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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter One
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c Solutions Manual Copyright 2005 by The McGrawHill Companies, Inc. !
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1.1 W = mg = 3(32.2) = 96.6 lb. 1.2 m = W / g = 100 / 9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) = 0.698 slug. 1.3 d = (50 + 5 / 12)(0.3048) = 15.37 m. 1.4 n = 1 / [60(1.341 × 10−3 )] = 12.43, or approximately 12 bulbs. 1.5 5(70 − 32) / 9 = 21.1◦ C 1.6
= 3000(2π ) / 60 = 314.16 rad/sec. Period P = 2π / = 60 / 3000 − 1 / 50 sec.
1.7 = 5 rad/sec. Period P = 2π / 0.796 Hz.
= 2π / 5 = 1.257 sec. Frequency f = 1 / P = 5 / 2π =
1.8
y − x = 5.66
x + y = 10.30
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1.9
Ae−1/τ 4.912 = = e2/tau 3.293 Ae−3/τ 2 4.912 = ln 3.293 2 =5 = ln(4.912 / 3.293) A = 4.912e1/τ = 6
1.10 y = [0, 7] and t = [0, 4(3)] = [0, 12] 1.11 y = [0, 6 + 12] = [0, 18]. Dominant time constant is 5. Thus t = [0, 4(5)] = [0, 20]. 1.12 10 sin 3t cos 2 + 10 cos 3t sin 2 = B sin 3t + C cos 3t
B = 10 cos 2 = −4.161
C = 10 sin 2 = 9.093
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1.13
A sin 6t cos A cos Thus
+ A cos 6t sin = −6.062 < 0
is in the third quadrant. = tan
−1
!
3.5 6.062
A=
#
"
= −6.062 sin 6t − 3.5 cos 6t
A sin
= −3.5 < 0
= 0.524 + π = 3.665 rad
(6.062)2 + (3.5)2 = 7
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1.14
y˙ = 0.5(7) cos(7t + 4) = 3.5 cos(7t + 4) y¨ = −0.5(7)2 sin(7t + 4) = −24.5 sin(7t + 4) Velocity amplitude is 3.5 m/s. Acceleration amplitude is 24.5 m/s2 .
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1.15
y (t) = A sin(3t + ) = A sin 3t cos
Thus
A sin
= 0.05 > 0
A cos
= 0.08 > 0
+ A cos 3t sin
is in the first quadrant. = tan
−1
!
0.05 0.08
"
= 0.559 rad
Thus
y (t) = 0.094 sin(3t + 0.559) The amplitude of the displacement is
A= For the velocity,
#
(0.05)2 + (0.08)2 = 0.094 ft
y˙ (t) = 3(0.094) cos(3t + 0.559) = 0.282 cos(3t + 0.559) The amplitude of the velocity is 0.282 ft/sec. For the acceleration,
y¨(t) = −3(0.282) sin(3t + 0.559) = 0.846 sin(3t + 0.559) The amplitude of the acceleration is 0.846 ft/sec2 .
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1.16 y˙ = 5(0.07) sin(5t + ). Thus y˙ (0) = 0.35 sin = sin
−1
!
0.041 0.35
"
= 0.041, and
= 0.117 or π − 0.117 = 3.02 rad
Without more information we cannot determine the quadrant of .
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1.17
y (0) = A sin y˙ (0) = 5A cos Thus
= 0.048 > 0 = 0.062 > 0
is in the first quadrant. −1
= tan
A=
$
%
0.048 = 1.318 rad (0.062 / 5)
#
(0.048)2 + (0.062 / 5)2 = 0.0496 m
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1.18 The radian frequency is 2π (10) = 20π . The displacement is y = A sin 20π t. The velocity and acceleration are y˙ = 20π A cos 20π t
Given the acceleration amplitude
y¨ = −(20π )2 A sin 20π t (20π )2 A = 0.7g
we solve for the displacement amplitude A as
A=
0.7g = 5.71 × 10−3 ft (20π )2
Thus the amplitude of the velocity is 20π A = 0.359 ft/sec.
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1.19 A = 0.15 mm. The acceleration is
x¨ = (2π f )2 A cos 2π f t Given the acceleration amplitude, (2π f )2 A = 0.6g we can solve for the frequency.
f=
&
0.6g = 0.997 Hz 4π 2 A
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1.20 y certainly lies within the range [−6, 6] and y disappears after about four time constants, 4(3) = 12.
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1.21
y¨ = −
A y˙ = − e−t/τ sin( t + ) + A e−t/τ cos( t + ) A 2
e−t/τ sin( t + )−
A
e−t/τ cos( t + )−
A
e−t/τ cos( t + )− A
2 −t/τ
e
sin( t + )
which reduces to
y¨ = A
!
1 2
−
2
"
e−t/τ sin( t + ) −
2A
e−t/τ cos( t + )
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1.22 Physical considerations require the model to pass through the origin, so we seek a model of the form f = kx. A plot of the data shows that a good line drawn by eye is given by f = 0.2x. So we estimate k to be 0.2 lb/in. Skipping ahead to Section 1.6, we can solve this problem using the least squares method, based on equation (1.6.3). The script file is x = f = num den k =
[4.7,7.2,10.6,12.9]4.7; [0,0.47,1.15,1.64]; = sum(x.*f); = sum(x.^2); num/den
The result is k = 0.1977.
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1.23 The script file is x = [0:0.01:1]; subplot(2,2,1) plot(x,sin(x),x,x),xlabel(#x (radians)#),ylabel(#x and sin(x)# ), . . . gtext(# x# ),gtext(#sin(x)# ) subplot(2,2,2) plot(x,sin(x)x),xlabel(#x (radians)# ),ylabel(#Error: sin(x)  x# ) subplot(2,2,3) plot(x,100*(sin(x)x). /sin(x)),xlabel(#x (radians)# ), . . . ylabel(#Percent Error# ),grid The plots are shown in the figure. 0
1
x and sin(x)
0.6
Error: sin(x) − x
x
0.8
sin(x)
0.4 0.2 0 0
0.5 x (radians)
1
0.5 x (radians)
1
−0.05 −0.1 −0.15 −0.2 0
0.5 x (radians)
1
Percent Error
0 −5 −10 −15 −20 0
Figure : for Problem 1.23. From the third plot we can see that the approximation sin x ≈ x is accurate to within 5% if x ≤ 0.5 radians.
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1.24 For
For
near π / 4,
near 3π / 4,
!
"!
π − 4
!
"!
3π − 4
π π f ( ) ≈ sin + cos 4 4
3π 3π + cos f ( ) ≈ sin 4 4
" "
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1.25 For
For
near π / 3,
near 2π / 3,
!
"!
π − 3
!
"!
2π − 3
π π f ( ) ≈ cos − sin 3 3
2π 2π − sin f ( ) ≈ cos 3 3
" "
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1.26 For h near 25,
f (h) ≈
√ 1 1 25 + √ (h − 25) = 5 + (h − 25) 10 2 25
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1.27 For r near 5, For r near 10,
f (r ) ≈ 52 + 2(5)(r − 5) = 25 + 10(r − 5) f (r ) ≈ 102 + 2(10)(r − 10) = 100 + 20(r − 10)
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1.28 For h near 16,
f (h) ≈
√ 1 1 16 + √ (h − 16) = 4 + (h − 16) 8 2 16
f (h) ≥ 0 if h > −16.
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1.29 Construct a straight line the passes through √ the two endpoints at p = 0 and p = 900. At p = 0, f (0) = 0. At p = 900, f (900) = 0.002 900 = 0.06. This straight line is
f (p) =
0.06 1 p= p 900 15, 000
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1.30 (a) The data is described approximately by the linear function y = 54x − 1360. The precise values given by the least squares method are y = 53.5x −1354.5 (see Problem 1.48a). (b) Only the loglog plot of the data gives something close to a straight line, so the data is best described by a power function y = bxm where the approximate values are m = −0.98 and b = 3600. The precise values given by the least squares method are y = 3582.1x−0.9764 (see Problem 1.48b). (c) Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx where the approximate values are m = −0.007 and b = 2.1 × 105 . The precise values given by the least squares method are y = 2.0622 × 105 (10)−0.0067x (see Problem 1.48c).
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1.31 With this problem, it is best to scale the data by letting x = y ear − 1990, to avoid raising large numbers like 1990 to a power. Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx . The approximate values are m = 0.035 and b = 9.98. Set y = 20 to determine how long it will take for the population to increase from 10 to 20 million. This gives 20 = 9.98(10)0.03x . Solve it for x: x = (log(20) − log(9.98)) / 0.035. The answer is 8.63 years, which corresponds to 8.63 years after 1990. More precise values are given by the least squares method (see Problem 1.49).
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1.32 (a) If C (t) / C (0) = 0.5 when t = 500 years, then 0.5 = e−5500b , which gives b = − ln(0.5) / 5500 = 1.2603 × 10−4 . (b) Solve for t to obtain t = − ln[C (t) / C (0)] / b using C (t) / C (0) = 0.9 and b = 1.2603 × 10−4 . The answer is 836 years. Thus the organism died 836 years ago. (c) Using b = 1.1(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 760 years. Using b = 0.9(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 928 years.
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1.33 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the temperature in degrees C and x is the time in seconds. The approximate values are m = −3.67 and b = 356. The alternate exponential form is y = be(m ln 10)x = 356e−8.451x . The time constant is 1 / 8.451 = 0.1183 s. The precise values given by the least squares method are y = 356.0199(10)−3.6709x (see Problem 1.51).
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1.34 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the bearing life thousands of hours and x is the temperature in degrees F. The approximate values are m = −0.007 and b = 142. The bearing life at 150 ◦ F is estimated to be y = 142(10)−0.007(150) = 12.66, or 12,600 hours. The alternate exponential form is y = be(m ln 10)x = 142e−0.0161x . The time constant is 1 / 0.0161 = 62.1 or 6.21 × 104 hr. The precise values given by the least squares method are y = 141.8603(10)−0.0070x (see Problem 1.52).
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1.35 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx where y is the voltage and x is the time in seconds. The first data point does not lie close to the straight line on the semilog plot, but a measurement error of ±1 volt would account for the discrepancy. The approximate values are m = −0.43 and b = 96. The alternate exponential form is y = be(m ln 10)x = 96e−0.99x . The time constant is 1 / 0.99 = 1.01 s. The precise values given by the least squares method are y = 95.8063(10)−0.4333x (see Problem 1.53).
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1.36 A semilog plot generated by the following script file shows that the exponential function T − 70 = bemt fits the data well. t = [0:300:3000]; temp = [207,182,167,155,143,135,128,123,118,114,109]; DT = temp70; semi logy(t ,DT, t ,DT,’o’) Fitting a line by eye gives the approximate values m = −4 × 10−4 and b = 125. The −4 corresponding function is T (t) = 70 + 125e−4×10 t . The precise values given by the least squares method are m = −4.0317 × 10−4 and b = 125.1276 (see Problem 1.54).
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1.37 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. The power function fitted by eye in terms of the height h is approximately f = 4h0.9 . Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye is approximately f = 3.2h. Using the least squares method gives more precise results: f = 4.1595h0.8745 and f = 3.2028h (see Problem 1.55)
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1.38 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. The power function fitted by eye in terms of the height h is approximately f = 4h0.9 . Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye is approximately f = 3.7h. Using the least squares method gives more precise results: f = 4.1796h0.9381 and f = 3.6735h (see Problem 1.56).
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1.39 Fitting a straight line by eye gives the approximate values m = 15 and b = 7. The precise values given by the least squares method are m = 15.0750 and b = 7.1500 (see Problem 1.57).
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1.40 Plot the data on a loglog plot. We must delete the first data point to avoid taking the logarithm of 0. The power function fitted by eye is approximately y = 7x3 . The precise values given by the least squares method are m = 2.9448 and b = 7.4053 (see Problem 1.58).
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1.41 Plot the data on a semilog plot. The exponential function fitted by eye is approximately y = 6e3x . The precise values given by the least squares method are y = 6.4224e2.8984x (see Problem 1.59).
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1.42 Fitting a straight line by eye through the origin gives the approximate values m = 17 and b = 0. The precise value given by the least squares method is m = 16.6071 (see Problem 1.60).
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1.43 Plot the data on a loglog plot. We must delete the first data point to avoid taking the logarithm of 0. The power function fitted by eye is approximately y = 7x3 . The precise value given by the least squares method is b = 7.4793 (see Problem 1.61).
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1.44 a) The equation for b is obtained as follows.
J=
n ' i=1
(bemxi − yi )2
n n ' ' ∂J = 2b e2mxi − 2 yi emxi = 0 ∂b i=1 i=1
( mx ye b = ( 2mx
e
b) Plot the data on a semilog plot. The exponential function fitted by eye is approximately y = 6e3x . The precise values given by the least squares method are y = 5.8449e3x (see Problem 1.62).
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1.45 The integral form of the sum of squares to fit the function y = 5x2 is
J=
)
0
4*
2
+2
5x − mx − b
dx =
)
4 0
,

25x4 − 10mx3 + (m2 − 10b)x2 + 2mbx + b2 dx
This evaluates to
J = 5120 − 640m + (m2 − 10b)
64 + 16mb + 4b2 3
Thus
128 ∂J = −640 + m + 16b = 0 ∂m 3 ∂J 640 =− + 16m + 8b = 0 ∂b 3 These give m = 20 and b = −13.333. Thus the line is y = 20x − 13.333.
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1.46 a) The integral form of the sum of squares to fit the function y = Ax2 + B x is
J= or
J=
)
L 0
)
0
L*
+2
Ax2 + B x − mx − b
dx
,

A2 x4 + (2AB − 2mA)x3 + (−2bA + B 2 + m2 − 2mB )x2 + (2mb − 2bB )x + b2 dx
This evaluates to
J = A2 Thus
L4 L2 L3 L5 + 2A(B − m) + (−2bA + B 2 + m2 − 2mB ) + 2b(m − B ) + b2 L 5 4 3 2 2 AL 4 2 ∂J =− + mL3 − B L3 + bL2 = 0 ∂m 2 3 3 2 ∂J = − AL3 + (m − B )L2 + 2bL = 0 ∂b 3
These give 4Lm + 6b = 4B L + 3AL2 3Lm + 6b = 2AL2 + 3B L which can be solved for m and b, given values of A, B , and L. b) With L = 2, A = 3, and B = 5, we obtain m = 11 and b = −2. The fitted straight line is y = 11x − 2.
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1.47 a) The integral form of the sum of squares to fit the function y = B eM x is
J= or
J=
)
0
This evaluates to
)
0
L*
+2
B e M x − mx − b
dx
L,

B 2 e2M x − 2mB xeM x − 2bB eM x + m2 x2 + 2bmx + b2 dx = + 2mB L + B 2 * 2M L 2mB * M L eM L + e −1 − e −1 2 2M M M 3 + 2bB * L + bmL2 + b2 L + 1 − eM L − m2 M 3
J =
Thus
+ 2mL3 2B L M L 2B * M L ∂J =− e + bL2 = 0 + 2 e −1 + ∂m M M 3 2B M L 2B ∂J =− e + mL2 + 2bL = 0 + ∂b M M
These give
+ 2B L M L 2B * M L 2L3 m + L2 b = e − 2 e −1 3 M M 2B M L 2B e L 2 m + 2L b = − M M which can be solved for m and b, given values of M , B , and L. b) With L = 1, M = −5, and B = 15, we obtain m = −10.973 and b = 8.466. The fitted straight line is y = −10.973x + 8.466.
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1.48 (a) The script file is x y p p
= [25:5:45]; = [5, 260, 480, 745, 1100]; = polyfi t(x,y,1) = 1.0e+003 * 0.0535 1.3545
The function is y = 53.5x − 1354.5.
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1.48 (b) Only the loglog plot of the data gives something close to a straight line, so the data is best described by a power function y = bxm . The script file to find the coe cients m and b is x y p m b
= = = = =
[2.5:0.5:6,7:10]; [1500,1220,1050,915,810,745,690,620,520,480,410,390]; polyfi t(log10(x), log10(y),1); p(1) 10^p(2)
The results are m = −0.9764 and b = 3582.1. Thus the power function is y = 3582.1x−0.9764 .
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1.48 (c) Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx . The script file to find the coe cients m and b is x y p m b m
= [550:50:750]; =[41.2,18.62,8.62,3.92,1.86]; = polyfi t(x, log10(y),1); = p(1) = 10^p(2) = 0.0067 b = 2.0622e+005 This gives the results m = −0.0067 and b = 2.0622 × 105 . Thus the exponential function is y = 2.0622 × 105 (10)−0.0067x . The results for the power function are obtained from p = polyfi t(log10(x), log10(y),1); m = p(1) b = 10^p(2) This gives the results m = −9.9949 and b = 1.0601 × 1029 . Thus the power function is y = 1.0601 × 1029 x−9.9949 . A plot of the data and the two functions shows that both functions describe the data well, but the exponential curve passes closer to the first data point than the power curve. In addition, we should be careful about using this power function because its coe cient b has a very large value. This large coe cient means that any predictions of y values made with this function will be very prone to error unless we use very precise values of x. Thus the exponential function is the best choice to describe this data.
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1.49 With this problem, it is best to scale the data by letting x = y ear − 1990, to avoid raising large numbers like 1990 to a power. Both the loglog and semilog plot (with the y axis logarithmic) give something close to a straight line, but the semilog plot gives the straightest line, so the data is best described by a exponential function y = b(10)mx . The script file to find the coe cients m and b is year = [1990:1995]; x = x = year1990; pop=[10,10.8,11.7,12.7,13.8,14.9]; p=polyfi t(x, log10(pop),1) p = 0.0349 0.9992 m=p(1); b=10^p(2) This gives the results m = 0.0349 and b = 9.9817. Thus the exponential function is y = 9.9817(10)0.0349x . Set y = 20 to determine how long it will take for the population to increase from 10 to 20 million. This gives 20 = 9.9817(10)0.0349x . Solve it for x: x = (log(20)−log(9.9817)) / 0.0349 = 8.6483 years, which corresponds to 8.6483 years after 1990.
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1.50 (a) If C (t) / C (0) = 0.5 when t = 500 years, then 0.5 = e−5500b , which gives b = − ln(0.5) / 5500. In Matlab this calculation is ( b = log(0.5)/5500 The answer is b = 1.2603 × 10−4 . (b) Solve for t to obtain t = − ln[C (t) / C (0)] / b using C (t) / C (0) = 0.9 and b = 1.2603 × 10−4 . In MATLAB this calculation is ( t = log(0.9)/b The answer is 836.0170 years. Thus the organism died 836 years ago. (c) Using b = 1.1(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 760 years. 0.9(1.2603 × 10−4 ) in t = − ln(0.9) / b gives 928 years.
Using b =
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1.51 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx . The script file to find the coe cients m and b is t ime = [0:0.1:0.6]; temp = [300,150,75,35,12,5,2]; p=polyfi t(t ime, log10(temp),1) m=p(1) b=10^p(2) This gives the results: m = −3.6709 and b = 356.0199. Thus the exponential function is y = 356.0199(10)−3.6709x , where y is the temperature in degrees C and x is the time in seconds. The alternate exponential form is y = be(m ln 10)x = 356.0199e−8.4526x . The time constant is 1 / 8.4526 = 0.1183 s.
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1.52 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx . The script file to find the coe cients m and b is temp = [100:20:220]; l ife = [28,21,15,11,8,6,4]; p = polyfi t(temp, log10(l ife),1); m = p(1) b = 10^p(2) This gives the results: m = −0.0070 and b = 141.8603. Thus the exponential function is y = 141.8603(10)−0.0070x , where y is the bearing life thousands of hours and x is the temperature in degrees F. The bearing life at 150 ◦ F is estimated to be y = 141.8603(10)−0.0070(150) = 12.6433, or 12,643 hours. The alternate exponential form is y = be(m ln 10)x = 141.8603e−0.0161x . The time constant is 1 / 0.0161 = 62.0421 or 6.20421 × 104 hr.
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1.53 Only the semilog plot of the data gives something close to a straight line, so the data is best described by an exponential function y = b(10)mx . The first data point does not lie close to the straight line on the semilog plot, but a measurement error of ±1 volt would account for the discrepancy. The script file to find the coe cients m and b is t ime = [0:0.5:4]; vol tage = [100,62,38,21,13,7,4,2,3]; p = polyfi t(t ime, log10(vol tage),1); m=p(1) b = 10^p(2) This gives the results: m = −0.4333 and b = 95.8063. Thus the exponential function is y = 95.8063(10)−0.4333x , where y is the voltage and x is the time in seconds. The alternate exponential form is y = be(m ln 10)x = 95.8063e−0.9977x . The time constant is 1 / 0.9977 = 1.002 s.
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1.54 The semilog plot generated by the following script file shows that the exponential function T − 70 = bemt fits the data well. t = [0:300:3000]; temp = [207,182,167,155,143,135,128,123,118,114,109]; DT = temp70; semi logy(t ,DT, t ,DT,’o’) p = polyfi t(t , log(DT),1) m = p(1) b = exp(p(2)) The results are m = −4.0317 × 10−4 and b = 125.1276. The time constant is 1 / 4.0317 × 10−4 = 2.4803 × 103 s, or 0.689 hr.
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1.55 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. The correct data are shown in the following script file. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try all both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. For the power function, we use (1.6.1) and (1.6.2) in terms of the variables x and Y = log y . t = [7,8,9,10,11,13,15,17,23]; x = [11:1:3]; y = 250. / t ; X = log10(x); Y = log10(y); a1 = sum(X.^2); a2 = sum(X); a3 = sum(Y.*X); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1) b = 10^solut ion(2) J = sum((b*x.^my).^2) S = sum((b*x.^mmean(y)).^2) r2 = 1  J/S The results are m = 0.8745, b = 4.1595, J = 5.3644, S = 493.5634, and r 2 = 0.9891. So the fitted function in terms of the height h is f = 4.1595h0.8745 . Note that the exponent is not close to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow. For the linear function f = mx, we use (1.6.3). m = sum(x.*y)/sum(x.^2) J = sum((m*xy).^2) S = sum((m*xmean(y)).^2) r2 = 1  J/S The results are m = 3.2028, J = 8.0247, S = 615.9936, and r 2 = 0.9870. The fitted function in terms of the height h is f = 3.2028h. Using only r 2 as the criterion, it is impossible to decide whether the linear or the power function is the best model. 148 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
1.56 In the first printing of the text, the time data was reversed (the largest height should have the smallest time). This misprint also occurs in the data table on page 28. The correct data are shown in the following script file. Plots of the data on a loglog plot and rectilinear scales both give something close to a straight line, so we try all both functions. (Note that the flow should be 0 when the height is 0, so we do not consider the exponential function and we must force the linear function to pass through the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate. The three lowest heights give the same time, so we discard the heights of 1 and 2 cm. For the power function, we use (1.6.1) and (1.6.2)in terms of the variables x and Y = log y . t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:1:3]; y = 250. / t ; X = log10(x); Y = log10(y); a1 = sum(X.^2); a2 = sum(X); a3 = sum(Y.*X); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1) b = 10^solut ion(2) J = sum((b*x.^my).^2) S = sum((b*x.^mmean(y)).^2) r2 = 1  J/S The results are m = 0.9381, b = 4.1796, J = 13.5531, S = 729.3505, and r 2 = 0.9814. So the fitted function in terms of the height h is f = 4.1796h0.9381 . For the linear function f = mx, we use (1.6.3). t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:1:3]; y = 250. / t ; m = sum(x.*y)/sum(x.^2) J = sum((m*xy).^2) S = sum((m*xmean(y)).^2) r2 = 1  J/S
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The results are m = 3.6735, J = 14.7546, S = 809.8919, and r 2 = 0.9818. The fitted function in terms of the height h is f = 3.6735h. Using only r 2 as the criterion, it is impossible to decide whether the linear or the power function is the best model.
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1.57 The script file is x = [0:2:6]; y = [4.5, 39, 72, 94]; a1 = sum(x.^2); a2 = sum(x); a3 = sum(y.*x); a4 = sum(y); n = length(x); A = [a1, a2; a2, n]; B = [a3; a4]; solut ion = A\B; m = solut ion(1) b = solut ion(2) J = sum((m*x+by).^2) S = sum((m*x+bmean(y)).^2) r2 = 1  J/S The results are m = 15.0750, b = 7.1500, J = 43.5750, S = 4.5451 × 103 , r 2 = 0.9904.
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1.58 Following the procedure shown in Example 1.6.1, we fit X = log x and Y = log y to a linear function. We must delete the first data point to avoid taking the logarithm of 0. The script file is x = [1:4]; y = [8, 50, 178, 490]; X = log10(x); Y = log10(y); a1 = sum(X.^2); a2 = sum(X); a3 = sum(Y.*X); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1) b = 10^solut ion(2) J = sum((b*x.^my).^2) S = sum((b*x.^mmean(y)).^2) r2 = 1  J/S The results are m = 2.9448, b = 7.4053, J = 2.7494×103 , S = 1.1218×105 , and r 2 = 0.9755.
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1.59 We fit x and Y = log y to the linear function Y = mx + B , where B = log b. The script file is x = [0:0.4:1.2]; y = [6.3, 22, 60, 215]; Y = log10(y); a1 = sum(x.^2); a2 = sum(x); a3 = sum(Y.*x); a4 = sum(Y); n = length(x); A = [a1, a2; a2, n]; C = [a3; a4]; solut ion = A\C; m = solut ion(1)/ log10(exp(1)) b = 10^solut ion(2) J = sum((b*exp(m*x)y).^2) S = sum((b*exp(m*x)mean(y)).^2) r2 = 1  J/S The results are m = 2.8984, b = 6.4244, J = 77.4488, S = 2.5496 × 104 , and r 2 = 0.9970. The fitted function is y = 6.4244e2.8984x .
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1.60 The equation for m, obtained from (1.6.3), is (
The script file is
xi yi m= ( 2 xi
x = [0:2:6]; y = [4.5, 39, 72, 94]; m = sum(x.*y)/sum(x.^2) J = sum((m*xy).^2) S = sum((m*xmean(y)).^2) r2 = 1  J/S The results are m = 16.6071, J = 116.6071, S = 5.5420 × 103 , and r 2 = 0.9790.
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1.61 The equation for b is obtained from equation (1) in Example 1.6.3 with m = 3. ( 3 x y b= ( 6
x
x = [0:4]; y = [1, 8, 50, 178, 490]; b = sum((x.^3).*y)/sum(x.^6) J = sum((b*x.^3y).^2) S = sum((b*x.^3mean(y)).^2) r2 = 1  J/S
The results are b = 7.4793, J = 799.4139, S = 1.6176 × 105 , and r 2 = 0.9951.
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1.62 a) The equation for b is obtained as follows.
J=
n ' i=1
(bemxi − yi )2
n n ' ' ∂J = 2b e2mxi − 2 yi emxi = 0 ∂b i=1 i=1
( mx ye b = ( 2mx
e
b) The script file is
x = [0:0.4:1.2]; y = [6.3, 22, 60, 215]; b = sum(y.*exp(3*x))/sum(exp(6*x)) J = sum((b*exp(3*x)y).^2) S = sum((b*exp(3*x)mean(y)).^2) r2 = 1  J/S The results are b = 5.8449, J = 27.7380, S = 2.7279 × 104 , and r 2 = 0.9990.
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Two
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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2.1 mv˙ = mg. Thus v(t) = gt = 32.2t
x(t) =
1 2 gt = 16.1t2 2
Thus t=
!
h(t) = 20 − x(t) = 20 − 16.1t2
20 − h(t) 16.1
For h = 10, t = 0.788 sec. For h = 0, t = 1.115 sec.
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2.2 t= x=
60/5280 3600 = 0.455 sec 90
1 2 gt = 16.1(0.455)2 = 3.326 ft 2
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2.3 Summing forces in the direction parallel to the plane gives mv˙ = f1 − mg sin φ − µmg cos φ Substituting the given values, 10v˙ = f1 − 98.1 sin 25◦ − 0.3(98.1) cos 25◦ = f1 − 68.132 Thus v˙ > 0 if f1 > 68.132. If f1 = 100 the block will continue to move up the plane. If f1 = 50, v˙ = −1.813 and the speed is given by v(t) = −1.813t + v(0) = −1.813t + 2 Thus v(0) = 0 at t = 1.103 s.
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2.4 Let d be the vertical distance dropped by the time the mass leaves the surface. See the following figure. Then L d= tan θ The speed v0 of the mass when it leaves the surface is found from conservation of energy: KE =
1 mgL mv02 = P E = 2 tan θ
Thus v0 =
!
2gL tan θ
(1)
The horizontal and vertical velocity components are vx = v0x sin θ v0y = v0 cos θ
(2)
Establish a coordinate system at the point where the mass leaves the surface, with x positive to the right and y positive down. In the vertical direction : y¨ = g and 1 y = gt2 + v0y t 2 The mass hits the ground when y = H, and timetohit tH is found from 1 2 gt + 2v0y tH = H 2 H or tH =
−2v0y ±
"
2 + 8gH 4v0y
2g
(3)
There will be one positive solution and one negative solution. Take the positive solution. In the horizontal direction: m¨ x = 0 and x = (v0 sin θ) t and D = (v0 sin θ) tH
(4)
The solution is given by (1) through (4).
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Figure : for Problem 4
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2.5 Let the origin of the launch point be at x = y = 0. Assuming the projectile is launched with a speed v0 at an angle θ from the horizontal, Newton’s law in the x and y directions gives x ¨=0 y¨ = −g vx = v0 cos θ
x = (v0 cos θ)t
vy = v0 sin θ − gt g y = − t2 + (v0 sin θ)t 2
These can be manipulated to show that vy2 = (v0 − sin θ)2 − 2gy which gives v0 sin θ =
"
vy2 − 2gy
Using this relation, the above can also be manipulated to show that vy = − Solve for x. x= where y is computed from
gx " 2 + vy − 2gy vx
vx " 2 vx vy vy − 2gy − g g y = R sin φ
(1) (2)
The desired distance D can be computed from D = x + R cos φ where x is computed from (1).
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2.6 a) From Newton’s law,
¨ = T − mg mh h˙ = h=
So the time to reach 10,000 ft is
#
#
T − mg m t=
!
$
T −g t m $
2
t2 = 10, 000 ft 2
10, 000m T − mg
Using the minimum thrust value, T = 11, 000 lb, and m = 100 slug, we obtain t = 16.03 sec. b) If T = 12, 000 ft and t = 16.03 sec, the maximum height will be h=
#
12, 000 − 100(32.2) 100
$
(16.03)2 = 11, 281 ft 2
c) Using t = 16.03 sec for the burn time, the fuel mass burned will be 0.5(16.03) = 8.015 slug. Thus the smallest possible launch mass is 100 − 8.015 = 91.985 slug, and the height reached will not be greater than h=
#
12, 000 − 91.985(32.2) 91.985
$
(16.03)2 = 12, 624 ft 2
So the maximum height will be in the range 11, 281 ≤ h ≤ 12, 624 ft.
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2.7 IG = 2mr 2 /5. Apply the parallel axis theorem. 2 IO = IG + mR2 = mr 2 + mR2 5
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2.8 a) Let point O be the pivot point and G be the center of mass. Let L be the distance from O to G. Treat the pendulum as being composed of three masses: 1) m1 , the rod mass above point O, whose center of mass is 1 ft above point O; 2) m2 , the rod mass below point O, whose center of mass is 1.5 ft below point O, and 3) m3 , the mass of the 10 lb block. Then, summing moments about G gives m1 g(L + 1) − m2 g(1.5 − L) − m3 g(3.5 − L) = 0 where
(1)
2 6 m1 g = 3 = lb 5 5 9 3 m2 g = 3 = lb 5 5 m3 g = 10 lb
From equation (1): 9 6 (L + 1) − (1.5 − L) − 10(3.5 − L) = 0 5 5 which gives L = 2.85 ft. b) Summing moments about the pivot point O gives IO θ¨ = −mgL sin θ where m is the total mass. From the parallelaxis theorem, treating the rod as a slender rod, we obtain IO =
1 12
# $
3 (5)2 + g
# $
3 (0.5)2 + g
#
$
10 (3.5)2 = 4.022 slug − ft2 g
and mgL = 13(2.85) = 37.05 ftlb. Thus the equation of motion is 4.022θ¨ = −37.05 sin θ or
θ¨ + 9.2118 sin θ = 0
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2.9 a) IA = mL22 + mC L21 IA θ¨ = −mC L1 g sin θ + mgL2 cos(β − θ)
b) Set θ¨ = 0 and solve for mg.
mg =
mC L1 g sin θ L2 cos(β − θ)
c) Substitute the given values to obtain mg =
5(0.2)g sin 20◦ = 2.315g = 22.713 N 0.15 cos 10◦
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2.10 The easiest way is to use the energyequivalence method. Let ωB the speed of pulley B (positive counterclockwise) and ωC the speed of pulley C (positive clockwise). The kinetic energy of the entire system is 1 1 1 1 1 2 2 2 2 2 KE = IB ωB + IC ω C + mL vC + mC vC = me vA 2 2 2 2 2 where me is the mass of the equivalent translational system. The potential energy is P E = (mL + mC )g(−sC ) = Fe sA where Fe is the equivalent gravity force. Thus Fe = −
mL + mC g 2
Using the kinematic relations: 1 sC = − sA 2
1 vC = − vA 2
ωB =
1 vA RB
ωC =
1 vC RC
the kinetic energy expression becomes 1 KE = 2
%
IC mL + mC IB + + 2 2 4 RB 4RC
Thus me =
&
2 vA =
1 2 me vA 2
IB IC mL + mC + + 2 2 RB 4RC 4
The model is me v˙ A = FA −
mL + mC g 2
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2.11 a) Let T be the tension in the cable attached to mass m2 . See the following figure. Then the cable force pulling up on m1 is T /2 because of the pulleys. Note also that because of the pulleys, x = 2y. Summing forces acting on m2 parallel to the plane, we obtain m2 y¨ = T − m2 g sin θ
(1)
Summing the vertical forces acting on m1 , we obtain
Since x = 2y, this becomes
1 m1 x ¨ = m1 g − T 2
(2)
1 2m1 y¨ = m1 g − T 2
(3)
T = 2m1 g − 4m1 y¨
(4)
Solve for T : Substitute this into (1) and collect the y¨ terms to obtain y = 2m1 g − m2 g sin θ (4m1 + m2 )¨
(5)
The mass m1 will lift m2 if y¨ > 0; that is, if 2m1 − m2 sin θ > 0 b) Follow the same procedure as in part (a) but include the friction force. Equation (1) becomes m2 y¨ = T − m2 g sin θ − µd m2 g cos θ (6) Equations (2) through (4) remain the same, but (5) becomes
y = 2m1 g − m2 g(µd cos θ + sin θ) (4m1 + m2 )¨
(7)
The mass m1 will lift m2 if y¨ > 0; that is, if 2m1 − m2 (µd cos θ + sin θ) > 0 For the case m1 = m2 /2, this becomes 1 − (µd cos θ + sin θ) > 0
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Figure : for Problem 11
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2.12 Refer to the following figure for definitions. This simplelooking system actually has complicated kinematics because the component of the cable tension T normal to the rod is a complicated function of θ, and because the acceleration x ¨ of m1 is a complicated function ˙ and θ. ¨ of θ, θ, Note that sin φ = cos θ and cos φ = sin θ. From the law of sines, H B = sin φ sin ψ
or
sin ψ =
H H sin φ = cos θ B B
(1)
From the law of cosines, B=
"
L2 + H 2 − 2LH cos φ =
Summing moments about point O,
'
L2 + H 2 − 2LH sin θ
(2)
T LH m2 g 1 L m2 L2 θ¨ = (T sin ψ) L − (m2 g cos θ) = cos θ − cos θ 12 2 B 2
(3)
Summing forces on m1 in the x direction: m1 x ¨ = m1 g − T ¨ because the cable is assumed Solving this for T , substituting into (3), and noting that x ¨=B inextensible, we obtain ( ) 1 m1 LH ¨ − m2 gL cos θ m2 L2 θ¨ = cos θ g − B 12 B 2
(4)
¨ To obtain the required expression for B ¨ as a function of θ and its derivaThis contains B. tives, differentiate (2) twice to obtain (
¨ = LH θ˙ 2 sin θ − θ¨ cos θ B +
)(
L2 + H 2 − 2LH sin θ
( )−3/2 LH ˙ θ cos θ −2LH θ˙ cos θ 2
(5)
)−1/2
The model consists of (2), (4), and (5).
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Figure : for Problem 12
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2.13 Let FR be the reaction force that acts on the contacting gear teeth. Then IG1 ω˙ 1 = T1 − r1 FR IG2 ω˙ 2 = T2 − r2 FR
If ω˙ 1 = ω˙ 2 = 0, these equation give T1 = r1 FR and T2 = r2 FR , and thus T1 = r1
T2 1 = T2 r2 N
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2.14 Let F be the contact force between the two gears. For gear 1, IG1 ω˙ 1 = T1 − r1 F For gear 2, If ω˙ 1 = ω˙ 2 = 0, or if IG1 = IG2 = 0,
IG2 ω˙ 1 = T2 − r2 F
T1 = r1 F which give T1 =
T 2 = r2 F r1 1 T2 = T 2 r2 N
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2.15 Let r2 be the radius of pulley 2. The equivalent inertia felt on shaft 1 is Ie = I1 +
1 1 1 I2 + 2 m2 r22 + 2 m3 r22 2 N N N
With N = 2, Ie = I1 + The equation of motion is Ie ω˙ 1 = T1 −
) 1( I2 + m2 r22 + m3 r22 4
m2 gr2 m3 gr2 gr2 + = T1 − (m3 − m2 ) N N 2
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2.16 The total kinetic energy is KE =
1 1 (Is + I) θ˙ 2 + mx˙ 2 2 2
KE =
) 1( Is + I + mR2 θ˙ 2 2
Substituting x = Rθ we obtain
Thus the equivalent inertia is
Ie = Is + I + mR2
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2.17 The total kinetic energy is KE =
1 1 1 (I1 + IS1 ) ω12 + (I2 + IS2 ) ω22 + mv 2 2 2 2
Substituting ω2 = r1 ω1 /r2 and v = r1 ω1 we obtain #
1 1 r1 ω 1 KE = (I1 + IS1 ) ω12 + (I2 + IS2 ) 2 2 r2 or
*
$2
1 + m (r1 ω1 )2 2
#
$2
+
#
$2
+ mr12
1 r1 KE = I1 + IS1 + (I2 + IS2 ) 2 r2
mr12
+
ω12
Thus the equivalent inertia is r1 Ie = I1 + IS1 + (I2 + IS2 ) r2
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2.18 a) With everything reflected to the load shaft (shaft 2), Newton’s law gives ω˙ 2 =
T2 + N T1 I2 + N 2 I1
b) To maximize ω˙ 2 , differentiate the above expression with respect to N , set the derivative equal to 0. This gives ∂ ω˙ 2 (I2 + N 2 I1 )T1 − 2I1 N (T2 + N T1 ) =0 = ∂N (I2 + N 2 I1 )2 This is true if the numerator is 0. Thus I1 T1 N 2 + 2I1 T2 N − I2 T1 = 0 The positive solution for N is T2 N =− + T1
! #
T2 T1
$2
+
I2 I1
This is the ratio that maximizes ω˙ 2 . (This can be confirmed to give a maximum rather than a minimum by showing that ∂ 2 ω˙ 2 /∂N 2 < 0). ' If the load torque2 T2 is 0, the optimal ratio for this case is denoted No and is: No = I2 /I1 , or I1 = I2 /No . This says that the ratio that maximizes the load acceleration is the ratio that makes the reflected load inertia (felt by the motor) equal to the motor’s inertia. This is the principle of inertia matching. If T2 = 0 and the actual ratio N differs from the optimal value No such that N = γNo , the efficiency E is actual ω˙ 2 2γ E= = max ω˙ 2 1 + γ2 Because E(γ) = E(1/γ), the efficiency when “overgearing” is the same as when “undergearing”. For example, if γ = 2 or γ = 1/2, E = 0.8, so the acceleration is 80% of the maximum possible.
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2.19 With I1 = I2 = I3 = 0, the total kinetic energy is 1 1 KE = I4 ω12 + I5 ω32 2 2 Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2 ω3 = 1.96ω3 , and I4 = 0.02, I5 = 0.1, we obtain 1 KE = (0.177)ω32 2 and the equivalent inertia is Ie = 0.177 kg·m2 . The equation of motion is Ie ω˙ 3 = (1.4)2 T , or 0.177ω˙ 3 = 1.96T
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2.20 The total kinetic energy is KE =
1 1 1 (I4 + I1 ) ω12 + I2 ω22 + (I3 + I5 ) ω32 2 2 2
Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2 ω3 = 1.96ω3 and the given values of the inertias, we obtain 1 KE = (0.203)ω32 2 and the equivalent inertia is Ie = 0.203 kg·m2 . The equation of motion is Ie ω˙ 3 = (1.4)2 T , or 0.203ω˙ 3 = 1.96T
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2.21 a)
ω4 ω3 ω2 73 ω4 3 = 2.1 = = ω1 ω3 ω2 ω1 65
b) The torque T1 felt on shaft 4 is T1 /2.1 and the equation of motion is I ω˙ 4 =
T1 2.1
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2.22 Using kinetic energy equivalence, 1 1 1 KE = mv 2 + Is ω 2 = Ie ω 2 2 2 2 The mass translates a distance x when the screw rotates by θ radians. When θ = 2π, x = L. ˙ ˙ we have Thus x = Lθ/2π and x˙ = v = Lθ/2π. Because ω = θ, KE =
#
$2
Ie =
mL2 + Is 4π 2
1 Lω m 2 2π
Solve for Ie to obtain
1 1 + Is ω 2 = Ie ω 2 2 2
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2.23 The expression for the kinetic energy is KE = KE of 2 rear wheels + KE of front wheel + KE of body # $ 1 1 1 = 2 Ir ωr2 + If ωf2 + mb v 2 2 2 2 But for the rear wheels, v = Rr ωr = 4ωr , and for the front wheel, v = Rf ωf = 2ωf . The inertias are calculated as follows. For the rear wheels: 1 1 500 2 4000 Ir = mr Rr2 = 4 = 2 2 g g For the front wheel: If =
1 1 800 2 1600 mf Rf2 = 2 = 2 2 g g
Also, mb = 9000/g. Thus KE =
4950 2 v g
and the equivalent mass is me = 2(4950)/g = 9000/g. The equation of motion is me v˙ = mg sin 10◦ , where m = 2mr + mf + mb = 2 Thus
500 800 9000 10, 800 + + = g g g g
9900 v˙ = 10, 800 sin 10◦ g
or v˙ = 0.1894. Thus v = 0.1894t ft/sec.
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2.24 In the first printing of the text, the first line of the problem refers to “mass m”. It should say “mass m1 ”. Using the equivalent mass approach, the equivalent mass referenced to the coordinate x is I me = m1 + m2 + 2 R where I is the inertia of the cylinder about its center. The force acting on me due to the weight of the cylinder is m1 g sin β. The force acting on me due to the weight of m2 is m2 g sin φ. See the following figure. The equation of motion of the equivalent system is me x ¨ = m1 g sin β − m2 g sin φ
Figure : for Problem 24
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2.25 Let y be the translational displacement of the cylinder to the right. Using the equivalent mass approach, the kinetic energy of the system is 1 1 1 KE = m2 y˙ 2 + I ω˙ 2 + m1 x˙ 2 2 2 2 where I = m2 R2 /2 is the inertia of the cylinder about its center. Because y˙ = 2x˙ and ω = ˙ y/R , we have ( ) 1 1 KE = m2 4x˙ 2 + 2 2
%
m2 R 2 2
&#
2x˙ R
$2
1 1 + m1 x˙ 2 = (6m2 + m1 ) x˙ 2 2 2
Thus the effective mass referenced to the coordinate x is me = 6m2 + m1 . The equation of motion is me x ¨ = m1 g or x = m1 g (6m2 + m1 )¨
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2.26 Summing moments about the pivot gives IO θ¨ = T − mgL sin θ where the effect of the motor torque is T = [2(1.5)]Tm = 3Tm and the inertia is IO = mL2 + Im [2(1.5)]2 + IG1 [2(1.5)]2 + IG2 (1.5)2 + IG3 (1.5)2 + IG4 = 1.936 kg · m2 Thus
1.936θ¨ = 3Tm − 29.43 sin θ
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2.27 a) The speed ratio of the sprocket drive at the driving shaft is the ratio of the sprocket diameters, which is 0.05/0.15 = 1/3. The equivalent inertia felt at the motor shaft is ,
Ie = I1 + I2 + IS1 + mc1 (0.05)2 +
,
IS2 + Id + 4Iw +
2mc2 rd2
 # 1 $2
+
10
mL rd2
 # 1 1 $2
3 10
where IS1 , IS2 , Id , and Iw are the inertias of the shafts 1 and 2, the drive shaft, and the drive wheels. The masses mc1 , mc2 , and mL are the masses of the sprocket chain, the drive chain, and the load. This expression evaluates to Ie = 0.0143 kg·m2 . The magnitude of the friction torque felt at the motor shaft is 54/[3(10)] = 1.8 N·m. The equation of motion is 0.0143ω˙ 1 = T1 − 1.8sgn(ω1 )
(1)
b) If T1 = 10, ω˙ 1 = 573.427 and ω1 = 573.427t rad/s. c) Solve (1) for T1 assuming ω1 > 0: T1 = 0.0143ω˙ 1 + 1.8 From the trapezoidal profile, 300
0 ≤ t ≤ 0.5 ω˙ 1 = 0 0.5 < t < 2.5 −300 2.5 ≤ t ≤ 3
Thus T1 =
300Ie + 1.8
1.8
1.8 − 300I e
0 ≤ t ≤ 0.5 0 ≤ t ≤ 0.5 6.09 0.5 < t < 2.5 = 1.8 0.5 < t < 2.5 2.5 ≤ t ≤ 3 −249 2.5 ≤ t ≤ 3
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2.28 a) The equation of motion for rotation is I ω˙ = Rft where I = mR2 /2 = 1.631 kg·m2 is the inertia of the cylinder about its center, and ft is the tangential force between the cylinder and the ground. The equation of motion for translation is I ω˙ m¨ x = f cos φ − ft = f cos φ − R Substituting x ¨ = Rω˙ we obtain (
)
mR2 + I ω ¨ = Rf cos φ
Noting that m = 800/9.81 = 81.549 kg, and substituting the given values, we obtain 16.636¨ ω = f cos φ b) Using the equivalent mass approach, the equivalent mass referenced to the coordinate x is I me = m + 2 R where I = mR2 /2 = 1.631 kg·m2 is the inertia of the cylinder about its center. The equation of motion is me x ¨ = f cos φ Substituting the given values, we obtain 83.18¨ x = f cos φ
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2.29 See the following figure for the coordinate definitions and the definition of the reaction force R. Let P be the point on the axle. Note that yP = 0. The coordinates of the mass center of the rod are xG = xP −
L sin θ 2
yG = −
L cos θ 2
Thus
L¨ L θ cos θ + θ˙ 2 sin θ 2 2 L L y¨G = θ¨ sin θ + θ˙ 2 cos θ 2 2 Let m be the mass of the rod. Summing forces in the x direction: ¨P − x ¨G = x
m¨ xG = f
#
$
L L m x ¨P − θ¨ cos θ + θ˙ 2 sin θ = f 2 2
or
(1)
Summing forces in the y direction: m¨ yG = R − mg
or
m
#
$
L¨ L θ sin θ + θ˙ 2 cos θ = R − mg 2 2
(2)
Summing moments about the mass center: L L IG θ¨ = f cos θ − R sin θ 2 2 Substituting for R from (2) and using the fact that IG = mL2 /12, we obtain mgL mL2 1 mL2 ˙ 2 fL mL2 θ¨ = cos θ − sin θ − sin2 θ θ¨ − θ sin θ cos θ 12 2 2 4 4
(3)
The model consists of (1) and (3) with m = 20 kg and L = 1.4 m.
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Figure : for Problem 29
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2.30 See the following figure for the coordinate definitions and the definition of the reaction forces Rx , Ry , N , and ft . Let P be the point on the axle. Note that yP = 0. The coordinates of the mass center of the rod are xG = xP −
L sin θ 2
yG = −
L cos θ 2
Thus
L¨ L θ cos θ + θ˙ 2 sin θ 2 2 L L y¨G = θ¨ sin θ + θ˙ 2 cos θ 2 2 Let m1 be the mass of the rod, m2 the mass of the wheel, and I the moment of inertia of the wheel about its center. Summing forces on the wheel in the x direction: x ¨G = x ¨P −
m2 x ¨P = f − Rx − ft
(1)
Summing forces on the wheel in the y direction: m2 y¨P = N − Ry − m2 g
(2)
Summing moments about the mass center of the wheel: Iω ¨ = Rft But x˙ P = Rω and x ¨P = Rω. ˙ Thus I x ¨ P = ft R2
(3)
Substitute this into (1) to obtain #
$
I m2 + 2 x ¨P = f − Rx R
(4)
Summing forces on the rod in the x direction: m1 x ¨ G = Rx
#
or
m1 x ¨P −
$
L¨ L θ cos θ + θ˙ 2 sin θ = Rx 2 2
(5)
Summing forces on the rod in the y direction: m1 y¨G = Ry − m1 g
or
m1
#
$
L¨ L θ sin θ + θ˙ 2 cos θ = Ry − m1 g 2 2
(6)
Summing moments about the mass center of the rod: L L IG θ¨ = Rx cos θ − Ry sin θ 2 2 235 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Substituting for Rx and Ry from (5) and (6), canceling terms using the identity sin2 θ + cos2 θ = 1, and using the fact that IG = m1 L2 /12, we obtain m1 L2 ¨ 1 m1 gL m1 L m1 L2 θ¨ = x ¨P cos θ − sin θ − θ 12 2 2 4 which can be rearranged as follows: m1 gL m1 L 1 m1 L2 θ¨ = x ¨P cos θ − sin θ 3 2 2
(7)
Substituting for Rx from (4) into (5) gives #
$ ) I m1 L ( ˙ 2 m1 + m2 + 2 x ¨P = F − θ sin θ − θ¨ cos θ R 2
(8)
The model consists of (7) and (8) with m1 = 20 kg, m2 = 3 kg, L = 1.4 m, R = 0.05 m, and I = m2 R2 /2.
Figure : for Problem 30
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2.31 Summing forces in the x direction: maGx = mg sin θ − F
(1)
Summing forces in the y direction: maGy = N − mg cos θ − F For no bounce, aGy = 0 and
N = mg cos θ
(2)
(3)
Summing moments about G: IG α = F r
(4)
From (1) and (4): maGx = mg sin θ − With slipping, aGx $= rα, but
F = µd N
IG α r
(5)
(6)
From (4) and (6): IG α = rµd N = rµd mg cos θ Thus, α=
(7)
rµd mg cos θ IG
From (1), aGx = g sin θ − µd g cos θ
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2.32 Follow the procedure of Example 2.4.2. Summing forces in the x direction: maGx = mg sin θ − F
(1)
Summing forces in the y direction: For no bounce, aGy = 0 and
maGy = N − mg cos θ − F N = mg cos θ
(2)
(3)
a) Summing moments about G: IG α = F r
(4)
From (1) and (4): maGx = mg sin θ −
For no slip,
IG α r
aGx = rα From (5) and (6): aGx =
(6)
mgr 2 sin θ mr 2 + IG
Since IG = mr 2 , aGx = and α=
g sin θ 2
(5)
(7)
(8)
aGx g sin θ = r 2r
b) Summing moments about P : MP = (mg sin θ)r = IG α + maGx r = IG Thus aGx =
aGx + maGx r r
mgr 2 sin θ g sin θ = mr 2 + IG 2
and
g sin θ aGx = r 2r c) The maximum friction force is Fmax = µs N = µs mg cos θ. From (4), (6), and (7), α=
F =
IG aGx IG mg sin θ = r2 mr 2 + IG
If Fmax > F there will be no slip; that is, if µs cos θ >
IG sin θ sin θ = 2 mr + IG 2
there will be no slip. 238 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
2.33 See the following figure. Summing forces in the x direction: maGx = µs NB − mg sin θ
(1)
Summing forces in the y direction: NA + NB = mg cos θ
(2)
Summing moments about G: NA LA − NB LB + µs NB H = 0
(3)
Solve (2) and (3): NA =
mg cos θ (µs H − LB ) 16, 677 cos θ (µs − 2.1) = µs H − LA − LB µs − 4.6
NB = −
mgLA cos θ 40, 025µs cos θ = µs H − LA − LB 4.6 − µs
The maximum acceleration is, from (1), aGx =
23.544µs cos θ − 9.81 sin θ 4.6 − µs
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Figure : for Problem 33
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2.34 a) Integrate the equation of motion as follows. v(t) =
2
v(t)
dv =
0
0
*
eat v(t) = 35 a 35 v(t) = a With a = 1 and g = 9.81,
2 t(
3#
#
1 t− a $
)
35teat − g dt $+t
0
− gt 4
1 at 1 t− e + − gt a a
,

v(t) = 35 (t − 1) et + 1 − 9.81t You can plot this to get an idea of the approximate value of t that makes v(t) = 300. It is approximately t = 2. To obtain a more precise value, create the following userdefined function file veldiff.m. function dv = veldiff(t) global a V dv = V  ((35/a)*((t1/a).*exp(a*t)+1/a)9.81*t); In the Command window, type %global a V %a = 1; V = 300; %T = fzero(# veldiff#,2) The answer returned is T = 2.0493 s.
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2.34 b) Create the following file vel2.m, using the time 2.05 s for a = 1 as the starting guess. global a V a = 1; T0 = fzero(# veldiff#,2.05); k = 0; for a = [1:0.01:5] k = k + 1; T(k) = fzero(#veldiff# ,T0); end plot([1:0.01:5],T) In the Command window, type %vel2 The plot is similar to that shown in Figure 2.5.1b, except that the maximum value is 2.05.
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2.35 Modify the program given in Example 2.5.2. a) % Set the values for initial speed, gravity, and angle. v0 = 40; g = 9.81; theta = 30*pi/180; % Compute the timetohit. t_hit = 2*v0*sin(theta)/g; % Compute the arrays containing time, height, and speed. t = [0:t_hit/100:t_hit]; y = v0*t*sin(theta)  0.5*g*t.^2; v = sqrt(v0^2  2*v0*g*sin(theta)*t + g^2*t.^2); % Determine when the height is no less than 15. u = find(y>=15); % Compute the corresponding times. t_1 = (u(1)1)*(t_hit/100) t_2 = u(length(u)1)*(t_hit/100) The results are 1.0194 and 3.0581 s. Thus the height is no less than 15 m for 1.0194 ≤ t ≤ 3.0581 s. b) % Set the values for initial speed, gravity, and angle. v0 = 40; g = 9.81; theta = 30*pi/180; % Compute the timetohit. t_hit = 2*v0*sin(theta)/g; % Compute the arrays containing time, height, and speed. t = [0:t_hit/100:t_hit]; y = v0*t*sin(theta)  0.5*g*t.^2; v = sqrt(v0^2  2*v0*g*sin(theta)*t + g^2*t.^2); % Determine when the height is no less than 15, % and the speed is no greater than 36. u = find(y>=15&v<=36); % Compute the corresponding times. t_1 = (u(1)1)*(t_hit/100) t_2 = u(length(u)1)*(t_hit/100) The results are 1.0601 and 3.0173 s. Thus the height is no less than 15 m and the speed is no greater than 36 m/s for 1.0601 ≤ t ≤ 3.0173 s.
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2.35 c) % Set the values for initial speed, gravity, and angle. v0 = 40; g = 9.81; theta = 30*pi/180; % Compute the timetohit. t_hit = 2*v0*sin(theta)/g; % Compute the arrays containing time, height, and speed. t = [0:t_hit/100:t_hit]; y = v0*t*sin(theta)  0.5*g*t.^2; v = sqrt(v0^2  2*v0*g*sin(theta)*t + g^2*t.^2); % Determine when the height is less than 5, % or the speed is greater than 35. u = find(y<5v>35); % Compute the corresponding times. t_1 = (u(1)1)*(t_hit/100) t_2 = u(length(u)1)*(t_hit/100) The results are 0 and 4.0775 s. Thus the height is less than 5 m or the speed is greater than 35 m/s for 0 ≤ t ≤ 4.0775 s. Note that the time to hit the ground is 4.0775 s.
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Three
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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3.1 a) Nonlinear because of the y y¨ term. b) Nonlinear because of the sin y term. c) √ Nonlinear because of the y term.
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3.2 a)
!
4
x
dx = 3
2
!
t
t dt 0
3 x(t) = 2 + t2 8
b)
!
5
x
dx = 2
3
!
t
e−4t dt
0
x(t) = 3.1 − 0.1e−4t
c) Let v = ˙ . x
3
!
v
dv = 5
7
!
t
t dt 0
dx 5 = 7 + t2 dt 6 # ! x ! t" 5 2 dx = 7 + t dt 6 2 0 5 x(t) = 2 + 7t + t3 18 v(t) =
d) Let v = x. ˙ 4
!
v 2
dv = 7
!
t 0
e−2t dt
23 7 −2t v(t) = − e 8 8 # ! x ! t" 23 7 −2t dx = dt − e 8 8 4 0 57 23 7 x(t) = + t + e−2t 16 8 16
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3.3 a)
!
Let
Thus
!
x
t dx = dt = t 2 3 25 − 5x 0 $√ %&x 1 5 + x && √ ln √ & 10 5 5 − x &3
√ 5+3 A = ln √ 5−3
√ √ 5+x = 10 5t + A ln √ 5−x √
√ eA e10 5t − 1 √ x(t) = 5 eA e10 5t + 1
b)
!
x
dx = 36 + 4x2
10
!
&
t
1 x &x tan−1 && = t 12 3 10
C = tan−1
x(t) = 3 tan(12t + C) c)
dt = y
0
!
x 4
dx = 5x + 25
&
!
10 3
t
dt 0
&x 1 ln(5x + 25)&& = t 5 4
ln(5x + 25) = 5t + ln 45 d)
x(t) = 9e5t − 5 !
x 5
dx =2 x
ln
xx5
!
t 0
e−4t dt &
&t 1 = − e−4t && 2 0
ln x = ln 5 −
( 1 ' −4t e −1 2 −4t −1)/2
x(t) = 5e−(e
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3.4 a) The roots are −3 and −5. The form of the free response is x(t) = A1 e−3t + A2 e−5t Evaluating this with the given initial conditions gives x(t) = 27e−3t − 17e−5t The steadystate solution is xss = 30/15 = 2. Thus the form of the forced response is x(t) = 2 + B1 e−3t + B2 e−5t Evaluating this with zero initial conditions gives x(t) = 2 − 5e−3t + 3e−5t The total response is the sum of the free and the forced response. It is x(t) = 2 + 22e−3t − 14e−5t The transient response consists of the two exponential terms. b) The roots are −5 and −5. The form of the free response is x(t) = A1 e−5t + A2 te−5t Evaluating this with the given initial conditions gives x(t) = e−5t + 9te−5t The steadystate solution is xss = 75/25 = 3. Thus the form of the forced response is x(t) = 3 + B1 e−5t + B2 te−5t Evaluating this with zero initial conditions gives x(t) = 3 − 3e−5t − 15te−5t The total response is the sum of the free and the forced response. It is x(t) = 3 − 2e−5t − 6te−5t The transient response consists of the two exponential terms.
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c) The roots are ±5j. The form of the free response is x(t) = A1 sin 5t + A2 cos 5t Evaluating this with the given initial conditions gives x(t) =
4 sin 5t + 10 cos 5t 5
The form of the forced response is x(t) = B1 + B2 sin 5t + B3 cos 5t Thus the entire forced response is the steadystate forced response. There is no transient forced response. Evaluating this function with zero initial conditions shows that B2 = 0 and B3 = −B1 . Thus x(t) = B1 − B1 cos 5t
Substituting this into the differential equation shows that B1 = 4 and the forced response is x(t) = 4 − 4 cos 5t The total response is the sum of the free and the forced response. It is x(t) = 4 + 6 cos 5t +
4 sin 5t 5
The entire response is the steadystate response. There is no transient response. d) The roots are −4 ± 7j. The form of the free response is x(t) = A1 e−4t sin 7t + A2 e−4t cos 5t Evaluating this with the given initial conditions gives x(t) =
44 −4t e sin 7t + 10e−4t cos 7t 7
The form of the forced response is x(t) = B1 + B2 e−4t sin 7t + B3 e−4t cos 7t The steadystate solution is xss = 130/65 = 2. Thus B1 = 2. Evaluating this function with zero initial conditions shows that B2 = −8/7 and B3 = −2. Thus the forced response is 8 x(t) = 2 − e−4t sin 7t − 2e−4t cos 7t 7
The total response is the sum of the free and the forced response. It is x(t) = 2 +
36 −4t e sin 7t + 8e−4t cos 7t 7
The transient response consists of the two exponential terms. 36 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
3.5 a) The root is s = 5/3, which is positive. So the model is unstable. b) The roots are s = 5 and −2, one of which is positive. So the model is unstable. c) The roots are s = 3 ± 5j, whose real part is positive. So the model is unstable. d) The root is s = 0, so the model is neutrally stable. e) The roots are s = ±2j, whose real part is zero. So the model is neutrally stable. f) The roots are s = 0 and −5, one of which is zero and the other is negative. So the model is neutrally stable.
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3.6 a) X(s) = b) X(s) =
2 10 + 3 s s
6 1 + 2 (s + 5) s+3
c) From Property 8, X(s) = −
dY (s) ds
where y(t) = e−3t sin 5t. Thus Y (s) =
5 5 = 2 (s + 3)2 + 52 s + 6s + 34
dY (s) 10s + 30 =− 2 ds (s + 6s + 34)2 Thus X(s) =
(s2
10s + 30 + 6s + 34)2
d) X(s) = e−5s G(s), where g(t) = t. Thus G(s) = 1/s2 and X(s) =
e−5s s2
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3.7 Thus
f (t) = 5us (t) − 7us (t − 6) + 2us (t − 14) F (s) =
5 e−6s e−14s −7 +2 s s s
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3.8 a)
5 5 = 3s + 7 3 5 x(∞) = lim s =0 s→0 3s + 7
x(0+) = lim s s→∞
b)
10 =0 + 7s + 4 10 x(∞) = lim s 2 =0 s→0 3s + 7s + 4
x(0+) = lim s s→∞
3s2
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3.9 a)
"
3 1 1 X(s) = − 2 s s+4 ( 3' x(t) = 1 − e−4t 2
b)
#
5 1 31 1 + 3s 3 s+3 5 31 x(t) = + e−3t 3 3
X(s) =
c)
13 1 1 1 + 3s+2 3 s+5 1 −2t 13 −5t x(t) = − e + e 3 3
X(s) = −
d) X(s) =
5 1 5 1 5/2 5 1 = + − 2 + 4) 8s 32 s 32 s + 4
s2 (s
x(t) = e)
5 5 5 t− + e−4t 8 32 32
2 1 13 1 13 1 − + 2 5s 25 s 25 s + 5 2 13 13 −5t x(t) = t + − e 5 25 25
X(s) =
f) X(s) = −
79 1 31 79 1 1 − + 2 4 (s + 3) 16 s + 3 16 s + 7
x(t) = −
31 −3t 79 −3t 79 −7t te + e − e 4 16 16
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3.10 a) X(s) =
7s + 2 5 s+3 = C1 + C2 2 2 2 2 (s + 3) + 5 (s + 3) + 5 (s + 3)2 + 52
or X(s) = −
19 s+3 5 +7 2 2 5 (s + 3) + 5 (s + 3)2 + 52
x(t) = − b) X(s) =
19 −3t e sin 5t + 7e−3t cos 5t 5
C1 4s + 3 5 s+3 = + C2 + C3 s[(s + 3)2 + 52 ] s (s + 3)2 + 52 (s + 3)2 + 52
or X(s) =
3 1 127 3 5 s+3 − + 2 2 34 s 170 (s + 3) + 5 34 (s + 3)2 + 52
x(t) =
3 3 127 −3t + e sin 5t − e−3t cos 5t 34 170 34
c) 4s + 9 [(s + + 52 ][(s + 2)2 + 42 ] 5 s+3 4 s+2 = C1 + C2 + C3 + C4 2 2 2 2 2 2 (s + 3) + 5 (s + 3) + 5 (s + 2) + 4 (s + 2)2 + 42
X(s) =
3)2
or 44 19 5 s+3 − 2 2 205 (s + 3) + 5 82 (s + 3)2 + 52 69 19 4 s+2 + 328 (s + 2)2 + 42 82 (s + 2)2 + 42
X(s) = − +
x(t) = −
44 −3t 19 69 −2t 19 e sin 5t − e−3t cos 5t + e sin 4t + e−2t cos 4t 205 82 328 82
d) X(s) = 2.625
1 1 1 − 18.75 + 21.125 s+2 s+4 s+6
x(t) = 2.625e−2t − 18.75e−4t + 21.125e−6t
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3.11 a) x˙ = 7t/5
b) x˙ = 3e−5t /4
c) x ¨ = 4t/7
d) x ¨ = 8e−4t /3
!
!
!
7 t t dt 5 0 3 7 x(t) = t2 + 3 10 x
dx =
!
3 t −5t dx = e dt 4 0 0 ( 3 ' x(t) = 1 − e−5t 20 x
!
4 t x(t) ˙ − x(0) ˙ = t dt 7 0 4 x(t) ˙ = t2 + 5 14 # ! x ! t" 4 2 t + 5 dt dx = 14 3 0 4 x(t) = t3 + 5t + 3 42 !
8 t −4t x(t) ˙ − x(0) ˙ = e dt 3 0 17 8 x(t) ˙ = − e−4t 3 12 # ! x ! t" 8 −4t 17 − e dx = dt 3 12 3 0 1 17 17 x(t) = t + e−4t + 3 6 6
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3.12 a) The root is −7/5 and the form is x(t) = Ce−7t/5 . With x(0) = 4, C = 4 and x(t) = 4e−7t/5 b) The root is −7/5 and the form is x(t) = C1 e−7t/5 + C2 . At steady state, x = 15/7 = C2 . With x(0) = 0, C1 = −15/7. Thus x(t) =
( 15 ' 1 − e−7t/5 7
c) The root is −7/5 and the form is x(t) = C1 e−7t/5 + C2 . At steady state, x = 15/7 = C2 . With x(0) = 4, C1 = 13/7. Thus x(t) = d)
( 13 ' 1 + e−7t/5 7
sX(s) − x(0) + 7X(s) = X(s) =
4 s2
5s2 + 4 4 4 249 −7t = 2− + e 2 s (s + 7) 7s 49 49
4 249 −7t 4 + e x(t) = t − 7 49 49
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3.13 a) The roots are −7 and −3. The form is x(t) = C1 e−7t + C2 e−3t Evaluating C1 and C2 for the initial conditions gives 9 25 x(t) = − e−7t + e−3t 4 4 b) The roots are −7 and −7. The form is x(t) = C1 e−7t + C2 te−7t Evaluating C1 and C2 for the initial conditions gives x(t) = e−7t + 10te−7t c) The roots are −7 ± 3j. The form is x(t) = C1 e−7t sin 3t + C2 e−7t cos 3t Evaluating C1 and C2 for the initial conditions gives x(t) =
20 −7t e sin 3t + 4e−7t cos 3t 3
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3.14 a) The roots are −3 and −7. The form is x(t) = C1 e−3t + C2 e−7t + C3 At steady state, x = 5/63 so C3 = 5/63. Evaluating C1 and C2 for the initial conditions gives 35 −3t 15 −7t 5 x(t) = − + + e e 252 252 63 b) The roots are −7 and −7. The form is x(t) = C1 e−7t + C2 te−7t + C3 At steady state, x = 98/49 = 2 so C3 = 2. Evaluating C1 and C2 for the initial conditions gives x(t) = −2e−7t − 16te−7t + 2 c) The roots are −7 ± 3j. The form is
x(t) = C1 e−7t sin 3t + C2 e−7t cos 3t + C3 At steady state, x = 174/58 = 3 so C3 = 3. Evaluating C1 and C2 for the initial conditions gives x(t) = −7e−7t sin 3t − 3e−7t cos 3t + 3
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3.15 a)
The root is s = −7/5. b)
15 X(s) = F (s) 5s + 7
X(s) 5 = 2 F (s) 3s + 30s + 63
The roots are s = −7 and s = −3. c)
The roots are s = −7 and s = −3. d)
The roots are s = −7 and s = −7. e)
The roots are s = −7 ± 3j. f)
4 X(s) = 2 F (s) s + 10s + 21
7 X(s) = 2 F (s) s + 14s + 49
6s + 4 X(s) = 2 F (s) s + 14s + 58
4s + 15 X(s) = F (s) 5s + 7
The root is s = −7/5.
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3.16 Transform each equation using zero initial conditions. 3sX(s) = Y (s) sY (s) = F (s) − 3Y (s) − 15X(s)
Solve for X(s)/F (s) and Y (s)/F (s).
X(s) 1 = 2 F (s) 3s + 9s + 15 3s Y (s) = 2 F (s) 3s + 9s + 15
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3.17 Transform each equation using zero initial conditions. sX(s) = −2X(s) + 5Y (s) sY (s) = F (s) − 6Y (s) − 4X(s)
Solve for X(s)/F (s) and Y (s)/F (s).
X(s) 5 = 2 F (s) s + 8s + 32 s+2 Y (s) = 2 F (s) s + 8s + 32
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3.18 a)
6 6 1 6 = − s(s + 5) 5s 5 s + 5 x(t) =
b)
4 1 4 1 4 = − s + 3)(s + 8) 5s+3 5s+8 x(t) =
c)
( 6' 1 − e−5t 5
( 4 ' −3t e − e−8t 5
1 27 1 43 1 8s + 5 8s + 5 = =− + 2s2 + 20s + 48 2 (s + 4)(s + 6) 4 s+4 4 s+6 x(t) = −
27 −4t 43 −6t + e e 4 4
d) The roots are s = −4 ± 10j. 4s + 13 4s + 13 + s2 + 8s + 116 (s + 4)2 + 102
x(t) = −
10 s+4 + C2 (s + 4)2 + 102 (s + 4)2 + 102 3 s+4 10 = − +4 10 (s + 4)2 + 102 (s + 4)2 + 102 = C1
3 −4t e sin 10t + 4e−4t cos 10t 10
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3.19 a)
1 1 1 1 1 1 3s + 2 = + − 2 + 10) 5s 50 s 50 s + 10
s2 (s
x(t) = b)
c)
1 1 1 5 − 0.3125 = 1.25 + 0.3125 2 2 (s + 4) (s + 1) (s + 4) s+4 s+1 x(t) = 1.25te−4t − 0.3125e−4t + 0.3125e−t 7 1 9 1 s2 + 3s + 5 5 1 9 1 + + = − 3 3 2 s (s + 2) 4s 16 s 64 s 64 s + 4 x(t) =
d)
( 1 1 ' t− 1 − e−10t 5 50
9 9 5 2 7 t + t+ − e−4t 8 16 64 64
1 1 1 1 11 1 1 s3 + s + 6 =3 4 − 3 + − + 4 2 s (s + 2) s s 2s 4s 4s+2 1 1 1 1 1 x(t) = t3 − t2 + t + − e−2t 2 2 2 4 4
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3.20 a) 5[sX(s) − 2] + 3X(s) = X(s) =
10 2 + 3 s s
10s3 + 10s2 + 2 2s3 + 2s2 + 2/5 2 1 10 1 140 1 86 1 = = − − + 3 3 3 2 5s (s + 3) s (s + 3/5) 3s 9 s 9 s 27 s + 3/5 1 10 140 86 −3t/5 − e x(t) = t2 − t + 3 9 27 27
b) 4[sX(s) − 5] + 7X(s) = X(s) = =
6 1 + 2 (s + 5) s+3
1 20s3 + 261s2 + 1116s + 1525 4 (s + 5)2 (s + 7/4)(s + 3) ) * 1 60 16904 14 1 474 1 1 1 + + − 4 13 (s + 5)2 169 s + 5 845 s + 7/4 5 s+3
x(t) = 1.1538te−5t − 0.7012e−5t + 5.0012e−7t/4 + 0.7e−3t c) This simplelooking problem actually requires quite a lot of algebra to find the solution, and thus it serves as a good motivating example of the convenience of using MATLAB. The algebraic complexity is due to a pair of repeated complex roots. First obtain the transform of the forcing function. Let f (t) = te−3t sin 5t. From Property 8, dY (s) F (s) = − ds where y(t) = e−3t sin 5t. Thus Y (s) =
5 5 = 2 2 2 (s + 3) + 5 s + 6s + 34
dY (s) 10s + 30 =− 2 ds (s + 6s + 34)2 Thus F (s) =
(s2
10s + 30 + 6s + 34)2
(1)
Using the same technique, we find that the transform of te−3t cos 5t is 1 2s2 + 12s + 18 − 2 2 2 (s + 6s + 34) s + 6s + 34
(2)
This fact will be useful in finding the forced response. 322 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
From the differential equation, 4[s2 X(s) − 10s + 2] + 3X(s) = F (s) = Solve for X(s). X(s) =
(s2
10s + 30 + 6s + 34)2
10s + 30 40s − 8 + 2 2 4s + 3 [(s + 3) + 25]2 (4s2 + 3)
The free response is given by the first fraction, and is √ √ 4 3 3 xfree (t) = − √ sin t + 10 cos t 2 2 3
(3)
The forced response is given by the second fraction, which can be expressed as 2.5s + 7.5 + 25]2 (s2 + 3/4)
(4) [(s √ The roots of this are s = ±j 3/2 and the repeated pair s = −3 ± 5j. Thus, referring to (1), (2), and (3), we see that the form of the forced response will be + 3)2
xforced (t) = C1 te−3t sin 5t + C2 te−3t cos 5t + C3 e−3t sin 5t + C4 e−3t cos 5t √ √ 3 3 + C5 sin t + C6 cos t (5) 2 2 The forced response can be obtained several ways. 1) You can substitute the form (5) into the differential equation and use the initial conditions to obtain equations for the Ci coefficients. 2) You can use (1) and (2) to create a partial fraction expansion of (4) in terms of the complex factors. 3) You can perform an expansion in terms of the six roots, of the form A1 (s + 3 + 5j)2
+ +
A3 A2 A4 + + 2 s + 3 + 5j (s + 3 − 5j) s + 3 − 5j √ A6 s 3A5 /2 + 2 2 s + 3/4 s + 3/4
4) You can use the MATLAB residue function. The solution for the forced response is xforced (t) = −0.0034te−3t sin 5t + 0.0066te−3t cos 5t
− 0.0026e−3t sin 5t + 2.308 × 10−4 e−3t cos 5t
+ 0.00796 sin 0.866t − 2.308 × 10−4 cos 0.866t The initial condition x(0) ˙ = 0 is not exactly satisfied by this expression because of the limited number of digits used to display it. 323 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
3.21 a) 7[sX(s) − 3] + 5X(s) = 4 X(s) =
25 25/7 = 7s + 5 s + 5/7
25 −5t/7 e 7 Note that this gives x(0+) = 25/7. From the initial value theorem x(t) =
x(0+) = lim s s→∞
25 25/7 = s + 5/7) 7
which is not the same as x(0). b) (3s2 + 30s + 63)X(s) = 5 5/3 5 1 5 1 5 = 2 = − + 30s + 63 s + 10s + 21 12 s + 3 12 s + 7 ' ( 5 e−3t − e−7t x(t) = 12 From the initial value theorem X(s) =
3s2
x(0+) = lim s s→∞
5/3 =0 s2 + 10s + 21
which is the same as x(0). Also x(0+) ˙ = lim s2 s→∞
s2
5 5/3 = + 10s + 21 3
which is not the same as x(0). ˙ c) s2 X(s) − 2s − 3 + 14[sX(s) − 2] + 49X(s) = 3 X(s) =
s2
1 2s + 34 1 = 20 +2 2 + 14s + 49 (s + 7) s+7 x(t) = 20te−7t + 2e−7t
From the initial value theorem x(0+) = lim s s→∞
2s + 35 =2 s2 + 14s + 49
which is the same as x(0). However, the initial value theorem is invalid for computing x(0+) ˙ and gives an undefined result because the orders of the numerator and denominator of sX(s) are equal (see the bottom of page 146). 324 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
d) s2 X(s) − 4s − 7 + 14[sX(s) − 4] + 58X(s) = 4
X(s) =
s2
4s + 67 4s + 67 3 s+7 = = 13 +4 2 2 2 2 + 14s + 58 (s + 7) + 3 (s + 7) + 3 (s + 7)2 + 32 x(t) = 13e−7t sin 3t + 4e−7t cos 3t
From the initial value theorem x(0+) = lim s s→∞
s2
4s + 67 =4 + 14s + 58
which is the same as x(0). However, the initial value theorem is invalid for computing x(0+) ˙ and gives an undefined result because the order of the numerator of sX(s) is greater than the denominator (see the bottom of page 146).
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3.22 a)
1 =4 s 25 25/7 X(s) = = 7s + 5 s + 5/7 25 −5t/7 x(t) = e 7
7[sX(s) − 3] + 5X(s) = 4s
From the initial value theorem
x(0+) = lim s s→∞
25 25/7 = s + 5/7 7
b)
1 6 7[sX(s) − 3] + 5X(s) = 4s + s s 1 25s + 6 6 1 83 25s + 6 1 = = + X(s) = s(7s + 5) 7 s(s + 5/7) 5 s 35 s + 5/7 6 83 x(t) = + e−5t/7 5 35 which gives x(0+) = 25/7. However, the initial value theorem is invalid for computing x(0+) and gives an undefined result because the orders of the numerator and denominator of X(s) are equal (see the bottom of page 146). c) 1 3[s2 X(s) − 2s − 3] + 30[sX(s) − 2] + 63X(s) = 4s = 4 s 1 6s + 73 55 1 31 1 X(s) = = − 3 (s + 3)(s + 7) 12 s + 3 12 s + 7 55 31 x(t) = e−3t − e−7t 12 12 From the initial value theorem 1 6s + 73 x(0+) = lim s =2 s→∞ 3 (s + 3)(s + 7) which is the same as x(0). However, the initial value theorem is invalid for computing x(0+) ˙ and gives an undefined result because the order of the numerator of sX(s) is greater than the denominator (see the bottom of page 146). d) 1 6 3[s2 X(s) − 4s − 7] + 30[sX(s) − 4] + 63X(s) = 4s + s s 1 12s2 + 145s + 6 1 1 1 X(s) = = 0.0952 + 8.9167 − 5.0119 2 3 s(s + 10s + 21) s s+3 s+7 x(t) = 0.0952 + 8.9167e−3t − 5.0119e−7t
The initial value theorem gives x(0+) = 4 but is invalid for computing x(0+) ˙ because the orders of the numerator and denominator of sX(s) are equal (see the bottom of page 146). 326 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
3.23 Transform each side of the equation. 5 4 6[sX(s) − 2] + X(s) = s s 2 2 3s + 1 = X(s) = 3 s2 + 5/6 3 2 x(t) = 3
+
$+
6 sin 5
,
s 6 5/6 +3 2 5 s2 + 5/6 s + 5/6
+
5 t + 3 cos 6
+
5 t 6
%
.
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3.24 Transform each equation. 3[sX(s) − 5] = Y (s) Solve for X(s) and Y (s).
sY (s) − 10 = 4 − 3Y (s) − 15X(s)
X(s) =
15s + 59 1 15s + 59 = + 9s + 15 3 s2 + 3s + 5
3s2
42s − 225 14s − 75 = 2 3s2 + 9s + 15 s + 3s + 5 The denominator roots are s = −1.5 ± 1.658j. Thus Y (s) =
)
1 1.658 s + 1.5 C1 + C2 3 (s + 1.5)2 + 2.75 (s + 1.5)2 + 2.75 s + 1.5 1.658 +5 = 7.3368 (s + 1.5)2 + 2.75 (s + 1.5)2 + 2.75
X(s) =
*
and x(t) = 7.3368e−1.5t sin 1.658t + 5e−1.5t cos 1.658t Also, 1.658 s + 1.5 + C2 (s + 1.5)2 + 2.75 (s + 1.5)2 + 2.75 1.658 s + 1.5 = −57.901 + 14 2 (s + 1.5) + 2.75 (s + 1.5)2 + 2.75
Y (s) = C1
and y(t) = −57.901e−1.5t sin 1.658t + 14e−1.5t cos 1.658t
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3.25 Transform each equation. sX(s) − 5 = −2X(s) + 5Y (s) sY (s) − 2 = −6Y (s) − 4X(s) + Solve for X(s) and Y (s). X(s) =
10 s
5s2 + 40s + 50 s3 + 8s2 + 32s
2s2 − 6s + 20 s3 + 8s2 + 32s The denominator roots are s = 0 and s = −4 ± 4j. Thus Y (s) =
X(s) = =
C1 4 s+4 + C2 + C3 s (s + 4)2 + 42 (s + 4)2 + 42 55 25 55 4 s+4 + + 2 2 16s 16 (s + 4) + 4 16 (s + 4)2 + 42
x(t) =
25 55 −4t 55 + e sin 4t + e−4t cos 4t 16 16 16
Also, Y (s) = =
C1 4 s+4 + C3 + C2 2 2 s (s + 4) + 4 (s + 4)2 + 42 33 5 11 4 s+4 − + 2 2 8s 8 (s + 4) + 4 8 (s + 4)2 + 42
y(t) =
5 33 −4t 11 − e sin 4t + e−4t cos 4t 8 8 8
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3.26 For zero initial conditions, the transform gives (s2 + 4)X(s) = or X(s) =
3 s2
3 C1 C2 2 s = 2 + + C4 2 + C3 2 s2 (s2 + 4) s s s +4 s +4
The solution form is thus x(t) = C1 t + C2 + C3 sin 2t + C4 cos 2t This form satisfies the differential equation if C1 = 3/4 and C2 = 0. From x(0) = 10, we obtain C4 = 10. From x(5) = 30, we obtain C3 = −63.675. Thus 3 x(t) = t − 63.675 sin 2t + 10 cos 2t 4
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3.27 Transform the equation. (s2 + 12s + 40)X(s) = 3
s2
5 + 25
The characteristic roots are s = −6 ± 2j. Thus 15 (s2 + 25)(s2 + 12s + 40) 5 s 2 s+6 + C2 2 + C3 + C4 = C1 2 s + 25 s + 25 (s + 6)2 + 4 (s + 6)2 + 4
X(s) =
or X(s) =
1 4 19 4 5 s 2 s+6 − + + 2 2 2 85 s + 25 85 s + 25 170 (s + 6) + 4 85 (s + 6)2 + 4
Thus x(t) =
4 19 −6t 4 1 sin 5t − cos 5t + e sin 2t + e−6t cos 2t 85 85 170 85
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3.28 From Example 3.7.6, the form A sin(ωt + φ) has the transform A
s sin φ + ω cos φ s2 + ω 2
For this problem, ω = 5. Comparing numerators gives s sin φ + 5 cos φ = 4s + 9 Thus A sin φ = 4
5A cos φ = 9
With A > 0, φ is seen to be in the first quadrant. φ = tan−1
sin φ 4/A 20 = tan−1 = tan−1 = 1.148 rad cos φ 9/5A 9
Because sin2 φ + cos2 φ = 1,
which gives A = 4.386. Thus
"
4 A
#2
+
"
9 5A
#2
=1
x(t) = 4.386 sin(5t + 1.148)
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3.29 Transform the equation. (s2 + 12s + 40)X(s) =
10 s
or, since the characteristic roots are s = −6 ± 2j, X(s) =
10 s[(s + 6)2 + 22 ]
(1)
From Example 3.7.6(b), the form Ae−at sin(ωt + φ) has the transform A
s sin φ + a sin φ + ω cos φ (s + a)2 + ω 2
For this problem, a = 6 and ω = 2. Thus X(s) = or X(s) =
10 C1 s sin φ + 6 sin φ + 2 cos φ = + C2 2 2 s[(s + 6) + 2 ] s (s + 6)2 + 22
C1 (s2 + 12s + 40) + C2 s2 sin φ + 6C2 s sin φ + 2C2 s cos φ s[(s + 6)2 + 22 ]
(2)
Collecting terms and comparing the numerators of equations (1) and (2), we have (C1 + C2 sin φ)s2 + (12C1 + 6C2 sin φ + 2C2 cos φ)s + 40C1 = 10 Thus comparing terms, we see that C1 = 1/4 and 1 + C2 sin φ = 0 4 3 + 6C2 sin φ + 2C2 cos φ = 0 So C2 sin φ = −
1 4
C2 cos φ = −
3 4
Thus φ is in the third quadrant and −1/4 = 0.322 + π = 3.463 rad −3/4
φ = tan−1 Because sin2 φ + cos2 φ = 1,
which gives C2 = 0.791. Thus
"
x(t) =
1 4C2
#2
+
"
3 4C2
#2
=1
1 + 0.791e−6t sin(2t + 3.463) 4
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3.30 Transform the equation. X(s) = Thus F (s) − X(s) = F (s) −
s2
F (s) + 8s + 1
F (s) s2 + 8s = F (s) s2 + 8s + 1 s2 + 8s + 1
Because F (s) = 6/s2 , F (s) − X(s) =
s2 + 8s 6 s+8 6 = 2 2 2 s + 8s + 1 s s + 8s + 1 s
From the final value theorem, fss − xss = lim s[F (s) − X(s)] = lim s s→0
s→0
s2
s+8 6 =8 + 8s + 1 s
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3.31 The roots are s = −2 and −4. Thus X(s) = Let
1 − e−3s (s + 2)(s + 4)
1 1 F (s) = = (s + 2)(s + 4) 2
so f (t) = From Property 6, x(t) =
"
1 1 − s+2 s+4
#
( 1 ' −2t e − e−4t 2
( 1/ 0 1 ' −2t e − e−4t − e−2(t−3) − e−4(t−3) us (t − 3) 2 2
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3.32 f (t) =
C 2C C tus (t) − (t − D)us (t − D) + (t − 2D)us (t − 2D) D D D
From Property 6, F (s) =
( C 2C −Ds C −2Ds C ' −Ds −2Ds − e + e = 1 − 2e + e Ds2 Ds2 Ds2 Ds2
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3.33 f (t) =
C C tus (t) − (t − D)us (t − D) − Cus (t − D) D D
From Property 6, F (s) =
C C −Ds C −Ds − e − e 2 Ds Ds2 s
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3.34 From Property 6,
f (t) = M us (t) − 2M us (t − T ) + M us (t − 2T ) F (s) =
2M −T s M −2T s M − e + e s s s
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3.35 From Property 6,
P (t) = 3us (t) − 3us (t − 5) P (s) = 1
3 3 −5s − e s s 2
3 1 − e−5s 3 1 − e−5s P (s) = = X(s) = 4s + 1 s(4s + 1) 4 s(s + 1/4) Let
Then
"
3 1 1 1 F (s) = =3 − 4 s(s + 1/4) s s + 1/4 '
f (t) = 3 1 − e−t/4 Since
'
(
X(s) = F (s) 1 − e−5s we have '
(
#
( /
0
x(t) = f (t) − f (t − 5)us (t − 5) = 3 1 − e−t/4 − 3 1 − e−(t−5)/4 us (t − 5)
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3.36 f (t) =
7 [tus (t) − (t − 10)us (t − 10) − (t − 30)us (t − 30)] 5
From Property 6, F (s) = Thus X(s) = Let
0 7 / −10s −30s −40s 1 − e − e + e 5s2
/ 0 7 F (s) −10s −30s −40s = 1 − e − e + e 2s + 1 10s2 (s + 1/2)
Y (s) = Then Y (s) =
s2 (s
1 + 1/2)
4 4 2 − + 2 s s s + 1/2
and Thus, from Property 6, x(t) =
y(t) = 2t − 4 + 4e−t/2
7 [y(t) − y(t − 10)us (t − 10) − y(t − 30)us (t − 30) + y(t − 40)us (t − 40)] 10
or x(t) = − +
/ 0 4 7 3 2t − 4 + 4e−t/2 − 2(t − 10) − 4 + 4e−(t−10)/2 us (t − 10) 10 0 4 7 3/ 2(t − 30) − 4 + 4e−(t−30)/2 us (t − 30) 10 0 4 7 3/ 2(t − 40) − 4 + 4e−(t−40)/2 us (t − 40) 10
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3.37 Let f (t) = t + Then F (s) =
t3 2t5 + 3 15
1 2 16 s4 + 2s2 + 16 + + = s2 s4 s6 s6
From the differential equation, X(s) = =
F (s) s4 + 2s2 + 16 = s+1 s6 (s + 1) 16 16 18 18 19 19 19 + − 5 + 4 − 3 + 2 − 6 s s s s s s s+1
Thus
2 5 2 4 t − t + 3t3 − 9t2 + 19t − 19 + 19e−t 15 3 On a plot of this and the solution obtained from the lowerorder approximation, the two solutions are practically indistinguishable. x(t) =
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3.38 a) [r,p,k] = residue([8,5],[2,20,48]) The result is r = [10.7500, 6.7500], p = [6.0000, 4.0000], and k = [ ]. The solution is x(t) = 10.75e−6t − 6.75e−4t b)
[r,p,k] = residue([4,13],[2,8,116]) The result is r = [1.0000  0.1701i, 1.0000 + 0.1701i], p = [2.0000 + 7.3485i, 2.0000  7.3485i], and k = [ ]. The solution is x(t) = (1 − 0.1701j)e(−2+7.3485j)t + (1 + 0.1701j)e(−2−7.3485j)t From (3.8.4) and (3.8.5), the solution is x(t) = 2e−2t (cos 7.3485t + 0.1701 sin 7.3485t) c) [r,p,k] = residue([3,2],[1,10,0,0]) The result is r = [ 0.2800, 0.2800, 0.2000], p = [10, 0, 0], and k = [ ]. The solution is x(t) = −0.28e−10t + 0.28 + 0.2t d)
[r,p,k] = residue([1,0,1,6],[1,2,0,0,0,0]) The result is r = [0.2500, 0.2500, 0.5000, 1.0000, 3.0000], p =[ 2, 0, 0, 0, 0], and k = [ ]. The solution is 1 1 x(t) = −0.25e−2t + 0.25 + 0.5t − t2 + t3 2 2
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e) [r,p,k] = residue([4,3],[1,6,34,0]) The result is r = [0.0441  0.3735i, 0.0441 + 0.3735i, 0.0882], p = [3.0000 + 5.0000i, 3.0000  5.0000i, 0], and k = [ ].The solution is x(t) = (−0.0441 − 0.3735j)e(−3+5j)t + (−0.0441 + 0.3735j)e(−3−5j)t + 0.0882 From (3.8.4) and (3.8.5), the solution is x(t) = 2e−3t (−0.0441 cos 5t + 0.3735 sin 5t) + 0.0882 f) [r,p,k] = residue([5,3,7],[1,12,144,48]) The result is r = [2.4759 + 1.3655i, 2.4759  1.3655i, 0.0482], p = 5.8286 +10.2971i, 5.8286 10.2971i, 0.3428], and k = [ ]. The solution is x(t) = (2.4759+1.3655j)e(−5.8286+10.2971j)t +(2.4759−1.3655j)e(−5.8286−10.2971j)t +0.0482e−0.3428t From (3.8.4) and (3.8.5), the solution is x(t) = 2e−5.8286t (2.4759 cos 10.29t − 1.3655 sin 10.29t) + 0.0482e−0.3428t
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3.39 a) [r,p,k] = residue(5,conv([1,8,16],[1,1])) The result is r = [0.5556, 1.6667, 0.5556], p = [4.0000, 4.0000, 1.0000], k = [ ]. The solution is x(t) = −0.5556e−4t − 1.6667te−4t + 0.5556e−t b) [r,p,k] = residue([4,9],conv([1,6,34],[1,4,20])) The result is r = [0.1159 + 0.1073i, 0.1159  0.1073i, 0.1159  0.1052i, 0.1159 + 0.1052i], p = 3.0000 + 5.0000i, 3.0000  5.0000i, 2.0000 + 4.0000i, 2.0000  4.0000i], and k = [ ]. The solution is x(t) = (−0.1159 + 0.1073j)e(−3+5j)t + (−0.1159 − 0.1073j)e(−3−5j)t + (0.1159 − 0.1052j)e(−2+4j)t + (0.1159 + 0.1052j)e(−2−4j)t
From (3.8.4) and (3.8.5), the solution is x(t) = 2e−3t (−0.1159 cos 5t − 0.1073 sin 5t) + 2e−2t (0.1159 cos 4t + 0.1052 sin 4t)
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3.40 a) sys = tf(1,[3,21,30]); step(sys) b) sys = tf(1,[5,20, 65]); step(sys) c) sys = tf([3,2],[4,32,60]); step(sys)
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3.41 a) sys = tf(1,[3,21,30]); impulse(sys) b) sys = tf(1,[5,20, 65]); impulse(sys)
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3.42 sys = tf(5,[3,21,30]); impulse(sys)
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3.43 sys = tf(5,[3,21,30]); step(sys)
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3.44 a) sys = tf(1,[3,21,30]); t = [0:0.001:1.5]; f = 5*t; [x,t] = lsim(sys,f,t); plot(t,x) b) sys = tf(1,[5,20,65]); t = [0:0.001:1.5]; f = 5*t; [x,t] = lsim(sys,f,t); plot(t,x) c) sys = tf([3,2],[4,32,60]); t = [0:0.001:1.5]; f = 5*t; [x,t] = lsim(sys,f,t); plot(t,x)
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3.45 a) sys = tf(1,[3,21,30]); t = [0:0.001:6]; f = 6*cos(3*t); [x,t] = lsim(sys,f,t); plot(t,x) b) sys = tf(1,[5,20,65]); t = [0:0.001:6]; f = 6*cos(3*t); [x,t] = lsim(sys,f,t); plot(t,x) c) sys = tf([3,2],[4,32,60]); t = [0:0.001:6]; f = 6*cos(3*t); [x,t] = lsim(sys,f,t); plot(t,x)
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3.46 The following is a modification of the code on pages 160161. C = 5; D1 = 2; D2 = 4; D3 = 6; D4 = 8; m1 = C/D1; m2 = C/(D3  D2); dt = 0.01; t1 = [0:dt:D1dt]; t2 = [D1:dt:D2dt]; t3 = [D2:dt:D3dt]; t4 = [D3:dt:D4]; t = [t1, t2, t3, t4]; f1 = m1*t1; f2 = m1*t2  m1*(t2D1); f3 = m1*t3m1*(t3D1)m2*(t3D2); f4 = m1*t4m1*(t4D1)m2*(t4D2)+m2*(t4D3); f = [f1,f2,f3,f4]; sys = tf(1, [3,21,30]); [x,t] = lsim(sys,f,t); plot(t,x)
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Four
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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4.1 For a helical coil spring k=
Gd4 (1/24)4 9 = 1.7 × 10 = 2.88 × 103 lb/ft 64nR3 64(6)(2/12)3
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4.2 Let D be the length of the dead space to the left of spring k2 . The plot follows.
Figure : for Problem 4.2
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4.3 a) Series. b) Let kb be the stiffness of the beam and kc be the stiffness of the cable. ke =
kb kc kb + kc
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4.4 Sum moments about the pivot. 0 = f L1 − k1 (L2 θ)L2 − k2 (L3 θ)L3 Thus f= But x = L1 θ, so f=
k1 L22 + k2 L23 θ L1
k1 L22 + k2 L23 x = ke x L21
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4.5 Series combination. ke = where
k1 k2 k1 + k2 !
EAi E Di = π ki = Li Li 2
"2
=
EπDi2 4Li
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4.6
!
EA ke = 4 L
"
=
4(2 × 1011 )π(0.03)2 4Eπ(d/2)2 = = 1.8π × 108 N/m L 4(1)
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4.7 The beam stiffness is kb =
4Ewh3 4(3 × 107 )(144)(1)(1/12)3 = = 3.7037 × 105 lb/ft L3 (3)3
a) The springs are in parallel, so ke = kb + k. We want ke = 2kb , so we must require that k = kb . b) # # ke 2(3.7037 × 105 ) = = 136.1 rad/sec ωn = m 40
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4.8 First reduce the system to the equivalent one shown in part (a) of the figure, where 1 1 3 1 = + = k1 2k k 2k Thus k1 = 2k/3. From part (b) of the figure, 1 k1 + 2k 1 1 = = + ke k k1 + k k(k1 + k) Thus, solving for ke and substituting for k1 , we obtain ke =
5k k(k1 + k) = k1 + 2k 8
Figure : for Problem 4.8
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4.9 The stiffnesses are in series. Thus kT e = where kT 1 = kT 2 =
kT 1 kT 2 kT1 + kT2
πG(D 4 − d4 ) π(8 × 1010 )(0.44 − 0.34 ) = = 6.874 × 107 32L 32(2)
π(8 × 1010 )(0.354 − 0.254 ) πG(D 4 − d4 ) = = 2.906 × 107 32L 32(3)
Thus kTe = 2.043 × 107 N · m/rad
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4.10 The plot is the following.
Figure : for Problem 4.10
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4.11 Since x1 and x2 are measured from the equilibrium positions, the gravity forces are canceled by the static spring forces in parts (a) and (b). Thus the equations of motion are the same for all three cases. They are ¨1 = −k1 x1 + k2 (x2 − x1 ) m1 x m2 x ¨2 = f − k2 (x2 − x1 )
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4.12 Sum moments about the pivot to obtain mL23 θ¨ = −(k1 L1 θ)L1 + k2 (x − L2 θ)L2 − mgL3 θ Collecting terms we obtain mL23 θ¨ + (k1 L21 + k2 L22 + mgL3 )θ = k2 L2 x
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4.13 Let T be the tension in the cable. See the figure. Since the pulley is considered massless, kx 2T = 2 For the mass m, kx + mg m¨ x = f − T + mg = f − 4 Thus kx m¨ x+ = f + mg 4 Note that the static spring force does not cancel the weight mg because x is not measured from the equilibrium position.
Figure : for Problem 4.13
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4.14 a) The buoyancy force is !
D B = (ρV )g = ρgπ 2 or
"2
x
πρgD 2 x 4 b) Summing forces in the vertical direction, B=
m¨ x = −B or m¨ x=−
πρgD 2 x 4
Thus, the natural frequency is ωn =
#
D πρgD 2 = 4m 2
c) The period is 2π 4π P = = ωn D or
4π P = 32.2(2)
$
$
$
πρg m
4π m = πρg gD
#
W πρ
1000 = 2.5 sec 1.94π
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4.15 a) Sum moments about the mass center G to obtain I θ¨ = −(B sin θ)h Because B = mg = W , the ship’s weight, I θ¨ + W h sin θ = 0 where I is the ship’s moment of inertia about G. b) For small angles, sin θ ≈ θ, and the equation becomes I θ¨ + W hθ = 0 %
%
The characteristic roots are s = ±j W h/I, and the roll period is given by 2π I/W h.
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4.16 Sum moments about the rotation axis. I θ¨ = k1 (φ − θ) − k2 θ Collecting terms we obtain The transfer function is
I θ¨ + (k1 + k2 )θ = k1 φ k1 Θ(s) = 2 Φ(s) Is + k1 + k2
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4.17 a) Refer to the figure on the following page. Summing forces in the x direction gives m¨ x = k(y − x) − ft
(1)
Summing moments about the mass center of the wheel gives I θ¨ = Rft But x = Rθ, and thus θ¨ = x ¨/R. Therefore ft =
I¨ I ¨ θ = 2x R R
(2)
Combine (1) and (2): m¨ x + kx = ky −
or
!
I x ¨ R2
"
I m+ 2 x ¨ + kx = ky R
where I = mR2 /2. Thus 1.5m¨ x + kx = ky
(3)
b) Substituting the values into (3) gives 15¨ x + 1000x = 1000y or 3¨ x + 200x = 200y %
The roots are s = ±j10 2/3. The response to a unitstep input is x(t) = 1 − cos 10
$
2 t 3
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Figure : for Problem 4.17
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4.18 Summing forces in the horizontal direction on each mass gives m1 x ¨1 = −k1 x1 + k2 (x2 − x1 ) m2 x ¨2 = −k3 x2 − k2 (x2 − x1 )
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4.19 Summing moments about the axis of rotation of each inertia gives I1 θ¨1 = −k1 θ1 − k2 (θ1 − θ2 ) I2 θ¨2 = T2 + k2 (θ1 − θ2 ) − k3 θ2
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4.20 Summing moments about each pivot point and assuming small angles, we obtain m1 L22 θ¨1 = −m1 gL2 θ1 − kL1 (θ1 − θ2 )L1 Collecting terms gives
m2 L22 θ¨2 = −m2 gL2 θ2 + kL1 (θ1 − θ2 )L1 m1 L22 θ¨1 + (m1 gL2 + kL21 )θ1 = kL21 θ2 m2 L22 θ¨2 + (m2 gL2 + kL21 )θ2 = kL21 θ1
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4.21 The torsional stiffness of the torsion bar is kT =
1.7 × 109 π(1.5/12)4 πGD 4 = = 104 lb − ft/rad 32L 32(4)
The characteristic equation is Is2 + kT = 0 where I = mL2 = (40/32.2)(2)2 = 4.9689. Thus the natural frequency is ωn =
#
kT = 44.86 rad/sec I
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4.22 The kinetic energy of the system is 1 1 1 1 KE = mx˙ 2 + Iω 2 = mx˙ 2 + I 2 2 2 2
!
x˙ R
"2
=
1 (1.5m)x˙ 2 2
since x˙ = Rω and I = mR2 /2. The potential energy is PE =
1 (k1 + k2 )x2 2
From conservation of mechanical energy, 1 1 KE + P E = (1.5m)x˙ 2 + (k1 + k2 )x2 = constant 2 2 Differentiating with respect to time gives 1.5mx¨ ˙ x + (k1 + k2 )xx˙ = 0 Canceling x˙ gives the equation of motion. 1.5m¨ x + (k1 + k2 )x = 0
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4.23 Because x = 0 at equilibrium, the static spring force cancels the constant weight m2 g. Taking the potential energy to be zero at x = 0, we obtain 1 P E = kx2 2 The kinetic energy of the system is KE =
1 1 1 m1 x˙ 2 + m2 y˙ 2 = (m1 + 4m2 )x˙ 2 2 2 2
since y = 2x. From conservation of energy, KE + P E = constant or
1 1 (m1 + 4m2 )x˙ 2 + kx2 = constant 2 2 Differentiating with respect to time and setting the derivative to 0, we obtain (m1 + 4m2 )x¨ ˙ x + kxx˙ = 0 Canceling x˙ gives the equation of motion. (m1 + 4m2 )¨ x + kx = 0
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4.24 Because x = 0 at equilibrium, the static spring force cancels the constant weight m1 g. Taking the potential energy to be zero at x = 0, and noting that the spring extension y from its equilibrium position is y = 2x, we obtain 1 P E = k(2x)2 = 2kx2 2 Note that y˙ = Rω = 2x˙ and the inertia of the cylinder is I = m2 R2 /2. The kinetic energy of the system is KE = = = =
1 1 1 m2 y˙ 2 + Iω 2 + m1 x˙ 2 2! 2 2 " 1 I 4m2 + 4 2 + m1 x˙ 2 2 R 1 (4m2 + 2m2 + m1 ) x˙ 2 2 1 (6m2 + m1 ) x˙ 2 2
From conservation of energy, KE + P E = constant or
1 (6m2 + m1 ) x˙ 2 + 2kx2 = constant 2 Differentiating with respect to time and setting the derivative to 0, we obtain (6m2 + m1 )x¨ ˙ x + 4kxx˙ = 0 Canceling x˙ gives the equation of motion. (6m2 + m1 )¨ x + 4kx = 0
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4.25 Because x = 0 at equilibrium, the static spring force cancels the constant weight mg. Taking the potential energy to be zero at x = 0, and noting that the spring extension y is y = x/2, we obtain 1 1 P E = ky 2 = kx2 2 8 Note that x˙ = Rω, where ω is the angular velocity of the pulley, and that the inertia of the cylinder is I = mR2 /2. The kinetic energy of the system is KE = = = =
1 1 mx˙ 2 + Iω 2 2 2 1 1 mR2 x˙ 2 mx˙ 2 + 2! 2 " 2 R2 1 m m+ x˙ 2 2 2 1 (1.5m)x˙ 2 2
Let A = xmax . Then x ˙ = ωn A for simple harmonic motion, and Rayleigh’s principle gives 1 1 (1.5m)ωn2 A2 = kA2 2 8 This gives # k ωn = 6m
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4.26 The kinetic energy of the system is 1 1 1 1 T = mx˙ 2 + Iω 2 = mx˙ 2 + I 2 2 2 2
!
x˙ R
"2
1 = (1.5m)x˙ 2 2
since x˙ = Rω and I = mR2 /2. The potential energy is 1 V = kx2 2 From the Rayleigh principle, Tmax = Vmax − Vmin . For simple harmonic motion, x˙ = ωn A, and this implies that 1 1 (1.5m)ωn2 A2 = kA2 2 2 Thus # k ωn = 1.5m
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4.27 The kinetic energy of the system is 1 1 L2 T = mv 2 = m 12 x˙ 2 2 2 L2 ˙ since v = L1 θ˙ and x˙ = L2 θ. The potential energy is From the Rayleigh principle, Tmax and this implies that
1 V = kx2 2 = Vmax − Vmin . For simple harmonic motion, x˙ = ωn A, 1 L21 2 2 1 2 m ω A = kA 2 L22 n 2
Thus L2 ωn = L1
#
k m
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4.28 See the figure. Assuming that θ is small, summing moments about the pivot gives IO θ¨ = fc L1 − k2 L2 x = fc L1 − k2 L2 (L2 θ) The equivalent mass at x due to the valve and spring is 1 m e = mv + ms 3 Since IO = Ir + me L22 , we have (Ir + me L22 )θ¨ = fc L1 − k2 L22 θ Thus the force on the cam is given by fc =
(Ir + me L22 )θ¨ + k2 L22 θ L1
¨ to use Knowing the cam profile and the cam rotation speed, we can compute θ(t) and θ(t) in the expression for fc .
Figure : for Problem 4.28
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4.29 The equivalent mass is me = 12 + 0.23(3) = 12.69 kg. From statics, k(0.02) = 12(9.81) or k = 5886 N/m.
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4.30 The equivalent mass is me = 3 + 0.38(1) = 3.38 slug. From statics, k(0.01) = 3(32.2) or k = 9660 lb/ft.
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4.31 The equation given in the problem statement is xy =
P y 2 (3L − y) 6EIA
(1)
The deflection xL at the end of the beam, where y = L, is found from (1) to be: xL =
P L3 3EIA
(2)
Comparing (1) and (2) shows that xy =
y 2 (3L − y) xL 2L3
(3)
Differentiate this equation with respect to time for a fixed value of y to obtain x˙ y =
y 2 (3L − y) x˙ L 2L3
(4)
Because (1) describes the static deflection, it does not account for inertia effects, and (4) is not exactly true. However, lacking another reasonable expression for x˙ y , we will use (4). The kinetic energy of a beam mass element dm at position y is x˙ 2y dm/2. Let ν be the beam’s mass per unit length. Then dm = ν dy, and the total kinetic energy KE in the beam is KE = = =
1 2
&
L 0
x˙ 2y dm &
ν 2 L 4 x˙ y (3L − y)2 dy 8L6 L 0 33νL 2 ν 2 33L7 = x˙ x˙ L 6 8L 35 2(140) L
Because the beam mass mb = νL, KE =
33mb 2 x˙ 2(140) L
If a mass me is located at the end of the beam, and moves with a velocity x˙ L , its kinetic energy is me x˙ 2L /2. Comparing this with the beam’s energy, we see that me = (33/140)mb ≈ 0.23mb . Thus the equivalent mass of the cantilever spring is 23% of the beam mass.
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4.32 a) See part (a) of the figure. The equivalent mass of the beam and winch is me = 0.23mb + mw . Assuming that x1 and x2 are measured from the equilibrium positions, the equation of motion for me is me x ¨1 = T − kx1 (1)
where T is the tension in the rope and the beam stiffness is k = Ewh3 /4L3 . For the hoisted mass, mh x ¨2 = −T (2) Solve this for T and substitute into (1) to obtain
me x ¨1 + mh x ¨2 = −kx1
(3)
If the rope does not stretch, then x1 = x2 and (3) becomes (me + mh )¨ x1 = −kx1 b) If the rope acts like a spring, then T = kr (x2 − x1 ), and we obtain the freebody diagrams shown in part (b) of the figure. For me , me x ¨1 = kr (x2 − x1 ) − kx1 For mh ,
mh x ¨2 = −kr (x2 − x1 )
These are the equations of motion.
Figure : for Problem 4.32
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4.33 a) Summing forces parallel to the plane, we obtain mv˙ = mg sin 30◦ − cv or 6v˙ + v = 29.43 b) v(t) = 29.43 − 25.43e−t/6
c) The steadystate speed is 29.43 m/s, and it takes approximately 4τ = 4(6) = 24 s to reach that speed. d) Because the steadystate speed can be found directly from the equation of motion, by setting v˙ = 0, it is independent of the initial speed.
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4.34 a) The characteristic equation is 40s2 + 680s + 1200 = 0, which has the roots s = −2 and s = −15. The form of the solution is x(t) = A1 e−2t + A2 e−15t + 6000/1200 For zero initial conditions, A1 = −75/13 and A2 = 10/13, so x(t) = −
75 −2t 10 −15t e + e +5 13 13
2 b) The √ characteristic equation is 40s + 400s + 1200 = 0, which has the roots s = −5 ± j 5. The form of the solution is √ √ x(t) = A1 e−5t sin 5t + A2 e−5t cos 5t + 6000/1200 √ For zero initial conditions, A1 = −5 5 and A2 = −5, so √ √ √ x(t) = −5 5e−5t sin 5t − 5e−5t cos 5t + 5
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4.35 a) Summing the horizontal forces on each mass gives m1 x ¨1 = −cx˙ 1 + k(x2 − x1 ) m2 x ¨2 = f − k(x2 − x1 )
b) Summing the horizontal forces on each mass gives
m1 x ¨1 = −kx1 + c(x˙ 2 − x˙ 1 ) m2 x ¨2 = f − c(x˙ 2 − x˙ 1 )
c) Summing the horizontal forces on each mass gives
m1 x ¨1 = −cx˙ 1 − k1 x1 + k2 (x2 − x1 ) m2 x ¨2 = f − k2 (x2 − x1 )
d) Summing the horizontal forces on each mass gives
m1 x ¨1 = −cx˙ 1 − k1 x1 + k2 (x2 − x1 ) + c2 (x˙ 2 − x˙ 1 ) m2 x ¨1 = f − k2 (x2 − x1 ) − c2 (x˙ 2 − x˙ 1 )
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4.36 a) See the diagram. Let F be the tension in the cable. Summing moments about the drum center gives I ω˙ = T − cT ω − F R (1) Summing vertical forces on the mass m gives
mv˙ = F − mg
(2)
Solve (2) for F , substitute for F in (1), and use the fact that v = Rω to obtain (I + mR2 )v˙ + cT v = T R − mgR2 b) Using the given values, we obtain 24v˙ + v = 243.3 The solution has the form v(t) = 243.3 + Ae−t/24 For v(0) = 0, A = −243.3, and
'
v(t) = 243.3 1 − e−t/24
(
Figure : for Problem 4.36
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4.37 Summing moments about the pivot, and assuming small, angles, we obtain '
(
I θ¨ = f L1 − c L2 θ˙ L2 − k (L3 θ) L3 or
I θ¨ + cL22 θ˙ + kL23 θ = f L1
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4.38 Summing horizontal forces gives m¨ x = −k2 x + k1 (y − x) + c(y˙ − x) ˙ Collect terms to obtain m¨ x + cx˙ + (k1 + k2 )x = k1 y + cy˙ The transfer function is
cs + k1 X(s) = 2 Y (s) ms + cs + k1 + k2
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4.39 Summing moments about the rotation axis of the pulley gives '
Ip θ¨p = kT (φ − θp ) − cT θ˙p − θ˙d
(
Summing moments about the center of mass of the damper gives '
Id θ¨d = cT θ˙p − θ˙d These two equations form the system model.
(
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4.40 Summing moments about the pivot, and assuming small, angles, we obtain '
(
IO θ¨ = −mgL3 θ − c L2 θ˙ L2 + k (z − L1 θ) L1 where IO = mL23 + I. Collecting terms gives
(mL23 + I)θ¨ + cL22 θ˙ + (kL21 + mgL3 )θ = kL1 z
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4.41 Summing horizontal forces on m gives m¨ x = f + k(xA − x) From statics, at point A, 0 = −k(xA − x) − cx˙ A
These can be combined into a single equation by using the Laplace transform. ms2 X(s) = F (s) + k [XA (s) − X(s)] 0 = −k [XA (s) − X(s)]) − csXA (s)
These can be solved for the transfer function relating the given output xA to the given input f. XA (s) k = 2 F (s) s(mcs + mks + ck) This is the required model.
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4.42 Summing horizontal forces on m gives m¨ x = c(x˙ A − x) ˙ − k2 x From statics, at point A, 0 = k1 (y − xA ) − c(x˙ A − x) ˙
These can be combined into a single equation by using the Laplace transform. ms2 X(s) = csXA (s) − csX(s) − k2 X(s) 0 = k1 [Y (s) − XA (s)]) − csXA (s) + csX(s)
These can be solved for the transfer function relating the given output x to the given input y. X(s) ck1 s = 3 2 Y (s) mcs + mk1 s + c(k1 + k2 )s + k1 k2 This is the required model.
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4.43 See the diagram. Let F be the force on the pinion due to the rack. Summing moments on the pinion gives (Im + Ip )θ¨ = T − RF (1) Summing horizontal forces on the rack gives
mr x ¨ = F − cx˙ − kx
(2)
Solving (2) for F , substituting in (1), and using the fact that x = Rθ, gives (Im + Ip + mr R2 )θ¨ + cR2 θ˙ + kR2 θ = T
Figure : for Problem 4.43
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4.44 We can represent the system as shown in the diagram, where I1e = N 2 I1
T1e = N T1
Summing moments on I1e gives I1e θ¨2 = T1e + kT (θ3 − θ2 )
(1)
Summing moments on I2 gives I2 θ¨3 = −kT (θ3 − θ2 ) − cT θ˙3
(2)
To obtain the model in terms of θ1 , substitute θ2 = θ1 /N into (1) and (2) to obtain I1e ¨ θ1 = T1e + kT N and
!
!
θ3 −
θ1 I2 θ¨3 = −kT θ3 − N
The system model consists of (3) and (4).
"
θ1 N
"
− cT θ˙3
(3)
(4)
Figure : for Problem 4.44
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4.45 a) Assume that the equilibrium position corresponds to θ = 0. Let φ = 45◦ + θ and δ be the static spring deflection at equilibrium. Summing moments about the pivot and assuming small angles, we obtain mL2 θ¨ = (mg sin φ)L − 2k(δ + L1 θ)L1 But
√ √ √ √ 2 2 2 2 sin φ = sin(45 + θ) = cos θ + sin θ ≈ + θ 2 2 2 2 for small angles. Thus )√ √ * 2 2 2¨ mL θ = mgL + θ − 2kL1 δ − 2k1 L21 θ 2 2 ◦
At equilibrium,
√
2 − 2kL1 δ = 0 2 and thus the equation of motion becomes ) * √ 2 2¨ 2 mL θ + mgL − 2k1 L1 θ = 0 2 mgL
The characteristic equation is mL s + mgL 2 2
√
2 − 2k1 L21 = 0 2
This will be neutrally stable if mgL and unstable if mgL
√
2 − 2k1 L21 ≥ 0 2
√
2 − 2k1 L21 < 0 2
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b) Assume that the equilibrium position corresponds to θ = 0. Let φ = 135◦ + θ and δ be the static spring deflection at equilibrium. Summing moments about the pivot and assuming small angles, we obtain mL2 θ¨ = (mg sin φ)L − 2k(δ + L1 θ)L1 But
√ √ √ √ 2 2 2 2 sin φ = sin(135 + θ) = cos θ − sin θ ≈ − θ 2 2 2 2 for small angles. Thus )√ √ * 2 2 2¨ mL θ = mgL − θ − 2kL1 δ − 2k1 L21 θ 2 2 ◦
At equilibrium,
√
2 − 2kL1 δ = 0 2 and thus the equation of motion becomes ) * √ 2 2¨ 2 mL θ + mgL + 2k1 L1 θ = 0 2 mgL
The characteristic equation is mL s + mgL 2 2
√
2 + 2k1 L21 = 0 2
This will always be neutrally stable since √ 2 + 2k1 L21 > 0 mgL 2
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4.46 Assume that the equilibrium position corresponds to θ = 0. Let φ = 135◦ + θ and δ be the static spring deflection at equilibrium. Let L be the distance from the pivot to the mass m. Let L1 be the distance from the pivot to the connection point of the spring and damper. Summing moments about the pivot and assuming small angles, we obtain ˙ 1 mL2 θ¨ = (mg sin φ)L − k(δ + L1 θ)L1 − c(L1 θ)L But
√ √ √ √ 2 2 2 2 cos θ − sin θ ≈ − θ sin φ = sin(135 + θ) = 2 2 2 2 for small angles. Thus )√ √ * 2 2 2¨ mL θ = mgL − θ − kL1 δ − k1 L21 θ − cL21 θ˙ 2 2 ◦
At equilibrium,
√
2 − kL1 δ = 0 2 and thus the equation of motion becomes ) * √ 2 + k1 L21 θ = 0 mL2 θ¨ + cL21 θ˙ + mgL 2 mgL
The characteristic equation is mL s + 2 2
cL21 s
+ mgL
√
2 + k1 L21 = 0 2
This will always be stable since mL2 > 0 cL21 > 0 and mgL
√
2 + k1 L21 > 0 2
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4.47 Let I be the inertia of the pendulum about the pivot point. Then summing moments about the pivot point, and assuming small angles, we give ˙ 2 I θ¨ = −k1 (L1 θ)L1 + k2 (y − L2 θ)L2 + c(y˙ − L2 θ)L Collecting terms we obtain I θ¨ + cL22 θ˙ + (k1 L21 + k2 L22 )θ = k2 L2 y + cL2 y˙
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4.48 Assuming that x1 , x2 , and x3 are measured from equilibrium, summing vertical forces on m1 gives ¨1 = k1 (x2 − x1 ) + c1 (x˙ 2 − x˙ 1 ) − k3 (x1 − x3 ) − c3 (x˙ 1 − x˙ 3 ) m1 x Summing vertical forces on m2 gives m2 x ¨2 = k2 (y − x2 ) − k1 (x2 − x1 ) − c1 (x˙ 2 − x˙ 1 ) Summing vertical forces on m3 gives m3 x ¨3 = k3 (x1 − x3 ) + c3 (x˙ 1 − x˙ 3 ) The system model consists of these three equations.
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4.49 Summing vertical forces acting on the rocket, we obtain 1485 v˙ = 18, 000 − 1485 − 0.56v 32.2 or 82.354v˙ + v = 29, 491 The steadystate speed is 29,491 ft/sec, and the solution form is v(t) = 29, 491 + Ae−t/82.354 If v(0) = 0, the solution is A = −29, 491 and '
(
'
v(t) = 29, 491 1 − e−t/82.354 = 29, 491 1 − e−0.0121t Integrate v(t) to obtain h(t).
h(t) = 29, 491t + or
(
( 29, 491 ' −0.0121t e −1 0.0121 '
(
h(t) = 29, 491t + 2.437 × 106 e−0.0121t − 1
At t = 2.5 sec, v = 878.7 ft/sec, ot 599 mph. At t = 2.5 sec, h = 1112 ft.
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4.50 Summing vertical forces on the rocket, we obtain mv˙ = −mg − cv or
c v = −32.2 − 0.0429v (1) m The characteristic root is s = −0.0429, and the solution has the form v˙ = −g −
v(t) = C1 e−at + C2 where a = 0.0429 and C2 is the steadystate speed, which can be found from (1) to be −750.58. Applying the initial condition v(0) = 6828 gives C1 = 7578.58. Thus v(t) = 7578.58e−0.0429t − 750.58
(2)
The peak height occurs when the speed becomes zero. This time tp is found from (2) by setting v(t) = 0: 750.58 1 ln = 53.89 sec tp = − 0.0429 7578.58 The height can be found by integrating v(t). h(t) =
&
tp
v(t) dt + h(0)
0
where h(0) = 26.4(5280) = 139, 392 ft. The result is h(t) =
&
tp 0
(C1 e−at + C2 ) dt + h(0) =
( C1 ' 1 − e−atp + C2 tp + h(0) a
At t = tp , this gives h(tp ) = 258, 098 ft, or 48.88 miles above the surface of the Earth.
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4.51 a) Summing horizontal forces on each mass gives m1 x ¨1 = f − k1 x1 − k2 (x1 − x2 ) m2 x ¨2 = k2 (x1 − x2 )
Substituting the parameter values and collecting terms, we obtain 20¨ x1 = f − 9 × 104 x1 + 6 × 104 x2 60¨ x2 = 6 × 104 (x1 − x2 )
b) Apply the Laplace transform to both sides of each equation, using zero initial conditions: ' ( 20s2 + 9 × 104 X1 (s) − 6 × 104 X2 (s) = F (s) '
(
−6 × 104 X1 (s) + 60s2 + 6 × 104 X2 (s) = 0
The solutions are
X1 (s) s2 + 103 = F (s) 20s4 + 1.1 × 105 s2 + 3 × 107 103 X2 (s) = F (s) 20s4 + 1.1 × 105 s2 + 3 × 107
c) The MATLAB session is
'sys = tf([1, 0, 1e+3],[20, 0, 1.1e+5, 0, 3e+7]); 'step(sys)
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4.52 a) Summing moments about the axis of rotation of each inertia gives I1 θ¨1 = −k1 θ1 − k2 (θ1 − θ2 ) I2 θ¨2 = T2 + k2 (θ1 − θ2 ) − k3 θ2
b) Substitute the given parameter values and collect terms to obtain I θ¨1 = −2kθ1 + kθ2 2I θ¨2 = T2 − 2kθ2 + kθ1
Apply the Laplace transform to both sides of each equation, using zero initial conditions: (Is2 + 2k)Θ1 (s) − kΘ2 (s) = 0 The solutions are
−kΘ1 (s) + (2Is2 + 2k)Θ2 (s) = T2 (s) k Θ1 (s) = 2 4 T2 (s) 2I s + 6kIs2 + 3k 2 Is2 + 2k Θ2 (s) = 2 4 T2 (s) 2I s + 6kIs2 + 3k 2
c) With I = 10 and k = 60, Θ1 (s) 60 = 4 T2 (s) 200s + 3600s2 + 10, 800 The MATLAB session is 'sys = tf(60,[200, 0, 3600, 0, 10800]); 'impulse(sys)
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4.53 Summing horizontal forces on each mass gives m1 x ¨1 = f − k1 x1 − k2 (x1 − x2 ) m2 x ¨2 = k2 (x1 − x2 )
Substituting the parameter values and collecting terms, we obtain 20¨ x1 = f − 9 × 104 x1 + 6 × 104 x2 60¨ x2 = 6 × 104 (x1 − x2 )
Apply the Laplace transform to both sides of each equation, using zero initial conditions: (20s2 + 9 × 104 )X1 (s) − 6 × 104 X2 (s) = F (s) −6 × 104 X1 (s) + (60s2 + 6 × 104 )X2 (s) = 0
The transfer functions can be found by using Cramer’s method. The session is 'D = conv([20, 0, 9e+4], [60, 0, 6e+4])[0, 0, 0, 0, 3.6e+9]; 'D1 = [60, 0, 6e+4]; 'D2 = 6e+4; 'sys1 = tf(D1, D) 'sys2 = tf(D2, D) The displayed transfer functions agree with the answers found by hand in Problem 4.51.
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4.54 Summing moments about the axis of rotation of each inertia gives I1 θ¨1 = −k1 θ1 − k2 (θ1 − θ2 ) I2 θ¨2 = T2 + k2 (θ1 − θ2 ) − k3 θ2
Substitute the given parameter values and collect terms to obtain 10θ¨1 = −120θ1 + 60θ2 20θ¨2 = T2 − 120θ2 + 60θ1
Apply the Laplace transform to both sides of each equation, using zero initial conditions: (10s2 + 120)Θ1 (s) − 60Θ2 (s) = 0 −60Θ1 (s) + (20s2 + 120)Θ2 (s) = T2 (s)
The transfer functions can be found by using Cramer’s method. The session is 'D = conv([10, 0, 120], [20, 0, 120])[0, 0, 0, 0, 60*60]; 'D1 = 60; 'D2 = [10, 0, 120]; 'sys1 = tf(D1, D) 'sys2 = tf(D2, D) The displayed transfer functions agree with the answers found by hand in Problem 4.52.
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4.55 a) Summing moments about each pivot point and assuming small angles, we obtain m1 L22 θ¨1 = −m1 gL2 θ1 − kL1 (θ1 − θ2 )L1 Collecting terms gives
m2 L22 θ¨2 = −m2 gL2 θ2 + kL1 (θ1 − θ2 )L1 m1 L22 θ¨1 + (m1 gL2 + kL21 )θ1 = kL21 θ2 m2 L22 θ¨2 + (m2 gL2 + kL21 )θ2 = kL21 θ1
b) Substitute the given values: m1 = 1, m2 = 4, L1 = 2, L2 = 5, and k = 10, we obtain 25θ¨1 + 89.05θ1 − 40θ2 = 0 400θ¨2 + 118.48θ2 − 40θ1 = 0
Apply the Laplace transform to both sides of each equation, using the given initial conditions: θ1 (0) = 0.1, θ2 (0) = 0, θ˙1 (0) = 0, and θ˙2 (0) = 0. (25s2 + 89.05)Θ1 (s) − 40Θ2 (s) = 2.5s −40Θ1 (s) + (400s2 + 118.48)Θ2 (s) = 0
The transfer functions can be found by using Cramer’s method. Note that we must append a 0 to the numerator polynomial D1, so that we can use the step function to obtain the free response. See Example 4.6.1. The session is 'D = conv([25, 0, 89.05], [400, 0, 118.48])  [0, 0, 0, 0, 40*40]; 'D1 = conv([2.5, 0, 0], [400, 0, 118.48]); 'sys1 = tf(D1, D); 'step(sys1)
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4.56 a) Summing horizontal forces on each mass gives m1 x ¨1 = −k1 x1 − k2 (x1 − x2 ) m2 x ¨2 = −k3 x2 − k2 (x1 − x2 )
b) Substitute the parameter values and collect terms to obtain x ¨1 = −3.2 × 104 x1 + 1.6 × 104 x2 2¨ x2 = −3.2x2 + 1.6 × 104 x2
Apply the Laplace transform to both sides of each equation, using the initial conditions: x1 (0) = 0.1, x2 (0) = 0, x˙ 1 (0) = 0, and x˙ 2 (0) = 0. (s2 + 3.2 × 104 )X1 (s) − 1.6 × 104 X2 (s) = 0.1s −1.6 × 104 X1 (s) + (2s2 + 3.2 × 104 )X2 (s) = 0
The transfer functions can be found by using Cramer’s method. Note that we must append a 0 to the numerator polynomial D1, so that we can use the step function to obtain the free response. See Example 4.6.1. The session is 'D = conv([1, 0, 3.2e+4], [2, 0, 3.2e+4])  [0, 0, 0, 0, 2.56e+8]; 'D1 = conv([0.1, 0, 0], [2, 0, 3.2e+4]); 'sys1 = tf(D1, D); 'step(sys1)
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Five
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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5.1 From the diagram: X(s) = Solve for the ratio:
1 {6 [F (s) − X(s)] + 4X(s)} s X(s) 6 = F (s) s+2
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5.2 From the problem figure: X(s) =
1 {G(s) + 10 [F (s) − X(s)]} s+3
Set G(s) = 0 and solve for the ratio: X(s) 10 = F (s) s + 13
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5.3 From the problem figure: X(s) =
!
1 4 −G(s) − 8X(s) + [F (s) − 6X(s)] s s
"
Set G(s) = 0 and solve for the ratio: X(s) 4 = 2 F (s) s + 8s + 7
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5.4 The following diagram is one of several possible ones:
Figure : for Problem 5.4
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5.5 The following diagram is one of several possible ones:
Figure : for Problem 5.5
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5.6 Define the following state variables: x1 = x and x2 = x. ˙ Then the state equations are x˙ 1 = x2
1 x˙ 2 = [−4x1 − 7x2 + f (t)] 5
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5.7 Define the following state variables: x1 = y, x2 = y, ˙ and x3 = y¨. Then the state equations are x˙ 1 = x2
x˙ 2 = x3
1 x˙ 3 = [f (t) − 7x1 − 4x2 − 5x3 ] 2
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5.8 Define the following state variables: x1 = x and x2 = x. ˙ Then the state equations are x˙ 1 = x2
x˙ 2 =
1 (4y − 4x1 − 5x2 ) 2
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5.9 From the transfer function we obtain: 3¨ y + 6y˙ + 10y = 6f (t) Define the following state variables: x1 = x and x2 = x. ˙ Then the state equations are x˙ 1 = x2
1 x˙ 2 = [6f (t) − 10x1 − 6x2 ] 3
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5.10 Define the following state variables: z1 = x1 , z2 = x˙ 1 , z3 = x2 , and z4 = x˙ 2 . Then the state equations are 1 z˙1 = z2 z˙2 = [f (t) − k1 z1 + k1 z3 ] m1 1 z˙4 = [k1 z1 − (k1 + k2 )z3 ] z˙3 = z4 m2
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5.11 Define the following state variables: z1 = x1 , z2 = x˙ 1 , z3 = x2 , and z4 = x˙ 2 . Then the state equations are z˙1 = z2 z˙3 = z4
z˙2 = z˙4 =
1 (−40z1 − 8z2 + 25z3 + 5z4 ) 10
1 [f (t) + 25z1 + 5z2 − 25z3 − 5z4 ] 5
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5.12 Define the following state variables: x1 = x and x2 = x. ˙ Then the state equations are 1 x˙ 2 = [4y(t) − 4x1 − 5x2 ] 2
x˙ 1 = x2 Then A=
#
0 1 −2 −5/2
C=
%
1 0
&
$
B=
#
0 2
$
D = [0]
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5.13 A=
#
C=
−5 3 0 −4
#
1 3 0 1
$
$
B= D=
#
#
2 0 0 6
2 0 0 0
$
$
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5.14 a) A=
#
C= b) A=
#
C=
−5 3 1 −4
#
1 0 0 1
−5 3 0 −4
#
1 0 0 0
$
$ $
$
B=
#
0 5
D=
#
0 0
$
B=
#
4 0 0 5
D=
#
0 0 0 0
$
$
$
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5.15 Define the following state variables: z1 = x1 , z2 = x˙ 1 , z3 = x2 , and z4 = x˙ 2 . Then the state equations are z˙1 = z2 z˙3 = z4
z˙2 = z˙4 =
1 (−40z1 − 8z2 + 25z3 + 5z4 ) 10
1 [f (t) + 25z1 + 5z2 − 25z3 − 5z4 ] 5
The matrices are
A=
0 1 0 0 −4 −4/5 5/2 1/2 0 0 0 1 5 1 −5 −1 C=
#
1 0 0 1
$
D=
#
B= 0 0
$
0 0 0 1/5
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5.16 First way: Isolate a 1 in the denominator: 6 + 7/s Y (s) = F (s) 1 + 3/s Thus
3 7 1 Y (s) = − Y (s) + 6F (s) + F (s) = [7F (s) − 3Y (s)] + 6F (s) s s s
Let X(s) =
1 [7F (s) − 3Y (s)] s
(1)
(2)
Thus, from (1), Y (s) = X(s) + 6F (s)
(3)
Substitute this into (2): X(s) =
1 1 {7F (s) − 3 [X(s) + 6F (s)]} = [−3X(s) − 11F (s)] s s
Thus x˙ = −3x − 11f (t)
where
y = x + 6f Thus x(0) = y(0) − 6f (0) = y(0) if we take f (0) = 0. Second way: Write the equation as Y (s) = (6s + 7) Let X(s) =
F (s) s+3
F (s) s+3
(1)
(2)
Then (s + 3)X(s) = F (s) which gives sX(s) = −3X(s) + F (s)
and From (1), (2), and (3),
(3)
x˙ = −3x + f (t)
Y (s) = 6sX(s) + 7X(s) = 6[−3X(s) + F (s)] + 7X(s) = −11X(s) + 6F (s) Thus y(t) = −11x(t) + 6f (t) 517 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
So x(0) = −
1 6 1 y(0) + f (0) = − y(0) 11 11 11
if we take f (0) = 0. Thus the state model is where
x˙ = −3x + f (t) x(0) = −
1 y(0) 11
518 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.17 First way: Divide by s2 to obtain a 1 in the denominator. 2s−2 + s−1 Y (s) = F (s) 1 + 4s−1 + 3s−2 Use the 1 in the denominator to solve for Y (s). Y (s) = =

.

.
2s−2 + s−1 F (s) − 4s−1 + 3s−2 Y (s) !
1 1 −4Y (s) + F (s) + [2F (s) − 3Y (s)] s s
"
(2)
This equation shows that Y (s) is the output of an integration. Thus y can be chosen as a statevariable. X1 (s) = Y (s) The term within square brackets in (2) is the input to an integration, and can be chosen as the second statevariable: X2 (s) =
1 1 [2F (s) − 3Y (s)] = [2F (s) − 3X1 (s)] s s
Then from equation (2) X1 (s) =
1 [−4X1 (s) + F (s) + X2 (s)] s
The state equations are x˙ 1 = −4x1 + x2 + f x˙ 2 = 2f − 3x1
(3) (4)
and the output equation is y = x1 . The matrices of the standard form are
−4 1
A=
−3 0
C=
%
1 0
&
B=
#
1 2
$
D = [0]
Because x1 = y, we have x1 (0) = y(0). From (3), x2 = x˙ 1 + 4x1 − f = y˙ + 4y − f . Thus, taking f (0) = 0, we obtain x2 (0) = y(0) ˙ + 4y(0). Second way: Express the model as Y (s) = Let the first state variable be X1 (s) =
s + 2 F (s) s+1s+3 F (s) s+3
(1)
519 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Then x˙ 1 = f − 3x1
Then
Y (s) =
(2)
s+2 X1 (s) s+1
(3)
Let the second state variable be X2 (s) =
X1 (s) s+1
(4)
This gives x˙ 2 = x1 − x2
From (3)
(5)
Y (s) = (s + 2)X2 (s) and y = x˙ 2 + 2x2 From (5), y = x1 − x2 + 2x2 = x1 + x2
Thus the model is
(6)
x˙ 1 = f − 3x1 x˙ 2 = x1 − x2 y = x1 + x2
For the initial conditions, note that (1) and (3) give y˙ + y = x˙ 1 + 2x1 = f − x1
or
x1 = f − y˙ − y
Taking f (0) = 0, this gives From (6),
x1 (0) = −y(0) ˙ − y(0)
(7)
x2 = y − x1
which gives
x2 (0) = y(0) − x1 (0) = 2y(0) + y(0) ˙
(8)
This initial conditions are given by (7) and (8). The state model matrices are
A=
−3
0
1
−1
C=
%
1 1
&
B=
#
1 0
$
D = [0]
520 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.18 a) The matrices are A=
#
C=
−5 3 0 −4
#
1 0 0 1
$
$
B=
#
0 5
#
0 0
$
D=
$
The MATLAB session is #A = [5, 3; 1, 4]; B = [0; 5]; #C = [1, 0; 0, 1]; D = [0;0]; #sys = ss(A,B,C,D); #systf = tf(sys) The resulting transfer functions displayed on the screen are for x1 and x2 in that order. 15 X1 (s) = 2 U (s) s + 9s + 17 X2 (s) 5s + 25 = 2 U (s) s + 9s + 17 b) The matrices are A=
#
C= The MATLAB session is
−5 3 1 −4
%
1 0
&
$
B= D=
%
#
4 0 0 5
0 0
$
&
#A = [5, 3; 1, 4]; B = [4, 0;0, 5]; #C = [1, 0]; D = [0, 0]; #sys = ss(A,B,C,D); #systf = tf(sys) The resulting transfer functions displayed on the screen are for u1 and u2 in that order. They are 4s + 16 X1 (s) = 2 U1 (s) s + 9s + 17 15 X1 (s) = 2 U2 (s) s + 9s + 17 521 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.19 a) The session is #sys1 = tf(1, [2, 5, 4, 7]); #sys2 = ss(sys1) The matrices displayed on the screen are
−2.5 −0.5 −0.2188 A= 4 0 0 0 4 0 C= b) The session is
%
0 0 0.25
&
0.125 B= 0 0 D = [0]
#sys1 = tf(6, [3, 6, 10]); #sys2 = ss(sys1) The matrices displayed on the screen are A=
#
$
−2 −0.4167 8 0
C=
%
0 0.5
&
B=
#
0.5 0
$
D = [0]
522 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.20 a) The session is #sys1 = tf([6, 7], [1, 3]); #sys2 = ss(sys1) The matrices, which are scalars in this case, displayed on the screen are A = −3
B=4
C = −2.75
So the model is
D=6
x˙ = −3x + 4f
y = −2.75x + 6f
b) The session is
#sys1 = tf([1, 2], [1, 4, 3]); #sys2 = ss(sys1) The matrices displayed on the screen are A=
#
−4 −0.375 8 0
C= So the model is
%
1 0.25
$ &
x˙ 1 = −4x1 − 0.375x2 + f
B=
#
1 0
$
D = [0] x˙ 2 = 8x1
y = x1 + 0.25x2
523 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.21 The matrices are A=
#
−5 3 1 −4
C= The session is
%
1 0
$
#
B=
&
D = [0]
0 5
$
#A = [5 ,3; 1, 4]; B = [0; 5]; C = [1, 0]; D = 0; #sys = ss(A,B,C,D); #initial(sys, [3, 5]) b) The session is #A = [5 ,3; 1, 4]; B = [0; 5]; C = [1, 0]; D = 0; #sys = ss(A,B,C,D); #step(sys) c) The session is #A = [5 ,3; 1, 4]; B = [0; 5]; C = [1, 0]; D = 0; #sys = ss(A,B,C,D); #t = [0:0.01:2]; #f = 3*sin(10*pi*t); #lsim(sys,f,t) This plots both the input f (t) and the output x1 (t) on the same plot. To plot only the output, replace the last line with #y = lsim(sys, f, t); plot(t, y)
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5.22 To obtain the roots, all we need is the A matrix, which is A=
#
−5 3 0 −4
$
The session is #A = [5, 3; 0, 4]; #poly(A) ans = 1 9 20 #eig(A) ans = 5 4 Thus the characteristic equation is s2 + 9s + 20 = 0 and the roots are s = −5 and s = −4.
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5.23 a) Let z1 = x1 , z2 = x˙ 1 , z3 = x2 , and z4 = x˙ 2 . Then the state equations are z˙1 = z2 z˙2 =
1 [−k1 z1 − c1 z2 + k1 z3 + c1 z4 ] m1 z˙3 = z4
1 [k1 z1 + c1 z2 − k1 z3 − c1 z4 + k2 y − k2 z3 ] m2 The following script file creates the state model. z˙4 =
m1 = 36;m2 = 240;k1 = 1.6e+5;k2 = 1.6e+4;c1 = 98; A = [0, 1, 0, 0; k1/m1, c1/m1, k1/m1, c1/m1; .... 0, 0, 0, 1; k1/m2, c1/m2, (k1+k2)/m2, c1/m2]; B = [0; 0; 0; k2/m2]; C = [1, 0, 0, 0; 0, 0, 1, 0]; D = [0; 0]; sys = ss(A,B,C,D); b), c) To compute the impulse response, add the following lines to the script file. impulse(sys) format long poly(A) eig(A) We must use the format long command to see the results adequately. The characteristic polynomial given by the poly(A) function is s4 + 3.13106s3 + 517777s2 + 181.481s + 2.96296 × 105 = 0 The roots given by the eig(A) function are s = −1.56527 ± 71.5364j
s = −3.07931 × 10−5 ± 7.60733j
The period of the first root pair is 2π/71.5364 = 0.0878 s, whereas the dominant time constant of the system is 1/(3.07931×10−5 ) = 3.247×104 s. So there will be approximately 4(3.247×104 )/0.0878 = 1.479×105 oscillations before the impulse response disappears. Thus the impulse(sys) will produce a plot on which the individual oscillations are impossible to discern. To remedy this, you can use the syntax [y, t] = impulse(sys); plot(t, y),axis([0 1000 10 10]). This will show several oscillations. d) Add the following line to the script file. tfsys = tf(sys)
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The following transfer functions are displayed on the screen. 181.5s + 2.963 × 105 X1 (s) = 4 Y (s) s + 3.131s3 + 5178s2 + 181.5s + 2.963 × 105 66.67s2 + 181.5s + 2.963 × 105 X2 (s) = 4 Y (s) s + 3.131s3 + 5178s2 + 181.5s + 2.963 × 105
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5.24 (a) The small displacement assumption implies that the suspension forces are nearly perpendicular to the centerline of the mass m, and thus are nearly vertical. To obtain the equations of motion, assume arbitrarily that y1 > x − L1 θ y˙ 1 > x˙ − L1 θ˙
y2 > x + L2 θ y˙ 2 > x˙ + L2 θ˙ We obtain the following moment equation about the mass center G. ˙ 1 − k1 (y1 − x + L1 θ)L1 IG θ¨ = − c1 (y˙ 1 − x˙ + L1 θ)L ˙ 2 + k2 (y2 − x − L2 θ)L2 + c2 (y˙ 2 − x˙ − L2 θ)L Rearranging gives IG θ¨ + (c2 L22 + c1 L21 )θ˙ + (k1 L21 + k2 L22 )θ − (c1 L1 − c2 L2 )x˙ − (k1 L1 − k2 L2 )x
= −c1 L1 y˙ 1 + c2 L2 y˙ 2 − k1 L1 y1 + k2 L2 y2
(1)
Summing forces in the vertical direction gives ˙ + k1 (y1 − x + L1 θ) + c2 (y˙ 2 − x˙ − L2 θ) ˙ + k2 (y2 − x − L2 θ) m¨ x = c1 (y˙ 1 − x˙ + L1 θ) Rearranging gives m¨ x + (c1 + c2 )x˙ + (k1 + k2 )x − (c1 L1 − c2 L2 )θ˙ − (k1 L1 − k2 L2 )θ = c1 y˙ 1 + c2 y˙ 2 + k1 y1 + k2 y2
(2)
Equations (1) and (2) are the desired model. b) The transformed equations of motion have the following forms: AΘ(s) + BX(s) = CY1 (s) + DY2 (s) EΘ(s) + F X(s) = GY1 (s) + HY2 (s) where A = IG s2 + (c2 L22 + c1 L21 )s + k1 L21 + k2 L22 B = (c2 L2 − c1 L1 )s + k2 L2 − k1 L1 C = −c1 L1 s + k1 L1 D = c2 L2 s + k2 L2
528 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
E = −B
F = ms2 + (c1 + c2 )s + k1 + k2 G = c1 s + k1 H = c2 s + k2 Because these models have input derivatives, we must be careful in deriving a state variable model. First obtain the transfer functions using Cramer’s method, and then use the MATLAB functions tf and ss to obtain the state model. For Cramer’s method, let
/ / / N1 = / / / / / N3 = / /
/
/ / A / M =/ / E
C B // / = CF − BG G F /
B F
/ / / / = AF − BE /
/
A C // / = AG − CE E G /
Then the transfer functions are found as follows
/ / / N2 = / / / / / N4 = / /
/
D B // / = DF − BH H F / /
A D // / = AH − DE E H /
Θ(s) N1 = Y1 (s) M
Θ(s) N2 = Y2 (s) M
N3 X(s) = Y1 (s) M
N4 X(s) = Y2 (s) M
We can implement Cramer’s method in MATLAB by using the conv function. To do this, note that A and F are secondorder polynomials, and that B, C, D, E, G, and H are firstorder polynomials. This distinction is important because will need to add leading zeros to subtract polynomials of different orders. Note that with a multiinput, multioutput (MIMO) model whose transfer functions all have the same denominator, we must store the numerators in a cell array. The MATLAB script file is shown on the following page.
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% Store the parameter values. k1 = 1100; k2 = 1525; c1 = 4; c2 = c1; L1 = 4.8; L2 = 3.6; m = 50; IG = 1000; % Create the coefficients of the transformed equations. A = [IG, c2*L2^2 + c1*L1^2, k1*L1^2 + k2*L2^2]; B = [c2*L2  c1*L1, k2*L2  k1*L1]; C = [c1*L1, k1*L1]; D = [c2*L2, k2*L2]; E = B; F = [m, c1 + c2, k1 + k2]; G = [c1, k1]; H = [c2, k2]; % Create the Cramer determinants. M = conv(A, F)  [0, 0, conv(E, B)]; N1 = conv(C, F)  [0, conv(G, B)]; N2 = conv(D, F)  [0, conv(B, H)]; N3 = conv(A, G)  [0, conv(C, E)]; N4 = conv(A, H)  [0, conv(D, E)]; % Create the transfer functions. % Store the numerators in a cell array. numerators = { N1, N2; N3, N4 }; tfsys = tf(numerators, M); % Create the state variable models. statesys = ss(tfsys)
The results are (note that MATLAB returns the state model matrices as lowercase symbols, a, b, c, d. We use these symbols to avoid confusion with the equation coefficients A, B,C, and D defined above.) a=
#
a1 a2 a3 a4
$
where
a1 =
−0.304 −6.102 −0.1151 −2.314 16 0 0 0 0 8 0 0 0 0 8 0 a3 = a2
a2 =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
a4 = a1
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c=
#
b=
2 0 0 0 0 0 0 0
0 0 0 0 2 0 0 0
−0.0096 −0.1651 −0.006891 −0.1376 0.0072 0.1716 0.006891 0.1376 0.04 0.6879 0.02814 0.4737 0.04 0.9534 0.02943 0.683 d=
#
0 0 0 0
$
$
c) With y1 = 0, we must redefine b as
b=
0 0 0 0 0 0 0 0
0 0 0 0 2 0 0 0
To do this, and to compute and plot the impulse response, add the following lines to the script file. [a, b, c, d] = ssdata(statesys) % Obtain unitimpulse response for y_1 = 0. b(:,1) = zeros(size(b(:,1))); statesys2 = ss(a,b,c,d); impulse(statesys2) Note that we do not need the state model to compute the impulse response, which can also be computed from the transfer functions N2 /M and N4 /M . The problem statement, however, specifically asks for the state space model.
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5.25 a) Create the following function file. function ydot = problem25(t,y) ydot = cos(t); Then type #[ta, ya] = ode45(" problem25",[0, 12], 6); b) The closedform solution is found as follows. 0
y 6
dy =
0
t
cos t dt
0
Thus y(t) − 6 = sin t
To compare the two solutions, continue the session as follows: #tb = [0:0.01:12]; #yb = 6 + sin(tb); #plot(ta,ya,tb,yb) The plots are identical.
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5.26 a) Create the following function file. function ydot = problem26(t,y) ydot = 5*exp(4*t); Then type #[ta, ya] = ode45(" problem26",[0, 1], 2); b) The closedform solution is found as follows. 0
y 2
dy = 5
0
t 0
Thus
e−4t dt = −
5 −4t //t e / 0 4
13 5 −4t − e 4 4 To compare the two solutions, continue the session as follows: y(t) =
#tb = [0:0.01:1]; #yb = 5*exp(4*tb); #plot(ta,ya,tb,yb) The plots are identical.
533 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.27 a) Create the following function file. function ydot = problem27(t,y) ydot = 5*exp(4*t)3*y; Then type #[ta, ya] = ode45(" problem27",[0, 1], 10); b) The closedform solution is found with the Laplace transform. sY (s) − y(0) + 3Y (s) = With y(0) = 10, this gives Y (s) =
5 s−4
10s − 35 65/7 5/7 = + (s + 3)(s − 4) s+3 s−4
and
65 −3t 5 4t e + e 7 7 To compare the two solutions, continue the session as follows: y(t) =
#tb = [0:0.01:1]; #yb = (65*exp(3*tb)+5*exp(4*tb))/7; #plot(ta,ya,tb,yb) The plots are identical.
534 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.28 a) Create the following function file. function ydot = problem28(t,y) ydot = sin(y); Then type #[ta, ya] = ode45(" problem28",[0, 4], 0.1); b) The closedform solution to the approximate equation y˙ = −y is found as follows. The root is s = −1, so the solution form is y(t) = Ae−t Because y(0) = 0.1, we obtain A = 0.1. Thus the solution is y(t) = 0.1e−t To compare the two solutions, continue the session as follows: #tb = [0:0.01:4]; #yb = 0.1*exp(tb); #plot(ta,ya,tb,yb) The plots are identical.
535 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.29 a) Create the following function file. function ydot = problem29(t,y) if t <= 2 f = 3*t; elseif t <= 5 f = 6; else f = 3*(t  5) + 6; end ydot = f  2*y; Then type #[ta, ya] = ode45(" problem29",[0, 7], 2);
536 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
5.30 A MATLAB solution is as follows. First create the following function file. function vdot = problem30(t,v) vdot = (800020*v0.05*v.^2)/50; Then type #[t,v] = ode45(" problem30",[0, 5], 0); #plot(t,v),xlabel("t (sec)"),ylabel("v (ft/sec)") The final time used (5 sec) was an estimate. The solution actually reaches steadystate in about 4.6 sec.
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5.31 Put the model into state variable form. Let x1 = y and x2 = y. ˙ Then x˙ 1 = x2 and 3 x˙ 2 = 9.81 − 180x1 − 340x1 . A MATLAB solution is as follows. First create the following function file. function xdot = problem31(t,x) xdot = [x(2);9.81180*x(1)340*x(1).^3; Then type #[ta,xa] = ode45("problem31",[0, 2], [0.06, 0]); #[tb,xb] = ode45("eqn" ,[0, 2], [0.1, 0]; #plot(ta,xa(:,1),tb,xb(:,1)),xlabel("t (s)" ),ylabel("y(m)" )
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5.32 The state variable form, with x1 = y and x2 = y, ˙ is x˙ 1 = x2 x˙ 2 = −x1 + b(1 − x21 )x2
Create the following function file:
function xdot = vander(t,x) global b xdot(1) = x(2); xdot(2) = b*(1x(1)^2)*x(2)  x(1); xdot = [xdot(1);xdot(2)]; Then use the following script file to solve the equation and plot x1 = y. For Case 1: global b b = 0.1; [t, x] = ode45(" vander", [0, 25], [1, 1]); plot(t,x(:,1)),xlabel("t" ),ylabel("y(t)" ) For Case 2: global b b = 0.1; [t, x] = ode45(" vander", [0, 25], [3, 3]); plot(t,x(:,1)),xlabel("t" ),ylabel("y(t)" ) For Case 3: global b b = 3; [t, x] = ode45(" vander", [0, 25], [1, 1]); plot(t,x(:,1)),xlabel("t" ),ylabel("y(t)" ) Note on Passing Additional Parameters to an ODE Function. Rather than using the global statement, the extended syntax of the ODE solvers using the options argument lets you pass any input parameters that follow the options argument to the ODE function and any function you specify in options. Since this syntax is not treated in the text and uses function handles, we have not used it in our solutions. For example, you can pass the parameter b directly to the van der Pol function, as follows: 539 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
function xdot = vander(t,x,b) xdot(1) = x(2); xdot(2) = b*(1x(1)^2)*x(2)  x(1); xdot = [xdot(1);xdot(2)]; Then pass the parameter b to the function vander by specifying it after the options argument in the call to the solver, using options = [ ] as a placeholder, as follows: [t,x] = ode45(@vander, [0, 25], [1, 1], [ ],b) The function handle @vander calls vander(t,x,b).
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5.33 Put the model in state variable form as follows. Let x1 = y and x2 = y. ˙ x˙ 1 = x2
Create the following function file.
x˙ 2 = −x1 + b(1 − x21 )x2
function xdot = vander(t,x); b = 1000; xdot = [x(2); x(1) + b*(1x(1)^2)*x(2)]; The following script file uses the ode23s function. [t, x] = ode23s("vander" , [0, 3000], [2, 0]); plot(t,x(:,1)),xlabel("t" ),ylabel("y" ) The ode23 function successfully solves the equation, but the functions ode45 and ode23 fail to converge to a solution for this problem. They can, however, obtain a solution for “nonstiff” values of b, such as b = 1.
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˙ is 5.34 The state variable form, with x1 = θ and x2 = θ, x˙ 1 = x2 a(t) cos θ − g sin θ L Express the acceleration a(t) as the linear function a(t) = mt + b. Thus for cases (a) and (b), m = 0 and b = 5. For case (c), m = 0.5 and b = 0. Create the following function file: x˙ 2 =
function xdot = accbase(t,x) global m b L = 1; g = 9.81; xdot(1) = x(2); xdot(2) = ((m*t + b)*cos(x(1)) g*sin(x(1)))/L; xdot = [xdot(1);xdot(2)]; To solve the equation and plot the solution, create the following script file. global m b m = 0; b = 5; [t, x] = ode45(" accbase", [0, 10], [0.5, 0]); plot(t,x(:,1)),xlabel("t (seconds)"),ylabel("theta(t) (radians)") For case (b), change the third line to [t, x] = ode45(" accbase", [0, 10], [3, 0]); For case (c), change the second line to m = 0.5; b = 0; Note on Passing Additional Parameters to an ODE Function. Rather than using the global statement, the extended syntax of the ODE solvers using the options argument lets you pass any input parameters that follow the options argument to the ODE function and any function you specify in options. Since this syntax is not treated in the text and uses function handles, we have not used it in our solutions. For example, you can pass the parameters L, g, m, and b directly to the accbase function, as follows: function xdot = accbase(t,x,L,g,m,b) xdot(1) = x(2); xdot(2) = ((m*t + b)*cos(x(1)) g*sin(x(1)))/L; xdot = [xdot(1);xdot(2)];
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Then pass the parameters L, g, m, and b to the function accbase by specifying them after the options argument in the call to the solver, using options = [ ] as a placeholder, as follows: [t,x] = ode45(@accbase, [0, 10], [3, 0], [ ], L, g, m, b) The function handle @accbase calls accbase(t,x,L,g,m,b).
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5.35 The cosine function is obtained with the Sine Wave Function block by using a phase shift of π/4 (pi/2). Make sure the Sine Wave blocks are set to use simulation time. Otherwise a Clock block will be required. Set the initial condition of the Integrator block to 1 and the initial condition of the Integrator1 block to 4.
Figure : for Problem 5.35.
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5.36 The time variable tout is automatically placed in the workspace by selecting Data Import/Export under the Configuration Parameters menu. Thus a Clock block is not needed. You can enter the initial condition x(0) ˙ as 100*cos(30*pi/180), with a similar expression for y(0). ˙ To plot the trajectory y versus x, type plot(simout(:,1),simout(:,2)),xlabel("x" ),ylabel("y" )
Figure : for Problem 5.36.
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5.37 The model is shown in the following figure. To plot the percent relative error in the series solution, type #t = tout; #x = (t.^3)/3t.^2+3*t3+3*exp(t); #plot(t,100*(simoutx)/simout),xlabel("t" ),ylabel("Percent Error" ),grid The plot shows that the percent relative error in the series solution is less than 1% for t < 0.75. It is about 6% at t = 1.
Figure : for Problem 5.37.
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5.38 The model is shown in the following figure. Set the Step time and Final value for the Step block to 0 and 4, respectively. Set the Step time and Final value for the Step1 block to 2 and 4, respectively. Set the Initial condition of the Integrator to 2. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.38.
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5.39 The model is shown in the following figure. Set the Step time and Final value for the Step block to 0 and 5, respectively. Set the Step time and Final value for the Step1 block to 2 and 5, respectively. Set the numerator of the Transfer Function block to [1] and the denominator to [2, 12, 10]. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.39.
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5.40 The Simulink model is shown in the figure. Set the Amplitude of the Sine Wave block to 5 and the Frequency to 0.8. Make sure the Initial conditions of the Integrator blocks are set to 0. The Save format of the To Workspace block has been selected as Array, and the simulation time tout has been saved to the workspace (Use the Workspace I/O tab on the Simulation Parameters menu under the Simulation menu to do this). Then the output can be plotted in MATLAB by typing #plot(tout,simout)
Figure : for Problem 5.40.
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5.41 The model is shown in the following figure. Set the Sine Wave block to use simulation time. Set the Amplitude to 10 and the Frequency to 3. In the Saturation block set the limits to ±8. Set the Initial condition of the Integrator to 2. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.41.
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5.42 The model is shown in the following figure. Set the Sine Wave block to use simulation time. Set the Amplitude to 10 and the Frequency to 4. In the Saturation block set the limits to ±5. Set the Initial condition of the Integrator to 0. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.42.
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5.43 The model is shown in the following figure. Set the Sine Wave block to use simulation time. Set the Amplitude to 2 and the Frequency to 4. In the Math Function block select the square function. Set the Initial condition of the Integrator to 1. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" ) Using a stop time of 3 results in an error message that indicates that Simulink is having trouble finding a small enough step size to handle the rapidly decreasing solution. By experimenting with the stop time, we find that, to two decimal places, a stop time of 1.38 will not generate an error. This model is unstable, and the output x → −∞ if the time span is long enough. We can see this by writing the equation as
Once x drops below negative values.
1
x˙ = 2 sin 4t − 10x2
2/10, x˙ remains negative and thus x continues to decrease through
Figure : for Problem 5.43.
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5.44 The model is shown in the following figure. Set the Start and End of the Dead Zone to −0.5 and 0.5 respectively. Set the Sine Wave block to use simulation time. Set the Amplitude to 2 and the Frequency to 4. In the Math Function block select the square function. Set the Initial condition of the Integrator to 1. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" ) Using a stop time of 3 results in an error message that indicates that Simulink is having trouble finding a small enough step size to handle the rapidly decreasing solution. By experimenting with the stop time, we find that, to two decimal places, a stop time of 1.49 will not generate an error. This model is unstable, and the output x → −∞ if the time span is long enough. We can see this by writing the equation as (not including the effect of the dead zone)
Once x drops below negative values.
1
x˙ = 2 sin 4t − 10x2
2/10, x˙ remains negative and thus x continues to decrease through
Figure : for Problem 5.44.
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5.45 The model is shown in the following figure. Set the Initial condition of Integrator to 0.5 and the Initial condition of Integrator1 to 0. In the Fcn block type 900*u(1)+1700*u(1)^3 for the expression. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.45.
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5.46 The model is shown in the following figure. Set the Initial condition of Integrator to 30*pi/180 and the Initial condition of Integrator1 to 0. In the Fcn block type 17500*cos(u(1))+626000*sin(1.33+u(1))/sqrt(2020+1650*cos(1.33+u(1)) for the expression. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.46.
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5.47 The model is shown in the following figure. Set the Step time and Final value for the Step block to 0 and f1 − mg cos(30π/180) = 5 − 3(9.8) cos(30π/180), respectively. Set the gain in the top gain block to µmg sin(30π/180) = 0.5(2)(9.8) sin(30π/180). Set the Initial condition of the Integrator to 3. The block comes to rest in about 0.35 s. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" )
Figure : for Problem 5.47.
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5.48 a) The model is shown in the following figure. Set the Step time and Final value for the Step block to 0 and 4, respectively. Set the Step time and Final value for the Step1 block to 2 and 4, respectively. Set the Initial condition of the Integrator to 2. You can plot the results by typing #plot(tout,simout),xlabel("t" ),ylabel("x" ) b) Delete the Step block and replace it with a Sine Wave block. Set the Amplitude to 10 and the frequency to 2.5.
Figure : for Problem 5.48.
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Six
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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6.1 See the following figure. From part (a) of the figure, R1 = R + From part (b) of the figure,
1 R
1 +
1 R
=
3R 2
1 1 1 = + R2 R R1
which gives R2 = 3R/5. From part (c) of the figure, Re = R + R2 = Thus vs =
8R 5
8R i 5
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Figure : for Problem 6.1
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6.2 From the following figure,
1 1 1 7 = + = R1 5 2 10
So R1 = 10/7. Also, R2 = 8 + 2 = 10. From conservation of charge, i1 + i2 + i3 = i
(1)
Kirchhoff’s voltage law applied to the three loops gives vs = 1i + 4i3
(2)
4i3 = R1 (i1 + i2 ) + 3i2 3i2 = R2 i1
(3)
(4)
Also, v1 = 2i1
(5)
Use (1) and (2) to eliminate i: i1 + i2 + 5i3 = vs
(6)
Collect terms, substitute the values of R1 and R2 , and rearrange (3) and (4): 10 31 i1 + i2 − 4i3 = 0 7 7 10i1 − 3i2 + 0i3 = 0
(7) (8)
Equations (6), (7), and (8) have the solution i1 = 0.1193vs . (These three equations can be solved with MATLAB by setting vs = 1.) From (5) we have the answer: v1 = 0.2386vs
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Figure : for Problem 6.2
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6.3 Let i1 be the current through R1 and R4 , and i2 be the current through R2 and R3 . Then vs = (R1 + R4 )i1 = (R2 + R3 )i2 and v1 = R4 i1 − R3 i2 = This gives v1 =
!
R4 R3 − R1 + R4 R2 + R3
"
vs
R2 R4 − R1 R3 vs (R1 + R4 )(R2 + R3 )
If the changes in R3 and R4 are small enough so that (R1 + R4 )(R2 + R3 ) ≈ constant = K, then R2 vs R1 vs v1 = R4 − R3 = αR4 − βR3 K K where α and β are approximately constant.
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6.4 The two resistors in series are equivalent to one 70 kΩ resistor, which is in parallel with the 80 kΩ resistor. Thus the total equivalent resistance R is found from 1 1 1 = + R 80 70 which gives R = 112/3 kΩ. Thus i= The power P is computed from
9 = 2.411 × 10−4 A 112/3 × 103
P = iv = (2.411 × 10−4 )9 = 2.17 × 10−3 W
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6.5 The model is v1 =
1 C
# !
is −
1 v1 R
"
dt
Differentiate both sides with respect to t to obtain a differential equation. RC
dv1 + v1 = Ris dt
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6.6 a) Kirchhoff’s voltage law gives vs = R1 i + vo + R2 i = (R1 + R2 )i + vo For the capacitor:
1 vo = C
which gives
#
(1)
i dt
dvo =i (2) dt Solve equation (1) for i and substitute into equation (2) to obtain the answer: C
C
dvo vs − vo = dt R1 + R2
which, in standard form, is (R1 + R2 ) C
dvo + vo = vs dt
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6.7 a) From Kirchhoff’s voltage law, vs = R2 i2 = R1 i1 + vo which gives i1 = For the capacitor, vo =
vs − vo R1
1 C
#
(1)
(2)
i1 dt
Differentiate (2) with respect to time, and substitute for i1 from (1): C
dvo vs − vo = i1 = dt R1
Thus the model is
dvo + vo = vs (3) dt Note the we did not need to find i2 or i3 , and that the model is independent of R2 because there is no element between vs and R2 , so the voltage input to the rightmost loop is vs . b) From (3) with vs = 0, we obtain the free response: R1 C
vo (t) = vo (0)e−t/R1 C If vs (t) = V us (t) and if vo (0) = 0, the response is $
vo (t) = V 1 − e−t/R1 C This is the forced response. So the total response is $
%
vo (t) = vo (0)e−t/R1 C + V 1 − e−t/R1 C
%
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6.8 a) Kirchhoff’s voltage law gives vs = R1 i1 + vo For the capacitor:
1 vo = R2 i2 = C
From conservation of charge,
#
(1)
i3 dt
(2)
i1 = i2 + i3 Substitute for the currents to obtain: vs − vo vo dvo = +C R1 R2 dt This gives R1 R2 C
dvo + (R1 + R2 )vo = R2 vs dt
b) The free response is vo (t) = vo (0)e−t/τ where τ= The forced response is vo (t) =
R1 R2 C R1 + R2
% R2 V $ 1 − e−t/τ R1 + R2
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6.9 a) The model of the circuit is Ri + L
di = vs dt
b) If vs (t) = V us (t) and i(0) = 0, then I(s) =
Vs (s) V V V 1 = = − Ls + R s(Ls + R) R R s + R/L
Thus i(t) =
% V $ 1 − e−Rt/L R
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6.10 Define i to be the current passing through R and L. From Kirchhoff’s voltage law, vs = Ri + v0 Solve for i: i=
vs − v0 R
For the inductor: v0 = L
di dt
Substitute for i to obtain v0 = L
d dt
!
vs − v0 R
"
=
L R
!
dvs dv0 − dt dt
"
Move the output terms (the v0 terms) to the left side, putting the highestorder derivative first, and move the input terms to the right side, to obtain the answer in a somewhat standard form: L dv0 L dvs + v0 = R dt R dt
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6.11 Define i to be the current passing through R1 , C, and R2 . Then from Kirchhoff’s voltage law: vs = R1 i + v0 (1) From the definition of v0 , v0 =
1 C
Differentiate this equation to obtain:
#
i dt + R2 i
(2)
dv0 1 di = i + R2 dt C dt Solve equation (1) for i: i=
(3)
vs − v0 R1
(4)
!
"
and differentiate to obtain: di 1 = dt R1
dvs dv0 − dt dt
(5)
Substitute equations (4) and (5) into equation (3) to obtain dv0 1 R2 = (vs − v0 ) + dt R1 C R1
!
dvs dv0 − dt dt
"
Putting this into standard form gives the answer: (R1 + R2 ) C
dv0 dvs + v0 = R2 C + vs dt dt
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6.12 From the voltage law,
di 1 vs = Ri + L + dt C Differentiate this with respect to time: LC
#
i dt
d2 i di dvs + RC + i = C 2 dt dt dt
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6.13 is = i1 + i2 vo = Ri1 Thus v1 = Also, or Thus
1 C
#
i2 dt =
1 C
#
(is − i1 ) dt =
di1 1 L = v1 − vo = dt C L dvo 1 = R dt C
#
L d2 vo 1 = 2 R dt C
Rearranging gives LC
#
(is −
(is − !
1 C
#
(is −
vo ) dt R
vo ) dt − vo R
vo ) dt − vo R
vo is − R
"
−
dvo dt
d2 vo dvo + RC + vo = Ris 2 dt dt
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6.14
di1 + Ri3 (1) dt # 1 Ri3 = i2 dt + v2 C
v1 = L
Thus
RC
dv2 di3 = i2 + C (2) dt dt i1 = i2 + i3 (3)
Transform (1) through (3) to obtain LsI1 (s) + RI3 (s) = V1 (s) I2 (s) − RCsI3 (s) = −CsV2 (s) I1 (s) − I2 (s) − I3 (s) = 0
These have the following solution
I1 (s) = I2 (s) =
(RCs + 1)V1 (s) − RCsV2 (s) D(s)
RCsV1 (s) − (RCs + LCs2 )V2 (s) D(s)
I3 (s) =
V1 (s) + LCs2 V2 (s) D(s)
where D(s) = LRCs2 + Ls + R The differential equation model is LRC LRC
d2 i1 di1 dv1 dv2 +L + Ri1 = RC + v1 − RC dt2 dt dt dt
d 2 i2 di2 dv1 dv2 d2 v2 +L + Ri2 = RC − RC − LC 2 2 dt dt dt dt dt LRC
d 2 i3 di3 d2 v2 + L + Ri = v + LC 3 1 dt2 dt dt2
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6.15 v1 = R1 i1 + L1 1 v3 = C which gives
di1 + v3 dt #
i3 = C v3 = L2
(1)
i3 dt
dv3 dt
di2 + v2 dt
i1 = i2 + i3 = i2 + C
(2) dv3 dt
(3)
Transform (1) through (3) to obtain (R + L1 s)I1 (s) + V3 (s) = V1 (s) L2 sI2 (s) − V3 (s) = −V2 (s)
I1 (s) − I2 (s) − CsV3 (s) = 0
These have the following solution
I1 (s) = I2 (s) =
(L2 Cs2 + 1)V1 (s) − V2 (s) D(s)
V1 (s) − (L1 Cs2 + RCs + 1)V2 (s) D(s)
V3 (s) =
L2 sV1 (s) + (L1 s + R)V2 (s) D(s)
where D(s) = L1 L2 Cs3 + RL2 Cs2 + (L1 + L2 )s + R The differential equation model is L1 L2 C L1 L2 C
d 3 i1 d 2 i1 di1 d2 v1 + RL C + (L + L ) + Ri = L C + v1 − v2 2 1 2 1 2 dt3 dt2 dt dt2
d 3 i2 d 2 i2 di2 d2 v2 dv2 + RL C + (L + L ) + Ri = v − L C − RC − v2 2 1 2 2 1 1 3 2 2 dt dt dt dt dt
L1 L2 C
d3 v3 d2 v3 dv3 dv1 dv2 + RL C + (L1 + L2 ) + Rv3 = L2 + L1 + Rv2 2 3 2 dt dt dt dt dt
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6.16 a) Choose v1 and i1 as the state variables because they describe the energy storage in the circuit. The basic circuit equations are 1 v1 = C v1 = L
#
(is − i1 ) dt
di1 + vo dt
vo = Ri1
These give the state equations C
dv1 = is − i1 dt
L
di1 = v1 − Ri1 dt
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6.17 Choose i1 and vc as the state variables because they describe the energy storage in the circuit (vc is the voltage drop across the capacitor). The basic circuit equations are v1 = L1 vc = These give the state equations L
di1 + Ri3 dt
1 C
#
Ri3 = vc + v2 i 1 = i2 + i3
i2 dt
di1 = v1 − v2 − vc dt
C
dvc 1 1 = i1 − vc − v2 dt R R
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6.18 a) Choose i1 , i2 , and v3 as the state variables because they describe the energy storage in the circuit. The basic circuit equations are 1 v3 = C
di1 v1 = R1 i1 + L1 + v3 dt v3 = L2
di2 + v2 dt
#
i3 dt
i1 = i2 + i3
These give the state equations L1
di1 = v1 − v3 − Ri1 dt C
L2
di2 = v3 − v2 dt
dv3 = i1 − i2 dt
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6.19 It can be seen that R2 and C are in parallel. Thus, their equivalent impedance Z(s) is found from 1 1 1 = + Z(s) 1/Cs R2 or R2 Z(s) = R2 Cs + 1 It can be seen that Z and R1 are in series. Thus, Vs (s) = Z(s) + R1 I(s) and Vo (s) = R1 I(s) Eliminating I(s) from the last two relations yields the desired transfer function. Vs (s) =
Vo (s) [Z(s) + R1 ] R1
or T (s) = =
Vo (s) R1 = Vs (s) Z(s) + R1 R1 R2 Cs + R1 R1 R2 Cs + R2 + R1
The network is seen to be a firstorder system with numerator dynamics.
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6.20 The impedance of R1 and C is their parallel combination: 1 R1
1 R1 = 1 R1 Cs + 1 + 1/Cs
The series combination of this impedance and R2 is: R2 +
R1 R1 R2 Cs + R1 + R2 Vs (s) = = R1 Cs + 1 R1 Cs + 1 I1 (s)
and the required transfer function is I1 (s) R1 Cs + 1 = Vs (s) R1 R2 Cs + R1 + R2
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6.21 Kirchhoff’s voltage law gives vs = Ri + vo Solve this for i: i=
vs − vo R
For the capacitor: vo = which gives C
1 C
#
vo = L i1 = From conservation of charge:
i2 dt
dvo = i2 dt
For the inductor: Solve this for i1 :
(1)
1 L
#
(2) di1 dt
vo dt
i = i1 + i2
(3) (4)
Substitute i, i1 , and i2 from equations (1), (2), and (3) into equation (4) to obtain: vs − vo 1 = R L
#
vo dt + C
dvo dt
Differentiate both sides to get 1 vo (v˙ s − v˙ o ) = + C v¨o R L Rearrange in standard form to get the answer: RLC v¨o + Lv˙ o + Rvo = Lv˙ s
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6.22 From conservation of charge: is = i1 + i2 + i3
(1)
R, L, and C have the same voltage drop vo across them, so di2 1 vo = L = Ri1 = dt C
#
i3 dt
Integrate the first equation for i2 to obtain: 1 i2 = L
#
Solve the second equation for i1 : i1 =
(2)
vo dt vo R
(3)
Solve the third equation for i3 :
dvo (4) dt Substitute equations (2), (3), and (4) into (1) to obtain: i3 = C
dvo vo 1 is = C + + dt R L
#
vo dt
Differentiate both sides to obtain dis d2 vo 1 dvo 1 =C 2 + + vo dt dt R dt L If desired, this can be “cleaned up” somewhat by multiplying both sides by RL to obtain RLC
d2 vo dvo dis +L + Rvo = RL dt2 dt dt
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6.23 Let Z1 (s) = Vs (s)/I(s) be the series impedance of L, C1 , C2 , and R. Z1 (s) = Ls +
1 1 LC1 C2 s2 + RC1 C2 s + C1 + C2 + +R = C1 s C2 s C1 C2 s
Also, Vo (s) =
1 1 Vs (s) C1 s I(s) = = Vs (s) C2 s C2 s Z1 (s) LC1 C2 s3 + RC1 C2 s2 + (C1 + C2 )s
and LC1 C2
d3 vo d2 vo dvo dvs + RC1 C2 2 + (C1 + C2 ) = C1 3 dt dt dt dt
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6.24 Let i1 be the current through R1 ; i2 be the current through the leftside C; i3 be the current through R2 ; i4 be the current through the rightside C; and i5 be the current to the output vo . Then from Kirchhoff’s voltage law, Vs (s) =
1 I2 (s) + R2 I2 (s) Cs R1 I1 (s) =
R2 I3 (s) =
1 I4 (s) + Vo (s) Cs
1 1 I2 (s) + I4 (s) Cs Cs
From conservation of charge, I2 (s) = I3 (s) + I4 (s)
I5 (s) = I1 (s) + I4 (s) = 0
Thus I4 (s) = −I1 (s). Use these equations to eliminate the current variables, and find the ratio Vo (s)/Vs (s). The answer is Vo (s) R1 R2 C 2 s2 + 2R2 Cs + 1 = Vs (s) R1 R2 C 2 s2 + (R1 + 2R2 )Cs + 1
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6.25 The impedance of the R1 C1 combination is Z1 (s) =
R1 R1 C1 s + 1
For the R2 C2 combination: Z2 (s) = R2 +
1 R2 C2 s + 1 = C2 s C2 s
Thus Vs (s) = Z1 (s)I(s) + Z2 (s)I(s) = [Z1 (s) + Z2 (s)] I(s) and Vo (s) = Z2 (s)I(s) = Thus
Z2 (s) Vs (s) Z1 (s) + Z2 (s)
Vo (s) Z2 (s) (R1 C1 s + 1)(R2 C2 s + 1) = = Vs (s) Z1 (s) + Z2 (s) R1 C2 s + (R1 C1 s + 1)(R2 C2 s + 1)
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6.26 There are many possible diagrams, depending on how the equations are arranged. Two diagrams are shown in the figure. The diagram in part (a) results from arranging the equations from Problem 6.14 in the following way. LsI1 (s) + RI3 (s) = V1 (s) I2 (s) = RCsI3 (s) + CsV2 (s) I1 (s) − I2 (s) = I3 (s)
The diagram in part (b) results from the following solution for I2 (s) from Problem 6.14. I2 (s) =
RCsV1 (s) − (RCs + LCs2 )V2 (s) D(s)
Figure : for Problem 6.26
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6.27 There are many possible diagrams, depending on how the equations are arranged. One such diagram is shown in the figure. It results from arranging the equations from Problem 6.15 in the following way. I1 (s) =
1 [V1 (s) − V3 (s)] R + L1 s
1 [V2 (s) − V3 (s)] + I1 (s) Ls CsV3 (s) = I3 (s) = I1 (s) − I2 (s)
I2 (s) =
Figure : for Problem 6.27
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6.28 The transfer function is
Tf (s) Vo (s) =− Vi (s) Ti (s)
where Ti (s) = R1 and Tf (s) = R2 + Thus
R2 Cs + 1 1 = Cs Cs
Vo (s) R2 Cs + 1 =− Vi (s) R1 Cs
and R1 C
dvo dvi = −R2 C − vi dt dt
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6.29 The transfer function is
where Tf (s) = R3 and
Tf (s) Vo (s) =− Vi (s) Ti (s) Ti (s) = Z(s) + R2 1 1 1 = + 1 Z(s) R1 Cs
Thus Z(s) = and Ti (s) = Thus
R1 R1 Cs + 1
R1 R1 R2 Cs + R1 + R2 + R2 = R1 Cs + 1 R1 Cs + 1 Vo (s) R3 (R1 Cs + 1) =− Vi (s) R1 R2 Cs + R1 + R2
and R1 R2 C
dvo dvi + (R1 + R2 )vo = −R1 R3 C − vi dt dt
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6.30 The transfer function is
Tf (s) Vo (s) =− Vi (s) Ti (s)
where Tf (s) = R2 +
1 R2 C2 s + 1 = C2 s C2 s
and Ti (s) = Z(s) + R3 1 1 1 = + 1 Z(s) R1 C1 s Thus Z(s) = and Ti (s) = Thus
R1 R1 C1 s + 1
R1 R3 R1 C1 s + R3 + R1 + R3 = R1 C1 s + 1 R1 C1 s + 1
Vo (s) (R2 C2 s + 1)(R1 C1 s + 1) =− Vi (s) C2 s(R3 R1 C1 s + R3 + R1 )
and R3 R1 C1 C2
d2 vo dvo d2 vi dvi + (R + R )C = −R R C C − (R1 C1 + R2 C2 ) − vi 3 1 2 2 1 1 2 2 2 dt dt dt dt
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6.31 Let Z1 (s) be the forward impedance, and Z2 (s) be the feedback impedance. Then 1 1 1 = + 1 Z1 (s) R1 C1 s Z1 (s) =
R1 R1 C1 s + 1
Z2 (s) =
R2 R2 C2 s + 1
Similarly,
and
Vo (s) Z2 (s) R2 C2 s + 1 R1 =− =− Vi (s) Z1 (s) R1 C1 s + 1 R2
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6.32 Let Z1 (s) be the impedance of the series R1 C1 connection, Z2 (s) the forward impedance, Z3 (s) the impedance of the series R3 C2 connection, and Z4 (s) the feedback impedance. Then 1 R1 C1 s + 1 Z1 (s) = R1 + = C1 s C1 s 1 1 1 (R1 C1 + R2 C1 )s + 1 = + = Z2 (s) R2 Z1 (s) R1 C1 s + 1 Z3 (s) = Z4 (s) = Thus
R3 C2 s + 1 C2 s
R3 C2 s + 1 (R3 C2 + R4 C2 )s + 1
Vo (s) Z4 (s) R3 C2 s + 1 (R1 C1 + R2 C1 )s + 1 =− =− Vi (s) Z2 (s) (R3 C2 + R4 C2 )s + 1 R1 C1 s + 1
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6.33 a) From equations (3) and (4) of Example 6.5.1, (Is2 + cs + KT )Θ(s) = nBLrI(s) (Ls + R)I(s) + nBLrsΘ(s) = Vi (s) Let A = nBLr. The solution for Θ(s) gives Θ(s) A A = = 2 2 3 2 Vi (s) (Is + cs + KT )(Ls + R) + A LIs + (RI + Lc)s + (LKT + Rc)s + RKT + A2 b) Assuming all the parameters are positive, the system is stable if and only if (RI + Lc)(LKT + Rc) > LI(RKT + A2 ) If the system is stable, the final value theorem can be applied. It gives θss = lim s s→0
LIs3
+ (RI +
Lc)s2
A Vi A = Vi 2 + (LKT + Rc)s + RKT + A s RKT + A2
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6.34 a) With vf = 0, if = 0, and IsΩ(s) = −cΩ(s) − TL (s) Thus
Ω(s) 1 =− TL (s) Is + c
b) I Thus
d2 θ dθ = K T if − c − T L 2 dt dt
$
%
Is2 + cs Θ(s) = KT If (s) − TL (s)
Also,
Vf (s) = (Lf s + Rf )If (s) and therefore
This gives
$
%
Is2 + cs Θ(s) = KT
Vf (s) − TL (s) (Lf s + Rf )
Θ(s) KT KT = = 2 2 Vf (s) (Lf s + Rf )(Is + cs) s[ILf s + (Rf I + Lf c)s + Rf c] and
Θ(s) 1 =− TL (s) s(Is + c)
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6.35 See the following figure. The equivalent inertia and damping at the motor shaft are Ie = Im +
IL N2
ce =
cL + cm N2
The transfer functions are Ωf (s) KT /N = Vf (s) (Lf s + Rf )(Ie s + ce ) Ωf (s) 1/N 2 = TL (s) Ie s + ce
Figure : for Problem 6.35
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6.36 Summing moments on I1 and I2 gives I1
d2 θ1 = T − kT (θ1 − θ2 ) = KT if − kT (θ1 − θ2 ) dt2 I2
d2 θ2 = kT (θ1 − θ2 ) − cθ˙2 dt2
For the field circuit,
dif + Rif dt Transforming these equations and solving for Θ2 (s) gives vf = Lf
Θ2 (s) =
s2 (L
kT K T 2 f s + Rf )[I1 I2 s + cI1 s + kT (I1 + I2 )]
The differential equation is I1 I2 Lf
d5 θ2 d4 θ2 d3 θ2 d2 θ2 +(cI1 I2 Lf +Rf ) 4 +[Lf kT (I1 +I2 )+Rf cI1 ] 3 +Rf kT (I1 +I2 ) 2 = kT KT vf 5 dt dt dt dt
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6.37 There are two circuits and one inertia in this system, and we must write an equation for each. For the generator circuit, Kirchhoff’s voltage law gives vf = Rf if + Lf
dif dt
For the motor circuit, Kirchhoff’s voltage law gives va = Ra ia + La
dia + Kb ω dt
where va = Kf if . For the inertia I, Newton’s law gives: I
dω = T − cω − TL dt
where T = KT ia . Substitute for va and T , and rearrange to obtain: Lf
dif + Rf if = vf dt
dω + cω = KT ia − TL dt dia La + R a ia = K f if − K b ω dt I
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6.38 Equations (6.6.5) and (6.6.6) give ia =
5 × 10−4 Va + 0.2TL 7.5 + 200TL = −4 2 (0.8)5 × 10 + (0.2) 40.4 ω=
0.2Va − 0.8TL 300 − 80TL = −2 4.04 × 10 4.04
(1)
(2)
Setting TL = 0 in (2) gives the noload speed ω = 300/4.04 = 74.26 rad/s or 709 rpm. From (1) we obtain the noload current: ia = 7.5/40.4 = 0.186 A. Setting ω = 0 in (2) gives the stall torque TL = 0.2(15)/0.8 = 3.75 N·m. From (1) we obtain the stall current: ia = [7.5 + 200(15/4)]/4.04 = 187.5 A.
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6.39 (a) The characteristic equation is La Is2 + (La c + Ra I)s + Ra c + Kb KT = 0 The roots are s=
−(La c + Ra I) ±
&
Ra2 I 2 − 2Ra ILa c + L2a c2 − 4La IKb KT 2La I
b) For c = 0 the roots are −219 and −47.5. The speed and the current will not oscillate. It will take about 4(1/47.5) = 0.08 second for the speed and the current to become constant. For c = 0.01 the roots are −196 ± 73.5i. The speed and current will oscillate with a frequency of 73.5 rad/s and a period of 0.085 s. It will take about 4(1/196) = 0.02 second for the speed and current to become constant. Because this time is less than the period, we will not see any oscillations in the plots. For c = 0.1 the roots are −1240 and −277. The speed and current will not oscillate. It will take about 4(1/277) = 0.014 second for them to become constant.
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6.40 Use the transfer functions given by (6.6.1) through (6.6.4). All four transfer functions have the same denominator, which is D(s) = 20 × 10−7 s2 + 4.02 × 10−4 s + 4 × 10−4 Current Response to Command Input Assuming va is a unit step input, we have from (6.6.1): 5 × 10−4 (s + 1) 250(s + 1) Ia (s) = = (1) 2 sD(s) s(s + 201s + 2020) The characteristic roots are s = −100.5 ± 100.5j. Thus the denominator can be expressed as (s + 100.5)2 + (100.5)2 . Expanding Ia (s) in a partial fraction expansion gives Ia (s) =
C1 100.5C2 (s + 100.5)C3 + + 2 2 s (s + 100.5) + (100.5) (s + 100.5)2 + (100.5)2
(2)
This gives a solution of the form: ia (t) = C1 + C2 e−100.5t sin 100.5t + C3 e−100.5t cos 100.5t Reducing (2) to a single fraction using the least common denominator s[(s + 100.5)2 + (100.5)2 ], we obtain Ia (s) =
C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s s[(s + 100.5)2 + (100.5)2 ]
(3)
Comparing the numerators of (1) and (3), we obtain C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s = 250(s + 1) or (C1 + C3 )s2 + (201C1 + 100.5C2 + 100.5C3 )s + 20, 200C1 = 250s + 250 Comparing like powers of s gives C1 = −C3 = 0.01238
C2 = 2.47518
Thus, after multiplying by 10, the magnitude of the step input, we have ia (t) = 0.1238 + 24.7518e−100.5t sin 100.5t − 0.1238e−100.5t cos 100.5t Speed Response to Command Input Assuming va is a unit step input, we have from (6.6.3): 0.2 105 Ω(s) = = (4) sD(s) s(s2 + 201s + 2020) Expanding Ω(s) in a partial fraction expansion gives Ω(s) =
C1 100.5C2 (s + 100.5)C3 + + 2 2 s (s + 100.5) + (100.5) (s + 100.5)2 + (100.5)2
(5)
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This gives a solution of the form: ω(t) = C1 + C2 e−100.5t sin 100.5t + C3 e−100.5t cos 100.5t Reducing (5) to a single fraction using the least common denominator s[(s + 100.5)2 + (100.5)2 ], we obtain Ω(s) =
C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s s[(s + 100.5)2 + (100.5)2 ]
(6)
Comparing the numerators of (5) and (6), we obtain C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s = 105 or (C1 + C3 )s2 + (201C1 + 100.5C2 + 100.5C3 )s + 20, 200C1 = 105 Comparing like powers of s gives C1 = −C3 = 4.9505
C2 = −C1 = −4.9505
(7)
Thus, after multiplying by 10, the magnitude of the step input, we have $
ω(t) = 49.505 1 − e−100.5t sin 100.5t − e−100.5t cos 100.5t
%
Current Response to Disturbance Input Comparing (6.6.2) with (6.6.3), and noting that Kb = KT , we see that Ia (s) Kb Ω(s) Ω(s) = = TL (s) KT Va (s) Va (s) Thus the response of ia to a unitstep disturbance TL will be the same as the response of ω to a unitstep voltage va . Thus we can use the coefficients given in (7) and multiply them by 0.2, the magnitude of TL , to obtain $
ıa (t) = 0.9901 1 − e−100.5t sin 100.5t − e−100.5t cos 100.5t
%
Speed Response to Disturbance Input To obtain the response of ω to TL , use (6.6.4). Assuming TL is a unit step input, we have from (6.6.4): Ω(s) =
4 × 10−3 s + 0.8 5 × 10−4 s + 0.1 = sD(s) s(s2 + 201s + 2020)
(8)
Expanding Ω(s) in a partial fraction expansion gives Ω(s) =
C1 100.5C2 (s + 100.5)C3 + + 2 2 s (s + 100.5) + (100.5) (s + 100.5)2 + (100.5)2
(9)
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This gives a solution of the form: ω(t) = C1 + C2 e−100.5t sin 100.5t + C3 e−100.5t cos 100.5t Reducing (9) to a single fraction using the least common denominator s[(s + 100.5)2 + (100.5)2 ], we obtain Ω(s) =
C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s s[(s + 100.5)2 + (100.5)2 ]
(10)
Comparing the numerators of (9) and (10), we obtain C1 [(s + 100.5)2 + (100.5)2 ] + [100.5C2 + (s + 100.5)C3 ]s = 5 × 10−4 s + 0.1 or (C1 + C3 )s2 + (201C1 + 100.5C2 + 100.5C3 )s + 20, 200C1 = 5 × 10−4 s + 0.1
Comparing like powers of s gives
C1 = −C3 = 4.9505 × 10−6
C2 = −4.187 × 10−7
Thus, after multiplying by 0.2, the magnitude of the step input, we have ω(t) = 9.901 × 10−7 − 8.37 × 10−8 e−100.5t sin 100.5t − 9.901 × 10−7 e−100.5t cos 100.5t
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6.41 Given Va = 20 and Istall = 25
ino
load
= 0.6
Thus Ra =
Va istall
ωno
=
load
=
2400 2π = 80π rad/s 60
20 = 0.8 Ω 25
From (6.6.6) and (6.6.6) with TL = 0 and Kb = KT , ino
load
= 0.6 =
ωno
load
= 80π =
20c 0.8c + KT2 20KT 0.8c + KT2
These two equations have the solution KT = Kb = 0.078 N·m/A and c = 1.85 × 10−4 N·m·s/rad.
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6.42 Refer to Table 6.6.1. θf = 3π/4 rad. Also, I = 0.215 t1 = 0.3
Td = 4.2 t2 = 1.7
tf = 2 R=4
KT = 0.3 = Kb Note that L is not needed. The calculated quantities are E = 35.7 J per cycle ωmax = 1.386 rad/s = 13.2 rpm Tmax = 5.19 N · m Trms = 4.24 N · m imax = 17.3 A irms = 14.1 A vmax = 69.6 V
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6.43 Refer to Table 6.6.1. θf = 11(2π) = 22π rad. Also, I = 0.05
Td = 3.6
tf = 3
t1 = 0.5
t2 = 2.5
R=3
KT = 0.4 = Kb Note that L is not needed. The calculated quantities are E = 872 J per cycle ωmax = 27.65 rad/s = 264 rpm Tmax = 6.36 N · m Trms = 3.94 N · m imax = 15.9 A irms = 9.85 A vmax = 58.8 V
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6.44 a) The transfer function is Y (s) s2 = T (s) = 2 Z(s) s + 18s + 100 With s = 120j we obtain ' ' ' ' −(120)2 ' ' M = T (120j) = ' ' = 0.9957 ' 100 − (120)2 + 18(120)j '
φ = # T (120j) = − tan−1
Thus the steady state response is
18(120) = −2.992 100 − (120)2
y(t) = 9.957 sin(120t − 2.992) b) The transfer function is Y (s) s2 = T (s) = 2 Z(s) s + 1800s + 106 With s = 120j we obtain ' ' ' ' −(120)2 ' ' M = T (120j) = ' 6 ' = 0.0144 ' 10 − (120)2 + 1800(120)j '
φ = # T (120j) = − tan−1
Thus the steady state response is
1800(120) = −0.2157 106 − (120)2
y(t) = 0.144 sin(120t − 0.2157) For case (a), the amplitude of the response (9.957) is almost identical to the amplitude of the displacement input (10). Thus the instrument functions as a vibrometer. For case (b), the amplitude of the response (0.144) is equal to 10−6 times the amplitude of the acceleration of the input, which is 10(120)2 = 0.144 × 106 . Thus the instrument functions as an accelerometer with a gain of 10−6 .
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6.45 Summing forces on the mass m gives m
d2 x dx +c + kx = fs + Kf i 2 dt dt
or (ms2 + cs + k)X(s) = Fs (s) + Kf I(s)
(1)
Since the applied voltage is zero, the circuit equation is L
di dx + Ri + Kb =0 dt dt
or (Ls + R)I(s) + Kb sX(s) = 0
(2)
Solving (1) for X(s) and substituting into (2), and rearranging, gives the transfer function. I(s) −Kb s = 3 2 Fs (s) mLs + (cL + mR)s + (kL + cR + Kb Kf )s + kR
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6.46 a) The roots of the secondorder model are s = −4333 ± 13 462j. b) The roots of the thirdorder model are s = −1643 ± 16 513j, which is the dominant pair, and s = −8715.
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6.47 The script file is KT = 0.2; Kb = 0.2; c = 5e4; Ra = 0.8; La = 4e3; I = 5e4; den = [La*I, Ra*I + c*La, c*Ra + Kb*KT]; % Speed transfer function sys1 = tf(KT, den); % Current transfer function sys2 = tf([I, c], den); [om, t1] = step(sys1); [ia, t2] = step(sys2); subplot(2,1,1) plot(t1, 10*om),xlabel($t (s)$ ),ylabel($\omega (rad/s)$ ) subplot(2,1,2) plot(t1, 10*ia),xlabel($t (s)$ ),ylabel($i_a (A)$ ) The plot is shown in the following figure. The peak current is approximately 8 A. The more accurate value of 8.06 A can be found with the step(sys2) function by rightclicking on the plot, selecting characteristics, then peak response, and multiplying the answer by 10. 60
! (rad/s)
50 40 30 20 10 0
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
10 8
ia (A)
6 4 2 0 −2
Figure : for Problem 6.47
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6.48 The script file is KT = 0.2; Kb = 0.2; c = 5e4; Ra = 0.8; La = 4e3; I = 5e4; den = [La*I, Ra*I + c*La, c*Ra + Kb*KT]; % Speed transfer function sys1 = tf(KT, den); % Current transfer function sys2 = tf([I, c], den); % Applied voltage t = [0:0.001:0.05]; va = 10*ones(size(t)); om = lsim(sys1, va, t); ia = lsim(sys2, va, t); subplot(2,1,1) plot(t, om),xlabel($t (s)$ ),ylabel($\omega (rad/s)$ ) subplot(2,1,2) plot(t, ia),xlabel($t (s)$ ),ylabel($i_a (A)$ ) The plot is shown in the following figure. The peak current is approximately 8 A. The more accurate value of 8.06 A can be found with the lsim(sys2, va, t) function by rightclicking on the plot, selecting characteristics, then peak response.
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60
! (rad/s)
50 40 30 20 10 0
0
0.005
0.01
0.015
0.02
0.025 t (s)
0.03
0.035
0.04
0.045
0.05
0
0.005
0.01
0.015
0.02
0.025 t (s)
0.03
0.035
0.04
0.045
0.05
10 8
ia (A)
6 4 2 0 −2
Figure : for Problem 6.48
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6.49 The resistance R was incorrectly given as 104 Ω. It should be 103 Ω. See Example 6.3.3 for the transfer functions. The script file is R = 1e+3; C = 2e6; L = 2e3; den = [L*R*C, L, R]; % Transfer function for v1: sys1 = tf(1, den); % Transfer function for v2: sys2 = tf([R*C, 0], den); % Response to unitstep voltage v1: t = [0:0.00001:0.025]; i31 = step(sys1, t); % Voltage v2: v2 = 4*sin(2*pi*60*t); % Response to voltage v2: i32 = lsim(sys2, v2, t); % Total response: i3 = 5*i31 + i32; plot(t, i3),xlabel($t (s)$ ),ylabel($i_3 (A)$ ) The plot is shown in the following figure. The steadystate response is a sine wave.
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0.016
0.014
0.012
3
i (A)
0.01
0.008
0.006
0.004
0.002
0
0
0.005
0.01
0.015
0.02
0.025
t (s)
Figure : for Problem 6.49
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6.50 See equations (6.6.1) through (6.6.4) for the transfer functions, where I and c must be replaced with the equivalent inertia and equivalent damping: Ie = I + 4 × 10−4 /N 2 and ce = c + 1.8 × 10−3 /N 2 . The script file is KT = 0.2; Kb = 0.2; c = 3e4; Ra = 0.8; La = 4e3; I = 4e4; N = 3; % Equivalent inertia: Ie = I + 1e3/N^2; % Equivalent damping: ce = c + 1.8e3/N^2; den = [La*Ie, Ra*Ie + ce*La, ce*Ra + Kb*KT]; % Speed command transfer function sys1c = tf(KT, den); % Current command transfer function sys2c = tf([I, c], den); % Speed disturbance transfer function sys1d = tf([La, Ra], den); % Current disturbance transfer function sys2d = tf(Kb, den); % Unit Command response: [omc, t1c] = step(sys1c); [iac, t2c] = step(sys2c); % Unit Disturbance response: omd = step(sys1d, t1c); iad = step(sys2d, t2c); % Total response: om = 20*omc + 0.04*omd; ia = 20*iac + 0.04*iad; subplot(2,1,1) plot(t1c, om),xlabel($t (s)$ ),ylabel($\omega (rad/s)$) subplot(2,1,2) plot(t2c, ia),xlabel($t (s)$ ),ylabel($i_a (A)$ ) The plot is shown in the following figure. The peak current is 12.73 A, which can be found by displaying the values in the array ia.
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120
! (rad/s)
100 80 60 40 20 0
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
0
0.01
0.02
0.03 t (s)
0.04
0.05
0.06
15
ia (A)
10
5
0
−5
Figure : for Problem 6.50
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6.51 The main script file is Problem6p51.m. % Program Problem6p51.m KT = 0.05; Kb = KT; c = 0; Ra = 0.8; La = 3e3; I = 8e5; trapezoid motortf current = lsim(currenttf, v, t); speed = lsim(speedtf, v, t); subplot(2,1,1) plot(t,current),xlabel($t (s)$ ),ylabel($Current (A)$ ) subplot(2,1,2) plot(t,speed),xlabel($t (s)$ ),ylabel($Speed (rad/s)$ ) performance
This program calls the following files. % trapezoid.m Trapezoidal voltage profile t1 = 0.5; t2 = 2; tfinal = 2.5; t3 = 4; max_v = 30; dt = t3/1000; t = [0:dt:t3]; for k = 1:1001 if t(k) <= t1 v(k) = (max_v/t1)*t(k); elseif t(k) <= t2 v(k) = max_v; elseif t(k) <= tfinal v(k) = (max_v/t1)*(tfinal  t(k)); else v(k) = 0; end end
% motortf.m Motor transfer functions for voltage input. % current: currenttf = tf([I,c],[La*I,Ra*I+c*La,c*Ra+Kb*KT]); % speed: speedtf = tf(KT,[La*I,Ra*I+c*La,c*Ra+Kb*KT]);
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% performance.m Computes motor performance measures. ia = current; dt = t(2)  t(1); E = trapz(t,Ra*ia.^2) + trapz(t,c*speed.^2) i_max = max(ia) i_rms = sqrt(trapz(t,ia.^2)/t3) T_max = KT*i_max T_rms = KT*i_rms speed_max = max(speed) v_max = Ra*i_max+Kb*speed_max The plots are shown in the following figure. The performance results are E = 2.8203 J/cycle
imax = 1.9200 A
Tmax = 0.0960 N · m
speedmax = 600 rad/s
irms = 0.9388 A
Trms = 0.0469 N · m vmax = 31.536 V
2
Current (A)
1
0
−1
−2
0
0.5
1
1.5
2 t (s)
2.5
3
3.5
4
0
0.5
1
1.5
2 t (s)
2.5
3
3.5
4
700
Speed (rad/s)
600 500 400 300 200 100 0
Figure : for Problem 6.51
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6.52 The equations are La
dia = va − Ra ia − Kb ω dt dθ = ω (2) dt
(1)
dω = KT ia − ce ω − 4.2 sin θ (3) dt Choose the three state variables to be x1 = ia , x2 = θ, and x3 = ω. The slew speed required for a rotation angle of 3π/4 rad is ωslew = 3π/4/t2 = 1.386 rad/s. Thus, letting A represent ωslew , the specified trapezoidal speed profile is Ie
0 ≤ t ≤ t1 ω(t) = A t 1 ≤ t ≤ t2 − A (t − t ) + A t ≤ t ≤ t 2 2 f inal t1 At t1
The specified angle θ(t) resulting from the trapezoidal speed profile can be found by integrating that profile. The result is θ(t) =
At2 2t1
A(t − t ) +
At1 2
1 − A(t−t2 )2 + A(t − t ) + At − 2 2 2t1
0 ≤ t ≤ t1 t1 ≤ t ≤ t2 t2 ≤ t ≤ tf inal
At1 2
The voltage required to move the arm at the desired angle θ(t) is found from (1): va = La
dia + R a ia − K b ω dt
(4)
We obtain ia from (3): 1 ia (t) = KT
!
dω Ie + ce ω + 4.2 sin θ dt
"
(5)
Note that dω/dt = α, the angular acceleration, where α(t) = Thus (5) can be expressed as ia (t) =
A t1
0
−A t1
0 ≤ t ≤ t1 t1 ≤ t ≤ t2 t2 ≤ t ≤ tf inal
1 (Ie α + ce ω + 4.2 sin θ) KT
(6)
Differentiate this to obtain dia /dt: dia 1 = dt KT
!
Ie
dα + ce α + 4.2θ˙ cos θ dt
"
(7)
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Note that θ˙ = ω and that dα/dt = 0. Also, for this problem, ce = 0. Thus (7) becomes dia 1 = 4.2ω cos θ dt KT
(8)
Thus, with ce = 0, (4) becomes va =
4.2La Ra ω cos θ + (Ie α + 4.2 sin θ) − Kb ω KT KT
(9)
In summary, va is computed from (9) using the expressions for α(t), ω(t) and θ(t). The computed values of va are then used with (2) and (3) to compute the actual values of ω(t) and θ(t). The mfiles required for this are shown below. % Program problem6p52.m global KT Kb La Ra Ie global t1 t2 tfinal A KT = 0.3; Kb = 0.3; Ra = 4; La = 3e3; Ie = 0.215; t1 = 0.3;t2 = 1.7; tfinal = 2; % A is the slew speed A = (3*pi/4)/t2; [t, x] = ode45($ problem6p52a$, [0, tfinal], [0, 0, 0]); subplot(3,1,1) plot(t, x(:,1)),ylabel($i_a (A)$ ) subplot(3,1,2) plot(t, x(:,2)),ylabel($\theta (rad)$ ) subplot(3,1,3) plot(t, x(:,3)),xlabel($t (s)$ ),ylabel($\omega (rad/s)$ )
function xdot = problem6p52a(t,x) % Differential equation file for program problem6p52.m global KT Kb La Ra Ie global t1 t2 tfinal A % Evaluate trapezoidal speed profile % om is the speed if t <= t1 om = (A/t1)*t; elseif t <= t2 om = A; else om = (A/t1)*(tt2) + A; end % Evaluate angle profile 662 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
% theta is the angle if t <= t1 theta = (A/(2*t1))*t.^2; elseif t <= t2 theta = A*(tt1)+A*t1/2; else theta = (A/(2*t1))*(tt2).^2 + A*(tt2) + A*t2A*t1/2; end % Evaluate angular acceleration profile % alpha is the acceleration if t <= t1 alpha = A/t1; elseif t <= t2 alpha = 0; else alpha = A/t1; end % Evaluate the voltage profile % va is the voltage va = 4.2*(La/KT)*om*cos(theta) + (Ra/KT)*(Ie*alpha + 4.2*sin(theta))  Kb*om; % x(1) is the current; x(2) is the angle; x(3) is the speed xdot = [(Ra*x(1)Kb*x(3)+va)/La;x(3);(KT*x(1)4.2*sin(x(2)))/Ie]; The simulation results are shown in the following figure. The angle θ does not quite reach the desired value of 3π/4. From the velocity plot we see that the velocity starts to fall below that specified by the trapezoidal profile. This is due to the approximation made by neglecting dα/dt at the corners of the velocity profile, at the time t1 and t2 . Because α, the slope of the profile, changes suddenly at these times, dα/dt is undefined, and thus our approximation does not represent a true solution of the differential equations. This problem is an example of openloop control in which the voltage is precalculated from the differential equations. Thus the effectiveness of this method depends on the accuracy of the differential equation model and on any approximation made to solve for the voltage. The method is also tedious, as this problem illustrates. A better way to control the arm angle is to use feedback control, which is discussed in Chapter 10.
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ia (A)
15 10 5 0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0
0.2
0.4
0.6
0.8
1 t (s)
1.2
1.4
1.6
1.8
2
" (rad)
3 2 1 0
! (rad/s)
2 1 0 −1 −2
Figure : for Problem 6.52
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6.53 From Example 6.3.1, I3 (s) =
V1 (s) + RCsV2 (s) LRCs2 + Ls + R
To avoid using a Transfer Function block containing the illegal transfer function RCs (which is a differentiator), we can write the equation as follows: I3 (s) =
RCsV1 (s) RCsV2 (s) + 2 RCs(LRCs + Ls + R) LRCs2 + Ls + R
The corresponding Simulink model is shown in the following figure. In the Sine block, set the Amplitude to 4 and the Frequency to 120*pi. In the Pulse Generator block, set the Amplitude to 5, the Period to 1, and the Pulse Width to 5. Set the Stop Time to whatever value you want, greater than 0.05 to see the full pulse but less than 1 so that the pulse will not repeat; say 0.075. In the To Workspace block, set the Save Format to array. Before running the model, type the following in the MATLAB Command window. R = 1e4; C = 2e6; L = 2e3; You can plot the response by typing plot(tout, simout(:,1)). Note that if you use a larger Stop time, the buffer for tout may overflow. The number of points to be retained in tout can be increased if necessary. This is done under the Configuration Parameters menu.
Figure : for Problem 6.53
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6.54 From Figure 6.3.2b we can create the corresponding Simulink model, which is shown in the following figure. In the Sine block, set the Amplitude to 3 and the Frequency to 120*pi. In the Step block, set the Step time to 0, the Initial value to 0, and the Final value to 12. Set the Stop Time to an appropriate value, say 0.5. In the To Workspace block, set the Save Format to Array. Before running the model, type the following in the MATLAB Command window. R = 2e4; C = 3e6; You can plot the response v1 (t) by typing plot(tout, simout(:,1)). You can plot the response v2 (t) by typing plot(tout, simout(:,2)).
Figure : for Problem 6.54
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6.55 The Simulink model is like that shown in Figure 6.9.5, with the disturbance torque Td set to 0. In the Step block, set the Step time to 0, the Initial value to 0, and the Final value to 10. Set the Stop Time to an appropriate value, say 0.1. In the To Workspace block, set the Save Format to Array. Before running the model, in the MATLAB Command window, type KT = 0.2; Kb = KT; c = 5e4; Ra = 0.8; La = 4e3; I = 5e4; For part (a), set the limits on the saturation block very high, say ±100. You can plot the torque response by typing plot(tout, simout(:,1)). You can plot the speed response by typing plot(tout, simout(:,2)). The following figure shows the two plots created with the subplot function. The maximum torque is 1.6 N·m. 2
Torque (N m)
1.5 1 0.5 0 −0.5
0
0.01
0.02
0.03
0.04
0.05 t (s)
0.06
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.05 t (s)
0.06
0.07
0.08
0.09
0.1
60
Speed (rad/s)
50 40 30 20 10 0
Figure : for Problem 6.55a
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For part (b), set the limits on the saturation block to ±0.8. The resulting responses are shown in the following figure. 0.8
Torque (N m)
0.6 0.4 0.2 0 −0.2
0
0.01
0.02
0.03
0.04
0.05 t (s)
0.06
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.05 t (s)
0.06
0.07
0.08
0.09
0.1
60
Speed (rad/s)
50 40 30 20 10 0
Figure : for Problem 6.55b
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6.56 The model is shown in the following figure. Set the Initial output to 0 in all four Ramp blocks. In the topmost Ramp block, set the Slope to 60 and the Start time to 0. In the Ramp1 block, set the Slope to −60 and the Start time to 0.5. In the Ramp2 block, set the Slope to −60 and the Start time to 2. In the Ramp3 block, set the Slope to 60 and the Start time to 2.5. Set the Stop Time to 4. In the To Workspace block, set the Save Format to array. Before running the model, type the following in the MATLAB Command window. KT = 0.05;Kb = 0.05;c = 0;Ra = 0.8;La = 3e3;I = 8e5; You can plot the current response by typing plot(tout, simout(:,1)) and the speed response by typing plot(tout, simout(:,2)).
Figure : for Problem 6.56
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Seven
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c Solutions Manual Copyright 2005 by The McGrawHill Companies, Inc. !
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7.1 Note that f1 = pA1 and f2 = pA2 . Thus f1 = (A1 /A2 )f2 = (10/30)60 = 20 lb. Also, A1 x1 = A2 x2 from conservation of fluid mass. Thus x1 = (A2 /A1 )x2 = (30/10)6 = 18 in. The work done is f2 x2 = 60(6) = 360 lbin.
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7.2 The crosssectional area is A = π(11/2)2 = 95.033 ft2 . The net inflow rate is (1000 − 800)(0.13368) = 26.736 ft3 /min The initial volume is 5A = 475.166 ft3 , and the volume after 5 hrs (300 minutes) is 475.166 + 26.736(300) = 8495.97 ft3 Thus the height after 5 hrs is h=
8495.97 8495.97 = = 89.4 ft A 95.033
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7.3 Summing forces in the horizontal direction and assuming zero acceleration, we obtain µmg = A(p1 − p2 ) or A=
µmg 0.6(1000) = = 0.003 m2 p1 − p2 (3 − 1) × 105
This corresponds to a radius of 30 mm.
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7.4 Let ft be the tangential force of the surface acting on the cylinder (positive to the left). Summing moments in the clockwise direction about the mass center of the cylinder gives I ω˙ = Rft
(1)
From kinematics, if there is no slipping, x˙ = Rω and thus x ¨ = Rω. ˙ Summing horizontal forces on the cylinder gives m¨ x = f − ft Thus ft = f − m¨ x. Substituting this into (1) gives I ω˙ = Rf − mR¨ x = Rf − mR(Rω) ˙ or (I + mR2 )ω˙ = Rf = R(p1 − p2 )A
With the given values, this becomes !
"
7 + 100(0.4)2 ω˙ = 0.4(3 × 105 )0.005 = 600
or
ω˙ = 26.087 Thus ω(t) = 26.087t
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7.5 Assuming no friction and summing horizontal forces, we obtain m¨ x = A(p1 − p2 ) or
600 3 x ¨= (10)144 = 10 32.2 144
or
x ¨ = 0.537 ft/sec2
Thus x˙ = 0.537t and x= The volume is V = Ax =
0.537 2 t = 0.268t2 2 3 0.067 = 0.001396 ft3 144
or 2.412 in3 .
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7.6 From conservation of water mass, qo = 10 + 2 = 12 m3 /s. From conservation of salt mass, d (300so ) = 2si − qo so dt or 300
dso = 2si − 12so dt
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7.7 From conservation of water mass, qo = 10 + 2 = 12 m3 /s. From conservation of salt mass, d (V so ) = 2si − qo so dt or
dso = 2si − 12so dt The time constant is τ = V /12. Taking the lag time to be four time constants, we have V
4τ = 4 which gives V =
V = 20 s 12
20(12) = 60 m3 4
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7.8 The capacitance can be computed from (7.2.3). C= The liquid surface area is Thus
A(h) g #
A(h) = 2L Dh − h2 C=
2L # Dh − h2 g
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7.9 The capacitance can be computed from (7.2.3). C=
A(h) g
The liquid surface area is A = D1 L if h < D2 and A= Thus C=
2L(h − D2 ) + D1 L tan φ $
D1 L g 2L(h−D2 ) D1 L g tan φ + g
h ≥ D2 h < D2 h ≥ D2
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7.10 From (7.2.5), dh = qmi − qmo dt here. From Problem 7.8, ρA(h)
where qmo = 0 and qm = qmi
#
A(h) = 2L Dh − h2 Thus
#
ρ2L Dh − h2
dh = qm dt
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7.11 From (7.2.5), ρA(h)
dh = qmi − qmo dt
where qmo = 0 here. From Problem 7.9, A(h) = Thus the model is
$
D1 L 2L(h−D2 ) + tan φ
h < D2 D1 L h ≥ D2
dh = qmi h < D2 dt % & 2L(h − D2 ) dh ρ + D1 L = qmi h ≥ D2 tan φ dt ρD1 L
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7.12 For the laminar resistance R=
128µL πρD 4
where L = 1 m, D = 10−3 m. Thus R=
128(1.58 × 10−5 )1 = 4.996 × 108 m−1 s−1 π(0.12885)(10−3 )4
The mass flow rate is qm =
p 1.0133 × 104 = = 2.028 × 10−5 kg/s R 4.996 × 108
The average velocity is found from v¯ =
qm 2.028 × 10−5 = = 20 m/s Aρ π(10−3 /2)2 (1.2885)
The Reynolds number is Ne =
ρ¯ vD 1.2885(20)10−3 = = 1634 µ 1.58 × 10−5
which is less than 2300, so the flow is laminar. The maximum entrance length Le is found from Ne = 0.06DNe = 0.06(10−3 )1634 = 0.098 m Since this length is much less than the pipe length of 1 m, most of the pipe has laminar flow.
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7.13 From (7.3.1) R= The flow rate is
'
dp dqm
(
qm =
)
where Ro =
(1)
r
p Ro
1 2ρCd2 A2o
Thus 2 p = R o qm
and from (1), R = 2Ro qmr = 2Ro But pr = ρghr , so R = 2Ro
*
)
pr Ro
ρghr 1 # = 2ghr Ro Cd Ao
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7.14 a) From (7.3.12), C
dp = qmi − qmo dt
where C= and qmo = Thus
A g
)
p R1
A dp = qmi − g dt
b) From conservation of mass, ρA
)
p R1
dh = qmi − qmo = qmi − dt
But p = ρgh, so dh ρA = qmi − − dt
*
)
p R1
ρgh R1
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7.15 a) The model is dh g =− h dt R where A = 20. The time constant is τ = RA/g. Taking the time to empty to be 4τ , we obtain τ = 200/4 = 50 s. Thus A
R=
τg = 24.525 m−1 s−1 A
b) The model is A Thus hss =
dh g =3− h dt R
3R 3(24.525) = = 7.5 m g 9.81
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7.16 a) The model is A The time constant is τ=
dh g =− h dt R
RA 150(2) = = 93.168 sec g 32.2
The time to empty is approximately 4τ , regardless of the initial height, and is 4τ = 372.671 sec. b) The model is g dh A = 0.1 − h dt R The time constant is τ = 93.168 and the steadystate height is hss = 0.1R/g = 0.466 ft. The response is + , h(t) = 0.466 1 − e−t/93.168 (1) Setting h(t) = hss /3 = 0.466/3 in (1) gives
1 = 1 − e−t/93.168 3 or t = −93.168 ln
2 = 37.776 sec 3
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7.17 The model is A At steady state,
# dh = qvi − Cd Ao 2gh dt #
qvi = Cd Ao 2gh so that
qvi Cd Ao = √ 2gh
(1)
Note that 1 cm is 10−2 m and that 1 liter/min =
10−3 = 1.667 × 10−5 m3 /s 60
So (1) in consistent units becomes qvi (1.667 × 10−5 ) qvi Cd Ao = # = 3.763 × 10−5 √ −2 2(9.81)h(10 ) h
(2)
where qvi is in liters/min and h is in cm. We will compute Cd Ao from (2) for each data pair in the table, and then average the results. The answers are Cd Ao = (6.7329 6.7350 6.9899 7.0619 7.1843 7.2870 7.3867 7.9023 8.6029 10.6456) × 104 The mean of the Cd Ao values is 7.6528 × 104 , and the standard deviation is 1.1936 × 104 , which is 16% of the mean.
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7.18 Let p1 − p3 denote the pressure drop from the bottom of the tank to the outlet of pipe 3. Denote the pressure drop over the length of pipe 1 by ∆p1 , that across pipe 3 by ∆p3 , and that across the component by ∆p2 . The turbulent resistance relation (7.3.3) is 2 = p − p . Thus, because the mass flow rate q Rqm 1 3 m is the same through each element, 2 2 2 R1 qm = ∆p1 , R2 qm = ∆p2 , and R3 qm = ∆p3 . The total pressure drop across all three elements is the sum of the drops across each element. Thus 2 2 2 p1 − p3 = ∆p1 + ∆p2 + ∆p3 = R1 qm + R 2 qm + R 3 qm
or 2 p1 − p3 = (R1 + R2 + R3 ) qm
and thus the total resistance is R = R1 + R2 + R3 , which shows that turbulent resistance obeys the series law. b) From conservation of mass, dh ρA = qmi − qmo = qmi − dt Because p1 − p3 = ρgh, the model becomes dh ρA = qmi − dt
*
)
p1 − p3 R
ρgh R
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7.19 From (7.3.3), dh ρA = qmi − dt
*
ρgh R1
where p = ρgh.
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7.20 If h < D, ρA If h ≥ D,
ρA
dh 1 = ps + qmi dt R1
dh 1 1 = ps + qmi − ρg(h − D) dt R1 R2
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7.21 a) The model is dh ρg = qmi − h dt R
ρA If there is no inflow,
dh g =− h dt R The time constant is τ = RA/g and the response is A
h(t) = h(0)e−t/τ Take the log of both sides: 1 ln h(t) = ln h(0) − t τ
(1)
Equation (1) has the form of the equation of a straight line: ln h(t) = mt + b where m = −1/τ and b = ln h(0). Use the leastsquares method to find m and b. The MATLAB code is t = [0:300:2400]; h = [20.2, 17.26, 14.6, 12.4, 10.4, 9, 7.6, 6.4, 5.4]; lnh = log(h); coeff = polyfit(t,lnh,1) where m is given by coeff(1) and b is given by coeff(2). The results are m = −5.48868 × 10−4 and b = 3.01. Thus, τ = 1.822 × 103 and R=
gτ = 9.778 × 103 ft−1 sec−1 A
b) If h(0) is known to be exactly 20.2 ft, then (1) becomes 1 ln h(t) − 3.006 = − t = mt τ
(2)
Using (1.6.3) from Chapter 1, we have m
9 i=1
Continue the code above as follows
t2i =
9 
ti ln h(ti )
i=1
m = sum(t.*(lnhlog(20.2)))/sum(t.^2) This gives m = −5.463 × 10−4 . Thus, R = 9.824 × 103 . 722 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.22 Applying conservation of mass to each tank gives ρA1 ρA2
dh1 ρg h1 = qmi − dt R1
ρg dh2 ρg h1 − h2 = dt R1 R2
Note that ρ cancels out in the second equation.
723 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.23 a) Applying conservation of mass to each tank gives ρA1 ρA2
dh1 ρg = qmi − (h1 − h2 ) dt R1
dh2 ρg ρg = (h1 − h2 ) − h2 dt R1 R2
Note that ρ cancels out in the second equation. b) Substituting the given values we obtain ρA
dh1 ρg = qmi − (h1 − h2 ) dt R
dh2 g g = (h1 − h2 ) − h2 dt R 3R Applying the Laplace transform with zero initial conditions, we obtain 4A
'
As +
(
g g Qmi (s) H1 (s) − H2 (s) = R R ρ '
(
g 4g − H1 (s) + 4As + H2 (s) = 0 R 3R Let b = g/RA. After dividing both equations by A, they can be expressed as (s + b) H1 (s) − bH2 (s) = '
Qmi (s) ρ
(
4 −bH1 (s) + 4s + b H2 (s) = 0 3 Using Cramer’s method to solve these equations, we obtain
where
Thus
. . s + b Q (s)/ρ . mi H2 (s) = . . −b 0
. . s+b −b . D=. . −b 4s + 4b/3
. . bQmi (s) . . /D = . ρD
. . 4 16 1 . bs + b2 . = (s + b)(4s + b) − b2 = 4s2 + . 3 3 3
H2 (s) b/ρ = 2 Qmi (s) 4s + (16/3)bs + (1/3)b2
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7.24 Applying conservation of mass to each tank gives ρA
dh1 ρg = − (h1 − h2 ) dt R
dh2 ρg ρg = qmi + (h1 − h2 ) − h2 dt R 3R If we divide both equations by ρA and let b = g/RA, these equations can be expressed as 2ρA
dh1 = −b(h1 − h2 ) dt 2
dh2 qmi b = + b(h1 − h2 ) − h2 dt ρ 3R
Applying the Laplace transform with zero initial conditions, we obtain (s + b) H1 (s) − bH2 (s) = 0 '
(
4 Qmi (s) −bH1 (s) + 2s + b H2 (s) = 3 ρ Using Cramer’s method to solve these equations, we obtain . . 0 . H1 (s) = . . Qmi (s)/ρ
where
Thus
. . s+b −b . D=. . −b 2s + 4b/3
−b 2s + 4b/3
. . bQmi (s) . . /D = . ρD
. ' ( . 4 10 1 . bs + b2 . = (s + b) 2s + b − b2 = 2s2 + . 3 3 3
H2 (s) b/ρ = 2 Qmi (s) 2s + (10/3)bs + (1/3)b2
The characteristic roots are s=
−10b/3 ±
#
'
(
100b2 /9 − 8b2 /3 5 1√ = − ± 19 b = −1.56b, −0.1069b 4 6 6
725 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.25 From Example 7.4.4 the damping constant is given by 128µLA2 πD 4 Substituting the given and desired values, we obtain c=
2000 =
128(0.9)LA2 πD 4
which gives
LA2 2000π = = 54.5415 (1) 4 D 0.9(128) So we have three parameters to select: L, A, and D. Let n be the ratio of the piston area A to the area Ao of the hole through the piston. A n= (2) Ao But ' (2 D Ao = π 2 and thus nπD 2 A= 4 From (1), LA2 n2 π 2 = L = 54.5416 D4 16 so 16(54.5415) 88.419 L= = 2 2 n π n2 Now we try various values for n to see if we obtain a reasonable value for the piston length L. Using n = 50 gives L = 0.035 m, which is 1.38 in., which is a reasonable length. Now we pick the piston area A. A cylinder diameter of 0.05 m (1.96 in.) gives '
0.05 A=π 2 So from (2), Ao =
(2
= 1.963 × 10−3 m2 '
A 1.963 × 10−3 D = = 3.9 × 10−5 = π n 50 2
Thus the hole diameter is
(2
D = 7.1 × 10−3 m
which is about 0.3 in. So one of many possible designs is
piston diameter = 0.05 m piston length, L = 0.035 m piston hole diameter, D = 7.1 × 10−3 m 726 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.26 (a) If m1 = 0, a force balance on the spool valve gives f (t) = k1 x + c1 x˙
(1)
Considering the masses m2 and m3 to constitute a single rigid body, Newton’s law gives (m2 + m3 )¨ y + c2 y˙ + k2 y = A(p1 − p2 )
(2)
where A is the piston area and (p1 − p2 ) is the pressure difference across the piston (refer to Figure 7.4.8). From equation (5) in Example 7.4.9, p1 − p2 = ps − 2∆p
(3)
In this problem, unlike Example 7.4.9, we cannot make the assumption that (m2 + m3 )¨(y) = 0 because the load inertia and the forces k2 y and c2 y˙ are not negligible. Thus p1 &= p2 here. Assuming that x, y, and ∆p are small deviations from equilibrium, then the deviation ps will be 0 if the supply pressure is constant, and (3) becomes p1 − p2 = −2∆p
(4)
and (2) becomes (m2 + m3 )¨ y + c2 y˙ + k2 y = −2A∆p
(5)
In addition, we can linearize the relation qv = f (x, ∆p) for the volume flow rate through the spool valve, as follows: qv = B1 x + B2 ∆p From conservation of mass, Ay˙ = qv = B1 x + B2 ∆p This gives ∆p =
Ay˙ − B1 x B2
Substitute this into (5) to obtain (m2 + m3 )¨ y + c2 y˙ + k2 y = −2A Collect terms:
/
2A2 (m2 + m3 )¨ y + c2 + B2
0
y˙ + k2 y =
Ay˙ − B1 x B2 2AB1 x B2
(6)
The system model is given by equations (1) and (6). (b) The total mass of the spool valve (considered to be rigid) is 2m1 . Newton’s law applied to the spool valve gives 2m1 x ¨ = f (t) − k1 x − c1 x˙ 727 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
or 2m1 x ¨ + c1 x˙ + k1 x = f (t)
(7)
The rest of the problem proceeeds as in part (a). The system model is given by equations (6) and (7).
728 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.27 The equivalent mass of the cylinder is me = m +
I R2
Use this instead of m in equation (9) of Example 7.4.6 to obtain '
I m+ 2 R
(
x ¨ = (R1 + R2 )ρA2 x˙ = A(p1 − p2 )
729 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.28 Let qv1 and qv2 be the volume flow flow rates into and out of the center section where the pressure is p. If the pressure drop ρgh in going up a height h from this point to the plate is small compared to the pressure p, then the pressure at the plate is approximately the same as p. From Newton’s law applied to the plate, m¨ x = pA − kx
(1)
where the force pA is that due to the liquid pressure p acting on the plate. Assuming that m¨ x of the plate is small, then the above equation shows that the pressure force equals the spring force. pA = kx (2) Thus x = pA/k. Differentiate this with respect to time to obtain. dx A dp = dt k dt
(3)
From conservation of volume, d dx (Ax) = A = qv1 − qv2 dt dt
(4)
The model for the dynamic behavior of the pressure as a function of the flow rates is obtained by substituting this expression for dx/dt into (3). The result is A2 dp = qv1 − qv2 k dt
(5)
Using the resistances upstream and downstream, the flow rates qv1 and qv2 can be expressed as functions of the upstream and downstream pressures. The mass flow rates in the sections are 1 ρqv1 = (p1 − p) R 1 ρqv2 = (p − p2 ) R and (5) becomes A2 dp 1 1 ρ = (p1 − p) − (p − p2 ) k dt R R or R ρA2 dp p1 + p2 +p= 2 k dt 2 This equation can be solved for p as a functions of time if we are given p1 and p2 as time functions.
730 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.29 A force balance on the plate gives kx = Ap. The volume swept out by the plate is V = Ax. With V = 30 in3 and p = 1.5 psi, we have that x=
30 A
and
k=
Ap 1.5A2 = x 30
We were given no indication of any size limits or available spring constants, so we are free to choose a reasonable value for A. For example, using a plate 6 inches in diameter, A = 9π, and this gives k = 40 lb/in. This gives a plate displacement of x = 3.75 in.
731 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.30 a) For the two resistances in series: qm2 =
1 ∆p R1 + R2
From the straight line on the graph, 30 = Since R2 = 400, we obtain
1 3 × 104 R1 + R2
R1 = 600 N · s/kg · m2
From equation (4) of Example 7.4.10,
ρgh = Thus h=
'
R2 ∆p R1 + R2 (
400 3 × 104 /9.81ρ = 1223/ρ 100
where ρ, in kg/m3 , is the mass density of the liquid (which was not specified). b) 1 δqm1 = − δ(∆p) r where −1/r is the slope of the straight line. Thus r = 700. We have d δh = −bδh dt where ' ( ' ( 1 R1 + R2 1 g 1 1000 1 9.81 b= + = + = 0.02978 r R2 R2 A 700 400 400 2 Thus
d δh = −0.02978δh dt The time constant is 1/0.02978 = 33.58 s.
732 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.31 From (7.2.5), ρA(h)
dh = qmi − qmo dt
where A(h) = (2L tan θ) h and qmo = Thus ρ (2L tan θ) h
1 ρgh R
dh 1 = qmi − ρgh dt R
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7.32 From (7.2.5), ρA(h)
dh = qmi − qmo dt
where A(h) = (2L tan θ) h and qmo = Ro =
*
ρgh Ro
1 2ρCd2 A2o
Thus
#
qmo = ρCd Ao 2gh and ρ (2L tan θ) h
# dh = qmi − ρCd Ao 2gh dt
734 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.33 From (7.2.5), ρA(h)
dh = qmi − qmo dt
where, from Problem 7.8, and
#
A(h) = 2L Dh − h2 qmo =
)
p = R
*
ρgh R
Thus #
dh 2ρL Dh − h2 = qmi − dt
*
ρgh R
735 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.34 (a) From conservation of mass and (7.3.9): #
#
ρAh˙ = ρqv − Cd Ao 2pρ = ρqv − ρCd Ao 2gh Thus ρ cancels out of the equation, and we obtain √ √ 100h˙ = qv − 0.5 64.4h = qv − 16.1h (1) √ (b) At steady state, h˙ = 0 and qv = 16.1h. With qv = 5, this gives h=
qv2 = 1.55 ft 16.1
(c) At h = 1.55, √
16.1h ≈
Then (1) becomes
1
16.1(1.55) +
. 1 1 . (16.1h) −1/2 . (h − 1.55) = 5 + (h − 1.55) h=1.55 2 10
1 (h − 1.55) 10 Let x = h − 1.55 and u = qv − 5 to obtain the linearized model: 100h˙ = qv − 5 +
100x˙ = u − 0.1x whose time constant is 100(0.1) = 10 sec.
736 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.35 a) The capacitance can be obtained from C= where
A(h) g '
h A(h) = π tan θ
Thus
π C= g
b) From (7.2.5), ρA(h)
'
h tan θ
(2
dh = qmi − qmo dt
where, from part (a),
'
h A(h) = π tan θ and
qmo = Thus πρ
'
h tan θ
(2
(2
(2
ρgh R
dh ρgh = qmi − dt R
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7.36 For an isothermal process, n = 1, and from (7.5.6), C=
V 20 = = 2.2 × 10−5 slug − ft2 /lb nRg T 1(1715)(70 + 460)
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7.37 If pi − p < 0, the flow will be out of the tank (and will be negative). Thus, C
dp = −f (pi − p) dt
739 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.38 For the lefthand tank, C1
d(δp1 ) 1 1 = (δpi − δp1 ) − (δp1 − δp2 ) dt R1 R2
For the righthand tank, C2
d(δp2 ) 1 = (δp1 − δp2 ) dt R2
740 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.39 a) b)
C = ρV cp = 1000(250 × 10−6 )4.18 × 103 = 1045 J/◦ C E = C(T − To ) = 1045(99 − 20) = 8.256 × 104 J
741 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.40 a) V = 15(10)(8) = 1200 ft3 b)
C = ρV cp = 0.0023(6.012 × 103 )1200 = 1.659 × 104 ft − lb/◦ F E = C(T − To ) = 1.659 × 104 (72 − 68) = 6.636 × 104 ft − lb
742 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.41 See Example 7.6.1, which with T (0) = 20, qv = 0.5, and V = 12 gives +
,
T (t) = 20e−t/24 + 1 − e−t/24 80
743 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.42 a) For conduction, the thermal resistance is given by R= Thus R1 = R2 = The total resistance is
L kA
10 × 10−3 = 7.958 ◦ C/W 400π(10−3 )2
5 × 10−3 = 7.074 ◦ C/W 400π((1.5/2) × 10−3 )2 R = R1 + R2 = 15.032 ◦ C/W
b) The heat flow rate is q=
30 30 = = 1.995 W R 15.032
744 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.43 The total resistance is R= R=
'
1 1 L + + h1 A kA h2 A
1 3/[16(12)] 1 + + 85 47 15
(
1 0.0788 = A A
745 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.44 The resistances are in series and thus they add. Using R = L/kA for conduction and R = 1/hA for convection, and letting x be the required thickness of the middle layer, we have 1 R= 30
/
1 10 × 10−3 x 20 × 10−3 + + + 30 0.2 0.04 0.1
where q=
0
=
'
x 1 0.047634 + 30 0.04
(
1 ∆T R
With q = 400 and ∆T = 40, we have '
40 1 x R= = 0.047634 + 400 30 0.04
(
This gives x = 3.54 m, which is rather large.
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7.45 a) The resistance formula is R= For the brick, Rbrick =
L kA
4/12 = 3.876 0.086(12)2 /144
For the concrete, Rconcrete =
4/12 = 1.852 0.02(36)2 /144
The segments are in parallel because they have the same temperature difference, so the total resistance is given by 1 1 1 = + R Rbrick Rconcrete which gives R = 1.253 ◦ F − sec/lb − ft
b) The heat flow rate is given by q = ∆T /R. For the brick, qbrick = For the concrete, qconcrete =
40 = 10.32 Rbrick 40 = 21.6 Rconcrete
The total heat flow rate is q = qbrick + qconcrete = 31.92 ft − lb/sec
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7.46 The solution procedure follows that of Example 7.7.4. a) Assuming that the temperature inside the pipe wall does not change with time, then the same heat flow rate occurs in the inner and outer convection layers and in the pipe wall. Thus the three resistances are in series and we can add them to obtain the total resistance. The inner and outer surface areas are Ai = 2πri L = 2π Ao = 2πro L = 2π The inner convective resistance is
' ('
1 2
' ('
3 4
(
1 10 = 2.618 ft2 12 (
1 10 = 3.927 ft2 12
Ri =
1 1 sec ◦ F = = 0.0239 hi A i 16(2.618) ft lb
Ro =
1 1 sec ◦ F = = 0.2315 ho A o 1.1(3.927) ft lb
The conductive resistance of the pipe wall is Rc =
ln
+
ro ri
,
2πLk
=
ln
+
3/4 1/2
,
2π(10)(10.1)
= 6.389 × 10−4
sec ◦ F ft lb
Thus the total resistance is R = Ri + Rc + Ro = 0.0239 + 6.389 × 10−4 + 0.2315 = 0.256
sec ◦ F ft lb
The heat loss from the pipe, assuming that the water temperature is a constant 120◦ along the length of the pipe, is qh =
1 1 ft lb ∆T = (120 − 70) = 195 R 0.256 sec
To investigate the assumption that the water temperature is constant, compute the thermal energy E of the water in the pipe, using the mass density ρ = 1.94 slug/ft3 and cp = 25, 000 ftlb/slug◦ F: E = mcp Ti = (πri2 Lρ)cp Ti = 3.1743 × 105 ft lb Assuming that the water flows at 1 ft/sec, a slug of water will be in the pipe for 20 sec. During that time it will lose 195(20) = 3900 ftlb of heat. Because this amount is approximately only 1% of E, our assumption that the water temperature is constant is confirmed.
748 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
7.47 The thermal energy E of the water in the pipe is E = mcp Ti =
(πri2 Lρ)cp Ti
'
1 =π 48
(2
6(1.94)(2.5 × 104 )Ti = 396.8Ti
From conservation of heat energy, dE 1 = − (Ti − 70) dt R or
dTi 1 = − (Ti − 70) dt R The resistance of the inside surface is mcp
Ri =
1 = 0.2123 6(0.785)
and the total resistance is R = Ri + Rv + Ro = 0.2123 + 2.15 × 10−4 + 0.77 = 0.925 Thus 396.8 or
dTi = −1.081 (Ti − 70) dt 367
dTi + Ti = 70 dt
For Ti (0) = 120, the solution is Ti (t) = 50e−t/367 + 70
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7.48 a)
C1 = mcp = ρV cp = 1.94(1000)2.5 × 104 = 4.85 × 107 ft − lb/◦ F
b) From conservation of heat energy,
d 1 (C1 T1 ) = − (T1 − To ) dt R1 or 4.85 × 107 R1
dT1 + T1 = To dt
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7.49 The steadystate temperature difference is 90 − 70 = 20◦ , and the temperature difference has decayed by 98% (to 70.4◦ ) in 4000 sec. So we take 4000 sec to be four time constants. Thus τ = 4000/4 = 1000 sec. This value is confirmed by noting from the data that it took 1000 sec for the temperature difference to decay by 63% (to 77◦ ). From Problem 7.48, the model is dT1 4.85 × 107 R1 + T1 = To dt so τ = 4.85 × 107 R1 = 1000 Thus
R1 = 2.06 × 10−5 ◦ F − sec/ft − lb
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7.50 From conservation of heat energy, d (C1 T1 ) = qi − q1 dt d (C2 T2 ) = q1 − qo dt where q1 =
1 (T1 − T2 ) R1
Thus C1 C2
qo =
1 (T2 − To ) R2
dT1 1 = qi − (T1 − T2 ) dt R1
dT2 1 1 = (T1 − T2 ) − (T2 − To ) dt R1 R2
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7.51 a) From conservation of heat energy, C1
1 dT1 1 = − (T1 − To ) + (T2 − T1 ) dt R1 R2 C2
dT2 1 = − (T2 − T1 ) dt R2
(1)
(2)
b) If C2 ≈ 0, (2) shows that T1 = T2 , and (1) becomes C1
dT1 1 = − (T1 − To ) dt R1
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7.52 a) The sphere model is
1 dT = − (T − To ) dt R where R = 1/hA. The time constant is τ = cp ρV /hA, and the response is cp ρV
T (t) = 22 + [T (0) − 22]e−t/τ which has the form ∆T (t) = ∆T (0)e−t/τ where ∆T = T − 22. The following MATLAB session computes the answer. t = [0:15:135,180:60:960]; DeltaT = [95,93,92,90,89,88,87,86,85,84,82,79,... 76,73,71,69,67,65,62,61,59,57,56,54]22; p = polyfit(t,log(DeltaT),1) tau = 1/p(1); cp = 500; rho = 7920;d = 0.025;r = d/2; V = (4/3)*pi*r^3; A = 4*pi*r^2; h = cp*rho*V/(A*tau) The regression coefficients are p = [0.0008 K). b) The Biot number is NB =
4.2533]. The answer is h = 13.69 J/(m2 s
hL hr/3 13.69(0.025/2)/3 = = = 1.426 × 10−4 k k 400
Because NB is much less than 0.1, the lumped parameter model can be considered accurate. c) Radiation heat transfer, which is dependent on T 4 , is thus more significant at higher temperatures, and does not give an exponential response. A plot of the data and the regression curve shows that the curve is a good fit. The greatest error occurs when T > 90◦ and is less than 3◦ . This indicates that radiation heat transfer is affecting the process, but only for the first 30 seconds or so.
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7.53 From conservation of heat energy, C where R=
dT 1 = (To − T ) dt R 1 hA
C = ρV cp
The surface area is and the volume is
A = 4πr 2 = 4π(30 × 10−3 )2 = 1.131 × 10−2 V =
4 3 4 πr = π(30 × 10−3 )3 = 1.13 × 10−4 3 3
The resistance is R=
1 = 0.2947 hA
The capacitance is Thus the model is
C = ρV cp = 8900(1.13 × 10−4 )385 = 387.5 387.5
or
dT 1 = (50 − T ) dt 0.2947
dT = 50 − T dt The time constant is τ = 114.2. The response is 114.2
T (t) = T (0)e−t/τ + To (1 − e−t/τ ) = 400e−t/τ + 50(1 − e−t/τ ) = 50 + 350e−t/τ The time to reach 130◦ is found from 50 + 350e−t/τ = 130 which gives t = −τ ln(0.2286) = 168.5 s
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7.54 Let T1 be the sphere tempewrature and T2 be the bath temperature. From conservation of heat energy dT1 1 C1 = (T2 − T1 ) dt R dT2 1 C2 = (T1 − T2 ) dt R where R is the surface convective resistance. The sphere’s surface area is A = 4πr 2 = 4π(30 × 10−3 )2 = 1.131 × 10−2 and the volume is
4 4 V1 = πr 3 = π(30 × 10−3 )3 = 1.13 × 10−4 3 3 R=
1 1 = = 0.2947 hA 300(1.131 × 10−2 )
C1 = ρV1 cp = 8900(1.13 × 10−4 )385 = 387.5 So
C2 = ρV2 cp = 7900(0.1)400 = 3.16 × 105 RC1 = 114.2
Thus
RC2 = 9.3125 × 104
dT1 = T2 − T1 dt dT2 9.3125 × 104 = T1 − T2 dt 114.2
Let τ1 = 114.2 and τ2 = 9.3125 × 104 . Applying the Laplace transform to each equation gives τ1 sT1 (s) − τ1 T1 (0) = T2 (s) − T1 (s) These have the solution
τ2 sT2 (s) − τ2 T2 (0) = T1 (s) − T2 (s) T1 (s) =
where a=
bs + c s(s + a)
τ1 + τ2 0.008767 τ2 τ2
b = T1 (0) = 400 c=
T1 (0) T2 (0) + = 0.4421 τ2 τ1 756
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The response is
c ab − c −at + e = 50.43 + 349.6e−0.008767t a a The temperature T1 will reach 130◦ at T1 (t) =
t=
1 ln 0.2276 = 168.8 s 0.008767
Contrast this result with the result of Problem 7.53, which is t = 168.5 s. Thus the assumption of a constant bath temperature in Problem 7.53 is very justified.
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7.55 The last three equations in Example 7.71 are: (R1 + R2 )T1 − R1 T2 = R2 Ti R3 T1 − (R2 + R3 )T2 + R2 T3 = 0 In matrix form these are
−R4 T2 + (R3 + R4 )T3 = R3 To
(R1 + R2 ) −R1 0 T1 R 2 Ti R −(R + R ) R T = 0 2 3 2 3 2 0 −R4 (R3 + R4 ) T3 R 3 To )
Once T1 is computed, qh can be computed from q = (T1 − Ti )/R1 . The script file is: R A b T q
= = = = =
[0.036,4.01,0.408,0.038];Ti = 20;To = 10; [R(1)+R(2),R(1),0;R(3),(R(2)+R(3)),R(2);0,R(4),R(3)+R(4)]; [R(2)*Ti;0;R(3)*To]; A\b (1/R(1))*(Ti  T(1))
The results are T = [19.7596, −7.0214, −9.7462] ◦ C and q = 6.6785 watts/m2 . Thus the total heat loss is 10(6.6785) = 66.785 W.
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7.56 (a)
1 (p1 − pb ) R2 1 q3 = (p1 − pc ) R3 q2 =
b) Rearrange the equations by bringing all the unknowns to the left side. R 1 q1 + p1 = pa R2 q2 − p1 = −pb
R3 q3 − p1 = −pc q1 − q2 − q3 = 0
These equations have the form Ax = b where
A=
R1 0 0 1 0 R2 0 −1 0 0 R3 −1 1 −1 −1 0
b=
(c) The script file is
pa −pb −pc 0
x=
q1 q2 q3 p1
pa = 30*144;pb = 25*144; pc = 20*144; R = [10000,14000,14000]; A = [R(1),0,0,1;0,R(2),0,1;0,0,R(3),1;1,1,1,0]; b = [pa;pb;pc;0]; format long x = A\b When this file is run it gives the following output. x = 1.0e+003 * 0.00006352941176 0.00000605042017 0.00005747899160 3.68470588235294 Thus q1 = 6.35 × 10−2 , q2 = 6.05 × 10−3 , q3 = 5.75 × 10−2 ft3 /sec, and p1 = 3685 lb/ft2 .
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7.57 For the given values Cd A = 0.5π(2 × 10−2 )2 = 2π × 10−4 and the differential equation is π(6h − h2 ) or
√ dh = −2π × 10−4 19.62h dt
√ dh 2 × 10−4 19.62h =− dt 6h − h2
(a) The greatest outflow rate occurs when the water level is the highest (h = 5). Thus using h = 5 in the differential equation, we can obtain an lower bound on the time required to drain the tank. When h = 5, √ dh 2 × 10−4 19.62h =− = −3.9618 × 10−4 dt 6h − h2 This implies that h(t) = −3.9618 × 10−4 t + 5, and h = 0 at t = 5/(3.9618 × 10−4 ) = 12, 620 s, or 210 minutes. Thus the tank will empty in no less than 210.34 minutes. Instead of a lower bound on the estimate, we can obtain a higher estimate by using the midpoint value for h; namely, h = 5/2 = 2.5. This gives √ dh 2 × 10−4 19.62h =− = −1.6008 × 10−4 dt 6h − h2 This implies that h(t) = −1.6008 × 10−4 t + 5, and h = 0 at t = 5/(1.6008 × 10−4 ) = 31, 234 s, or 521 minutes. (b) To use the ode45 solver, solve for the derivative: √ dh 2 × 10−4 19.62h =− dt 6h − h2
and create the following function file:
function hdot = tank(t,h) hdot = (0.0002*sqrt(19.62*h))/(6*hh^2); Then use the ode45 solver in the following script file. [t, h] = ode45(’tank’, [0, 25200], 5); plot(t,h),xlabel(’t (seconds)’),ylabel(’ h (feet)’)
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Start with a final time of something more than 12,620 s, and run the file until the plot shows the height approaching zero. The time to empty, which is 25,200 s or 420 minutes, was found this way. The estimate of 521 minutes obtained with the midpoint height is not much different, and establishes confidence in the numerical result. Note that if you choose a final time somewhat larger than 25,200 s, the denominator in the expression for dh/dt becomes zero because h = 0, and the expression for dh/dt becomes undefined. This causes difficulties for the numerical algorithm. Thus it is best to start with a small value for the final time, and increase it. The plot is shown in the figure. 5 4.5 4 3.5
h (feet)
3 2.5 2 1.5 1 0.5 0 0
0.5
1
1.5 t (seconds)
2
2.5
3 4
x 10
Figure : for Problem 7.57.
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7.58 (a) Write the equation as
2y dy =4− dt 10 + 2t Then create the following function file. function ydot = salt(t,y) ydot = 42*y/(10+2*t); The following file solves the problem using the ode45 solver. [t, h] = ode45(’salt’, [0, 10], 0); plot(t,h),xlabel(’Time t’),ylabel(’Salt Mass y’) The plot is shown in the figure. 30
25
Salt Mass y
20
15
10
5
0 0
1
2
3
4
5 Time t
6
7
8
9
10
Figure : for Problem 7.58.
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(b) The variable coefficient 2/(10 + 2t) varies from 2/10 to 2/30 as t varies from 0 to 10. Its midpoint value is 4/30. Using this value the model becomes dy 4 + y=4 dt 30 The step response is y(t) =
, + , 4 + 1 − e−30t/4 = 30 1 − e−30t/4 4/30
This equation predicts that y(10) = 30. The plot shows that the numerical solution gives y(10) = 27 approximately. Thus we can have confidence that the numerical solution is correct.
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7.59 The relation between height h and the volume inflow rate r is 100 Thus h(t) =
dh =r dt 1 100
The MATLAB mfile is
8
t
r dt 0
t = [0:10]; r = [0,80,130,150,150,160,165,170,160,140,120]; for k=2:11 h(k) = (1/100)*trapz(t(1:k),r(1:k)); end plot(t,h) The answer for the final height is given by h(11) and is 13.65 ft. The plot is shown below. 14
12
10
h (ft)
8
6
4
2
0
0
1
2
3
4
5 t (sec)
6
7
8
9
10
Figure : for Problem 7.59
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7.60 This problem requires both analytical and numerical methods. (a) Let ' (2 ' ( 1 R 1 1.5 2 3π b= π = π = 3 H 3 4 64 Then V = bh3 . When the cup is full, h = 4 in, and the water volume is V = b(4)3 = 64b = 3π. Let q be the flow rate (q = 2 cubic in/sec). From mass conservation, dV =q dt and V (t) =
8
t
q(t) dt =
0
8
t
2 dt = 2t
0
Equating the two expressions for the water volume gives 2t = 64b, or t = 32b = 3π/2 = 4.7 s. So we do not need MATLAB for this part of the problem. (b) If q(t) = 2(1 − e−2t ), then V (t) =
8
t
q(t) dt =
0
8
t 0
9
2e−2t 2(1 − e−2t ) dt = 2t − −2
:.t . . . = 2t + e−2t − 1 . 0
The time to fill is found by equating the two expressions for the volume: 2t + e−2t − 1 = 64b = 3π
We can solve this for t by using the fzero function. First define the function cup: function f = cup(t) f = 2*t+exp(2*t)1  3*pi; Use the fzero function with the answer from part (a) as the starting guess: (fzero(’cup’,4.7) ans = 5.2124 Thus it will take about 5.2 sec to fill the cup.
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7.61 a) The model is just like that shown in Figure 7.10.2 except that the two SSR blocks are not present. Create a subsystem block and save it. b) Create a Simulink model like that shown in Figure 7.10.3, using the subsystem block created in part (a). c) In the MATLAB Command window, type the following parameter values. (A_1 = 2;A_2 = 5;R_1 = 400;R_2 = 600; (rho = 1000; g = 9.81; q_1 = 50; (h10 = 1.5;h20 = 0.5; Then run the simulation. A Stop time of 2000 s shows the complete response.
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7.62 Using the values given, the equation can be reduced to √ √ dh Cd A 2gh h −4 =− = −8.8589 × 10 2 dt π(2rh − h ) (6h − h2 ) The model is shown in the following figure. Set the Initial condition of Integrator to 5. In the Fcn block type 8.8589*10^(4)*sqrt(u(1))/(6*u(1)u(1)^2) for the expression. You can plot the results by typing (plot(tout,simout),xlabel(#t# ),ylabel(#x# ) By experimenting with the Stop time, we find that the height is essentially 0 after 25,230 s.
Figure : for Problem 7.62
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7.63 From conservation of fluid mass, 100h˙ = q, where q is the flow rate. Thus 1 h(t) = 100
8
t
q(t) dt 0
The model is shown in the following figure. We assume that the flow rate remains constant at the previous value for 1 min. Thus in the Lookup Table block, we select the Lookup method to be Use Input Below. In this block, the Vector of input values is [0:10] and the Vector of output values is [0,80,130,150,50,160,165,170,160,140,120]. The height is approximately 13 ft after 10 min.
Figure : for Problem 7.63
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7.64 From conservation of mass, the rate of change of water mass in the cup is d(ρV ) = ρq dt or
dV ) = ρq dt where ρ is the mass density of the water. We see that ρ cancels out of the equation, which can then be expressed as ' (2 dV R dh =π h2 =q dt H dt ρ
Using the values given for R and H and solving for dh/dt we obtain dh q q = + ,2 = + , 9 dt R π 64 h2 π H h2
a) With q = 2 the equation becomes
dh 2 = + , 9 dt π 64 h2
The model is shown in the following figure. Set the Initial condition of the Integrator to a small positive number, say 0.01, to avoid a singularity (because h appears in the denominator of dh/dt. In the Fcn block type 2/pi*(9/64)*u(1)^2) for the expression. You can plot the results by typing (plot(tout,simout),xlabel(#t# ),ylabel(#x# ) By experimenting with the Stop time, we find that the height is essentially equal to 4 in. after 4.7 sec.
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Figure : for Problem 7.64a
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b) With q a function of time, we must modify the model in part (a). The equation becomes dh q(t) = + , dt π 9 h2 64
The model is shown in the following figure. Set the Initial condition of the Integrator to a small positive number, say 0.01, to avoid a singularity (because h appears in the denominator of dh/dt. In the Fcn block type pi*(9/64)*u(1)^2 for the expression. In the Fcn1 block type 2*(1exp(2*u(1))) for the expression. The Clock block provides the −2t input+time , t to compute the expression 2(1 − e ). We use the Divide block to divide q(t) 9 by π 64 h2 . In the Divide block enter the number of inputs as */ (to multiply by q(t) and to divide by π
+
9 64
,
h2 ). You can plot the results by typing
(plot(tout,simout),xlabel(#t# ),ylabel(#x# ) By experimenting with the Stop time, we find that the height is essentially equal to H = 4 after 5.21 sec.
Figure : for Problem 7.64b
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7.65 Use the model developed in Example 7.10.1 and change the Relay settings.
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7.66 The model is shown in the following figure. In the Relay block, set the Switch on point to 5.5 and the Switch off point to 4.5. Set the Output when on to 0 and the Output when off to 50. Set the Initial condition of the Integrator to 1. Set the gain of the Gain block to 1/ρA = 1/2000. Set the gain of the other block to ρg/R. Set the Stop time to 1000. Using a value of R = 400, the height does not reach the desired band of 4.5 to 5.5 ft because the inflow cannot compensate for the loss due to the small resistance. However, for a larger value, say R = 4000, the height oscillates between the desired values. You can plot the results by typing (plot(tout,simout),xlabel(#t# ),ylabel(#x# )
Figure : for Problem 7.66
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Eight
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8.1 From Table 8.1.1, v(t) = Because x˙ = v, x(t) = Set x(t) = 2000.
#
" ! " 2700 ! 1 − e−4t = 112.5 1 − e−4t 24 t 0
!
v(t) dt = 112.5t + 28.125 e−4t − 1 !
2000 = 112.5t + 28.125 e−4t − 1 Combine terms and solve the following equation for t.
"
"
112.5t + 28.125e−4t − 2028.125 = 0 This can be solved by plotting or by iteration, for example. The answer is t = 17.8 sec.
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8.2 τ = 0.25 sec. Thus the speed takes approximately 4(0.25) = 1 sec to reach constant speed. Because 0.04 sec is small compared to 1 sec, the answer is yes.
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8.3 From Table 8.1.1, a) τ = 8/7, x(t) = 6e−7t/8 b) τ = 12/5, x(t) = 3e−5t/12 c) τ = 13/6, x(t) = −2e−6t/13 d) No time constant is defined because the model is unstable. x(t) = 9e5t/7
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8.4 a) xss = 10, t = 4τ = 8 b) xss = 10, t = 4τ = 8 c) xss = 200, t = 4τ = 8
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8.5 Note that the initial condition does not affect the steadystate response. a) xss = 4, t = 4τ = 24/5 b) xss = 4, t = 4τ = 4/5 c) No steadystate response exists because the model is unstable.
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8.6 From Table 8.1.1, a)
!
x(t) = 4 1 − e−5t !
"
"
b) x(t) = e−5t + 4 1 − e−5t = 4 − 3e−5t c)
x(t) = −2e6t/13 −
" 18 ! 1 − e6t/13 = −3 + e6t/13 6
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8.7 The time constant of the rotational system is τ1 = 100/5 = 20 s. The time constant of the field circuit is τ2 = 0.002/4 = 5 × 10−4 s. The steadystate current is if ss = 12/4 = 3 A, and thus the steadystate torque is Tss = KT if ss = 15(3) = 45 N·m. It takes about 4τ2 = 2 × 10−3 s for the torque to reach steady state, and since this time is much smaller than τ1 , we may treat the torque as a step input to the rotational system. The steadystate speed is 45/5 = 9 rad/s, and it takes about 4τ1 = 80 s to reach the steadystate speed.
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8.8 RC v˙ + v = vs , where RC = 3 s. From Table 8.1.1, !
v(t) = 6e−t/3 + 12 1 − e−t/3
"
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8.9 (AR/g)h˙ + h = (R/g)qv . With the given values, this becomes 93.1677h˙ + h = 1.8634qv From Table 8.1.1,
!
h(t) = 2e−t/93.1677 + 18.634 1 − e−t/93.1677 Set h(t) = 15 ft and solve for e−t/93.1677 :
"
e−t/93.1677 = 0.2184 which gives t = 142 sec.
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8.10 mcp RT˙ + T = Tb , where mcp R = 100(500)0.09 = 4500 s. From Table 8.1.1, !
T (t) = 20e−t/4500 + 80 1 − e−t/4500
"
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8.11 Transforming the first model, 2v˙ + v = ˙ (t) g + g(t), gives 2[sV (s) − v(0)] + V (s) = (s + 1)G(s) =
10(s + 1) s
With v(0) = 5, this becomes V (s) = 10
s + 0.5 10 = s(s + 0.5) s
Thus v(t) = 10us (t). Transforming the second model, 2v˙ + v = g(t), gives 2[sV (s) − v(0)] + V (s) = G(s) =
10 s
With v(0) = 5, this becomes V (s) = 5
s+1 10 5 = − s(s + 0.5) s s + 0.5
Thus v(t) = 10us (t) − 5e−0.05t .
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8.12 Note that g˙ = 0 for −∞ ≤ t ≤ ∞. Thus the model is equivalent to 5v˙ + v = g. From Table 8.1.1, v(t) = 10 − 5e−0.2t for both cases.
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8.13 a) Transforming the equation gives (6s + 3)V (s) = (s + 1)G(s) = or V (s) =
s+1 s
s+1 1 s+1 11 1 1 = = − s(6s + 3) 6 s(s + 0.5) 3 s 6 s + 0.5
Thus v(t) =
1 1 −0.5t − e 3 6
Note that v(0+) = 1/6. b) Transforming the equation gives (6s + 3)V (s) = (s + 1)G(s) = (s + 1) or V (s) =
$
%
1 1 s+1 − =5 s s+5 s(s + 5)
5 s+1 11 5 1 4 1 = − − 6 s(s + 5)(s + 0.5) 3 s 27 s + 0.5 27 s + 5
Thus v(t) =
1 5 4 − e−0.5t − e−5t 3 27 27
Note that v(0+) = 0.
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8.14 Transforming the equation gives (2s + 1)V (s) = or V (s) =
2s2 (s
5 s2
5 5 10 10 = 2− + + 0.5) s s s + 0.5
Thus v(t) = 5t − 10 + 10e−0.5t
The transient response is 10e−0.5t . The steadystate response is 5t − 10.
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8.15 Transforming the equation gives (9s + 3)V (s) = or V (s) =
9s2 (s
7 s2
7 7 7 7 = 2− + + 1/3) 3s s s + 1/3
Thus
7 v(t) = t − 7 + 7e−t/3 3 The steadystate response is 7t/3 − 7, which is not parallel to the input 7t.
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8.16 The equation reduces to x ¨ + 4x = 0, whose roots are s = ±2j. Thus the oscillation frequency is 2 rad/sec for both cases, and x(t) = B sin(2t + φ) and x(0) = B sin φ x(0) ˙ = B cos φ a) B sin φ = 5 and B cos φ = 0. Thus, φ = π/2 rad and the amplitude is B=
5 =5 sin π/2
b) B sin φ = 0 and B cos φ = 5. Thus, φ = 0 and the amplitude is B=
5 =5 cos 0
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8.17 From Table 8.2.2, a) s = −2 ± 2j. x(t) = 0.5e−2t sin 2t. b) s = −6,−2. x(t) = 0.25e−2t − 0.25e−6t . c) s = −2, −2. x(t) = te−2t .
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8.18 a) The roots are s = −2 ± 2j, so we use the trial solution x(t) = Be−2t sin(2t + φ) + C At steady state, xss = 2/8 = 0.25, so C = 0.25. We obtain B and φ from the initial conditions. The solution is $ & ' % 1 √ −2t 5π x(t) = 2e sin 2t + +1 4 4
b) The roots are s = −6, −2, so we use the trial solution x(t) = C1 e−6t + C2 e−2t + C3 At steady state, xss = 2/12 = 1/6, so C3 = 1/6. We obtain C1 and C2 from the initial conditions. The solution is 1 x(t) = 6
&
1 −6t 3 −2t e − e +1 2 2
'
c) The roots are s = −2, −2, so we use the trial solution x(t) = C1 e−2t + C2 te−2t + C3 At steady state, xss = 2/4 = 0.5, so C3 = 0.5. We obtain C1 and C2 from the initial conditions. The solution is x(t) =
) 1( 1 − (2t + 1)e−2t 2
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8.19 Using Table 8.2.2, a) s = −2, −5. b) s = −2, −2. c) s = −2 ± 5j.
x(t) =
4 −2t 1 −5t e − e 3 3
x(t) = (1 + t)e−2t x(t) = 1.02e−2t sin (5t + 1.373)
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8.20 a) s = −2, −5.
x(t) =
b) s = −2, −2. c) s = −2 ± 5j.
1 30
x(t) =
x(t) =
&
2 −5t 5 −2t e − e +1 3 3
'
) 1 ( 1 − (2t + 1)e−2t 20
$ % 1 1 √ −2t 29e sin (5t + 4.33) + 1 58 5
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8.21 a) The roots are s = −2, −5. 4 4 X(s) = 2 2 = 3s (s + 7s + 10) 3
&
0.1 0.07 0.0833 0.0133 − − + s2 s s+2 s+5
'
Thus
" 4! 0.1t − 0.07 + 0.0833e−2t − 0.0133e−5t 3 b) The roots are s = −2, −2.
x(t) =
4 4 X(s) = 2 2 = 5s (s + 4s + 4) 5
&
0.25 0.25 0.25 0.25 − + + 2 2 s s (s + 2) s+2
'
Thus
" 4! 0.25t − 0.25 + 0.25te−2t + 0.25e−2t 5 c) The roots are s = −2 ± 5j.
x(t) =
X(s) = Thus
$
5 5 29 4 21 s+2 = − − +4 2s2 [(s + 2)2 + 25] 2(292 ) s2 s (s + 2)2 + 25 (s + 2)2 + 25 &
5 21 x(t) = 29t − 4 + e−2t sin 5t + 4e−2t cos 5t 1682 5
%
'
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8.22 X(s) =
a s2 (ms2 + cs + k)
Let e(t) be the difference between the input and the response: e(t) = f (t) − x(t). Then a E(s) = F (s) − X(s) = 2 s
&
1 1− 2 ms + cs + k
'
a = 2 s
=
,
*
ms2 + cs + k − 1 ms2 + cs + k
+
From the final value theorem, ess
a = lim s 2 s→0 s
*
ms2 + cs + k − 1 ms2 + cs + k
+
∞ if k = ' 1 ac if k = 1
The response will be parallel to the input only if k = 1. In that case, the difference between the input and the response is ac.
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√ √ 8.23 a) ζ = cos[tan−1 (6/2)] = 0.316. τ = 1/2. ωd = 6. ωn √ = 4 + 36√ = 40. b) Unstable so ζ and τ are not defined. ωd = 5. ωn = 1 + 25 = 26. c) ζ = 1 because the roots are real and equal. τ = 1/10. ωd and ωn are not defined because the free response is not oscillatory. d) ζ is not defined because the system is first order. τ = 1/10. ωd and ωn are not defined because the free response is not oscillatory.
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8.24 a) τ = 0.5, but ζ, ωn , and ω√d do not apply because the dominant root is real. b) τ = 0.5, ζ = 0.707, ωn = 2 2, ωd = 2.
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8.25 The dominant root pair is s = −2 ± 4j. For this root, ζ = cos[tan−1 (4/2)] = 0.447, τ = 1/2, and ωd = 4.
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8.26 a) The RouthHurwitz criterion implies that the system is stable if and only if −(µ + 2) > 0 and 2µ + 5 > 0, which gives −2.5 < µ < −2 for stability. Neutral stability occurs if either the s term or the constant term is missing in the characteristic equation. This occurs if µ + 2 = 0 (the roots are s = ±j) or if 2µ + 5 = 0 (the roots are s = 0, −0.5).
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8.27 We have 3¨ x + cx˙ + 27x = 0. Using Table 8.2.2, a) (c = 0) x(t) = cos 3t b) (c = 9) c) (c = 18) d) (c = 22)
x(t) = e−1.5t cos 2.6t − 0.5774e−1.5t sin 2.6t x(t) = −3te−3t + e−3t x(t) = 1.3695e−5.774t − 0.3695e−1.558t
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8.28 a) Stable if and only if 6d > 0 and 25d2 > 0 (from the RouthHurwitz criterion). Thus, stable if and only if d > 0. b) The damping ratio is 6d 3 ζ= √ = < 1 for all d > 0 2 10 2 100d Thus there will be damped oscillations for any d > 0.
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8.29 a) Stable if and only if 6b > 0 and 5b − 10 > 0 (from the RouthHurwitz criterion). Thus, stable if and only if b > 2. b) The roots are s = −3b ± 9b2 − 5b + 10
There will be imaginary parts if f (b) = 9b2 − 5b + 10 < 0. Solving df /db = 0 gives b = 5/18. At b = 5/18, df 2 /db2 = 18 > 0, and thus f (b) has a minimum value of f (5/18) = 9.305. Therefore there is no real value of b for which 9b2 − 5b + 10 < 0, and thus no decaying oscillations.
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8.30 Transforming each equation gives Ω(s) =
T (s) 50s + 10
T (s) = 25If (s) 1 If (s) = V (s) 0.0001s + 5 Thus Ω(s) 25 1 500 500 = = = 2 V (s) 50s + 10 0.001s + 5 (s + 0.2)(s + 5000) s + 5000.2s + 1000 The damping ratio is
5000.2 ζ= √ = 79.06 2 1000
The time  constants are τ1 = 5 and τ2 = 1/5000 s. The undamped natural frequency is ωn = 1000/1 = 31.62 rad/s.
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8.31 The characteristic equation becomes 24 × 10−8 s2 + (3.6 × 10−5 + 4c × 10−3 )s + 0.6c + 0.01 = 0 The damping ratio is 3.6 × 10−5 + 4c × 10−3 0.09 + 10c ζ= =−8 2 24 × 10 (0.6c + 0.010 6(0.6c + 0.01)
The undamped natural frequency is ωn =
.
0.6c + 0.01 = 5000 24 × 10−8
/
0.6c + 0.01 6
If ζ < 1, the real part of the roots is −
3.6 × 10−5 + 4c × 10−3 900 + c × 105 = − 24 × 10−8 6
Thus the time constant is
τ=
6 900 + c × 105
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8.32 From the quadratic formula, s=
−12 ±
√ 0 144 − 12k = −2 ± 4 − k/3 6
For k ≤ 12 the roots are real, the dominant root lies between 0 and −2, and the dominant time constant lies between 1/2 and ∞. For k > 12 the roots are complex, and the real part is −2. Thus the dominant time constant is 1/2. The root locus plot is identical to Figure 8.2.7. The smallest possible dominant time constant is 1/2 and is obtained for any k ≥ 12.
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8.33 From the quadratic formula, s=
−c ±
√
c2 − 144 6
For c ≥ 12 the roots are real, the dominant root lies between 0 and −2, and the dominant time constant lies between1/2 and ∞. For c < 12 the roots are complex, and the undamped natural frequency is ωn = 12/3 = 2. This means that the complex roots lie on a semicircle of radius 2 centered at the origin. The root locus plot is identical to Figure 8.2.8. The smallest possible dominant time constant is 1/2 and is obtained for c = 12.
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8.34 Dividing by the highest coefficient and factoring out c gives the standard form (8.222): s2 + 666.7s + 2.78 × 104 + 8333c(s + 666.7) = 0 where α = β = 666.7, γ = 2.78 × 104 , and µ = 8333c. From the results of (8.223) and (8.224) we see that the root locus is a circle centered at s = −666.7 with a radius of 166.7, and is similar to Figure 8.29(a) with the ×’s at s = −44.7 and s = −622, and the ! at s = −667. The starting points with c = 0 are at the ×’s. As c → ∞ one root approaches s = −667 and the other root approaches s = −∞. The plot illustrates the sometimes counterintuitive behavior of dynamic systems. One would think that increasing the damping c would increase the time constant and thus slow the response. However, as c is increased up to the value where the two roots meet at s = −833, the dominant time constant decreases, and we see that the smallest possible dominant time constant is τ = 1/833 s.
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8.35 Let µ = (m − 2)/2. Thus m = 2µ + 2 and the characteristic equation becomes (2µ + 2)s2 + 12s + 10 = 0 Divide by 2: (µ + 1)s2 + 6s + 5 = 0 From the quadratic formula,
√ −3 ± 4 − 5µ s= µ+1
For 0 ≤ µ ≤ 4/5, the roots are real. For µ = 0, the roots are s = −1 and s = −5. For µ = 4/5, the roots are equal at s = −1.67. For µ > 4/5 the roots are complex and approach s = 0 as µ → ∞. The plot is similar to that in Figure 8.2.10 with the ×’s located at s = −1 and s = −5, and with the circle intersecting the real axis at s = −1.67.
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√ 8.36 ζ = 1/ 2 = 0.707 which implies from Figure 8.3.6a that the maximum percent overshoot is ≈ 5%. Because xss = 2/8 = 0.25, the overshoot is 0.05(0.25) ≈ 0.01. √ For ζ √ = 0.707, Figure 8.3.6c shows that ωn tr ≈ 3.2. Because ωn = 8/1 = 2 2, tr = 3.2/2 2 = 1.1. √ √ For ζ = 0.707, Figure 8.3.6a shows that ωn tp ≈ 4.6. Because ωn = 2 2, tp = 4.6/2 2 = 1.6. √ For ζ = 0.707 and ωn = 2 2, Table 8.3.2 gives td = 0.53. The roots are −2 ± 2j, so the time constant is τ = 0.5. The 2% settling time is 4τ = 2.
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8.37 ωn = 4/1 = 2. For tr ≤ 3, ωn tr ≤ 2(3) = 6. From Figure 8.3.6c, we see that this requires that ζ ≤ 0.9. Thus ζ must be no greater than 0.9 if tr is to be no greater than 3. Percent overshoot ≤ 20% implies from Figure 8.3.6a that ζ must be ≥ 0.43. Thus ζ must be in the range 0.43 ≤ ζ ≤ 0.9 to satisfy both specifications. To minimize the overshoot, Figure 8.3.6a shows that we should choose ζ as large as possible (within the above √ range). Thus we choose ζ = 0.9. The damping c is found from the definition of ζ: ζ = c/2 4 = c/4. Thus c = 0.9(4) = 3.6.
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8.38 The approach is similar to that of Problem 8.37, but the two specifications cannot both be satisfied, so we give  priority to the overshoot specification, as directed. The calculations are as follows. ωn = 4/9 = 2/3 For tr ≤ 3, ωn tr ≤ 2(3)/3 = 2. From Figure 8.3.6c, we see that this requires that ζ ≤ 0.35. Thus ζ must be no greater than 0.35 if tr is to be no greater than 3. Percent overshoot ≤ 20% implies from Figure 8.3.6a that ζ must be ≥ 0.43. Thus ζ cannot satisfy both specifications. So we set ζ = 0.43, giving priority to the overshoot specification, as directed. (We could choose ζ > 0.43 to minimize the overshoot, but this would cause the rise time to become much larger than desired.) Thus we choose ζ = 0.43. The damping c is found from the definition of ζ: ζ = c/2 9(4) = c/12. Thus c = 0.43(12) = 5.16.
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8.39 tp =
ωn
π 1 − ζ2


But the imaginary part of the roots is ωd = ωn 1 − ζ 2 . Thus tp = π/ωd , and thus tp depends only on the imaginary part of the roots.
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8.40 The equation of motion is m¨ y = mg sin 30◦ − c1 y˙ − c2 y˙ − ky or 3¨ y + 30y˙ + 500y = 3(9.81) sin 30◦ or y¨ + 10y˙ + 166.67y = 4.905 The roots are s = −5 ± 11.9025j and 10 ζ= √ = 0.3873 2 166.67 From Table 8.3.1, (
y(t) = 0.0294 1.085e−5t sin(11.9t + 4.315) + 1
)
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8.41 For the system in part (a) of the figure, m¨ x + cx˙ + kx = cy˙ + ky With the given values, 3¨ x + 18x˙ + 10x = 18y˙ + 10y Note that Table 8.3.1 cannot be used because of the input derivative term y. ˙ Transforming this equation with zero initial conditions gives X(s) =
18s + 10 6s + 10/3 Y (s) = 3s2 + 18s + 10 s(s + 0.6195)(s + 5.3805)
This expands to X(s) =
1 0.13 1.13 + − s s + 0.6195 s + 5.3805
Thus x(t) = 1 + 0.13e−0.6195t − 1.13e−5.3805t
For the system in part (b) of the figure,
m¨ x + cx˙ + kx = ky With the given values, 3¨ x + 18x˙ + 10x = 10y Note that Table 8.3.1 can be used here. This result is x(t) = 1 + 0.13e−0.5.3805 − 1.13e−0.6195t The responses are shown in the following plot. The two responses are very different, due to the numerator dynamics of the model in part (a) of the figure.
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1.4
1.2 (a) 1
x(t)
0.8
(b) 0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5 t
3
3.5
4
4.5
5
Figure : for Problem 8.41
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8.42 For part (a) of the figure, X(s) 18s + 10 = 2 Y (s) 3s + 18s + 10 In MATLAB, type sysa = tf([18,10],[3,18,10]); When the plot appears, right click on it and select “Characteristics”. From there you can select the desired items to display on the plot. The results are: maximum per cent overshoot = 7%, peak time = 0.924, 2% settling time = 3.02, and rise time = 0.298. Note that you cannot use the formulas in Table 8.3.2 because this model has numerator dynamics. For part (a) of the figure, X(s) 10 = 2 Y (s) 3s + 18s + 10 In MATLAB, type sysb = tf(10,[3,18,10]); When the plot appears, right click on it and select “Characteristics”. From there you can select the desired items to display on the plot. The results are: maximum per cent overshoot = 0% (there is no overshoot), so the peak time does not apply. Also, 2% settling time = 6.51 and rise time = 3.6. Note how much larger the settling time at the rise time are, as compared to the model having numerator dynamics. Thus numerator dynamics can reduce response time but can cause overshoot.
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8.43 For the circuit, L or
di + Ri = vi (t) dt
di + 10i = vi (t) dt Assuming that the applied voltage is 12 volts (as in the text example), the impulse strength is 12(0.3) = 3.6. So we model vi (t) as 3.6δ(t), and thus 5
(5s + 10)I(s) = 3.6 which gives I(s) =
3.6 0.72 = 5s + 10 s+2
and i(t) = 0.72e−2t
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8.44 From impulsemomentum, (m + 5m)v(0+) − [mv1 + 5m(0)] = 0 Thus v(0+) = v1 /6. The equation of motion is 6m¨ x + kx = 0, and its solution with x(0+) = 0, x(0+) ˙ = v(0+) = v1 /6 is . . v(0+) x(t) = sin ωn
k v1 t= 6m 6
/
6m sin k
k t 6m
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8.45 With m1 = m and m2 = 5m, (8.4.7) becomes v3 =
m − 5m 2 v1 = − v1 m + 5m 3
The change in momentum is &
'
2 m − v1 − mv1 = 3 Thus
#
0+ 0
#
0+
f (t) dt
0
5 f (t) dt = − mv1 3
The linear impulse applied to m2 is (5/3)mv1 , so 5 5m¨ x + kx = mv1 δ(t) 3 or where ωn =

1 x ¨ + ωn2 x = v1 δ(t) 3
k/5m. The response is given by v1 v1 x(t) = sin ωn t = 3ωn 3
/
5m sin k
.
k t 5m
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8.46 The free response of the circuit is vC (t) = vC (0)e−t/τ where τ = RC = 3 × 106 C. A plot of ln vC (t) versus t gives a straight line whose slope −1/τ is computed from 1 ln 12 − ln 6.2 − = = −0.033 τ 0 − 20 Thus τ = 1/0.033 = 30.286 and C = τ /R = 10−5 F.
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8.47 The free response is ∆T (t) = ∆T (0)e−t/τ A plot of ln ∆T (t) versus t gives a straight line whose slope −1/τ is computed from −
1 ln(178 − 68) − ln(82 − 68) = = −6.8714 × 10−4 τ 0 − 3000
Thus τ = 1455 sec. Note that ∆T (0) = 178 − 68 = 110. The model is
T (t) = 68 + 110e−t/1455 For T = 135, we have 135 = 68 + 110e−t/1455 or t = −1455 ln
135 − 68 = 721 sec 110
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8.48 From the graph we can easily identify the following three characteristics: the steadystate response, xss = 0.025 m, the peak time, tp = 0.13 s, and the maximum per cent overshoot, 0.0395 − 0.025 M% = 100 = 58% 0.025 From Table 8.3.2, 100 R = ln = 0.5447 M% Thus ζ=√ 
R = 0.171 + R2
π2
Thus 1 − ζ 2 = 0.985. Because the applied force is 1000 N, xss = 1000/k, and thus k = 1000/0.025 = 40 000 N/m. From π tp = ωn 1 − ζ 2
we obtain
ωn = so that m= Because we have
tp
π = 24.53 rad/s 1 − ζ2

k 40 000 = = 66.476 kg ωn2 (24.53)2 c ζ= √ 2 mk
0 √ c = 2ζ mk = 2(0.171) 66.476(40 000) = 557.7 N · s/m
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8.49 From the log decrement relations (8.5.12) and (8.5.14), δ= Thus δ= Also,
1 B1 ln n Bn+1
ζ=√
1 ln 5 = 0.0536 30
δ 4π 2
+ δ2
ζ = 0.0085
c c c ζ= √ = √ = √ = 0.0085 2 mk 2 100k 20 k
√ Thus c = 0.17 k. We are told the time to complete the 30 cycles is 60 s, so we can compute the period P for one cycle from P = 60/30 = 2 s. Then, k = mωn2 =
mωd2 m(2π/P )2 100(2π/2)2 = = = 987 1 − ζ2 1 − ζ2 1 − (0.0085)2
N/m
√ and thus c = 0.17 k = 5.34 N s/m.
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8.50 a) The roots are s = −1, −10, and −11. The steadystate response is xss = 1/110 = 0.0091. The dominant root is s = −1, and the dominant time constant is 1. Because all roots are real, there will be no oscillations in the response, and the steadystate response will be reached at approximately t = 4. b) We can use the Laplace transform or the trial substitution method. Using the latter, we try x(t) = xss + C1 e−t + C2 e−10t + C3 e−11t For zero initial conditions, x(0) = 0.0091 + C1 + C2 + C3 = 0 x(0) ˙ = −C1 − 10C2 − 11C3 = 0 x ¨(0) = C1 + 100C2 + 121C3 = 0 The solution is C1 = 0.0111, C2 = −0.0111, C3 = 0.0091. So x(t) = 0.0091 + 0.0111e−t − 0.0111e−10t + 0.0091e−11t
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8.51 The roots are s = −1, −2, and −3. The steadystate response is xss = 1/6 = 0.1667. The dominant root is s = −1, and the dominant time constant is 1. Because all roots are real, there will be no oscillations in the response, and the steadystate response will be reached at approximately t = 4. We can use the Laplace transform or the trial substitution method. Using the latter, we try x(t) = xss + C1 e−t + C2 e−10t + C3 e−11t For the given initial conditions, x(0) = 0.1667 + C1 + C2 + C3 = 2 x(0) ˙ = −C1 − 2C2 − 3C3 = −4 x ¨(0) = C1 + 4C2 + 9C3 = 1 The solution is C1 = −4, C2 = 9.5, C3 = −3.667. So x(t) = 0.1667 − 4e−t + 9.5e−2t − 3.667e−3t
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8.52 The factor of the given root pair is (s + 3)2 + 25 = s2 + 6s + 34. To find the third root s = −A, expand the equation as follows: (s2 + 6s + 34)(s + A) = s3 + (6 + A)s2 + (34 + 6A)s + 34A Comparing this with the characteristic polynomial s3 + (6 + µ)s2 + (34 + 6µ)s + 34µ we see that the third root is s = −µ. Thus, as long as µ > 3, the root pair s = −3 ± 5j is dominant, and the dominant time constant is τ = 1/3. Thus the settling time is 4τ = 4/3. −1 The damping ratio of the dominant pair √ is ζ = cos[tan (5/3)] = 0.514, and its undamped √ natural frequency is ωn = 9 + 25 = 34. Thus the maximum percent overshoot is √ 2 M% = 100e−πζ/ 1−ζ = 15% The peak time is tp =
ωn
The rise time is
π = 0.628 1 − ζ2

2π − φ = 0.422 ωn 1 − ζ 2
tr = where
−1
φ = tan

1 − ζ2 + π = 4.17 ζ
When µ = 30, the root separation factor is 30/3=10. When µ = 6, the root separation factor is 6/3 = 2. When µ = 3, the root separation factor is 1. The above calculations are based on the dominant root approximation, which is less accurate when the root separation factor is small. Thus the calculations for µ = 30 have the most accuracy.
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8.53 The steadystate response is yss = 5000/4680 = 1.068. The dominant root pair is s = −3 ± 6j. For the dominant root pair, $
ζ = cos tan
−1
M% = 100e−πζ/ ωn =
& '%
√
6 3
= 0.447
1−ζ 2
= 20.8%

32 + 62 = 6.708 π = 0.524 tp = ωn 1 − ζ 2
φ = tan
−1
tr =
*
1 − ζ2 ζ
+
+ π = 4.249
2π − φ = 0.339 ωn 1 − ζ 2
The 2% settling time is estimated to be 4/3 = 1.33. Simulation of the fourthorder model gives the following characteristics: M% = 13% tp = 0.748 tr = 0.324 ts = 1.39
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8.54 The roots are s = −12.68 ± 54.28j, and −0.323 ± 1.3j. The dominant root pair is s = −0.323 ± 1.3j, for which $
ζ = cos tan
−1
&
1.3 0.323 √
M% = 100e−πζ/ ωn =
'%
1−ζ 2
= 0.241
= 46%

0.3232 + 1.32 = 1.34 π = 2.42 tp = ωn 1 − ζ 2
φ = tan
−1
tr =
*
1 − ζ2 ζ
+
+ π = 4.47
2π − φ = 1.4 ωn 1 − ζ 2
The steadystate response is found by setting the derivatives to zero to obtain 11, 202yss = 50, 000, or yss = 4.46. The form of the response based on the dominant root pair is y(t) = Ae−0.323t sin(1.3t + θ) + 4.46 The dominant time constant is τ = 1/0.323 = 3.2, and the settling time is 4τ = 12.38. Simulation of the fourthorder model gives the following characteristics: M% = 46% tp = 2.4 tr = 0.936 ts = 12.2
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8.55 Using the given parameter values with (8.6.12), we obtain D(s) = (3s2 + 35s + 5)(s2 + 20s + 3) − (20s + 3)2 or D(s) = 3s4 + 90s3 + 314s2 + 85s + 6 Thus X1 (s) = X2 (s) =
20s + 2 F (s) D(s)
3s2 + 35s + 6 F (s) D(s)
The characteristic roots are s = −0.1442 ± 0.00364j, −3.692, and −26.019. With F (s) = 1/s, we have X1 (s) =
20s + 2 0.00364C1 + C2 (s + 0.1442) C3 C4 C5 = + + + sD(s) (s + 0.1442)2 + (0.00364)2 s + 3.692 s + 26.019 s
Using the partial fraction expansion or the MATLAB residue function, we obtain x1 (t) = 2e−0.1442t (−0.178 cos 0.00364t + 3.0672 sin 0.00364t) + 0.0231e−3.692t − 0.0004e−26.019t + 0.333 Similarly, for x2 , with F (s) = 1/s, we have X2 (s) =
3s2 + 35s + 6 0.00364C1 + C2 (s + 0.1442) C3 C4 C5 = + + + 2 2 sD(s) (s + 0.1442) + (0.00364) s + 3.692 s + 26.019 s
Using the partial fraction expansion or the MATLAB residue function, we obtain x2 (t) = 2e−0.1442t (−0.5137 cos 0.00364t − 3.4987 sin 0.00364t) + 0.0264e−3.692t + 0.001e−26.019t + 1
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8.56 Using (8.6.7), (8.6.11), and (8.6.12), and substituting F (s) = 1, we have D(s) = 3s4 + 14s2 + 6 X1 (s) =
3s4
3 3 F (s) = 4 2 + 14s + 6 3s + 14s2 + 6
3s2 + 5 3s2 + 5 F (s) = 3s4 + 14s2 + 6 3s4 + 14s2 + 6 The roots are s = ±2.0468j and s = ±0.6909j. Thus X2 (s) =
X1 (s) =
1 [s2
+
(2.0468)2 ][s2
+
(0.6909)2 ]
=
(1) (2)
cC1 + C2 s dC3 + C4 s + s2 + c2 s2 + d2
where c = 2.0468 and d = 0.6909. The inverse transform can be found with the MATLAB residue function or with algebra. Using the least common denominator and comparing numerators, we find that C2 = −C4 = 0 and 1 C1 = = −0.1316 2 c(d − c2 ) C3 = − Thus
d(d2
1 = 0.3899 − c2 )
x1 (t) = C1 sin ct + C3 sin dt = −0.1316 sin 2.0468t + 0.3899 sin 0.6909t Also, X2 (s) =
s2 + 5/3 cD1 + D2 s dD3 + D4 s = + [s2 + (2.0468)2 ][s2 + (0.6909)2 ] s2 + c2 s2 + d2
Using the least common denominator and comparing numerators, we find that D2 = −D4 = 0 and 5 − 3c2 D1 = = 0.9961 c(d2 − c2 ) D3 =
1 − cD1 = −1.5036 d
Thus x2 (t) = D1 sin ct + D3 sin dt = 0.9961 sin 2.0468t − 1.5036 sin 0.6909t
Transforming the differential equations with nonzero initial conditions gives (3s2 + 5)X1 (s) − 3X2 (s) = 3sx1 (0) + 3x˙ 1 (0) −3X1 (s) + (s2 + 3s)X2 (s) = sx2 (0) + x˙ 2 (0) + F (s) 858
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The solutions are X1 (s) =
[3sx1 (0) + 3x˙ 1 (0)](s2 + 3s) + 3sx2 (0) + 3x˙ 2 (0) + 3F (s) D(s)
X2 (s) =
[sx2 (0) + x˙ 2 (0) + F (s)](3s2 + 5) + 9sx1 (0) + 9x˙ 1 (0) D(s)
(3) (4)
So we see from the latter equation that x˙ 2 (0) = 1 and F (s) = 0 is equivalent to having F (s) = 1 and x˙ 2 (0) = 0. In order for (1) and (2) to match (3) and (4) with F (s) = 0, we must have all zero initial conditions except for x˙ 2 (0) = 1.
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8.57 The equations of motion are m1 x ¨1 = f (t) − k1 x1 − c1 x˙ 1 + k2 (x2 − x1 ) + c2 (x˙ 2 − x˙ 1 ) m2 x ¨2 = f (t) − k2 (x2 − x1 ) − c2 (x˙ 2 − x˙ 1 )
Transform each equation and collect terms: (
)
m1 s2 + (c1 + c2 )s + k1 + k2 X1 (s) − (c2 s + k2 )X2 (s) = F (s) −(c2 s + k2 )X1 (s) + (m2 s2 + c2 s + k2 )X2 (s) = 0
Cramer’s determinant D(s) for this set of equations is the same as that given by (8.6.7). (
)
D(s) = m1 s2 + (c1 + c2 )s + k1 + k2 (m2 s2 + c2 s + k2 ) − (c2 s + k2 )2 The solutions are
X1 (s) m 2 s2 + c2 s + k2 = F (s) D(s) X2 (s) c2 s + k2 =− F (s) D(s)
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8.58 The approach is outlined in Example 8.6.4. The transfer function expressions are given in Problem 8.57. They are X1 (s) m 2 s2 + c2 s + k2 = F (s) D(s) X2 (s) c2 s + k2 =− F (s) D(s) where
(
)
D(s) = m1 s2 + (c1 + c2 )s + k1 + k2 (m2 s2 + c2 s + k2 ) − (c2 s + k2 )2
For this problem the program is as follows.
% Parameter values. m1 = 3; m2 = 1; c1 = 15; c2 = 20; k1 = 2; k2 = 3; % Form the determinant. D = conv([m1,c1+c2,k1+k2],[m2,c2,k2]conv([0,c2,k2],[0,c2,k2]) % Form the numerators of the transfer functions. N1 = [m2,c2,k2]; N2 = [c2,k2]; % Form and display the transfer functions. sys1 = tf(N1,D) sys2 = tf(N2,D) % Compute and plot the step response. step(sys1,sys2,’’) The variable x1 will be plotted with a solid line and x2 with a dashed line.
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8.59 The transfer function is Y (s) as + 1 = 4 3 F (s) s + 24s + 225s2 + 900s + 2500 The MATLAB file is den = [1,24,225,900,2500]; numa = [0.2,1]; sysa = tf(numa,den); numb = [1,1]; sysb = tf(numb,den); numc = [10,1]; sysc = tf(numc,den); Type step(sysa) at the prompt to see the response for the case where a = 0.2. Then right click on the plot and select “Characteristics” to determine the overshoot, etc. Doing the same thing for the other two cases, we can construct the following table.
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a 0.2 1 10
% Overshoot 29.6 190 2250
Table : Values for Problem 8.59 Peak Time Rise Time Settling Time 0.745 0.457 1.87 0.497 0.195 2.38 0.469 0.07 2.3
Compare these values with the following predicted from the dominant root s = −2 + 4j: Overshoot = 20.8%, peak time = 0.79, rise time = 0.508, and settling time = 2.
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8.60 The standard form is 1+
c s =0 5 s2 + 9
where the locus parameter is K = c/5. Type rlocus([1,0],[1,0,9]) to plot the root locus.
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8.61 The equation of motion is 4¨ x + 8x˙ + kx = f (t) and the characteristic equation is 4s2 + 8s + k = 0 The standard form is 1+
k 1 =0 2 4 s + 2s
where the locus parameter is K = k/4. Type rlocus(1,[1,2,0]) to plot the root locus.
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8.62 The equation of motion is 4¨ x + cx˙ + 64x = f (t) and the characteristic equation is 4s2 + cs + 64 = 0 The standard form is 1+
s c =0 2 4 s + 16
where the locus parameter is K = c/4. Type rlocus([1,0],[1,0,16]) to plot the root locus.
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8.63 The equation of motion is 2¨ x + 8x˙ + (26 + k2 )x = k2 y(t) and the characteristic equation is 2s2 + 8s + 26 + k2 = 0 The standard form is 1+
k2 1 =0 2 2 s + 4s + 13
where the locus parameter is K = k2 /2. Type rlocus(1,[1,4,13]) to plot the root locus. The plot shows that the largest attainable value of ζ for k2 > 0 is ζ = cos[tan−1 (3/2)] = 0.55.
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8.64 The equation of motion is 2¨ x + (8 + c2 )x˙ + 26x = c2 vi (t) and the characteristic equation is 2s2 + (8 + c2 )s + 26 = 0 The standard form is 1+
s c2 =0 2 2 s + 4s + 13
where the locus parameter is K = c2 /2. Type rlocus([1,0],[1,4,13]) to plot the root locus. The plot shows that the smallest attainable value of τ for c2 > 0 occurs when K = 3.21 at s = −3.61. This gives τ = 1/3.61 and c2 = 2K = 6.42.
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8.65 The characteristic equation in standard form is 1+K
s3
+
13s2
1 =0 + 52s + 60
Type rlocus(1,[1,13,52,60]),sgrid([0.5,0.707],[3,5]) to plot the root locus with a (ζ, ωn ) grid. The plot shows that the specifications are achievable with K in the range 38.6 ≤ K ≤ 83.8.
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8.66 a) The characteristic equation in standard form is 1+
K 1 =0 2 s3 + 6s2 + 8s
where the locus parameter is K/2. Type rlocus(1,[1,6,8,0]),sgrid(0.707,[]) to plot the root locus with a (ζ, ωn ) grid. b) The plot shows that the specifications are achievable with K/2 ≈ 5.18; that is, with K = 10.36. c) In the first printing, the transfer function was not given. It is X(s) K = 3 2 F (s) 2s + 12s + 16s + K The step response with K = 10.36 is obtained by typing sys = tf(10.36,[2,12,16,10.36]);step(sys) Right click on the plot and select “Characteristics”, then “Peak response”. The maximum overshoot is 4.03% at t = 4.4. With K = 10.36, the roots are s = −4.4694 and s = −0.7653 ± 0.7572j. The dominant time constant is 1/0.7653 = 1.31. The secondary time constant is 1/4.4694 = 0.2237. It is much smaller than the dominant time constant and thus has little effect on the response. This is illustrated by the fact that the overshoot predicted from the dominant roots with ζ = 0.707 is 4.33%, which is almost identical to that of the thirdorder system.
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8.67 Isolate the parameter c as follows. La Is2 + Ra Is + Kb KT + c(La s + +Ra ) = 0 Put this into the form of (8.7.1): 1+
c s + Ra /La =0 2 I s + (Ra /La )s + Kb KT /La I
where K = c/I. The zero is s = −Ra /La = −666.67. The poles are the roots of s2 + (Ra /La )s + Kb KT /La I = s2 + 666.67s + 8.3333 × 104 = 0 and are s = −500 and s = −166.7. Type rlocus([1, 666.67], [1, 666.67, 8.3333e+4]), axis equal to plot the root locus. The locus is a circle centered at s = −666.67 with a radius of 288.5. The smallest possible time constant occurs when both roots are repeated at s = −955. This occurs when K = 1240, or when c = 1240I = 0.0496.
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8.68 a) From (8.6.7), D(s) = s4 + (16 + c1 )s3 + (9 + 8c1 )s2 + (8 + 4c1 )s + 4 = 0 Isolate the parameter c1 as follows. s4 + 16s3 + 9s2 + 8s + 4 + c1 (s3 + 8s2 + 8s) = 0 Put this into the form of (8.7.1): 1 + c1
s3 + 8s2 + 8s =0 s4 + 16s3 + 9s2 + 8s + 4
where K = c1 . The zeros are the roots of s3 + 8s2 + 8s = 0 and are s = 0, s = −6.8284, and s = −1.1716. The poles are the roots of s4 + 16s3 + 9s2 + 8s + 4 = 0 and are s = −15.4499, s = −0.0107 ± 0.6997j, and s = −0.5287. Type rlocus([1, 8,8,0], [1,16,9,8,4]), sgrid(0.707,[]) to plot the root locus with a gridline corresponding to ζ = 0.707. b) By moving the cursor along the plot we can determine that a root exists at s = −0.973 + 0.966j, which gives ζ = 0.71, which is close enough to 0.707. The value of c1 required to give this root is determined as we move the cursor, and is the “gain”, which is 3.69. However, this root is not the dominant root. This can be seen by typing K = 3.69; roots([1,(16+K),(9+8*K),(8+8*K), 4]). The roots are s = −17.6245, s = −0.9725 ± 0.9673j, and s = −0.1206. The latter is the dominant root. So it is not possible to achieve a dominant root with a damping ratio of 0.707. c) Type rlocus([1,8,8,0], [1,16,9,8,4]),sgrid([],0.25) to plot the root locus with a gridline corresponding to ωn = 0.25. This will give a circular gridline that intersects the real axis at the point corresponding to τ = 4. By moving the cursor along the plot we can determine that a root exists at s = −0.252, which gives τ close enough to 4. The value of c1 required to give this root is determined as we move the cursor, and is the “gain”, which is 1.51. To see if this is the dominant root, type K = 1.51; roots([1,(16+K),(9+8*K),(8+8*K), 4]). The roots are s = −16.2908, s = −0.4835 ± 0.8601j, and s = −0.2522. The latter is the dominant root. So it is possible to achieve a dominant real root with a time constant of 4. d) We obtain the two transfer functions from (8.6.11) and (8.6.12), using c1 = 1.51. X1 (s) 8s + 4 8s + 4 = 4 = 4 3 2 3 D(s) s + (16 + c1 )s + (9 + 8c1 )s + (8 + 4c1 )s + 4 s + 17.51s + 21.08s2 + 14.04s + 4 X2 (s) s2 + 9.51s + 5 = 4 D(s) s + 17.51s3 + 21.08s2 + 14.04s + 4 The unitstep response plot for x1 is found by typing sys1 = tf([8,4],[1,17.51,21.08,14.04,4]); step(sys1)
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The unitstep response plot for x2 is found by typing sys2 = tf([1,9.51,5],[1,17.51,21.08,14.04,4]); step(sys2)
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8.69 The Simulink digram is identical to that shown in Figure 8.8.6. Edit the Signal Builder and the LookUp Table blocks to reflect the new data. Change the mfile damper.m to read as follows. function f = damper(v) if v <= 0 f = 500*(abs(v)).^(1.2); else f = 50*(abs(v)).^(1.2); end A Stop Time of 15 seconds gives a suitable plot. The maximum overshoot is 0.522−0.3 = 0.222 m at approximately t = 0.57 s. The maximum undershoot is 0 − 0.1765 = −0.1765 m at approximately t = 4.73 s.
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8.70 The state variable representation is much easier to do because it does not require the algebra needed to find the transfer functions. However, since much of that algebra has been done already here —see (8.6.7), (8.6.11), and (8.6.12)— then the two methods are about equal in difficulty, for this particular problem. The state space matrices are
A=
0 1 0 0 −5 −4 4 1 0 0 0 1 4 1 −4 −1
C=
7
1 0 0 0 0 0 1 0
8
B= D=
7
0 0
8
0 0 0 1
The transfer functions are X1 (s) s+4 = 4 3 F (s) s + 5s + 12s2 + 13s + 4 X2 (s) s2 + 4s + 5 = 4 F (s) s + 5s3 + 12s2 + 13s + 4 The next figure shows the Simulink model using both representations. For each you must create the following MATLAB function. function f = func(t) if t <= 1 f = t; elseif t < 2 f = 2  t; else f = 0; end
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Figure : for Problem 8.70
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Nine
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c Solutions Manual Copyright 2005 by The McGrawHill Companies, Inc. !
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9.1 a)
b)
c)
13 13 yss = ! 15 = ! 15 = 14.1 2 2 (7ω) + 9 [7(1.5)]2 + 92 15ω 15(2) yss = ! 6= ! 6 = 24.96 2 2 [9ω] + 4 (9(2))2 + 42 yss
d) yss
√ (ω)2 + 502 104 + 502 =! 2 2= √ 2 = 1.24 2 (ω) + 150 104 + 1502 !
√ √ 33 ω 2 + 1002 33 502 + 1002 √ √ = 8= 8 = 2.46 200 ω 2 + 332 200 502 + 332
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9.2 a) In standard form of (9.1.2), the transfer function is T (s) =
15 15 1 = 6s + 2 2 3s + 1
Here τ = 3 and the multiplicative factor 15/2 shifts the m curve up by 20 log 15 2 = 17.501 dB. The m plot looks like that in Figure 9.1.6, but shifted up by 17.501 dB. Because ! (15/2) = 0◦ , the phase curve is identical to that shown in Figure 9.1.6. b) In standard form of τ s/(τ s + 1), the transfer function is T (s) =
9s 9 2s = 8s + 4 2 2s + 1
Here τ = 2 and the multiplicative factor 9/2 shifts the m curve up by 20 log 92 = 13.064 dB. The m plot looks like that in Figure 9.1.9, but shifted up by 13.064 dB. Because ! (9/2) = 0◦ , the phase curve is identical to that shown in Figure 9.1.9. c) In standard form of (9.1.13), the transfer function is T (s) = 6
14s + 7 7 2s + 1 = 10s + 2 2 5s + 1
Here τ1 = 2 and τ2 = 5. The multiplicative factor 7/2 shifts the m curve up by 20 log 72 = 10.881 dB. The m plot looks like that in part (a) of Figure 9.1.12 (because τ1 < τ2 ), but shifted up by 10.881 dB. Note that Figure 9.1.12 applies to the case where K = 1 in (9.1.13). Because ! (7/2) = 0◦ , the phase curve is given by φ(ω) = ! (1 + 2ωj) − ! (1 + 5ωj) At low frequencies, φ ≈ 0◦ but negative. At ω = 1/τ2 = 1/5 and at ω = 1/τ1 = 1/2, φ = −23◦ (this identical value is a coincidence because here τ1 = 1/τ2 ). As ω → ∞, φ → 0◦ through negative values.
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9.3 The input is hi (t) = 10 + 3 sin
2π t 12
We are given that the steadystate response is h(t) = 10 + 2 sin
"
2π t+φ 12
#
Thus the amplitude ratio is M = 2/3. The model is 1 Ah˙ = − (h − hi ) R so that the time constant is τ = RA. Thus the model can be written as τ h˙ + h = hi The amplitude ratio for this model is M=√
1 2 = 2 2 3 1+ω τ
where ω = 2π/12 = π/6. Solving for τ gives τ=
2√ 5 = 1.423 hr π
The phase shift is φ = − tan−1 (ωτ ) = −0.641 rad
and the time shift is φ/ω = 1.22 hr.
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9.4 In the first printing of the text, the value of R was given incorrectly. The correct value is R = 4.5 × 10−3 . The total wall surface area is A = 4(5×3) = 60 m2 , and the air volume is V = 3(5×5) = 75 m3 . The thermal capacitance of the room air is C = mcp = ρV cp = 1.289(75)(1004) = 9.704 × 104 The outside temperature is To = 15 + 5 sin ωt where ω=
2π = 7.272 × 10−5 rad/sec 24(3600)
The model is C or
dT 1 = (To − T ) dt RT
dT + T = To dt where RT is the total wall resistance, which is RT C
RT = AR = 60(4.5 × 10−3 ) = 0.27 The time constant is The amplitude ratio is
τ = RT C = 2.62 × 104 s M=√
1 = 0.4647 1 + ω2 τ 2
Thus the amplitude of oscillation of the indoor temperature is 5(0.4647) = 2.32◦ C. The phase shift is φ = − tan−1 ωτ = −1.087 rad
Thus
T (t) = 15 + 2.32 sin(7.272 × 10−5 t − 1.087)
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9.5 The model is I ω˙ + cω = T , or 2ω˙ + 4ω = T . The transfer function is Ω(s) 0.25 1 = = T (s) 2s + 4 0.5s + 1 and the amplitude ratio is M (ω) = √
0.25 1 + 0.25ω 2
The phase shift is φ(ω) = − tan−1 (ωτ ) = − tan−1 (0.5ω). Because there are three terms in the forcing function, there are three corresponding terms in the steadystate response. These are ωss (t) = A1 M1 + A2 M2 sin(3t + φ2 ) + A3 M3 cos(5t + φ3 ) From the M (ω) and φ(ω) equations we obtain M1 = M (0) = 0.25
M2 = M (3) = 0.1387
φ2 = φ(3) = − tan−1 (1.5) = −0.983
M3 = M (5) = 0.0928
φ3 = φ(5) = − tan−1 (2.5) = −1.19
Thus, noting from the input that A1 = 30, A2 = 5, and A3 = 2, we have ωss (t) = 7.5 + 0.6935 sin(3t − 0.983) + 0.1856 cos(5t − 1.19)
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9.6 The model is
1 Ah˙ = − h + qvi R
or
RAh˙ + h = Rqvi
With the given values,
6000π h˙ + h = 1500qvi
The time constant is τ = 6000π = 1.885 × 104 The amplitude ratio is M (ω) = √
1500 1 + ω2 τ 2
where ω = 0.002. The phase shift is φ(ω) = − tan−1 (ωτ )
Since the input consists of the sum of two terms, the steadystate response has the form hss (t) = A1 M1 + A2 M2 sin(0.002t + φ2 ) where A1 = 0.2, A2 = 0.1, and M1 = M (0) = 1500
Thus
M2 = M (0.002) = 39.775
φ2 = φ(0.002) = − tan−1 [0.002(1.885 × 104 )] = −1.544 hss (t) = 300 + 3.9775 sin(0.002t − 1.544)
The lag is given by φ2 /ω = 772 sec.
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9.7 The transfer function given in the text is T (s) =
R1 R2 Cs + R2 R 2 τ1 s + 1 = R1 R2 Cs + R1 + R2 R 1 + R 2 τ2 s + 1
where τ1 = R1 C
τ2 =
R1 R2 C R1 + R2
For the circuit to be a lowpass filter, the corner frequency of the numerator must be greater than the corner frequency of the denominator; that is, 1 1 > τ1 τ2 This implies that τ2 > τ1 , which says R1 R2 C > R1 C R1 + R2 Canceling R1 C gives
R2 >1 R1 + R2
which is impossible to satisfy. Thus the circuit cannot be a lowpass filter.
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9.8 a)
b)
20 5 ! ! yss = ! 5= ! 20 = 34.92 2 2 2 (10ω) + 1 (4ω) + 1 [10(0.2)] + 1 [4(0.2]2 + 1 0.5 0.5 yss = ! 32 = ! 32 = 0.178 2 2 2 (100 − ω ) + (10ω) (100 − 52 )2 + [10(5)]2
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9.9 a)
4 jω(100 − ω 2 + 10ωj)
T (jω) = M (ω) = Thus M (9) = 0.0048.
4 ω (100 − ω 2 )2 + 100ω 2 !
φ(ω) = −! jω − ! [(100 − ω 2 ) + 10ωj] So φ(9) = π/2 − tan−1 (90/19) = 0.2081. Thus yss (t) = 6M (9) sin[9t + φ(9)] = 0.0288 sin(9t − 0.2081) b)
10 + ωj)
T (jω) =
−ω 2 (1
M (ω) =
10 √ ω2 1 + ω2
Thus M (2) = 1.118. φ(ω) = −! (−10/ω 2 ) − ! [(1 + ωj) So φ(2) = π − tan−1 (2) = 2.0344. Thus yss (t) = 3M (2) sin[2t + φ(2)] = 3.354 sin(2t + 2.0344) c) ωj (1 + 2ωj)(1 + 5ωj) ω √ M (ω) = √ 2 1 + 4ω 1 + 25ω 2 T (jω) =
Thus M (0.7) = 0.1118. φ(ω) = ! ωj − ! (1 + 2ωj) − ! (1 + 5ωj) So φ(0.7) = π/2 − tan−1 (1.4) − tan−1 (3.5) = −0.6722. Thus yss (t) = 5M (0.7) sin[0.7t + φ(0.7)] = 0.559 sin(0.7t − 0.6722)
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d) T (jω) =
−ω 2 (1 + 2ωj)(1 + 5ωj)
M (ω) = √
ω2 √
1 + 4ω 2 1 + 25ω 2
Thus M (0.7) = 0.0782. φ(ω) = ! (−ω 2 ) − ! (1 + 2ωj) − ! (1 + 5ωj) So φ(0.7) = −π/2 − tan−1 (1.4) − tan−1 (3.5) = −3.8138. Thus yss (t) = 5M (0.7) sin[0.7t + φ(0.7)] = 0.391 sin(0.7t − 3.8138)
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9.10 The resonant frequency and peak magnitude are given by $
ωr = ωn 1 − 2ζ 2
Mr =
√ Note that ωn = 2. Thus a) For ζ = 0.1, ωr = 1.4 and Mr = 5.025. b) For ζ = 0.3, ωr = 1.28 and Mr = 1.747.
1 2ζ 1 − ζ 2 !
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9.11 The maximum amplitude occurs at ω = ωr and is 22Mr , where Mr = The damping ratio is
1 2ζ 1 − ζ 2 !
c c ζ= √ = √ 2 50 10 2
Thus 22Mr =
22 =3 2ζ 1 − ζ 2
Solve for ζ by squaring each side to obtain
!
4ζ 4 − 4ζ 2 − 13.44 = 0
√ which has the positive root ζ 2 = 4.2, or ζ = 2.05. Thus c = 10 2(2.05) = 28.99.
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9.12 The transfer function is T (s) =
13s2
From (9.2.20)
1 + 2s + k
$
ωr = ωn 1 − 2ζ 2
where
ωn =
%
k 13
2 1 ζ= √ =√ 2 13k 13k We want ωr = 4. Thus we must solve the following for k. 4=
%
k 13
&
1−
2 1√ = 13k − 2 13k 13
This gives k = 208 lb/ft. This gives ζ = 0.0192. From (9.2.21), 1 Mr = ! = 26.295 2ζ 1 − ζ 2
Thus the amplitude of the steadystate response is 26.295(10) = 262.95 The phase shift of the response is (with ω = 4) φ = −! [(k − 13ω 2 ) + 2ωj] = −! [−8 + 8j] = −2.356 Thus the steadystate response is xss (t) = 262.95 sin(4t − 2.356)
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9.13 a) T (jω) = M (ω) =
7 ωj(58 − ω 2 + 6ωj)
7 ω (58 − ω 2 )2 + 36ω 2 !
Since M (0) = ∞, the resonant frequency is at ω = 0. b) 7 T (jω) = 2 (174 − 3ω + 18ωj)(58 − 2ω 2 + 8ωj)
7 ! M (ω) = ! (174 − 3ω 2 )2 + 324ω 2 (58 − 2ω 2 )2 + 64ω 2
A plot of M versus ω shows that M has a single peak near ω = 4.97, which is the resonant frequency. We might have expected two peaks since there are two quadratic factors in the denominator. These quadratic factors have resonant frequencies of 4.58 and 6.32. Because these frequencies are so close in value, their two peaks merge into a single peak at ω = 4.97.
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9.14 Applying the voltage law to each loop gives v1 − Ri1 − L
di3 =0 dt
'
1 di3 i2 dt − L =0 C dt From conservation of charge, i3 = i1 + i2 . Take the Laplace transform of each equation using zero initial conditions. Solve the first two for I1 (s) and I2 (s): V1 (s) − LsI3 (s) I1 (s) = R v2 −
I2 (s) = CsV2 (s) − LCs2 I3 (s)
Substitute these into the third equation to obtain
(RLCs2 + Ls + R)I3 (s) = V1 (s) + RCsV2 (s) The two transfer functions are I3 (s) 1 1 = = −5 2 2 V1 (s) RLCs + Ls + R 10 s + 0.1s + 100 I3 (s) RCs 10−4 s = = V2 (s) RLCs2 + Ls + R 10−5 s2 + 0.1s + 100 The roots are s = −1127 and s = −8873. Thus the corner frequencies are ω1 = 1127 rad/s and ω2 = 8873 rad/s. First consider the transfer function I3 (s)/V1 (s). At low frequencies the m curve approaches 20 log(105 /[(1127)(8873)] = −40 dB. The m curve breaks down at ω1 = 1127 and again at ω2 = 8873 rad/s. So the system acts like a low pass filter for the input voltage v1 , and it tends to filter out frequency components in v1 that are higher than ω1 = 1127 rad/s. The low frequency gain is M = 10m/20 = 10−40/20 = 0.01. Now consider the transfer function I3 (s)/V2 (s). At low frequencies the m curve approaches 20 log(0) = −∞. At ω = 10, "
m = 20 log √
10(10) √ 2 1127 + 102 88732 + 102
#
= −100 dB
The m curve has a slope of 20 dB/decade until ω1 = 1127, when the slope becomes approximately zero. The curve breaks downward at ω2 = 8873 rad/s, and the slope becomes −20 dB/decade for higher frequencies. So the system acts like a band pass filter for the input voltage v2 , and it passes frequency components in v2 that are between ω1 = 1127 rad/s and ω2 = 8873 rad/s.
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9.15 In the first printing of the text, Figure P9.15 was incorrect. The damper should be between the mass and the support. The displacement y should act on the spring. a) The equation of motion is m¨ x = k(y − x) − cx˙ or x ¨ + cx˙ + 600x = 600y The transfer function is
X(s) 600 = 2 Y (s) s + cs + 600 √ √ √ This has the same form as (9.2.14) with ωn = 600 = 10 6 and ζ = c/(20 6). So, provided that c is such that ζ ≤ 0.707, we can use (9.2.20) and (9.2.21) to obtain T (s) =
$ √ $ ωr = ωn 1 − 2ζ 2 = 10 6 1 − 2ζ 2 rad/s
Mr =
1 2ζ 1 − ζ 2 !
√ Since c = 20ζ 6, ζ will be no greater than 0.707 if 0 ≤ c ≤ 34.64 N·s/m. For 0 ≤ c ≤ 34.64, a plot of ωr versus c shows that the resonant frequency varies from 3.6 (for c = 34.64) to 24.5 rad/s (for c = 0). For 0 ≤ c ≤ 34.64, a plot of Mr versus c shows that Mr varies from ∞ (for c = 0) to 1 rad/s (for c = 34.64). For values of ζ > 0.707 (for c > 34.64), there is no resonant peak and thus no resonant frequency. b) The transfer function is T (s) =
X(s) k k/m ωn2 = = = Y (s) ms2 + cs + k s2 + (c/m)s + k/m s2 + 2ζωn s + ωn2
This has the same form as (9.2.14), so provided ζ ≤ 0.707, we can use (9.2.20) and (9.2.21) to compute ωr and Mr . For values of ζ > 0.707, there is no resonant peak and thus no resonant frequency.
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9.16 Applying the voltage law gives vs −
1 C
'
i dt − L
di − vo = 0 dt
where v0 = Ri. Use the latter equation to eliminate i: LC
d2 vo dvo dvs + RC + vo = RC 2 dt dt dt
For the given values, 5 × 10−8 The transfer function is
For R = 10,
d2 vo dvo dvs + 10−5 R + vo = 10−5 R dt2 dt dt Vo (s) 103 Rs = 2 Vs (s) 5s + 103 Rs + 108 Vo (s) 104 s = 2 Vs (s) 5s + 104 s + 108
The roots are s = −1000 ± 4359j. For R = 100, Vo (s) 105 s = 2 Vs (s) 5s + 105 s + 108 The roots are s = −100.05 and s = −199, 900. The plots are shown below. The peak is approximately the same for each. It is −0.000185 dB, or M = 10−0.000185/20 = 1 at 5170 rad/s. Thus an increase in the value of R by a factor of 100 does not significantly change the peak value or the peak frequency, but it does change the spread of the plot about the peak.
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0
−10
R = 1000
−20
Magnitude (dB)
−30
−40
R = 10 −50
−60
−70
−80
−90 0 10
1
10
2
10
3
10
4
10
5
10
6
10
7
10
Frequency (rad/sec)
Figure : Log magnitude plots for Problem 9.16
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9.17 The equations of motion are I1 ω˙ 1 = T1 − c1 (ω1 − ω2 ) I2 ω˙ 2 = c1 (ω1 − ω2 ) − c2 ω2
Applying the Laplace transform for zero initial conditions, and eliminating Ω1 (s), we obtain the transfer function: Ω2 (s) 100 = 2 T1 (s) s + 5s + 2 From this we obtain the magnitude ratio and the phase angle. 100 M (ω) = ! (2 − ω 2 )2 + (5ω)2
5ω 2 − ω2 Evaluation of M and φ for ω = 0, 1.5, and 2 gives the following results. φ(ω) = − tan−1
M (0) = 50 φ(0) = 0
M (1.5) = 13.3259
φ(1.5) = −1.604 rad
Thus the steady state response is
M (2) = 9.8058 φ(2) = −1.763 rad
ω2ss = 50(4) + 13.3259(2) sin(1.5t − 1.604) + 9.8058(0.9) sin(2t − 1.763) or ω2ss = 200 + 26.6518 sin(1.5t − 1.604) + 8.8252 sin(2t − 1.763)
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9.18 The forcing frequency is ω = 5.2 and the natural frequency is ωn = beat period is 2π 2π = = 31.4159 ω − ωn  5.2 − 5 The vibration period is
!
75/3 = 5. The
4π 4π = = 1.232 ω + ωn 10.2
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9.19 From (9.3.4) with fo = 0.2, k = 64, and ωn = 8, "
sin 8t x(t) = 0.0125 − t cos 8t 8
#
The plot is shown below. It takes approximately 8.2 sec for x(t) to become greater than 0.1 ft. 0.1
0.08
0.06
0.04
x(t) (ft)
0.02
0
−0.02
−0.04
−0.06
−0.08
−0.1
0
1
2
3
4
5 t (sec)
6
7
8
9
10
Figure : Plot for Problem 9.19
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9.20 From the given data
3400(2π) = 356 rad/s 60 √ c = 2ζ mk = 548 N · s/m
ω=
Also
Thus
Ft (s) cs + k = Fr (s) ms2 + cs + k √ k + cωj k 2 + c2 ω 2 Ft (ω) = Fr  = Fr  ! 2 k − mω + cωj (k − mω 2 )2 + c2 ω 2
For rotating unbalance
Fr  = mu 'ω 2 = 0.1(0.02)(356)2 = 253.5 N Substituting the values of k, c, m, and ω, we obtain Ft  = 253.5(0.1027) = 26 N
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9.21 For base motion,
√ X k 2 + c2 ω 2 =! Y (k − mω 2 )2 + c2 ω 2
Plotting this expression versus ω for the given values of m, c, and k, we find that the ratio X/Y is never less than 1. Therefore, X will never be greater than 2 mm as long as Y is never greater than 2 mm.
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9.22 The vibration frequency of the base motion as a function the vehicle velocity v is ω=
"
5280 20
#"
#
1 (2π)v = 0.4608v 3600
From the results of Example 8.7.1, D3 = (m1 s2 +c1 s+k1 )(m2 s2 +c1 s+k1 +k2 )−(c1 s+k1 )2 = m1 m2 s4 +(m1 k1 +m1 k2 +m2 k1 )s2 +k1 k2 where we have taken c1 = 0. Using the given values, we obtain 114.472s4 + 4.0838 × 105 s2 + 8 × 107 = 0 The roots are s = ±14.4j and s = ±57.96j. Thus the resonant frequencies are 14.4 and 57.96 rad/s, and the resonant speeds are v=
14.4 = 31.25 ft/sec 0.4608
and
v=
57.96 = 125.8 ft/sec 0.4608
or v = 21.3 mph and 85.8 mph. Note that the amplitude of the road surface variation, although given in the problem statement, was not needed.
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9.23 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be mb = 2388(2)(0.3)(0.03) = 42.98 kg The equivalent system mass is me = 50 + 0.23mb = 59.886 kg The stiffness of the beam is k=
Ewh3 2 × 1011 (0.3)(0.03)3 = = 1.6875 × 106 N/m 4L3 4(2)3
The forcing frequency is ω=
3400(2π) = 356 rad/s 60
and the amplitude of the unbalance force is Fr  = mu 'ω 2 = 0.02(356)2 = 2535 N Thus the displacement amplitude is, since c = 0,
or 0.428 mm.
Fr  Fr  X=! = = 4.29 × 10−4 m 2 2 k − me ω 2  (k − me ω )
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9.24 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be mb = 2388(2)wh = 4776wh where w is the beam width and h is the beam thickness. The equivalent system mass is me = 50 + 0.5mb = 50 + 2388wh The stiffness of the beam is k=
16Ewh3 = 4 × 1011 wh3 L3
The forcing frequency is ω=
3400(2π) = 356 rad/s 60
and the amplitude of the unbalance force is Fr  = mu 'ω 2 = 0.02(356)2 = 2535 N Thus the displacement amplitude is, since c = 0, Fr  Fr  X = 0.01 = ! = 2 2 k − me ω 2  (k − me ω )
(1)
This expression is a complicated function of w and h since both k and me are functions of w and h. So let us temporarily neglect the beam mass. Then me ≈ 50 kg and the previous expression becomes X = 0.01 =
2535 k − 6.3368 × 106 
This gives k = 6.5903 × 106 N/m. Thus wh3 =
6.5903 × 106 = 1.6 × 10−5 4 × 1011
There is an infinite number of solutions to this equation, but we should choose one that results in an equivalent beam mass that is small compared to 50 kg, because we neglected the beam mass in obtaining this solution. One solution is h = 0.04 m, which gives w = 0.25 m, both reasonable dimensions. These give an equivalent beam mass of 0.5(47.76) = 23.88 kg and a total equivalent mass of me = 73.88 kg. Using this value in equation (1) we obtain X=
2535 = 9.14 × 10−4 m 6.59 × 106 − 73.88(356)2 
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9.25 We √ are given m = 1500 kg, k = 20 000 N/m, ζ = 0.04, and Y = 0.01 m. Thus c = 2ζ mk = 438.18. From (9.3.9), with s = jω, !
At resonance, from (9.2.20),
mω 2 k 2 + (cω)2 Ft  = Y  ! (k − mω 2 )2 + (cω)2 $
ω = ωr = ωn 1 − 2ζ 2 = 0.998
%
k = 3.6442 m
Thus Ft  = 2500 N
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9.26 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also, ω = 40 Hz = 251 rad/s ωn =
%
k = m
%
66 667 = 57.2 rad/s (200/9.81)
From (9.3.7) with c = 0, X k =! = 0.055 Y  (k − mω 2 )2
Thus 5.5% of the airframe motion is transmitted to the module.
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9.27 a) Neglect damping in the isolator, and determine its required stiffness k. From (9.3.7) with c = 0, X k k/m ωn2 ! ! =! = = = 0.1 Y  (k − mω 2 )2 (k/m − ω 2 )2 (ωn2 − ω 2 )2
which gives (ω/ωn )2 = 11. Thus ωn2 =
ω2 [3000(2π)/60]2 (314)2 = = 11 11 11
and
2 (314)2 = 556.7 lb/ft 32.2 11 √ b) Let r = ω/ωn . From part (a), ωn = 314/ 11, we have k = mωn2 =
r1 =
2500(2π)/60 √ = 2.77 314/ 11
r2 =
3500(2π)/60 √ = 3.87 314/ 11
and
From (9.3.7) with c = 0,
X 1 = Y  1 − r 2 
Thus the highest percentage of motion will be transmitted at the lowest r value, which is r1 , the value corresponding to 2500 rpm. For 2500 rpm, X 1 = = 0.15 Y  1 − r12  For 3500 rpm,
X 1 = = 0.07 Y  1 − r22 
Thus at most, 15% of the crane motion will be transmitted to the module.
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9.28 The transmitted force is given by (9.3.14) and (9.3.17): Ft  = mu 'ω If we let
!
k 2 + (cω)2 (k − mω 2 )2 + (cω)2
2!
r = ω/ωn then we can express the previous equation as Ft  = mu 'ω
2
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. ' = 0.1.12 = 0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus r= Thus
ω 104.7 =! = 5.83 ωn 500/1.55
Ft  = (0.00155)(0.0083)(104.7)2
%
1 + 136ζ 2 = 0.141 1088 + 136ζ 2
%
1 + 136ζ 2 1088 + 136ζ 2
a) For ζ = 0.05, Ft = 0.0049 lb. b) For ζ = 0.7, Ft = 0.034 lb.
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9.29 The transmitted force is given by (9.3.14) and (9.3.17): Ft  = mu 'ω If we let
!
k 2 + (cω)2 (k − mω 2 )2 + (cω)2
2!
r = ω/ωn then if c = 0, we can express the previous equation as Ft  = mu 'ω 2
1 1 − r 2 
Here Ft = 15 and ωR = 200(2π)/60 = 20.9 rad/s. Thus r= Thus
ω 20.9 20.9 =! =! = 3.62 ωn k/m 2500/75 Ft  = mu '(20.9)2 (0.0826) = 15
Solve for mu ' to obtain mu ' = 0.416 kg·m.
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9.30 For a vibrometer we require that the natural frequency ωn be much less than the forcing frequency ω. Here ω = 2π(200) = 400π rad/s and ωn = So we require that
%
%
k 0.1
k ) 400π 0.1
or k ) 1.579 × 105 N/m
Suppose that k = 1.6×104 . Then the static deflection will be δst = mg/k = 6.1×10−5 m. The designer must verify that there will be enough space in the instrument to accommodate this deflection.
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9.31 In the first printing of the text, Figure P9.31 was incorrect. The damper should be between the mass and the support. The displacement y should act on the spring. The equation of motion is m¨ x = k(y − x) − cx˙ or
The transfer function is T (s) =
200¨ x + 2000x˙ + 2 × 104 x = 2 × 104 y X(s) 2 × 104 100 = = 2 2 4 Y (s) 200s + 2000s + 2 × 10 s + 10s + 100
This has the same form as (9.2.14) with ωn = 10 and ζ = 0.5. So we can use (9.2.20) and (9.2.21) to obtain $ ωr = ωn 1 − 2ζ 2 = 7.07 rad/s Mr =
1 = 1.155 2ζ 1 − ζ 2 !
For the bandwidth, equation (2) of Example 9.4.2 gives r=
&
1−
2ζ 2
$
± 2ζ 1 − ζ 2 = 1.167, 0.605j
Because there is only one positive, real solution, the bandwidth extend from ω = 0 to ω = 1.167ωn = 11.69 rad/s.
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9.32 The transfer function becomes T (s) =
10−6 s 3 × 10−4 s2 + 10−2 s + 1
From the frequency response plot shown below we can tell that the bandwidth is between 45 and 76 rad/s. You can obtain this plot in MATLAB by typing: sys = tf([1e6,0],[3e4,1e2,1]); bodemag(sys)
Bode Diagram −80
−85
−90
Magnitude (dB)
−95
−100
−105
−110
−115
−120
−125 0 10
1
2
10
10
3
10
Frequency (rad/sec)
Figure : Plot for Problem 9.32
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9.33 a) The transfer function for the circuit in part (a) of the figure was derived in Chapter 6. It is Vo (s) G = 2 2 2 Vs (s) R C s + 2RCs + 1 Its natural frequency is ωn = Its damping ratio is
&
1 R2 C 2
=
1 RC
2RC ζ= √ =1 2 R2 C 2
From Equation (1) in Example 9.4.2, with r = ω/ωn = RCω, 1 kM = ! (1 − r 2 )2 + 4r 2 √ which has a maximum of 1 at r = 0. Setting kM = 1/ 2 and solving for r gives r = 0.644. Thus the bandwidth corresponds to 0 ≤ r ≤ 0.644 or 0≤ω≤
0.644 RC
b) The transfer function for the circuit in part (b) of the figure was derived in Chapter 6. It is Vo (s) G = 2 2 2 Vs (s) R C s + 3RCs + 1 Its natural frequency is ωn = Its damping ratio is
&
1 R2 C 2
=
1 RC
3RC ζ= √ = 1.5 2 R2 C 2
From Equation (1) in Example 9.4.2, with r = ω/ωn = RCω, 1 kM = ! (1 − r 2 )2 + 9r 2 √ which has a maximum of 1 at r = 0. Setting kM = 1/ 2 and solving for r gives r = 0.374. Thus the bandwidth corresponds to 0 ≤ r ≤ 0.374 or 0≤ω≤
0.374 RC
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9.34 The circuit equation is LC v¨1 + RC v˙ 1 + v1 = vs or 10−7 v¨1 + 10−6 Rv˙ 1 + v1 = vs Its natural frequency and damping ratio are √ √ ωn = 107 = 103 10 10R 5R √ ζ= √ = 1000 10 2 107 The resonant frequency is $ √ ωr = ωn 1 − 2ζ 2 if ζ ≤ 1/ 2
Thus there is a resonant peak and a resonant frequency only if 5R 1 √ ≤√ 1000 10 2 which implies that
√ 1000 10 √ R≤ = 447 Ω 5 2
If there is a resonant peak, its value is Mr =
1 2ζ 1 − ζ 2 !
A plot of Mr versus R for 0 ≤ R ≤ 447 shows that Mr = 3.2026 for R = 100. As R is increased to 447, Mr decreases to Mr = 1. For R > 447, the maximum value of M occurs at ω = 0. The following plots show the variation of M versus r = ω/ωn , for R = 100, 250, 400, 500, 750, and 1000 Ω. The plots show that as R is increased, the system changes from a bandpass filter to a low pass filter. For R ≤ 447, increasing R decreases the bandwidth. For R > 447, increasing R decreases the bandwidth.
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3.5 3
100
2.5 M
2 1.5
250
1 400
0.5 0
0
0.2
0.4
0.6
0.8
1 r
1.2
1.4
1.6
1.8
2
1.2
1.4
1.6
1.8
2
1 0.8
500
750 M
0.6 0.4
1000
0.2 0
0
0.2
0.4
0.6
0.8
1 r
Figure : Plot for Problem 9.34
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9.35 The magnitude ratio is M (ω) = √ and the phase angle is
1 1 =√ 1 + R2 C 2 ω 2 1 + 36 × 10−8 ω 2
φ(ω) = − tan−1 RCω = − tan−1 6 × 10−4 ω The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1667 rad/s. Thus the cos 720πt term and the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0, 240π and 480π gives the following results. M (0) = 1 φ(0) = 0
M (240π) = 0.9111
φ(240π) = −0.425 rad
Thus the steady state response is voss =
M (480π) = 0.7415 φ(480π) = −0.735 rad
20 40 40 − 0.9111 cos(240πt − 0.425) − 0.7415 cos(480πt − 0.735) π 3π 15π
or voss = 6.3662 − 3.8664 cos(240πt − 0.425) − 0.6294 cos(480πt − 0.735)
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9.36 The magnitude ratio is M (ω) = √
1 1 =√ 1 + R2 C 2 ω 2 1 + 10−6 ω 2
and the phase angle is φ(ω) = − tan−1 RCω = − tan−1 10−3 ω The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1000 rad/s. Thus the sin 360πt term and the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0 and 120π gives the following results. M (0) = 1 φ(0) = 0 Thus the steady state response is
M (120π) = 0.9357 φ(120π) = −0.361 rad
voss = 5 + 0.9357
20 sin(120πt − 0.361) π
or voss = 5 + 5.9569 sin(120πt − 0.361)
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9.37 The equation of motion is m¨ x + cx˙ + kx = cy˙ + ky or x ¨ + 98x˙ + 4900x = 98y˙ + 4900y The transfer function is
The magnitude ratio is
and the phase angle is
X(s) 98s + 4900 = 2 Y (s) s + 98s + 4900 √ 49002 + 982 ω 2 M (ω) = ! (4900 − ω 2 )2 + 982 ω 2 φ(ω) = tan−1
98ω 98ω − tan−1 4900 4900 − ω 2
A plot √ of M versus ω shows that M has a peak of 1.2764 at ω = 55 rad/s. Noting that 1.2764/ 2 = 0.9026 and that M (0) = 1, we see that the lower bandwidth frequency is 0. The plot also shows that M (111) = 0.906 which is close to 0.9026. So the upper bandwidth frequency is 111 rad/s. The circuit is a low pass filter with the bandwidth of 0 ≤ ω ≤ 111 rad/s. So we see that only the terms up to and including the sin 30πt term lie within the bandwidth. Evaluation of M and φ for ω = 0, 10π, 20π, and 30π gives the following results. M (0) = 1 φ(0) = 0
M (10π) = 1.1623
φ(10π) = −0.1057 rad
Thus the steady state response is xss =
M (20π) = 1.263 φ(20π) = −0.5187 rad
M (30π) = 1.0395 φ(30π) = 2.2467 rad
1 1 1 1 −1.1623 sin(10πt−0.1057)−1.263 sin(20πt−0.5187)−1.0395 sin(30πt+2.2467) 20 10π 20π 30π
or xss = 0.05 − 0.037 sin(10πt − 0.1057) − 0.0201 sin(20πt − 0.5187) − 0.011 sin(30πt + 2.2467)
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9.38 a) The transfer function is T (s) =
1 0.1s + 1
The time constant is τ = 0.1 sec. and the bandwidth is 1/τ = 10 rad/sec. b) The magnitude ratio and phase angle are given by M (ω) = √
1 0.01ω 2 + 1
φ(ω) = − tan−1 (1 + 0.1ωj)
The only components of the input that lie within the bandwidth are sin 4t and sin 8t. Thus we evaluate M (ω) and φ(ω) at ω = 4 and ω = 8.
Thus, or
M (4) = 0.928
φ(4) = −0.381 rad
M (8) = 0.781
φ(8) = −0.675 rad
y(t) ≈ 0.5(0.928) sin(4t − 0.381) + 2(0.781) sin(8t − 0.675) y(t) ≈ 0.464 sin(4t − 0.381) + 1.562 sin(8t − 0.675)
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9.39 a) The transfer function is T (s) =
X(s) 0.2 1 = = 2 2 F (s) 0.05s + 0.4s + 5 0.01s + 0.08s + 1
From the frequency response plot we can √ determine that the bandwidth is from 0 to 12 rad/sec. (The peak is Mr = 0.27, and Mr / 2 = 0.19, which occurs at approximately 2 rad/sec). b) Only the first two terms in the expansion have frequencies within the bandwidth. Thus 1 f (t) ≈ −0.2(sin 3t + sin 9t) 3 and at steady state: x(t) ≈ 0.2(−0.2)M1 sin(3t + φ1 ) + 0.2(−0.2/3)M2 sin(9t + φ3 ) where M (3) = 1.063 M (9) = 1.343 Thus
φ(3) = −0.258 rad φ(9) = −1.313 rad
x(t) ≈ −0.0425 sin(3t − 0.258) − 0.0179 sin(9t − 1.313)
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9.40 First normalize the vo data by dividing by the amplitude of the input. Let v=
vo 20
Then compute m = 20 log v dB and log ω, and plot m versus log ω. The plot looks like a first order system plot (see Figure 9.1.6). The peak value of m is −11.25 dB, and m is approximately 3 dB below that value at ω = 0.8. Thus we estimate the time constant to be τ = 1/0/8 = 1.25 s. Thus we surmise that the form of the transfer function is Vo (s) K = Vs (s) 1.25s + 1 To estimate K, note that for low frequencies (ω << 0.8), vo  = Kv. So from the first data point, we estimate that K = vo /20 = 5.48/20 = 0.274. Thus we estimate the transfer function to be Vo (s) 0.274 = Vs (s) 1.25s + 1 The MATLAB code to generate this analysis is shown below. omega = [0.1:0.1:1,1.5,2:7]; vo = [5.48,5.34,5.15,4.92,4.67,4.4,4.14,3.89,3.67,... 3.2,2.59,2.05,1.42,1.08,0.87,0.73,0.63]; v = vo/20; m = 20*log10(v); logom = log10(om); plot(logom,m) This result can be checked by comparing the m plot of the transfer function with the plot of the data, by continuing the above session as follows: [M, ph, w]=bode(sys); M = M(:); plot(log10(w),20*log10(M),logom,m,’o’) The plots agree very well.
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9.41 First convert the data to meters and then divide by 15, the amplitude of the input. This gives data for the magnitude ratio M (ω). M = 10−3
x 15
Then create the log magnitude ratio plot (see the following MATLAB program). The plot shows the presence of a denominator term s and a breakpoint frequency somewhere between 1 and 2 rad/s. This suggests a transfer function of the form X(s) K = F (s) s(τ s + 1) This transfer function form agrees with the physical situation, because it can be due to a mass attached to a damper, with no stiffness. This would have the following equation of motion: m¨ x + cx˙ = f (t) and a transfer function
X(s) 1 1/m = = F (s) s(ms + c) s(s + c/m)
The breakpoint frequency is at ω = c/m. From the data, we see that c/m is somewhere between 1 and 2. The magnitude ratio is M (ω) =
1/m ω ω 2 + (c/m)2 !
To estimate the mass m, we can choose one of the data points. Choosing the first point, for which M = 0.0139 and ω = 0.1, we obtain m=
1
$
0.0139(0.1) ωb2 + (0.1)2
where ωb is the estimate of the breakpoint frequency. The choice ωb = 2 gives a better fit to the data that ωb = 1. Thus the solution is c =2 m m=
1
!
0.0139(0.1) 22 + (0.1)2
= 359
from which we obtain c = 2m = 718. The estimated transfer function is X(s) 1 = F (s) s(359s + 718) The following program generates the plot, which shows good agreement with the data. 945 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
% File for problem 9.41 x = (1e3)*[209,52,28,19,7,2,1]/15; % Converted x data w = [0.1,0.4,0.7,1,2,4,6]; % Frequency data m = 20*log10(x); % Decibel conversion wb = 2; % Estimated breakpoint frequency M0 = x(1); % First converted data point w0 = w(1); % First frequency data point mass = 1/(M0*w0*sqrt(wb^2+w0^2)) % Estimated mass value w1 = [0.1:0.01:6]; % Frequency vector for plotting m1 = 20*log10((1/mass)./(w1.*(sqrt(w1.^2+wb^2)))); semilogx(w,m,’o’,w1,m1)
−35
−40
−45
−50
m (dB)
−55
−60
−65
−70
−75
−80
−85 −1 10
0
1
10 ! (rad/s)
10
Figure : Plot for Problem 9.41
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9.42 The equation of motion is m¨ x + cx˙ + kx = mu 'ω 2 sin ωt The frequency transfer function for this system, with x normalized by m/mu ' is
where
r2 mx =! mu ' (1 − r 2 )2 + (2ζr)2 r = ω/ωn
ωn =
%
(1)
k m
First plot the raw data after converting the frequency data to rad/sec and the displacement data to meters. From this plot you can estimate that the resonant frequency is 19 rad/sec. Assuming that the damping is small enough so that the resonant frequency is close to ωn , we estimate that ωn = 19, and thus k = mωn2 = 100(192 ) = 36100 N/m. Next, normalize the frequency data to find r: r = ω/19. To plot the normalized displacement mx/mu ' vs. r, we need to estimate mu '. To do this we need to plot the theoretical curve from (1), and this requires an estimate of ζ. After several attempts, using different values of mu ' and ζ, the estimates mu ' = 0.005 and ζ = 0.1 gave a good fit to the data. √ ! With ζ = 0.1 and k = 36100, we obtain c = 2 mkζ = 2 100(36100)(0.1) = 380 N·s/m.
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9.43 The equations of motion are m1 x ¨1 = −k1 (x1 − x2 ) m2 x ¨2 = k1 (x1 − x2 ) + k2 (y − x2 )
Convert these to state variable form by dividing by m1 and m2 . The state vector is [x1 , x˙ 1 , x2 , x˙ 2 ]. This gives the following state and input matrices for the given parameter values. 0 1 0 0 0 −64 0 0 64 0 A= B= 0 0 0 0 1 64 0 −128 0 64
Since the output is given to be x1 , the output matrices are C=
.
1 0 0 0
The MATLAB program is the following.
/
D = [0]
A = [0,1,0,0;64,0,64,0;0,0,0,1;64,0,128,0]; B = [0;0;0;64]; C = [1,0,0,0]; D = 0; sys1 = ss(A,B,C,D); sys2 = tf(sys1) eig(A) bode(sys1) The transfer function is X1 (s) 4096 = 4 Y (s) s + 192s2 + 4096 The roots are s = ±4.9443j and s = ±12.9443j. Thus the resonant frequencies in radian units are 4.9443 and 12.9443. The Bode plots are shown in the following figure. The resonant peaks are very tight, so the bandwidths are very small. You can type bodemag(sys1) to obtain just the m plot, which can be expanded about the resonant peaks. After doing this it is still difficult to identify the bandwidths, which appear to be about ±0.1 rad/unit time for each peak
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Bode Diagram 150
Magnitude (dB)
100
50
0
−50
−100 −540
Phase (deg)
−585
−630
−675
−720 −1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure : Plot for Problem 9.43
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9.44 The equations of motion are I1 θ¨1 = kT (φ − θ1 ) − cT 1 (θ˙1 − θ˙2 ) I2 θ¨2 = cT 1 (θ˙1 − θ˙2 ) − cT 2 θ˙2
Convert these to state variable form by dividing by I1 and I2 . The state vector is [θ1 , θ˙1 , θ2 , θ˙2 ]. This gives the following state and input matrices for the given parameter values.
A=
0 1 0 0 −1 −0.1 0 0.1 0 0 0 1 0 0.1 0 −0.2
Since the output is given to be x1 , the output matrices are C=
.
1 0 0 0
The MATLAB program is the following.
/
B=
0 1 0 0
D = [0]
A = [0,1,0,0;1,0.1,0,0.1;0,0,0,1;0,0.1,0,0.2]; B = [0;1;0;0]; C = [1,0,0,0]; D = 0; sys1 = ss(A,B,C,D); sys2 = tf(sys1) eig(A) bode(sys1) The transfer function is X1 (s) s + 0.2 = 3 Y (s) s + 0.3s2 + 1.01s + 0.2 The roots are s = 0, s = −0.202, and s = −0.049 ± 0.9939j. Thus the resonant frequency in radian units is 0.9939. The Bode plots are shown in the following figure. You can type bodemag(sys1) to obtain just the m plot. The bandwidth can be computed by sliding the cursor along the plot and noting the frequencies corresponding to 3 dB below the peak. These are ω = 0.942 and ω = 1.04, so the bandwidth is 1.04 − 0.942 = 0.098.
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Bode Diagram 40 30
Magnitude (dB)
20 10 0 −10 −20 −30 −40 0
Phase (deg)
−45
−90
−135
−180 −1
10
0
1
10
10
Frequency (rad/sec)
Figure : Plot for Problem 9.44
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9.45 The equation of motion is m¨ x + cx˙ + kx = ky(t) Thus the transfer function is X(s) k 5000 100 = = = 2 2 2 Y (s) ms + cs + k 50s + 200s + 5000 s + 4s + 100 The MATLAB code to generate the m, plot is [mag,phase,w]=bode(100,[1,4,100]); mag = mag(:); plot(w,20*log10(mag)),axis([6 15 0 10]),ylabel(’Magnitude (db)’),... xlabel(’Frequency (rad/sec)’),grid From the plot we can determine that the resonant frequency is approximately 9.6 rad/s, and the peak in m is approximately 8 db. Thus Mr = 10mr /20 = 2.51 The bandwidth frequencies are found from the plot to be from 7.2 rad/sec to 11.5 rad/sec.
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9.46 a) The MATLAB code is KT = 0.04;Ke = 0.04;c = 7e5;R = 0.6;L = 2e3;Im = 2e5;IL = 4e5; I = Im + IL; den = [I*L,c*L+I*R,c*R+Ke*KT]; num1 = [KT]; sys1 = tf(num1,den); bode(sys1) [mag,phase,w] = bode(sys1); mag = mag(:); To obtain the plots for the disturbance transfer function, type num2 = [L,R]; sys2 = tf(num2,den); bode(sys2) [mag,phase,w] = bode(sys2); mag = mag(:); You can examine the data by typing [w,20*log10(mag)]. This enables you to determine the resonant frequency and the bandwidth more precisely than from the plot. The plots show that there is no resonant frequency. The bandwidth for the voltage input is from 0 to 37 rad/s. The bandwidth for the disturbance torque input is from 0 to 50 rad/s. b) From the data generated by the above code, for the voltage input, m(0) = 27.733 and φ(0) = 0. This m value corresponds to M = 1027.733/20 = 24.36. Thus the constant term 10 in the input will produce a term 10(24.36) = 243.6 in the output. Also, from the data, M (130) = 8.4853 and φ(130) = −94.7◦ = −1.653 rad. Thus the steadystate output is ωss (t) = 243.6 + 8.4853(2) sin(130t − 1.653) = 243.605 + 16.9706 sin(130t − 1.653)
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9.47 The program is given below, along with the resulting plot. The resonant frequency is 2.51 × 104 rad/s. By moving the cursor along the plot we can determine the points at which the curve is 3 dB below the peak. These frequencies are 2.39 × 104 and 2.61 × 104 , so the bandwidth is (2.61 − 2.39) × 104 = 2.2 × 103 rad/sec. m = 0.002;k = 1e6;Kf = 20;Kb = 15;R = 10;L = 0.001; sys = tf(Kf,[m*L,m*R,k*L+Kf*Kb,k*R]) roots([m*L,m*R,k*L+Kf*Kb,k*R]) bode(sys)
Bode Diagram −100
Magnitude (dB)
−120
−140
−160
−180
−200
Phase (deg)
−220 0
−90
−180
−270 2
10
3
10
4
5
10
10
6
10
Frequency (rad/sec)
Figure : Plot for Problem 9.47
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9.48 Transform the equations and arrange them as follows. (m1 s2 + c1 s + k1 )X1 (s) − (c1 s + k1 )X2 (s) = 0 −(c1 s + k1 )X1 (s) + (m2 s2 + c1 s + k1 + k2 )X2 (s) = k2 Y (s)
Use Cramer’s method to find the transfer functions. You can use MATLAB to perform the algebra after inserting the given numerical values for the parameters. See Example 8.6.4 for an example of how this can be done. The general results are X1 (s) k2 (c1 s + k1 ) = Y (s) D(s) X2 (s) k2 (m1 s2 + c1 s + k1 ) = Y (s) D(s) where D(s) = m1 m2 s4 + c1 (m1 + m2 )s3 + (k1 m2 + k1 m1 + k2 m1 )s2 + c1 k2 s + k1 k2 The MATLAB code is m1 = 250;m2 = 40;k1 = 15000;k2 = 150000;c1 = 1000; D = [m1*m2,c1*(m1+m2),k1*m2+k1*m1+k2*m1,c1*k2,k1*k2]; num1 = k2*[c1,k1]; num2 = k2*[m1,c1,k1]; bode(num1,D) [mag,phase,w] = bode(num1,D,[5:.05:9]); mag = mag(:); plot(w,mag),axis([0 20 0 3]),xlabel(’Frequency (rad/sec)’),ylabel(’M’),grid The plots show that the major peak has a resonance frequency of 7.3 rad/sec and a bandwidth of 5.3 to 8.8 rad/sec. The secondary resonance is at 62 rad/sec. b) The peak amplitude occurs at ω = 7.3 rad/sec. With a period of 10 m, this frequency corresponds to a speed of 7.3(1) v= = 11.62 m/s 2π or 41.8 kilometers per hour.
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Ten
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10.1 a) A traffic light may be either closed or open loop, depending on whether or not it uses a sensor to detect the presence of vehicles in the roadway. b) Most washing machines are open loop, but some now have sensors that detect the amount of dirt in the wash water and adjust their cycle accordingly. These would be closedloop devices. c) If the toaster uses a timer, it is open loop. If it uses a sensor to tell when the desired temperature has been reached, it is closed loop. d) Cruise control is closed loop because it uses a measurement of the vehicle speed to control the engine. e) A aircraft autopilot is closed loop because it uses measurements of a variety of variables (speed, altitude, angle, etc.) to adjust the control surfaces (ailerons, rudder, elevators, etc.). f) Closed loop because it reacts to changes in the environment to keep the temperature near 98.6◦ .
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10.2 The diagram is like Figure 10.1.7 with Gc (s) = KP Ga (s) = 1 1 4s2 + 6s + 3 The command transfer function is found from the diagram as follows. Gp (s) =
C(s) = This gives
1 [Gf (s)R(s) + KP R(s) − KP C(s)] 4s2 + 6s + 3 Gf (s) + KP C(s) = 2 R(s) 4s + 6s + 3 + KP
The system will give perfect response if C(s)/R(s) = 1. This requires that Gf (s) = 4s2 + 6s + 3 The difficulty with implementing this design is that it requires the first and second derivatives of the input, r˙ and r¨, to be computed. This would be difficult to accomplish, especially for step inputs.
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10.3 The diagram is a combination of Figures 10.1.7 and 10.1.8 with Gc (s) = KP Ga (s) = 1 10 s Gf (s) = Kf Gp (s) =
Gd (s) = Kd The command transfer function is found from the diagram as follows. C(s) =
10 [−D(s) + Kd D(s) + Kf R(s) + KP R(s) − KP C(s)] s
or C(s) =
10(Kf + KP ) 10(Kd − 1) R(s) + D(s) s + 10KP s + 10KP
The disturbance will have no effect if Kd = 1. The command transfer function is C(s) 10(Kf + KP ) = R(s) s + 10KP If R(s) = 1/s, the steadystate response is css =
10(Kf + KP ) 10KP
which will be 1 if Kf + KP = KP , which is true if Kf = 0. Thus Kf is not needed to satisfy the given specifications. The time constant expression is 1 τ= 10KP Setting KP = 0.05 will give the desired time constant τ = 2. So the design is KP = 0.05, Kd = 1, and Kf = 0.
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10.4 From the diagram, C(s) =
6 [M (s) − D(s)] 15s + 2
(1)
4K E(s) 3s + 1
(2)
E(s) = R(s) − C(s)
(3)
M (s) =
To obtain the output C(s) in terms of R(s) and D(s), eliminate M (s) and E(s) from equations (1), (2), and (3) to obtain C(s) 24K = 2 R(s) 45s + 21s + 2 + 24K C(s) 6(3s + 1) =− D(s) 45s2 + 21s + 2 + 24K To obtain the error E(s) in terms of R(s) and D(s), eliminate C(s) and M (s) from equations (1), (2), and (3) to obtain E(s) 45s2 + 21s + 2 = R(s) 45s2 + 21s + 2 + 24K E(s) 6(3s + 1) = D(s) 45s2 + 21s + 2 + 24K To obtain M (s) in terms of R(s) and D(s), eliminate C(s) and E(s) from equations (1), (2), and (3) to obtain M (s) 4K(45s2 + 21s + 2) 4K(15s + 2) = = 2 2 R(s) (3s + 1)(45s + 21s + 2 + 24K) 45s + 21s + 2 + 24K M (s) 24K = 2 D(s) 45s + 21s + 2 + 24K The characteristic polynomial is the denominator of the transfer functions: 45s2 + 21s + 2 + 24K.
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10.5 The command transfer function is C(s) KP = 2 R(s) 5τa s + (5 + τa )s + 1 + KP If τa = 0.05, the characteristic roots are s = −11.1 ±
!
98.01 − 4KP
The roots are real if KP ≤ 98.01/4 = 24.5. If KP > 24.5 the roots are complex and the closedloop time constant is 1/11.1 = 0.09. If we neglect τa by setting it equal to zero, the characteristic equation becomes a firstorder equation: 5s + 1 + KP = 0. Thus the root is always real and the closedloop time constant is 5 τ= 1 + KP which goes to zero as we increase KP . So if we neglect the actuator time constant, we are led to believe that the step response will not oscillate and that we can make the closedloop time constant as small as we like by increasing KP . However, in fact the time constant can be made no smaller that 0.09 and the response will oscillate if KP > 24.5.
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10.6 a) Summing forces parallel to the plane gives mv˙ = f − F − Wx b) The block diagram is shown following part (c). The control algorithm is Gc (s), and the motor constants are KT and Rf . The second diagram is a simplified form where G1 (s) = Gc (s) G2 (s) =
Ka KT Rf
1 ms
c) From the second diagram, "
T (s) V (s) = G2 (s) − D(s) R T (s) = G1 (s)E(s) E(s) = Vr (s) − V (s)
#
(1) (2) (3)
To obtain the output V (s) in terms of Vr (s) and D(s), eliminate T (s) and E(s) from equations (1), (2), and (3) to obtain V (s) G1 (s)G2 (s) Ka KT Gc (s) = = Vr (s) R + G1 (s)G2 (s) mRf Rs + Ka KT Gc (s) V (s) RG2 (s) RRf =− =− D(s) R + G1 (s)G2 (s) mRf Rs + Ka KT Gc (s) To obtain the error E(s) in terms of Vr (s) and D(s), eliminate T (s) and V (s) from equations (1), (2), and (3) to obtain E(s) mRf Rs = Vr (s) mRf Rs + Ka KT Gc (s) E(s) RRf = D(s) mRf Rs + Ka KT Gc (s) To obtain the torque T (s) in terms of Vr (s) and D(s), eliminate E(s) and V (s) from equations (1), (2), and (3) to obtain T (s) mKa KT RsGc (s) = Vr (s) mRf Rs + Ka KT Gc (s) T (s) Ka KT RGc (s) = D(s) mRf Rs + Ka KT Gc (s) 107 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Figure : for Problem 10.6
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10.7 a) The diagram is shown in the following figure. Since no load torque is mentioned, we assume that TL = 0. b) 1 1 1 1 Ie = I1 + 2 I2 + I3 = I1 + I2 + I3 2 [2(3)]2 4 36 Ne = 2(3) = 6
Figure : for Problem 10.7
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10.8
KI K D s2 + K P s + K I + KD s = s s Thus KD = 15, KP = 6, and KI = 4, and KP +
TI =
KP 6 = = 1.5 KI 4
TD =
KD 15 = = 2.5 KP 6
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10.9 From Figure 10.4.4,
Rf =4 Ri 1 KI = = 0.08 Ri C KP =
Using C = 10−6 , we obtain Ri =
1 = 1.25 × 107 Ω 0.08 × 10−6 Rf = 4Ri = 5 × 107 Ω
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10.10 a) From Figure 10.4.6, KP =
R =2 R1 + R2
TD = R 2 C = 2 α=
(1) (2)
R1 = 0.1 R1 + R2
(3)
The denominator 1 + αTD s causes the m curve to level off for ω > 1/αTD . So we choose 1 =5 αTD which gives α = 0.2/TD = 0.2/2 = 0.1. Using C = 10−6 , we obtain from (2): R2 = 2 × 106 Ω. We then solve (3) for R1 : R1 = and solve (1) for R: R= b) The transfer function is
2 × 106 Ω 9
40 × 106 Ω 9
Vo (s) 2(1 + 2s) =− Vi (s) 1 + 0.2s
The MATLAB code to obtain the frequency response plot is sys = tf([4,2],[0.2,1]); bodemag(sys),grid The plot is shown in the following figure.
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Bode Diagram 30
25
Magnitude (dB)
20
15
10
5 −2 10
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure : for Problem 10.10
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10.11 From Figure 10.4.7, KP = β
RC + R2 C1 = 10 R2 C
(1)
β = 1.4 R2 C
(2)
KD = βRC1 = 4
(3)
KI =
β=
R2 R1 + R2
(4)
From (2), R2 =
β 1.4C
R=
4 βC1
From (3),
From (4), R1 = R2
(5)
(6)
1−β 1−β = β 1.4C
(7)
Substituting (2), (5), and (6) into (1) gives 1.4
$
4C βC1 + βC1 1.4C
%
= 10
This can be rearranged as follows: (βC1 )2 − 10C(βC1 ) + 5.6C 2 = 0 Choosing C = 10−6 we obtain (βC1 )2 − 10−5 (βC1 ) + 5.6 × 10−12 = 0 which has the solutions βC1 = 5.9546 × 10−7 , 9 × 10−6
Choosing the first solution we obtain
R = 6.7175 × 106 Ω R2 =
0.4253 C1
R1 = 7.1429 × 105 −
0.4253 C1
(8)
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From this, since we require that R1 > 0, we see that C1 must satisfy C1 > 5.9546 × 10−7
(9)
To limit the response above 100 rad/s, we require that 1 ≥ 100 βR1 C1 or
1 ≥ 100 5.9546 × 10−7 R1
This requires that R1 ≤ 1.679 × 104 . Combining this with (8) and (9) shows that C1 must lie in the range 5.9546 × 10−7 < C1 ≤ 6.3 × 10−7
Choosing C1 = 6 × 10−7 F gives R1 = 5407 Ω, R2 = 7.089 × 105 Ω, and R = 4/βC1 = 6.7175 × 106 Ω. The second solution, βC1 = 9 × 10−6 , gives a much smaller allowable range for C1 .
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10.12 a) Y (s) =
6 14 84 = 7s + 3 s (7s + 3)s
Thus yss = lim s s→0
84 84 = = 28 (7s + 3)s 3
b) Y (s) = Thus yss = lim s s→0
7s − 3 5 + 6s + 9 s
10s2
7s − 3 5 15 5 =− =− 10s2 + 6s + 9 s 9 3
c) Y (s) = Thus
3s + 5 12 s2 − 9 s
3s + 5 s2 − 9 The roots of the denominator are s = ±3. Since one root is positive, the final value theorem cannot be applied. d) 4s + 3 8 Y (s) = 2 s + 2s − 7 s Thus 4s + 3 sY (s) = 8 2 s + 2s − 7 One of the roots of the denominator is positive, so the final value theorem cannot be applied. sY (s) = 12
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10.13 For each case, the error transform is E(s) = R(s) − C(s) = R(s) − T (s)R(s) = [1 − T (s)]R(s) where T (s) = a)
C(s) R(s)
3s 6 18 1 = 3s + 1 s2 3s + 1 s 18s 1 18 = lim s = lim = 18 2 s→0 3s + 1 s s→0 3s + 1 E(s) =
ess b)
3s − 4 6 3s + 1 s2 3s − 4 6 = lim s =∞ s→0 3s + 1 s2 E(s) =
ess c)
3s2 + 5s 12 3s + 5 12 = 2 2 2 3s + 5s + 4 s 3s + 5s + 4 s2 3s + 5 12 3s + 5 = lim s 2 = lim 12 2 = 15 2 s→0 3s + 5s + 4 s s→0 3s + 5s + 4 E(s) =
ess d)
E(s) = ess = lim s s→0
2s2 + 4s − 5 8 2s2 + 4s + 5 s2
2s2 + 4s − 5 8 2s2 + 4s − 5 8 = lim =∞ 2s2 + 4s + 5 s2 s→0 2s2 + 4s + 5 s
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10.14 Divide the equation by 3 and obtain the characteristic equation: s2 − (b + 2)s + 2b + 5 = 0 The RouthHurwitz criterion implies that the system is stable if and only if −(b + 2) > 0 and 2b + 5 > 0, which gives −2.5 < b < −2 for stability. Neutral stability occurs if either the s term or the constant term is missing in the characteristic equation. This occurs if b + 2 = 0 (the roots are s = ±j) or if 2b + 5 = 0 (the roots are s = 0,−0.5.).
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10.15 From the RouthHurwitz criterion it is necessary and sufficient that K > 0 and 9(26) − K > 0. Thus, the system is stable if and only if 0 < K < 234.
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10.16 The time constant requirement means that no root can lie to the right of s = −2. Translate the origin of the s plane to s = −2 by substituting s = p−2 into the characteristic equation. This gives (p − 2)3 + 9(p − 2)2 + 26(p − 2) + K = 0 Expanding and collecting terms gives
p3 + 3p2 + 2p + K − 24 = 0 We can apply the RouthHurwitz criterion to this equation to determine when all roots p have negative real parts (and thus when all s roots lie to the left of s = −2). This occurs when K − 24 > 0 and 3(2) − (K − 24) > 0. Thus K must be in the range 24 < K < 30.
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10.17 a) From the RouthHurwitz criterion, a>0
K>0
b>0
b>0
K>0
and 2aK > 2b, or aK > b. b) From the RouthHurwitz criterion, a>0 and 5aK > 25K, or ab > 5K. c) From the RouthHurwitz criterion, 4+K >0 and 12(12) > 4(4 + K). Thus −4 < K < 32
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10.18 We require that
ωLss = 0.85 ωr
From equation (5) of the example, KP =
0.07(0.85) = 0.3967 1 − 0.85
From equation (3), 5.407 × 10−6 s2 + 1.623 × 10−3 s + 2.8 × 10−3 + 0.04KP = 0 which gives s = −11.9803 and s = −288.186. So the system is stable. The dominant time constant is 1/11.9803 = 0.08347 s. After approximately 4(0.08347) = 0.3339 s, the response will have reached steadystate. From equation (2) of the example, ∆ω = −
0.6(0.3)/1.5 = −6.428 rad/s 1.5(0.6)4.44 × 10−4 + 1.5(0.04)2
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10.19 In the first printing of the text, the reader is directed to Figure 10.3.9. It should be Figure 10.3.8. Referring to Example 10.3.5, we obtain the following transfer function. Ω(s) KP KT = Ω(s) D(s) where D(s) is given by equation (3) in that example. The equivalent inertia is Ie = Im +
IL + It = 17.75 × 10−4 N2
The equivalent damping is ce = cm +
cL = 10−3 N2
The characteristic polynomial is D(s) = 1.42 × 10−5 s2 + 2.848 × 10−3 s + 0.0816 + 0.2KP The steadystate requirement is ωLss 0.2KP = = 0.9 ωr 0.0816 + 0.2KP This gives KP = 3.672. The resulting roots are s = −100.282 ± 217.735j So the time constant is 1/100.282 = 0.00997 s and the damping ratio is 0.418. The disturbance transfer function is ΩL (s) (La s + Ra )/N =− TL D(s) The steadystate deviation caused by the disturbance torque is ωLss = −
Ra /N 0.8/2 TL = − 0.2 = −0.49(0.2) = 0.098 N R a ce + N K T K b + K P K T 0.816
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10.20 The steadystate offset error is offset error = ωr − ωss = 1 −
KP = 0.2 3 + KP
if KP = 12. The time constant for this value of KP is τ=
I 2 2 = = c + KP 3 + KP 15
The steadystate response due to the disturbance is disturbance response =
−1 −1 = = −0.067 c + KP 15
If both inputs are applied, the actual steadystate speed is ωss = command response + disturbance response = (1 − 0.2) + (−0.067) = 0.733 Note that we can make the offset error, the disturbance response, and the time constant smaller only by making KP larger than 12, but this increases the maximum required torque and probably the cost of the system.
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10.21 Substitute I = 2, c = 3, and KP = 12 into equations (1) and (2) of Example 10.6.1 to obtain: Ω(s) 12 = Ωr (s) 2s + 15 Ω(s) 1 =− Td (s) 2s + 15 Using Td (s) = 1/s2 with the final value theorem, we find the steadystate disturbance response to be −1 1 ωss = lim s = −∞ s→0 2s + 15 s2 Thus the controller cannot keep the output near its desired value if the ramp disturbance lasts too long. To investigate the command response, use Ωr (s) = 1/s2 with the first transfer function to obtain 12 1 4 8 8 Ω(s) = = 2− + 2s + 15 s2 5s 75s 75(s + 15/2) Thus the response is
4 8 8 t− + e−15t/2 5 75 75 Since the slope is 4/5 and is less than the slope of the input, we can see that the speed ω(t) never catches up with the command input ωr (t), and that the steadystate error is infinite. We could have also obtained this result from the final value theorem, using the following expression for the error when no disturbance is present. ω(t) =
$
E(s) = Ωr (s) − Ω(s) = Ωr (s) 1 − Thus, with Ωr (s) = 1/s2 , ess = lim s s→0
1 s2
$
12 2s + 15
2s + 3 2s + 15
%
%
= Ωr (s)
$
2s + 3 2s + 15
%
=∞
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10.22 The transfer functions are Ω(s) KI = 2 Ωr (s) 20s + cs + KI Ω(s) s =− Td (s) 20s2 + cs + KI The steadystate unitstep response for the command input is ωss =
KI =1 KI
which is perfect. The steadystate deviation caused by a unitstep disturbance is ∆ωss = 0 which is also perfect. We also have
c ζ= √ =1 2 20KI
which gives KI = Since ζ = 1 we can write τ= So, if c = 10,
c2 80
2(20) 40 = c c
100 = 1.25 80 40 τ= =4 10
KI =
If c = 0.2,
0.04 = 0.0005 80 40 τ= = 200 0.2
KI =
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10.23 (a) The command transfer function is Ω(s) KP s + KI = 2 Ωr (s) 4s + (4 + KP )s + KI
(1)
For both cases (1) and (2), the time constant is τ=
8 = 0.2 4 + KP
Therefore, KP = 36 for both cases. For case (1), 40 ζ= √ = 0.707 2 4KI Thus, KI = 200. For case (2), the same procedure gives KI = 100. (b) Because ζ > 1 is required, there will be two real roots, and the dominant root must be s = −1/τ = −5. For this to be the dominant root, the second root must be to the left of s = −5. Choosing an arbitrary separation factor of 10, we place the secondary root at s = −50. The characteristic polynomial must therefore be (s + 5)(s + 50) = s2 + 55s + 250 = 0 Multiply this equation by 4 before comparing it with the denominator of equation (1), so that the highest coefficients will be the same in each polynomial. Thus the required characteristic polynomial must be 4s2 + 220s + 1000 Comparing its coefficients with those of the transfer function denominator gives the equations for the required gain values. 4 + KP = 220 KI = 1000 Thus, KP = 216. For a unitstep command input, a MATLAB file to compute the unitstep response for the first case, where KP = 216 and KI = 1000, is KP = 216; KI = 1000; sys = tf(4*[KP,KI],[4,4+KP,KI]); Rightclick on the resulting plot, select “Characteristics”, then “Peak Response” to determine the maximum percent overshoot, and the rise time. The following table summarizes the important response characteristics. The overshoot decreases as ζ increases, as expected. But the 1090% rise time is the smallest for ζ = 1.74. A common misconception is that 1027 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
response is sluggish for ζ > 1, but this is clearly not the case here. The gains KP = 216, KI = 1000 give a good response with a very small rise time and a small overshoot. However, high gain values are required, and this might make the physical system expensive. The fast response with ζ = 1.74 is due to the numerator dynamics, whose effect is increased by the high gain values used in the third case. This effect can be seen by comparing the actual overshoot and rise time with those predicted by the secondorder model without numerator dynamics. These values are given in the following table.
Case 1 2 3
KP 9 9 54
KI 50 25 250
ζ 0.707 1 1.74
Overshoot % 16 8 5
Rise Time 10 − 90% 0.14 0.18 0.03
Without Numerator Dynamics Overshoot Rise Time % 10 − 90% 4 0.36 0 0.66 0 0.44
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10.24 With the disturbance torque absent, the motor torque expression can be found as follows. From the block diagram, $
KI T (s) = KP s
%
E(s) =
$
KP s + KI s
%
[Ωr (s) − Ω(s)]
Using the transfer function Ω(s)/Ωr (s), we have T (s) =
$
KP s + KI s
or T (s) = 4
%
$
KP s + KI Ωr (s) 1 − 2 4s + (4 + KP )s + KI
KP s2 + (KP + KI )s + KI Ωr (s) 4s2 + (4 + KP )s + KI
%
(1)
This can be used to obtain the unitstep responses for the various gain values computed above. For a unitstep command input, a MATLAB file to compute the actuator response for the first case, where KP = 36 and KI = 200, is KP = 36; KI = 200; sys = tf(4*[KP,KP+KI,KI],[4,4+KP,KI]); Rightclick on the resulting plot, select “Characteristics”, then “Peak Response” to determine the peak value. For Case 1 it is 36. Repeating this for the other cases, we obtain peak values of 36 for case 2 and 216 for case 3. Note that the maximum torque occurs at t = 0 for all three cases, that the maximum torque equals KP , and thus is much greater for the case 3, which has the largest gain values. So we see that the fast rise time of case 3 is obtained at the expense of having to provide a motor with a larger torque capability. In the absence of a disturbance, the maximum torque occurs at t = 0 for both the proportional control system and the PI control system, if the command is a step function. The maximum torque is KP M , where M is the step magnitude. This is because the step input is a sudden command, and the proportional term responds instantaneously to try to reduce the error, whereas the integral term takes time to build up. Thus the maximum error, and maximum torque, occur at t = 0. At t = 0, the speed is zero and the torque is T (0) = KP e(0) = KP [ωr (0) − ω(0)] = KP ωr (0) = KP M This reasoning assumes that the maximum error occurs at t = 0. This is not true for some systems, such as the system shown in Figure 10.6.7. For such systems, there is no simple formula to compute the maximum torque.
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10.25 The error equation is E(s) =
4s2 + 4s s Ωr (s) + 2 Td (s) 2 4s (4 + Kp )s + KI 4s (4 + Kp )s + KI
Apply the final value theorem using Ωr (s) = Td (s) = 1/s2 . For the command error, ess =
4 KI
• For ζ = 0.707, KI = 200, and ess = 4/200 • For ζ = 1, KI = 100, and ess = 4/100 • For ζ = 1.74, KI = 1000, and ess = 4/1000 For the disturbance error, ess =
1 KI
• For ζ = 0.707, KI = 200, and ess = 1/200 • For ζ = 1, KI = 100, and ess = 1/100 • For ζ = 1.74, KI = 1000, and ess = 1/1000
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10.26 From the figure,
Ω(s) KP s + KI = 2 Ωr (s) 2s + (2 + KP )s + KI
So the characteristic polynomial is 2s2 + (2 + KP )s + KI
(1)
a) 1. The characteristic polynomial must be factored as 2(s + 10)(s + 8) = 2s2 + 36s + 160 Comparing coefficients with the characteristic polynomial (1), we obtain KP = 34
KI = 160
2. The characteristic polynomial must be factored as 2(s + 10)(s + 20) = 2s2 + 60s + 400 Comparing coefficients with the characteristic polynomial (1), we obtain KP = 58
KI = 400
3. The characteristic polynomial must be factored as 2(s + 10)(s + 50) = 2s2 + 120s + 1000 Comparing coefficients with the characteristic polynomial (1), we obtain KP = 118
KI = 1000
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10.27 From the figure,
Ω(s) KI = 2 Ωr (s) 4s + (4 + K2 )s + KI
So the characteristic polynomial is 4s2 + (4 + K2 )s + KI and the damping ratio is
4 + K2 4 + K2 ζ= √ = √ 2 4KI 4 KI
a) 1. Since ζ ≤ 1, we can write an expression for the time constant: τ=
8 = 0.2 4 + K2
which gives K2 = 36. Thus ζ=
4 + K2 10 √ =√ = 0.707 4 KI KI
which gives KI = 200. 2. Since ζ ≤ 1, we can write an expression for the time constant: τ= which gives K2 = 36. Thus ζ=
8 = 0.2 4 + K2
4 + K2 10 √ =√ =1 4 KI KI
which gives KI = 100. 3. The characteristic polynomial must be factored as 4(s + 5)(s + 50) = 4s2 + 220s + 1000 Comparing coefficients with the characteristic polynomial, we obtain K2 = 216
KI = 1000
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10.28 The actuator equation is T (s) =
(4s + 4)KI K2 s + KI Ωr (s) + 2 Td (s) 4s2 + (4 + K2 )s + KI 4s + (4 + K2 )s + KI
For a unitstep command input, a MATLAB file to compute the actuator response for the first case, where K2 = 36 and KI = 200, is K2 = 36; KI = 200; sys = tf(KI*[4,4],[4,4+K2,KI]); Rightclick on the resulting plot, select “Characteristics”, then “Peak Response” to determine the peak value. For Case 1 it is 14.5. Repeating this for the other cases, we obtain peak values of 8.58 for case 2 and 16.1 for case 3.
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10.29 The error equation is E(s) =
4s2 + (4 + K2 )s s Ωr (s) + 2 Td (s) 2 4s (4 + K2 )s + KI 4s (4 + K2 )s + KI
Apply the final value theorem using Ωr (s) = Td (s) = 1/s2 . For the command error, ess =
4 + K2 KI
• For ζ = 0.707, K2 = 36, KI = 200, and ess = 40/200 • For ζ = 1, K2 = 36, KI = 100, and ess = 40/100 • For ζ = 1.74, K2 = 216, KI = 1000, and ess = 220/1000 For the disturbance error, ess =
1 KI
• For ζ = 0.707, KI = 200, and ess = 1/200 • For ζ = 1, KI = 100, and ess = 1/100 • For ζ = 1.74, KI = 1000, and ess = 1/1000
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10.30 From the figure,
Ω(s) KI = 2 Ωr (s) 2s + (2 + K2 )s + KI
So the characteristic polynomial is 2s2 + (2 + K2 )s + KI
(1)
a) 1. The characteristic polynomial must be factored as 2(s + 10)(s + 8) = 2s2 + 36s + 160 Comparing coefficients with the characteristic polynomial (1), we obtain K2 = 34
KI = 160
2. The characteristic polynomial must be factored as 2(s + 10)(s + 20) = 2s2 + 60s + 400 Comparing coefficients with the characteristic polynomial (1), we obtain K2 = 58
KI = 400
3. The characteristic polynomial must be factored as 2(s + 10)(s + 50) = 2s2 + 120s + 1000 Comparing coefficients with the characteristic polynomial (1), we obtain K2 = 118
KI = 1000
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10.31 The compensated system diagram is shown in the following figure.
Figure : for Problem 10.31 Command compensation cannot affect the characteristic roots or the disturbance response, so we set D(s) = 0 here. The error equation is E(s) =
Is2
s(Is + c − Kf ) R(s) + (c + KP )s + KI )
For a unitstep input, the final value theorem gives ess = 0 as long as the system is stable. Thus ess = 0 even if there is no command compensation (Kf = 0). For a unitramp input, the final value theorem gives c − Kf ess = =0 KI if Kf = c. Thus the command compensation improves the ramp response without affecting the step response.
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10.32 The requirements are (for unit step inputs) 1. the steadystate command error must be zero. 2. the magnitude of the steadystate disturbance error must be ≤ 0.1 3. ζ = 0.707
Part (a). We can do P control and PD control at the same time by finding the error equation for PD control, and then setting KD = 0 to obtain the results for P control. From Figure 10.7.4 with I = 20 and c = 10, the error equation can be written as follows (after doing some algebra): (1) E(s) = Θr (s)−Θ(s) =
20s2 + 10s 1 Θr (s)+ Td (s) 20s2 + (10 + KD )s + KP 20s2 + (10 + KD )s + KP
The characteristic equation is obtained from the denominator: (2)
20s2 + (10 + KD )s + KP = 0
For a unit step command, the steady state error is found by setting Θr (s) = 1/s and Td (s) = 0 in equation (1), and using the final value theorem. ess
&
20s2 + 10s = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
'
1 =0 s
if the system is stable (that is, if 10 + KD > 0 and KP > 0). For a unit step disturbance, the steady state error is found by setting Td (s) = 1/s and Θr (s) = 0 in equation (1), and using the final value theorem. ess
"
1 = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
#
1 1 = s KP
if the system is stable. Thus the steady state errors are the same for P and PD control. So they both satisfy specification #1, and specification #2 if KP ≥ 10. The difference between the two lies in their transient performance. P control: For P control, the characteristic equation is (set KD = 0 in equation (2)): 20s2 + 10s + KP = 0 Thus
10 ζ= √ 2 20KP 1037
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To satisfy specification #3 (ζ = 0.707), KP must be 2.5, which is smaller than the value required to satisfy specification #2. Thus P control cannot satisfy simultaneously both specification #2 and specification #3. PD control. For PD control, since ζ < 1 we can write the following formula for the time constant [the real part of the roots of equation (1) is −(10 + KD )/40]: (3) The formula for ζ is (4)
τ=
40 10 + KD
10 + KD ζ= √ = 0.707 2 20KP
A solution that exactly satisfies specification #3 is KP = 10. From equation (4) we find that KD = 10. From equation (3) we see that τ = 40/20 = 2. Other solutions are possible for values of KP greater than 10. Part (b). In summary 1. With P control, the requirements on damping ratio and disturbance error cannot be met simultaneously. 2. With PD control, one solution is KP = 10, KD = 10 Thus P control does not work. PD control works but has numerator dynamics that increases the overshoot.
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10.33 Because ζ = cos β, the value ζ = 0.707 corresponds to β = 45◦ , which is a 45◦ line on the complex plane. Thus ζ = 0.707 corresponds to a pair of roots whose real and imaginary parts have the same magnitude; that is, s = −a ± ja. The time constant of these roots is τ = 1/a. Thus the roots must be s = −1 ± j if τ = 1 and ζ = 0.707. These roots correspond to the polynomial equation (s + 1 − j)(s + 1 + j) = (s + 1)2 + 1 = s2 + 2s + 2 = 0 or 10s2 + 20s + 20 = 0 Compare this with the system’s characteristic equation obtained from the denominator of the transfer function: 10s2 + (3 + KD )s + KP = 0 Thus KP = 20 and 3 + KD = 20, or KD = 17.
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10.34 From the figure we obtain Θ(s) =
KP 1 Θr (s) − Td (s) 20s2 + (10 + K2 )s + KP 20s2 + (10 + K2 )s + KP
The steadystate unitstep response for the command input is θss =
KP =1 KP
which is perfect. The steadystate deviation caused by a unitstep disturbance is ∆θss = −
1 KP
Thus KP ≥ 10 is required to meet the specification. If ζ ≤ 1, 40 τ= = 0.1 10 + K2 which gives K2 = 390. Thus
10 + K2 200 ζ= √ =√ ≤1 2 20KP 20KP
if KP ≥ 2000. So an acceptable design is KP = 2000, K2 = 390, which gives ζ = 1, τ = 0.1, and ∆θss = −
1 2000
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10.35 (a) With the specific parameter and gain values used in Problem 10.33, the resulting command transfer function is Θ(s) 20 + 17s = 2 Θr (s) 10s + 20s + 20
(1)
A MATLAB file to compute the unitramp response is the following. systheta = tf([17,20],[10,20,20]); t = [0:0.001:4]; input = t; lsim(systheta,input,t) Even though the system is underdamped, we do not see oscillations in the response to a unit ramp command because the oscillation period is 2π = 6.28, which is greater than 4τ = 1. Thus the oscillations die out before one period has occurred. The steadystate error can be found using the final value theorem. From the block diagram in Figure 10.7.4, with the specific parameter and gain values used here, we obtain $
20 + 17s E(s) = Θr (s) − Θ(s) = Θr (s) 1 − 10s2 + 20s + 20
%
= Θr (s)
(
10s2 + 3s 10s2 + 20s + 20
)
With Θr (s) = 1/s2 , we have ess
1 = lim sE(s) = lim s 2 s→0 s→0 s
(
10s2 + 3s 10s2 + 20s + 20
)
=
3 20
From Figure 10.7.4 with the specific parameter and gain values used here and Θr (s) = 1/s2 , we obtain the following actuator equation: T (s) =
170s3 + 251s2 + 60s Θr (s) 10s2 + 20s + 20
Note that the MATLAB tf function cannot be used to compute this transfer function because the order of the numerator is greater than that of the denominator. We can, however, work around this difficulty as follows. With Θr (s) = 1/s2 , and canceling an s term from the numerator and denominator, we obtain 170s2 + 251s + 60 1 T (s) = 10s2 + 20s + 20 s This is equivalent to the unitstep response of the following transfer function T (s) 170s2 + 251s + 60 = Ωr (s) 10s2 + 20s + 20 A MATLAB file to compute the unitramp response is the following. 1041 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
systorque = tf([170,251,60],[10,20,20]); step(systorque) The torque approaches the constant value of 3 needed to counteract the damping torque 3θ˙ = 3(1) = 3. (b) Actual disturbance inputs are not always “clean” functions like steps, ramps, and sine waves. We often do not know the exact functional form of the disturbance. It is often a random function, such as the disturbance torque due to wind gusts of a rotating radar antenna. The analysis of random inputs is beyond the scope of this text. However, although frequency response plots strictly speaking describe only the steadystate response for periodic inputs, the plots are used to obtain a rough idea of the system’s transient response to fluctuating inputs that are not periodic. The disturbance transfer function for the system in question is Θ(s) −1 = Td (s) 10s2 + 20s + 20
(1)
The frequency response plot can be obtained with the following MATLAB file. sys=tf(1,[10, 20,20]); bodemag(sys) Rightclick on the plot, select “Characteristics”, then “Peak Response”, to see the peak value and the corresponding frequency. They are m = −26 dB at ω = 0 rad/s. The plot shows that the system responds more to slowly varying disturbances, and tends to reject disturbances whose frequencies lie above the frequency ω = 1.4 radians per unit time. For disturbance frequencies within the bandwidth of 0 ≤ ω ≤ 1.4, the system attenuates the input by approximately m = −26 db, or by a multiplicative factor of M = 10−26/20 = 0.05.
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10.36 a) Because this system is third order, we do not have formulas to use for the damping ratio and the time constant. In addition, we must interpret the specifications to apply to the dominant root or root pair, and must choose the secondary root somewhat arbitrarily. The values ζ = 0.707 and τ = 1 correspond to the dominant root pair s = −1 ± j. The third root must be less than −1 so that it will not be dominant. We arbitrarily select the third root to be s = −2. This choice can be investigated later if necessary. These three roots correspond to the polynomial equation *
+
(s + 2)(s + 1 − j)(s + 1 + j) = (s + 2) (s + 1)2 + 1 = s3 + 4s2 + 6s + 4 = 0 To compare this with the system’s characteristic equation, we multiply it by 10. 10s3 + 40s2 + 60s + 40 = 0 Compare this with the system’s characteristic equation: 10s3 + (3 + KD )s2 + KP s + KI Thus KP = 60, KI = 40, and 3 + KD = 40, or KD = 37. b) The resulting disturbance transfer function for this system is Θ(s) −s = 3 Td (s) 10s + 40s2 + 60s + 40 The frequency response plot can be obtained with the following MATLAB file. sys=tf([1,0],[10, 40,60,40]); bodemag(sys) Rightclick on the plot, select “Characteristics”, then “Peak Response”, to see the peak value and the corresponding frequency. They are m = −33.7 dB at ω = 1.14 rad/s. The plot shows that the system attenuates disturbance inputs by a factor of m = −33.7 dB or more. This corresponds to an amplitude reduction of M = 10−33.7/20 = 0.0207. The system rejects by an even greater amount any disturbances whose frequencies lie below or above the frequency ω = 1.14. Compare this performance with the PD control system of Problem 10.35, which does not have as great an attenuation and responds more to low frequency disturbances than high frequency ones.
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10.37 The requirements are (for unit step inputs) 1. the steadystate command error must be zero. 2. the magnitude of the steadystate disturbance error must be ≤ 0.1 3. the time constant must be 0.1 We can do P control and PD control at the same time by finding the error equation for PD control, and then setting KD = 0 to obtain the results for P control. From Figure 10.7.4 with I = 20 and c = 10, the error equation can be written as follows (after doing some algebra): (1) E(s) = Θr (s)−Θ(s) =
20s2 + 10s 1 Θr (s)+ Td (s) 20s2 + (10 + KD )s + KP 20s2 + (10 + KD )s + KP
The characteristic equation is obtained from the denominator: (2)
20s2 + (10 + KD )s + KP = 0
For a unit step command, the steady state error is found by setting Θr (s) = 1/s and Td (s) = 0 in equation (1), and using the final value theorem. ess
&
20s2 + 10s = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
'
1 =0 s
if the system is stable (that is, if 10 + KD > 0 and KP > 0). For a unit step disturbance, the steady state error is found by setting Td (s) = 1/s and Θr (s) = 0 in equation (1), and using the final value theorem. ess
"
1 = lim sE(s) = lim s s→0 s→0 20s2 + (10 + KD )s + KP
#
1 1 = s KP
if the system is stable. Thus the steady state errors are the same for P and PD control. So they both satisfy specification #1, and specification #2 if KP ≥ 10. The difference between the two lies in their transient performance. P control. For P control, we have 10 ζ= √ 2 20KP Thus, if KP ≥ 10, ζ ≤ 0.35. Because ζ < 1, the time constant gives τ=
40 =4 10
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Thus specification #3 is not satisfied. PD control. For PD control, specifications #1 and #2 are met if KP ≥ 10. There are two ways to meet specification #3: i) set ζ ≤ 1 or ii) set ζ > 1. Case (i). For case (i), (ζ ≤ 1), the time constant is given by τ=
40 = 0.1 10 + KD
if KD = 390. For this value, the damping ratio is found from equation (4) to be 10 + 390 400 ζ= √ =√ 2 20KP 20KP Thus ζ ≤ 1 only if KP ≥ 8000. Thus one solution is KP = 8000 and KD = 390. This gives a damping ratio ζ = 1. Case (ii). For case (ii), (ζ > 1), the dominant root must have a real part equal to −10 for the time constant to be 0.1. The other root must be placed to the left of s = −10. Thus there are an infinite number of solutions, depending on where the second root is placed. For example, if the second root is placed at s = −b, the characteristic equation will factor as 20(s + 10)(s + b) = 20s2 + (200 + 20b)s + 200b. Comparing this to the characteristic equation shows that KP = 200b and KD = 190 + 20b. Thus the larger b is, the greater the gains KP and KD . For example, using an arbitrary root separation factor of 10, one solution is b = 100, KP = 20, 000 and KD = 2190. Generally, placing the second root far to the left requires higher gains, which is not desirable. In this case, additional specifications are needed to arrive at a unique solution. Part (b). In summary, 1. Using P control, specifications #2 and #3 cannot be met simultaneously. 2. Using PD control, a solution with ζ ≤ 1 is: KP = 8000, KD = 390. A solution with ζ > 1 is: KP = 20, 000, KD = 2190. Other solutions are possible.
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10.38 From the following diagram with Td (s) = 0 we obtain Θ(s) =
1 {Kf Θr + (KP + KD s) [Θr (s) − Θ(s)]} s(Is + c)
Thus
Kf + KP + KD s Θ(s) = 2 Θr (s) Is + (c + KD )s + KP
The steadystate unitstep response for the command input is θss =
Kf + KP =1 KP
only if Kf = 0. The error equation is E(s) = Θr (s) − Θ(s) =
Is2 + cs − Kf Θr (s) Is2 + (c + KD )s + KP
For a unitramp command, ess = lim sE(s) = ∞ s→0
regardless of the value of Kf . So Kf does not improve the steadystate response for a step or a ramp input.
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Figure : for Problem 10.38
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10.39 a) Try Iaction: G(s) = KI /s. Thus C(s) KI = 2 R(s) 20s + 0.2s + KI and the error equation with a unitramp input is 1 E(s) = R(s) − C(s) = 2 s
(
)
20s2 + 0.2s 20s2 + 0.2s + KI √ Thus ess = 0.2/KI = 0.01 if KI = 20. This gives ζ = 0.2/(2 20KI ) = 0.005 and τ = 2(20)/0.2 = 200. b) Try PIaction: G(s) = KP + KI /s. Thus C(s) KP s + KI = 2 R(s) 20s + (0.2 + KP )s + KI and the error equation with a unitramp input is 1 E(s) = R(s) − C(s) = 2 s
(
20s2 + 0.2s 20s2 + (0.2 + KP )s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. This gives ζ=
0.2 + KP 0.2 + KP √ = =1 40 2 20KI
if KP = 38.8. This gives τ = 2(20)/(0.2 + KP ) = 1. c) Try PIDaction: G(s) = KP + KI /s + KD s. Thus C(s) K D s2 + K P s + K I = R(s) (20 + KD )s2 + (0.2 + KP )s + KI and the error equation with a unitramp input is 1 E(s) = R(s) − C(s) = 2 s
(
20s2 + 0.2s (20 + KD )s2 + (0.2 + KP )s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. The damping ratio is 0.2 + KP 0.2 + KP ζ= ! = ! =1 2 (20 + KD )KI 2 20(20 + KD )
Because ζ = 1, the expression for the time constant is τ=
2(20 + KD ) = 0.1 0.2 + KP
The solution is KP = 3.8 and KD = −19.8. 1048 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
10.40 The error equation is: (1)
E(s) =
s2 + s 1 Θr (s) + 2 Td (s) 2 s + (1 + KD )s + KP s + (1 + KD )s + KP
For a unit ramp command, the steady state error is found by setting Θr (s) = 1/s2 and Td (s) = 0 in equation (1), and using the final value theorem. ess
(
s2 + s = lim sE(s) = lim s 2 s→0 s→0 s + (1 + KD )s + KP
)
1 1 = 2 s KP
if the system is stable (that is, if 1 + KD > 0 and KP > 0). For a unit ramp disturbance, the steady state error is found by setting Td (s) = 1/s2 and Θr (s) = 0 in equation (1), and using the final value theorem. ess = lim sE(s) = lim s s→0
s→0
$
1 2 s + (1 + KD )s + KP
%
1 =∞ s2
if the system is stable.
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10.41 The system’s closedloop transfer function can be found from the block diagram. It is C(s) KP + KD s = 2 R(s) 2s + (2 + KD )s + KP Thus the characteristic equation is 2s2 + (2 + KD )s + KP = 0. Because ζ is specified to be less than 1, we can write the following formula for the time constant. τ=
2(2) =1 2 + KD
This gives KD = 2. The damping ratio is given by 2 + KD 2+2 2 ζ= √ = √ =√ = 0.9 2 2KP 2 2KP 2KP This gives KP = 2.469. If the command is a step input with a magnitude m, then R(s) = m/s, and the Final Value Theorem gives css
$
KP + KD s = lim sC(s) = lim s 2 s→0 s→0 2s + (2 + KD )s + KP
%
m =m s
Thus the steady state error is zero, so all three specifications are satisfied.
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10.42 a) The output equation is Θ(s) =
K D s2 + K P s + K I s Θr (s) − Td (s) 3 2 3 10s + (2 + KD )s + KP s + KI 10s + (2 + KD )s2 + KP s + KI
The characteristic equation 10s3 + (2 + KD )s2 + KP s + KI = 0
(1)
For case 1, the desired characteristic equation can be expressed as 10(s + 0.5)[(s + 5)2 + 25] = 10s3 + 105s2 + 550s + 250 = 0
(2)
Comparing coefficients in equations (1) and (2), we obtain KP = 550
KI = 250
KD = 103
For case 2, the desired characteristic equation can be expressed as 10(s + 0.5)(s + 1)(s + 2) = 10s3 + 35s2 + 35s + 10 = 0
(3)
Comparing coefficients in equations (1) and (3), we obtain KP = 35
KI = 10
KD = 33
b) The MATLAB file is KP1 = 550; KD1 = 103; KI1 = 250; systheta1 = tf([KD1,KP1,KI1],[10,2+KD1,KP1,KI1]); KP2 = 35; KD2 = 33; KI2 = 10; systheta2 = tf([KD2,KP2,KI2],[10,2+KD2,KP2,KI2]); step(systheta1,systheta2) Rightclick on the plot, select “Characteristics”, then “Peak Response” to determine the maximum percent overshoot and the peak time. For Case 1, the maximum percent overshoot is 22%; for Case 2, it is15%. For the gains given in Example 10.7.7, it is 10%. The separation factors are 10, 2, and 10, respectively. On the basis of this example we would conclude that a large separation factor is better, provide that the dominant root is complex. b) The disturbance transfer function is Θ(s) s =− 3 Td (s) 10s + (2 + KD )s2 + KP s + KI The MATLAB file, which is a continuation of the previous file, is 1051 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
sysdist1 = tf([1,0],[10,2+KD1,KP1,KI1]); sysdist2 = tf([1,0],[10,2+KD2,KP2,KI2]); bodemag(sysdist1,sysdist2) Rightclick on the plot, select “Characteristics”, then “Peak Response” to determine the peak response and the corresponding frequency. For Case 1, the maximum response is m = −54.2 dB at ω = 2.44; for Case 2, it is −30 dB at ω = 0.585. For the gains given in Example 10.7.7, it is m = −34.1 dB at ω = 0.7. The separation factors are 10, 2, and 10, respectively. On the basis of this example we would conclude that a large separation factor gives more attenuation, provided that the dominant root is real.
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10.43 With I action only, the transfer function is C(s) KI K = 2 R(s) τ s + s + KI K and With DI action,
and
1 ζ= √ 2 τ KI K C(s) K(KD s2 + KI ) = R(s) (τ + KKD )s2 + s + KI K 1 ζ= ! 2 (τ + KKD )KI K
If KD > 0 the derivative action makes ζ smaller, and thus the response is more oscillatory. So D action does not improve the response.
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10.44 The compensated system diagram is shown in the following figure.
Figure : for Problem 10.44 Command compensation cannot affect the characteristic roots or the disturbance response, so we set D(s) = 0 here. The error equation is E(s) =
Is2 + cs − Kf R(s) Is2 + cs + KP
For a unitstep input, the final value theorem gives ess = −Kf /KP as long as the system is stable. Thus ess = 0 only if there is no command compensation (Kf = 0)! For a unitramp input, the final value theorem gives ess = ∞. Thus the command compensation does not improve the ramp response. Therefore, command compensation for this system degrades the step response, and does not improve the ramp response.
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10.45 a) The closedloop transfer function is C(s) 3KP + 3KD s = 2 R(s) s + 3KD s + 3KP − 4 The damping ratio is
The time constant is
3KD ζ= √ = 0.707 2 3KP − 4 τ=
2 = 0.1 3KD
These two conditions give KP = 68 and KD = 20/3. b) The closedloop transfer function with the negative rate feedback gain K1 is C(s) 3KP = 2 R(s) s + 3K1 s + 3KP − 4 This denominator has the same form as the transfer function in part (a) with KD replaced by K1 . Thus KP = 68 and K1 = 20/3. c) For part (a), C(s) 20s + 204 = 2 R(s) s + 20s + 200 This has numerator dynamics and will overshoot more than the design in part (b), for which C(s) 204 = 2 R(s) s + 20s + 200
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10.46 Since the desired value of θ is 0, the error signal is e = θr − θ = 0 − θ = −θ. Trying ˙ Substituting this into the equation of PD control we use f = KP e + KD e˙ = −KP θ − KD θ. motion we obtain M Lθ¨ − (M + m)gθ = f = −KP θ − KD θ˙ The characteristic equation is
M Ls2 + KD s + KP − (M + m)g = 0 Using M = 40, m = 8, L = 20, and g = 32.2 this becomes 800s2 + KD s + KP − 1545.6 = 0 A 2% settling time of 10 seconds corresponds to 4τ = 10 or τ = 2.5 seconds. Because ζ < 1, we can use the following formula for the time constant. τ=
2(800) = 2.5 KD
This gives KD = 640 lb/rad/sec. The formula for the damping ratio is KD 640 ζ= ! = ! = 0.707 2 800(KP − 1545.6) 2 800(KP − 1545.6)
Solve this for KP to obtain KP = 1801.6 lb/rad.
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10.47 The closedloop transfer function is Θ(s) 2412 + 1200s = Θr (s) 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 The characteristic equation is 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 = 0 From the RouthHurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498, so the case where τ = 1 is unstable. If τ = 0.1, the roots are s = −0.3418 ± 0.4761j
The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specifications call for a dominant time constant of 2.5 and a damping ratio of 0.707.
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10.48 The closedloop transfer function is Θ(s) (2412 + 1200s)(τ s + 1) = Θr (s) 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 The characteristic equation is 1500τ s3 + 1500s2 + (1200 − 1932τ )s + 480 = 0 From the RouthHurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498, so the case where τ = 1 is unstable. If τ = 0.1, the roots are s = −0.3418 ± 0.4761j
The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specifications call for a dominant time constant of 2.5 and a damping ratio of 0.707.
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10.49 In the first printing of the text, Figure P10.49 is incorrect. The feedback gains should be ce and ke , and the gain in the control algorithm block should be me , not m. a) From Figure P10.49, (ms2 +cs+k)X(s) = Fd (s)+(ce s+ke )X(s)+me
,
K D s2 + K P s + K I s2 Xr (s) + [Xr (s) − X(s)] s

If me = m, ce = c, and ke = k, this equation becomes (ms3 + mKD s2 + mKP s + mKI )X(s) = sFd (s) + (ms3 + mKD s2 + mKP s + mKI )Xr (s) Thus
X(s) =1 Xr (s)
which is perfect, and X(s) s = 3 2 Fd (s) ms + mKD s + mKP s + mKI For a step disturbance, Fd (s) = 1/s, Xss = 0. For a unitramp disturbance, Fd (s) = 1/s2 , Xss = 1/mKI . Since the characteristic equation is third order, there is no expression for the damping ratio or time constant. We can, however, obtain a dominant root pair that meets the requirements that ζ = 1 and τ = τd . This means that the dominant root pair must be two identical roots at s = −1/τd . Thus the third root will be real. Denote it by s = −b. Then the characteristic equation can be factored as $
1 ms +mKD s +mKP s+mKI = m 2 + τd 3
2
%2
&
$
2 (s+b) = m s + b + τd 3
%
s + 2
(
1 2b + τd2 τd
)
b s+ 2 τd
'
Comparing coefficients, we see that the gains must be chosen so that KP = b + KP =
2 τd
1 2b + τd τd2
KI =
b τd2
The value of b must be selected such that the root at s = −b is not the dominant root. This requires that b > 1/τd . b) In addition to requiring accurate estimates of m, c, and k, this scheme requires that the second derivative, x ¨r , of the command input must be computed in real time. This can be difficult to do for noisy inputs or inputs having discontinuities in xr or x˙ r , such as steps and ramps. 1059 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
10.50 a) From the diagram, ΩL (s) =
1 KT [Vm (s) − Kb Ωm (s)] N (Ie s + ce ) La s + Ra
and Ωm (s) = N ΩL (s). Thus ΩL (s) KT = 2 Vm (s) N Ie La s + N (ce La + Ra Ie )s + ce Ra N + N KT Kb With the given values, ΩL (s) 1.4 × 105 4.997 = = Vm (s) 4.368s2 + 894.4s + 2.8015 × 104 1.5592 × 10−4 s2 + 0.0319s + 1 Since the two roots are real, this can be expressed as ΩL (s) K = Vm (s) (τ1 s + 1)(τ2 s + 1) where K = 4.997, τ1 = 2.591 × 10−2 and τ2 = 6.02 × 10−3 . b) Using PI control with the above transfer function results in a thirdorder system, for which there are no convenient design formulas. Since τ2 ( τ1 , we will neglect τ2 and express the transfer function as ΩL (s) K = Vm (s) τ1 s + 1 With PI control, the closedloop transfer function is ΩL (s) K(KP s + KI ) = 2 Ωr (s) τ1 s + (1 + KKP )s + KKI Since we will set ζ < 1, we can express the closedloop time constant as τ= Thus KP = The damping ratio is
2τ1 = 0.05 1 + KKP
2τ1 − 0.05 = 7.284 × 10−3 0.05K
1 + KKP 20τ1 ζ= √ =√ 2 τ1 KKI τ1 KKI
Thus KI =
400τ12 ζ 2 τ1 K
Choosing ζ = 0.707, we obtain KI = 4.1481 1060 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
The resulting thirdorder transfer function is ΩL (s) 0.0364s + 20.73 = Ωr (s) 0.0001559s3 + 0.03193s2 + 1.036s + 20.73 The roots of the thirdorder system are s = −170.39 and s = −17.21 ± 22j. Thus the dominant time constant is 1/17.21 = 0.058, which is close to the desired value. The damping ratio of the dominant root pair is ζ = cos[tan−1 (22/17.21)] = 0.616, which is within the required range.
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10.51 The value from Problem 10.19 is KP = 3.6772. From Example 10.8.6, Tm (s) N KP KT (Ie s + ce ) = Ωr (s) D(s) and thus, since ia = TM /KT , Ia (s) N KP (Ie s + ce ) = Ωr (s) D(s) where
*
D(s) = N La Ie s2 + (Ra Ie + ce La )s + Ra ce + KT Kb + KP KT Note that the term N in D(s) cancels N in the numerator of Ia (s)/Ωr (s). The MATLAB program is the following.
+
KT = 0.2; Kb = 0.2; Ie = 1.775e3;ce = 1e3; Ra = 0.8;La = 4e3; KP = 3.672; num = KP*[Ie, ce]; den = [La*Ie, Ra*Ie+ce*La, Ra*ce+KT*Kb+KP*KT]; sys = tf(num, den); step(209.4*sys) Note that we can use the step function by multiplying the transfer function by magnitude of the step input. The plot is shown next. The maximum current is 389 A! This unattainable value can be reduced by using a modified step input that increases slowly (see Problem 10.53 for a discussion of this type of input).
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Step Response 400
300
Current ia (A)
200
100
0
−100
−200
0
0.01
0.02
0.03
0.04
0.05
0.06
Time (sec)
Figure : for Problem 10.51.
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10.52 The new transfer functions are Ω(s) KP s + KI = Ωr (s) 0.5s3 + 5.2s2 + (2 + KP )s + KI 0.5s2 + 5.2s + 2 T (s) KP s + KI = Ωr (s) 0.1s + 1 0.5s3 + 5.2s2 + (2 + KP )s + KI Ω(s) −s(0.1s + 1) = Td (s) 0.5s3 + 5.2s2 + (2 + KP )s + KI The three cases are: 1. KP = 18, KI = 40; 2. KP = 18, KI = 20; 3. KP = 108, KI = 200; a) The MATLAB program is KP = [18, 18, 108]; KI = [40, 20, 200]; % Command transfer function k = 1; sysa1 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 2; sysa2 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 3; sysa3 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]); % Actuator transfer function k = 1; numb1 = conv([KP(k), KI(k)],[0.5, 5.2, 2]); denb1 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]); sysb1 = tf(numb1, denb1) k = 2; numb2 = conv([KP(k), KI(k)],[0.5, 5.2, 2]); denb2 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]); sysb2 = tf(numb2, denb2) k = 3; numb3 = conv([KP(k), KI(k)],[0.5, 5.2, 2]); denb3 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]); sysb3 = tf(numb3, denb3) subplot(2,1,1), step(sysa1, sysa2, sysa3) subplot(2,1,2), step(sysb1, sysb2, sysb3) The characteristic roots are as follows: KP KI Intended ζ Intended Roots Actual Roots 18 40 0.707 −2 ± 2j (dominant) −2.2353 ± 2.9147j (ζ = 0.6086), −5.9294 18 20 1 −2, −2 (dominant) −1.5016, −4.4492 ± 2.6159j 108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168 ± 13.6248j The plots are shown in the following figure. The peak actuator values are 16.2, 14.2, and 52. 1064 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Step Response 1.5
3
1
Speed (rad/sec)
2 1
0.5
0
0
0.5
1
1.5
1
1.5
Time (sec)
Step Response
Actuatot Response
60
3
40
20
1 2
0
−20
0
0.5 Time (sec)
Figure : for Problem 10.52a.
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b) The MATLAB program is KP = [18, 18, 108]; KI = [40, 20, 200]; % Disturbance transfer function k = 1; sysc1 = tf([0.1, 1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 2; sysc2 = tf([0.1, 1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]); k = 3; sysc3 = tf([0.1, 1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]); bodemag(sysc1, sysc2, sysc3) The plots are shown in the following figure. The peak values are −23.3, −24.7, and −30.8 dB. c) The peak actuator values are less than in Example 10.6.4 because the actuator responds more slowly in this problem (it has a time constant of 0.1, whereas in the example the actuator responded instantaneously). The peak disturbance response values are slightly higher than in the example. Bode Diagram −20
1
−25
2 −30
3 −35
Magnitude (dB)
−40
−45
−50
−55
−60
−65
−70 −1 10
0
1
10
10
2
10
Frequency (rad/sec)
Figure : for Problem 10.52b.
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10.53 a) The transfer functions are Ω(s) KP s + KI = 2 Ωr (s) 5s + (2 + KP )s + KI T (s) (KP s + KI )(5s + 2) = 3 Ωr (s) 5s + (2 + KP )s + KI The three cases are: 1. KP = 18, KI = 40; 2. KP = 18, KI = 20; 3. KP = 108, KI = 200; a) The MATLAB program is KP = [18, 18, 108]; KI = [40, 20, 200]; % Command transfer function k = 1; sysa1 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]); k = 2; sysa2 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]); k = 3; sysa3 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]); % Actuator transfer function k = 1; sysb1 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)]) k = 2; sysb2 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)]) k = 3; sysb3 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)]) tc = [0:.001:2]; rc = 1  exp(20*tc); subplot(2,1,1),lsim(sysa1,sysa2,sysa3,rc,tc) ta = [0:.001:0.8]; ra = 1  exp(20*ta); subplot(2,1,2),lsim(sysb1,sysb2,sysb3,ra,ta) The plots are shown in the following figure. The peak actuator values are 14.5, 13.4, and 40.1. The peak actuator values are less than in Example 10.6.4 because the command input does not act as fast as the step input used in the example. This gives the system more time to respond, and thus makes less demand on the actuator.
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Linear Simulation Results 1.4
1
1.2
Amplitude
1
3
Command
2
0.8 0.6 0.4 0.2 0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (sec)
Linear Simulation Results
Amplitude
40
3
30
20
1 10
0
2 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time (sec)
Figure : for Problem 10.53.
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10.54 The new transfer functions are Ω(s) KI = 3 2 Ωr (s) 0.5s + 5.2s + (2 + K2 )s + KI T (s) KI (5s + 2) = Ωr (s) 0.5s3 + 5.2s2 + (2 + K2 )s + KI Ω(s) −s(0.1s + 1) = 3 Td (s) 0.5s + 5.2s2 + (2 + K2 )s + KI The three cases are: 1. K2 = 18, KI = 40, ζ = 0.707; 2. K2 = 18, KI = 20, ζ = 1; 3. K2 = 108, KI = 200, ζ = 1.74; a) The MATLAB program is K2 = [18, 18, 108]; KI = [40, 20, 200]; % Command transfer function k = 1; sysa1 = tf(KI(k),[0.5, 5.2, 2 + K2(k), k = 2; sysa2 = tf(KI(k),[0.5, 5.2, 2 + K2(k), k = 3; sysa3 = tf(KI(k),[0.5, 5.2, 2 + K2(k), % Actuator transfer function k = 1; sysb1 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + k = 2; sysb2 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + k = 2; sysb3 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + subplot(2,1,1), step(sysa1, sysa2, sysa3) subplot(2,1,2), step(sysb1, sysb2, sysb3)
KI(k)]); KI(k)]); KI(k)]); K2(k), KI(k)]); K2(k), KI(k)]); K2(k), KI(k)]);
The characteristic roots are as follows: K2 KI Intended ζ Intended Roots Actual Roots 18 40 0.707 −2 ± 2j (dominant) −2.2353 ± 2.9147j (ζ = 0.6086), −5.9294 18 20 1 −2, −2 (dominant) −1.5016, −4.4492 ± 2.6159j 108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168 ± 13.6248j The plots are shown in the following figure. The peak actuator values are 8.64, 4.87, and 11.3.
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Step Response 1.4
" = 0.707
1.2
!(t)
1
" = 1.74
0.8
"=1
0.6 0.4 0.2 0
0
0.5
1
1.5
2
2.5
3
3.5
2
2.5
3
3.5
Time (sec)
Step Response 12
" = 1.74
10
" = 0.707 T(t)
8 6
"=1
4 2 0
0
0.5
1
1.5 Time (sec)
Figure : for Problem 10.54a.
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b) The MATLAB program is K2 = [18, 18, 108]; KI = [40, 20, 200]; % Disturbance transfer function k = 1; sysc1 = tf([0.1, 1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]); k = 2; sysc2 = tf([0.1, 1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]); k = 3; sysc3 = tf([0.1, 1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]); bodemag(sysc1, sysc2, sysc3) The plots are shown in the following figure. The peak values are −23.3, −24.7, and −30.8 dB. c) The peak actuator values are greater than in Example 10.6.5. The peak disturbance response values are slightly higher than in the example. Bode Diagram −20
" = 0.707 −30
" = 1.74
"=1 −40
Magnitude (dB)
−50
−60
−70
−80
−90
−100
−110 −2 10
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure : for Problem 10.54b.
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10.55 The characteristic equation is 10s3 + (2 + KD )s2 + KP s + KI = 0 With KP = 55 and KD = 58, we have 10s3 + 60s2 + 55s + KI = 0 or 1+
KI 1 =0 3 10 s + 6s2 + 5.5s
The root locus gain is K = KI /10. In MATLAB type sys = tf(1, [1, 6, 5.5, 0]); rlocus(sys) The plot follows. In Example 10.7.7, KI = 25. This gives ζ = 0.707 and τ = 2. The plot shows that to reduce the error by increasing KI , the dominant time constant will become larger and the damping ratio will decrease, making the system slower and more oscillatory. If we decrease the error by half by doubling KI to 50, the plot shows that the new damping ratio will be approximately 0.4 and the new time constant will be approximately 1/0.435 = 2.3.
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Root Locus 10
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−10 −14
−12
−10
−8
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 10.55.
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10.56 The characteristic equation is s3 + KD s2 + (KP − 4)s + KI = 0 With KP = 604 and KD = 40, we have s3 + 40s2 + 600s + KI = 0 or 1 + KI
1 =0 s3 + 40s2 + 600s
The root locus gain is K = KI . In MATLAB type sys = tf(1, [1, 40, 600, 0]); rlocus(sys) The plot follows. In Example 10.8.3, KI = 4000. This gives ζ = 0.707 and τ = 0.1. The plot shows that to reduce the error by increasing KI , the dominant time constant will become larger and the damping ratio will decrease, making the system slower and more oscillatory. If we decrease the error by half by doubling KI to 8000, the plot shows that the new damping ratio will be approximately 0.3 and the new time constant will be approximately 1/5.68 = 0.18.
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Root Locus 40
30
20
Imaginary Axis
10
0
−10
−20
−30
−40 −60
−50
−40
−30
−20
−10
0
10
Real Axis
Figure : for Problem 10.56.
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10.57 From equations (2) and (3) in the example, ΩL (s) (0.002s + 0.6)/1.5 =− −6 2 TL (s) 5.407 × 10 s + 1.623 × 10−3 s + 2.8 × 10−3 + 0.04KP The program is given below. KP = 0.63; num = [0.002, 0.6]/1.5; den = [5.407e6, 1.623e3, 2.8e3 + 0.04*KP]; sys = tf(num, den); t1 = [0:0.001:0.05]; t2 = [0.051:0.001:0.3]; t = [t1, t2]; TL1 = 40*t1; TL2 = 2*ones(size(t2)); TL = [TL1, TL2]; lsim(sys, TL, t) The plot is shown below. The steadystate speed deviation is 28.4 rad/s, or 271 rpm, below the desired speed of 1000 rpm.
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Linear Simulation Results 5
Load Torque (N m)
0
−5
Amplitude
−10
Speed Deviation (rad/s) −15
−20
−25
−30
0
0.05
0.1
0.15
0.2
0.25
0.3
Time (sec)
Figure : for Problem 10.57.
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10.58 Let K= where Ie = I1 + and K = 12.7547. The transfer function is
Kpot Ka KT 6Ie
"
#
1 1 I2 + I3 = 0.0157 4 9
Θ(s) K(KP s + KD ) = Θr (s) Ls3 + Rs2 + KKD s + KKP a) For L = 0,
Θ(s) K(KP s + KD ) = 2 Θr (s) Rs + KKD s + KKP
To obtain τ = 0.5 and ζ = 1, τ = 0.5 = and
2R KKD
KKD ζ=1= √ 2 RKKP
These give
4R = 0.0941 K Parts (b), (c), and (d) are done with the following MATLAB program. KD = KP =
I1 = 0.01; I2 = 5e4; I3 = 0.2; Ka = 1; KT = 0.6; Kpot = 2;R = 0.3; Ie = I1+(1/4)*(I2+(1/9)*I3) K = Kpot*Ka*KT/(6*Ie) KD = 4*R/K KP = 4*R/K sys1 = tf([K*KD,K*KP],[R,K*KD,K*KP]) L = 0.015; sys2 = tf([K*KD,K*KP],[L,R,K*KD,K*KP]) step(sys1,sys2) The resulting plot is shown on the next page. With L = 0, the overshoot is 13.5% and the settling time is 2.7 s. With L = 0.015, the overshoot is 16.2% and the settling time is 2.59 s. So the response predicted by the second order model, which makes the calculation of the gains KP and KD much easier, is close to that of the third order system. This will not be true if L is much larger than 0.015. 1078 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Step Response 1.4
1.2
L = 0.015 L=0
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
Time (sec)
Figure : for Problem 10.58.
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10.59 Since the acceleration time is 4 s, we set τ = 1 so that the controllers can follow the ramp input. The time constant expressions for the three controllers are: 1. τ = I/(c + KP ) = 10/(3 + KP ) for P control. 2. τ = I/(c + KP ) = 10/(3 + KP ) for PI control, if ζ ≤ 1. 3. τ = I/(c + K2 ) = 10/(3 + K2 ) for Modified I control, if ζ ≤ 1. Requiring, for example, a maximum steady state error of 10% of the slew speed means that ess ≤ 0.1. The slope of the input is 1/4, and the steady state errors for a ramp command of slope 1/4 is 1. ess = 0.25/(c + KP ) = 0.25/(3 + KP ) for P control. 2. ess = 0.25c/KI = (0.25)3/KI for PI control. 3. ess = 0.25(c + K2 )/KI = 0.25(3 + K2 )/KI for Modified I control. Combining these requirements and satisfying the requirement on τ exactly, we obtain the following gains: 1. KP = 7 for P control. 2. KP = 7 , KI = 7.5 for PI control. 3. K2 = 8 , KI = 110 for Modified I control. The following program is a modification of the programs command_tf.m, actuator_tf.m, and trap1.m given in Section 10.9. The program calls on the files named plot_command and plot_actuator listed in Table 10.9.3. KPa = 120; KPb = 7; KIb = 7.5; K2 = 8;KIc = 110; I = 10;c = 3; % Create the command transfer functions. sysa = tf(KPa,[I,c+KPa]); sysb = tf([KPb,KIb],[I,c+KPb,KIb]); sysc = tf(KIc,[I,c+K2,KIc]); % Create the actuator transfer functions sysaACT = tf(KPa*[I,c],[I,c+KPa]); sysbACT = tf(conv([KPb,KIb],[I,c]),[I,c+KPb,KIb]); syscACT = tf(KIc*[I,c],[I,c+K2,KIc]); % A specific trapezoidal profile t = [0:0.01:19]; for k = 1:length(t) 1080 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
if t(k) <= 4 r(k) = (1/4)*t(k); elseif t(k) <= 10 r(k) = 1; elseif t(k) <= 14 r(k) = 14/4(1/4)*t(k); else r(k) = 0; end end subplot(2,1,1),plot_command subplot(2,1,2),plot_actuator The value KP = 7 for P control was found to be too small, so KP was increased to 120 for P control only. This improved response is shown in the following figure. 1.2
PI
I Modified
Command Response
1 P
0.8 0.6 0.4 0.2 0 −0.2
0
2
4
6
8
10 Time
12
14
16
18
20
6
8
10 Time
12
14
16
18
20
6 I Modified
PI
Actuator Output
4 P 2 0 −2 −4
0
2
4
Figure : for Problem 10.59.
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10.60 From Figure 10.5.2, we can see that the maximum speed error occurs at t = 0 and is 1000 rpm, or 104.7 rad/s. From (6.6.7), the energy consumption from t = 0 to t = T is E=
.
T 0
Ra i2a (t) dt +
The rms current is irms =
/
The rms speed error is ωrms =
/
1 T
.
T 0
1 T
.
T 0
.
T 0
ce ωL2 (t) dt
i2a (t) dt
(1)
(2)
[104.7 − ωL (t)]2 (t) dt
(3)
We can perform these computations either 1) by obtaining the step response numerically and then performing the integrations numerically, or 2) by first obtaining the closedform expressions for the current and speed, and then evaluating the integrals. These integrals can be evaluated in closed form. Both methods require the transfer functions. From the example, we have the expression for the speed: ΩL (s) KP KT = Ωr (s) D(s) where D(s) = N La Ie s2 + N (Ra Ie + ce La )s + N Ra ce + N KT Kb + KP KT The expression for the current is Ia (s) KP (Ie s + ce ) = Ωr (s) D(s) The following MATLAB program does the computations. N = 1.5;La = 2e3;Ra = 0.6; Ie = 1.802e3;ce = 4.444e4; KP = 0.63;KT = 0.04;Kb = 0.04; D = [N*La*Ie,N*(Ra*Ie+ce*La),N*Ra*ce+N*KT*Kb+KP*KT]; sysia = tf(KP*[Ie,ce],D); syswL = tf(KP*KT,D); [ia,t1] = step(104.7*sysia); [wL,t2] = step(104.7*syswL); integrand1 = Ra*ia.^2; integrand2 = ce*wL.^2; E = trapz(t1,integrand1) + trapz(t2,integrand2) max_current = max(ia) 1082 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
rms_current = sqrt((1/max(t1))*trapz(t1,ia.^2)) rms_speed_error = sqrt((1/max(t2))*trapz(t2,(104.7wL).^2)) The results are: Energy consumed = E = 97.0283 J The maximum current is 64.2584 A, and the rms current is 23.1163 A. The rms speed error is 34.9 rad/s, or 333.271 rpm. For those not wishing to use MATLAB, we give the following expressions for the speed and current. ωL (t) = 94.041 + 6.5643e−282t − 100.6053e−18.4t ia (t) = 1.0448 − 83.3224e−282t + 82.2776e−18.4t
The integrals in equations (1), (2), and (3) can be evaluated in closed form or numerically.
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10.61 From Figure 10.6.5, we see that the only case where the saturation block with an upper limit of 20 will have any effect is the case for which ζ = 1.74. This corresponds to KP = 108 and KI = 200. The Simulink model is shown in the following figure. In the To Workspace block, specify the Save format as Array. Type KP = 108; KI = 200; in MATLAB and then run the model. Then in MATLAB type subplot(2,1,1),plot(tout,simout(:,2)), subplot(2,1,2),plot(tout,simout(:,1)) The resulting plot, which has been edited, is shown on the next page. As compared to the case with no limit on the actuator, the overshoot in ω(t) is larger (18% vs approximately 5%).
Figure : for Problem 10.61.
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1.2 1
!(t)
0.8 0.6 0.4 0.2 0
0
0.5
1
1.5
1
1.5
t 25 20
T(t)
15 10 5 0
0
0.5 t
Figure : for Problem 10.61.
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10.62 The Simulink model is shown in the following figure. We must use the PID Controller block instead of a Transfer Function block, because the latter does not allow the denominator to have an order greater than the numerator. In the Block Parameters window of the PID Controller block, enter 55 for Proportional, 25 for Integral, and 58 for Derivative. We must use the Transfer Function (with initial outputs) block because the initial position is not zero. In the To Workspace block, specify the Save format as Array.
Figure : for Problem 10.62. Run the model and then in MATLAB type plot(tout,simout) The resulting plot, which has been edited, is shown below.
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3
2.5
#(t)
2
1.5
1
0.5
0
1
2
3
4
5 t
6
7
8
9
10
Figure : for Problem 10.62.
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10.63 a) The Simulink model and the responses are shown in the following figures.
Figure : for Problem 10.63. b) Comparison with Figure 10.7.7 shows that the response with a limited actuator is much slower. However, the response shown in Figure 10.7.7 is misleading because it is caused by an actuator response that contains an impulse, something that is not physically possible. To see this, derive the expression for the actuator output, as follows. From the diagram in Figure 10.7.6, we can derive the following expression. T (s) =
IKD s4 + (IKP + cKD )s3 + (IKI + cKP )s2 + cKI s Θr (s) Is3 + (c + KD )s2 + KP s + KI
If the command is a unit step, then Θr (s) = 1/s, and, after canceling the s terms in the numerator and denominator, we obtain T (s) =
IKD s3 + (IKP + cKD )s2 + (IKI + cKP )s + cKI Is3 + (c + KD )s2 + KP s + KI
Using synthetic division, this can be expressed as T (s) = KD +
N (s) Is3 + (c + KD )s2 + KP s + KI
where N (s) is a secondorder polynomial. Thus the constant term KD will produce an impulse of strength KD in T (t). 1088 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
6
#(t)
4 2 0
0
5
10
15 t
20
25
30
0
5
10
15 t
20
25
30
0
5
10
15 t
20
25
30
T(t)
20 10 0
10000
T(t)
5000 0 −5000
Figure : for Problem 10.63. The top graph shows the command response, which is much slower than when the actuator output is unlimited. The second of the three graphs shows the limited actuator output. The third of the three graphs shows the actuator output if unlimited. It has a very large positive value initially and then goes negative. So the limiter causes the very slow command response. Note that Simulink cannot deal with an impulse in the actuator response, but rather computes the response numerically. This effect causes the simulated, unlimited actuator response to have some approximation error.
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10.64 a) The Simulink model and the responses are shown in the following figures. We must use the PID Controller block instead of a Transfer Function block, because the latter does not allow the denominator to have an order greater than the numerator. In the Block Parameters window of the PID Controller block, enter 55 for Proportional, 25 for Integral, and 58 for Derivative.
Figure : for Problem 10.64. b) The maximum overshoot in the command response is 42%, as compared to 10% in the example. The 2% settling time is 2.5 as compared with 2.89 in the example.
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1.5
#(t)
1
0.5
0
0
0.5
1
1.5 t
2
2.5
3
0
0.5
1
1.5 t
2
2.5
3
300 200
T(t)
100 0 −100 −200
Figure : for Problem 10.64.
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10.65 In the first printing of the text, the value of Ie was misprinted. The correct value is Ie = 2.08 × 10−3 slugft2 . a) The Simulink model is shown in the following figure. Note that since N = 1, we do not need a gain block for the gain 1/N .
Figure : for Problem 10.65a. In MATLAB, type Kb Ie Ra KP
= = = =
0.199; KT = 0.14; 2.08e3; ce = 3.6e4; 0.43; La=2.1e3; 7.284e3; KI = 4.1481;
Then run the model and type the following in MATLAB. subplot(2,1,1),plot(tout,simout(:,1)),subplot(2,1,2),plot(tout,simout(:,2)) This produces the following plots, which have been edited. b) The Simulink model is shown in the following figure. Note that since N = 1, we do not need a gain block for the gain 1/N . In MATLAB, type 1092 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
2
!L(t) (rad/sec)
0 −2 −4 −6 −8 −10
0
0.5
1
1.5
2
2.5 t (sec)
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5 t (sec)
3
3.5
4
4.5
5
10 8
ia(t) (A)
6 4 2 0 −2
Figure : for Problem 10.65a. Kb Ie Ra KP
= = = =
0.199; KT = 0.14; 2.08e3; ce = 3.6e4; 0.43; La=2.1e3; 7.284e3; KI = 4.1481;
Then run the model and type the following in MATLAB. subplot(2,1,1),plot(tout,simout(:,1)),subplot(2,1,2),plot(tout,simout(:,2)) This produces the following plots, which have been edited.
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Figure : for Problem 10.65b.
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0.05 0.04
L
Speed Error, ! (t) − !(t) (rad/sec)
0.06
0.03 0.02 0.01 0
0
0.2
0.4
0.6
0.8
1 t (sec)
1.2
1.4
1.6
1.8
2
0
0.2
0.4
0.6
0.8
1 t (sec)
1.2
1.4
1.6
1.8
2
0.02
ia(t) (A)
0.015
0.01
0.005
0
Figure : for Problem 10.65b.
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10.66 Modify the Simulink model from Problem 10.65a as shown below.
Figure : for Problem 10.66. In MATLAB, type Kb Ie Ra KP
= = = =
0.199; KT = 0.14; 2.08e3; ce = 3.6e4; 0.43; La=2.1e3; 7.284e3; KI = 4.1481;
Then run the model and type the following in MATLAB. subplot(2,1,1),plot(tout,simout(:,1)),subplot(2,1,2),plot(tout,simout(:,2)) This produces the following plots, which have been edited.
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!L(t) (rad/sec)
150
100
50
0
0
0.5
1
1.5
2
2.5 t (s)
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5 t (sec)
3
3.5
4
4.5
5
10
ia(t) (A)
5 0 −5 −10
Figure : for Problem 10.66.
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Eleven
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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11.1 a) The equation can be expressed as 1+
3p 1 =0 6 s(s + 4/3)
so K = 3p/6 = p/2. b) The equation can be expressed as 1+
p s+2 =0 2 3 s + 2s + 5/3
so K = p/3. c) The equation can be expressed as 1+
4p s2 + 1/4 =0 4 s(s2 + 0.5)
so K = 4p/4 = p.
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11.2 a) Write Gc (s)Gp (s) as Gc (s)Gp (s) = KP TD
!
s2 +
1 TD s
+
1 TD TI
"
(s + 10)
s(s + 1)(s + 2)
= 0.5KP
(s2 + 2s + 10)(s + 10) s(s + 1)(s + 2)
The poles are s = 0, −1, and −2. The zeros are s = −10 and −1 ± 3j. b) K = TD KP = 0.5KP c) The closedloop transfer function is C(s) Gc (s)Gp (s) = R(s) 1 + Gc (s)Gp (s) When KP = 10, this becomes C(s) 5s3 + 60s2 + 150s + 500 = 3 R(s) 6s + 63s2 + 152s + 500 The closedloop poles are the roots of the denominator, and are s = −8.688 and −0.9059 ± 2.962j. The closedloop zeros are the roots of the numerator, and are s = −10 and −1 ± 3j.
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11.3 Part (a) • Poles: s = 0 and s = −5 • Zeros: None • Two paths must leave the plot, so there will be two asymptotes. • The locus exists on the real axis in the following interval: −5 < Re(s) < 0. • To find the breakaway and breakin points: K = −(s2 + 5s) dK = −(2s + 5) = 0 ds The solution is s = −2.5, which must be a breakaway point. The value of K at the breakaway point is # # K = − (s2 + 5s)# = 6.25 s=−2.5
• Angles of the asymptotes:
θ=
(2n + 1)180◦ = 90◦ , 270◦ 2−0
• Intersection point of the asymptotes: σ=
$
$
sp − sz 0−5−0 = = −2.5 2−0 2
• Crossover point: None for K ≥ 0 • The root locus plot is shown in the following figure.
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Root Locus 2
1.5
1
Imaginary Axis
0.5
0
−0.5
−1
−1.5
−2
−5
−4.5
−4
−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
Real Axis
Figure : for Problem 11.3a
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11.3 Part (b) • Poles: s = 0, s = −7, and s = −9 • Zeros: None • Three paths must leave the plot, so there will be three asymptotes. • The locus exists on the real axis in the following intervals: − 7 < Re(s) < 0
Re(s) < −9 • To find the breakaway and breakin points:
K = −(s3 + 16s2 + 63s) dK = −(3s2 + 32s + 63) = 0 ds The solutions are s = −8.0618 and s = −2.6049. The point s = −8.0618 corresponds to a breakaway or breakin point for K < 0. The value of K at the breakaway point s = −2.6049 is # # K = − (s3 + 16s2 + 63s)# = 73.22 s=−2.6049
• Angles of the asymptotes:
θ=
(2n + 1)180◦ = 60◦ , 180◦ , 300◦ 3−0
• Intersection point of the asymptotes: σ=
$
$
sp − sz 0−7−9−0 16 = = − = −5.333 3−0 3 3
• The crossover point is found as follows. Substitute s = jω into the characteristic equation s3 + 16s2 + 63s + K = 0. −jω 3 − 16ω 2 + 63ωj + K = 0 which gives (63 − ω 2 )ω = 0
and K − 16ω 2 = 0
√ The first equation gives ω = 0, which is not of interest, and ω = ± 63. From the second equation, K = 16(63) = 1008 at the crossover points. • The root locus plot is shown in the following figure. 116 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus 20
15
10
Imaginary Axis
5
0
−5
−10
−15
−20
−35
−30
−25
−20
−15
−10
−5
0
5
10
15
Real Axis
Figure : for Problem 11.3b
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11.3 Part (c) • Poles: s = −1.5 ± 1.6583j • Zero: s = −3 • One path must leave the plot, so there will be one asymptote, which is 180◦ . • The locus exists on the real axis in the following interval: Re(s) < −3 • To find the breakaway and breakin points: K=−
s2 + 3s + 5 s+3
dK (s + 3)(2s + 3) − (s2 + 3s + 5) s2 + 6s + 4 =− = =0 ds (s + 3)2 (s + 3)2 The solutions are s = −5.2361 and s = −0.7639. The point s = −0.7639 corresponds to a breakaway or breakin point for K < 0. The value of K at the breakaway point s = −5.2361 is # s2 + 3s + 5 ## K=− = 7.47 # s + 3 #s=−5.2361
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s2 + 3s + 5 + K(s + 3) = 0. −ω 2 + 3jω + 5 + K(jω + 3) = 0 which gives (K + 3)ω = 0
and 5 + 3K − ω 2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because it does not lie on the root locus for K > 0, and 2) K = −3. With K = −3, the √ second equation gives ω = ± 5 − 9 = ±4j, which means that s = −4 and is thus not a solution of interest. • The root locus plot is shown in the following figure.
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Root Locus
3
2
Imaginary Axis
1
0
−1
−2
−3
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
Real Axis
Figure : for Problem 11.3c
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11.3 Part (d) • Poles: s = 0 and s = −4 • Zero: s = −5 • One path must leave the plot, so there will be one asymptote, which is 180◦ . • The locus exists on the real axis in the following intervals and
Re(s) < −5
− 4 < Re(s) < 0
• To find the breakaway and breakin points: K=−
s2 + 4s s+5
dK (s + 5)(2s + 4) − (s2 + 4s) s2 + 10s + 20 =− = =0 ds (s + 5)2 (s + 5)2 The solutions are s = −7.2361 and s = −2.7639. The point s = −2.7639 corresponds to the breakaway point, and s = −7.2361 corresponds to the breakin point. Because each point is symmetrically placed a distance 2.2361 from the zero, this indicates that the locus is a circle of radius 2.2361 off the real axis. • The value of K at the breakaway point is #
s2 + 4s ## K=− = 1.5279 # s + 5 #s=−2.7639
The value of K at the breakin point is
#
s2 + 4s ## K=− = 10.4721 # s + 5 #s=−7.2361
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s2 + 4s + K(s + 5) = 0. −ω 2 + 4jω + K(jω + 5) = 0 which gives (K + 4)ω = 0
and
5K − ω 2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because it does not lie on the root √ locus for K > 0, and 2) K = √ −4. With K = −4, the second equation gives ω = ±j 20, which means that s = − 20 and is thus not a solution of interest. • The root locus plot is shown in the following figure. 1110 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus
3
2
Imaginary Axis
1
0
−1
−2
−3
−10
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
Real Axis
Figure : for Problem 11.3d
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11.3 Part (e) • Poles: s = 0 and s = −1.5 ± 1.6583j • Zeros: None • Three paths must leave the plot, so there will be three asymptotes. • The locus exists on the real axis in the following interval: Re(s) < 0 • To find the breakaway and breakin points: K = −(s3 + 3s2 + 5s) dK = −(3s2 + 6s + 5) = 0 ds The solutions s = −1 ± 0.8165j are complex, so there are no breakaway or breakin points. • Angles of the asymptotes: θ=
(2n + 1)180◦ = 60◦ , 180◦ , 300◦ 3−0
• Intersection point of the asymptotes: σ=
$
$
sp − sz 0 − 1.5 − 1.6583j − 1.5 + 1.6583j − 0 3 = = − = −1 3−0 3 3
• The crossover point is found as follows. Substitute s = jω into the characteristic equation s3 + 3s2 + 5s + K = 0. −jω 3 − 3ω 2 + 5ωj + K = 0 which gives (5 − ω 2 )ω = 0
and K − 3ω 2 = 0
√ The first equation gives ω = 0, which is not of interest, and ω = ± 5. From the second equation, K = 3(5) = 15 at the crossover points. • The root locus plot is shown in the following figure.
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Root Locus 5
4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−5
−8
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 11.3e
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11.3 Part (f ) • Poles: s = 0, s = −3 and s = −7 • Zero: s = −4 • Two paths must leave the plot, so there will be two asymptotes. • Angles of the asymptotes: θ=
(2n + 1)180◦ = 90◦ , 270◦ 3−1
• Intersection point of the asymptotes: σ=
$
$
sp − sz 0−3−7+4 = = −3 3−1 2
• The locus exists on the real axis in the following intervals: and
−7 < Re(s) < −4
− 3 < Re(s) < 0
• To find the breakaway and breakin points: K=−
s3 + 10s2 + 21s s+4
dK (s + 4)(3s2 + 20s + 21) − (s3 + 10s2 + 21s) 2s3 + 22s2 + 80s + 84 =− = − =0 ds (s + 4)2 (s + 4)2 The only real solution is s = −1.7812. This point corresponds to a breakaway point. The value of K at the breakaway point is #
s3 + 10s2 + 21s ## K=− = 5.1062 # # s+4 s=−1.7812
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s3 + 10s2 + 21s + K(s + 4) = 0. −ω 3 − 10ω 2 + 21jω + K(jω + 4) = 0 which gives (K + 21 − ω 2 )ω = 0
and 4K − 10ω 2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because it does not lie on the root locus for K%> 0, and 2) K = ω 2 − 21. Substituting this into % the second equation gives ω = ±j 7/3, which means that s = − 7/3 and is thus not a solution of interest. • The root locus plot is shown in the following figure. 1114 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus 10
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−10
−15
−10
−5
0
5
Real Axis
Figure : for Problem 11.3f
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11.4 The root locus equation is 1 + Gc (s)Gp (s) = 1 + KP or 1 + KP TD
s2 +
&
'
1 s + 10 1+ + TD s =0 TI s (s + 2)(s + 5)
1 TD s
+
1 TI TD
s
s + 10 =0 (s + 2)(s + 5)
Substitute the given values to obtain 1 + 0.5KP
(s2 + 2s + 10)(s + 10) =0 s(s + 2)(s + 5)
The root locus parameter is K = 0.5KP . • Poles: s = 0, s = −2 and s = −5 • Zeros: s = −10 and s = −1 ± 3j
• No paths must leave the plot, so there will be no asymptotes. • The locus exists on the real axis in the following intervals: −10 < Re(s) < −5
and
− 2 < Re(s) < 0
• To find the breakaway and breakin points: K=−
s3 + 7s2 + 10s s3 + 12s2 + 30s + 100
dK 5s4 + 60s3 + 390s2 + 1400s + 1000 =− =0 ds (s3 + 12s2 + 30s + 100)2 The only relevant solution is s = −0.9187, which is a breakin point for K > 0. The value of K at this breakin point is #
# s3 + 7s2 + 10s # K=− 3 = 0.0496 # s + 12s2 + 30s + 100 #s=−0.9187
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s3 + 10s2 + 21s + K(s + 4) = 0. which gives
−ω 3 − 10ω 2 + 21jω + K(jω + 4) = 0 (K + 21 − ω 2 )ω = 0
and 4K − 10ω 2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because it 2 does not lie on the root locus for K % > 0, and 2) K = ω − 21. Substituting this into the second equation gives ω = ±j 7/3, which means that s = −sqrt7/3 and is thus not a solution of interest. • The root locus plot is shown in the following figure. 1116 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus 10
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−10
−15
−10
−5
0
5
Real Axis
Figure : for Problem 11.3f
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11.5 a) The equation can be expressed as 1+
5p s2 + 6s + 8 =0 4 s[s2 + (25/4)s + 4]
so K = 5p/4. b) The poles are the roots of s[s2 + (25/4)s + 4] = 0, which are s = 0, −5.5262, and −0.7238. The zeros are the roots of s2 + 6s + 8 = 0, which are s = −2 and −4. One path must leave the plot, so there will be one asymptote, which is 180◦ . The locus exists on the real axis in the following intervals: Re(s) < −5.5262
− 4 < Re(s) < −2
− 0.7238 < Re(s) < 0
To find the breakaway and breakin points: K=− dK ds
s[s2 + (25/4)s + 4] s3 + (25/4)s2 + 4s = − s2 + 6s + 8 s2 + 6s + 8
(s2 + 6s + 8)(3s2 + (25/2)s + 4) − (s3 + (25/4)s2 + 4s)(2s + 6) (s2 + 6s + 8)2 s4 + 12s3 + 57.5s2 + 100s + 32 =0 (s2 + 6s + 8)2
= − =
The solutions are s = −4.4115 ± 2.9797j, s = −2.7693, and s = −0.4077. The breakaway point is at s = −0.4077 and the breakin point is at s = −2.7693. The value of K at the breakaway point is #
s3 + (25/4)s2 + 4s ## K=− = 0.1153 # # s2 + 6s + 8 s=−0.4077
The value of K at the breakin point is
#
s3 + (25/4)s2 + 4s ## K=− = 16.49 # # s2 + 6s + 8 s=−2.7693
There is no crossover point for K ≥ 0. The root locus plot is shown in the following figure.
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Root Locus
3
2
Imaginary Axis
1
0
−1
−2
−3
−10
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
Real Axis
Figure : for Problem 11.5
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11.6 a) The poles are s = 0, 0, and −9. The zero is s = −1. The plot is shown in the following figure. Root Locus 4
3
2
Imaginary Axis
1
0
−1
−2
−3
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
1
Real Axis
Figure : for Problem 11.6a
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11.6 b) K=−
s2 (s + 9) s+1
dK (s + 1)(3s2 + 18s) − (s3 + 9s2 ) =− =0 ds (s + 1)2 which gives s = 0 and s = −3 for the breakaway points. For the breakaway point of interest, s = −3, # s2 (s + 9) ## K=− = 27 # s + 1 #s=−3 From Guide 10,
(
roots = −3 − 3 + r3 = −9
Thus the third root is r3 = −3. So when K = 27, all three roots are at s = −3. c) The smallest possible dominant time constant occurs at the breakaway point s = −3 and is τ = 1/3.
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11.7 a) The root locus equation is 1+K
s(s + 0.5)(s + 1.5) (s + 2.5)(s2 + 7s + 24.5)
The poles are s = −2.5, and −3.5 ± 3.5j. The zeros are s = 0, −0.5, and −1.5. The plot is shown in the following figure. b) The two points on the root locus that have a time constant of τ = 0.5 can be read off the plot. They are s = −2 and s = −2 ± 3.35j. You can use the graphical method to find the value of K at each point, and then use Guide 10 to determine the locations of the other roots. For the first point, s = −2, K = 4.92, and the other roots are s = −0.64 ± 2.19j, which are the dominant roots. Thus the dominant time constant is not 0.5. For the second point, s = −2 ± 3.35j, K = 0.73, and the third root is s = −2.32, which is not dominant. Thus the solution is K = 0.73. You can solve this problem very easily by using the rlocus and rlocfind functions of MATLAB to find the values of K and the root locations. Root Locus 4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−6
−5
−4
−3
−2
−1
0
1
2
3
Real Axis
Figure : for Problem 11.7
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11.8 a) Separate the equation as follows: 9s3 + 6s2 + 2 − 5ps = 0 The root locus form of the equation is &
5p 1+ − 9
'
s =0 s3 + (2/3)s2 + 2/9
The root locus parameter is K = −5p/9 ≤ 0. b) Separate the equation as follows: 4s3 + 2s + 7 − ps2 = 0 The root locus form of the equation is &
1+ −
p 4
'
s2 =0 s3 + (1/2)s2 + 7/4
The root locus parameter is K = −p/4 ≤ 0. c) Separate the equation as follows: s2 + 3s + 4 − p(s − 4) = 0 The root locus form of the equation is 1 + (−p)
s2
s−4 =0 + 3s + 4
The root locus parameter is K = −p ≤ 0.
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11.9 Part (a) • Poles: s = 0 and s = −5 • Zeros: None • Two paths must leave the plot, so there will be two asymptotes. • The locus exists on the real axis in the following intervals: Re(s) < −5 • Angles of the asymptotes:
θ=
and
Re(s) > 0
m360◦ = ±180◦ 0−2
• The two previous results lead us to conclude that there are no breakaway or breakin points. • Crossover point: None for K ≤ 0 • The root locus plot is shown in the following figure.
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Root Locus
6
4
Imaginary Axis
2
0
−2
−4
−6
−12
−10
−8
−6
−4
−2
0
2
4
6
8
Real Axis
Figure : for Problem 11.9a
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11.9 (b) • Poles: s = −1.5 ± 0.866j • Zero: s = −3 • One path must leave the plot, so there will be one asymptote. • The locus exists on the real axis in the following interval: Re(s) > −3 • Angles of the asymptote:
θ=
m360◦ = 0◦ 1−2
• The breakaway and breakin points are found as follows. K=−
s2 + 3s + 3 s+3
dK (s + 3)(2s + 3) − (s2 + 3s + 3 s2 + 6s + 6 =− = =0 ds (s + 3)2 (s + 3)2 The solutions are s = −4.7321 and s = −1.2679. The point is at s = −4.7321 is a breakaway or breakin point for K > 0. The point s = −1.2679 is a breakin point for K < 0. The value of K at the breakaway point is #
s2 + 3s + 3 ## K=− = −0.4641 # s + 3 #s=−1.2679
• The crossover point obviously occurs at s = 0, at which K = −1. • The root locus plot is shown in the following figure.
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Root Locus
1.5
1
Imaginary Axis
0.5
0
−0.5
−1
−1.5
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
Real Axis
Figure : for Problem 11.9b
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11.9 (c) • Poles: s = 0, s = −1.5 ± 0.866j • Zeros: None • Three paths must leave the plot, so there will be three asymptotes. • The locus exists on the real axis in the following interval: Re(s) > 0 • Angles of the asymptotes: θ=
m360◦ = 0◦ , −120◦ , −240◦ 0−3
which are equivalent to θ = 0◦ , −120◦ , 60◦ . σ=
$
$
sp − sz 0 − 1.5 − 0.866j − 1.5 + 0.866j − 0 = = −1 2−0 3
• The breakaway and breakin points are found as follows. K = −(s3 + 3s2 + 3s) dK = −(3s2 + 6s + 3) = 0 ds The solutions are s = −1 and s = −1, which do not lie on the locus for K < 0. Thus there are no breakaway or breakin points for K < 0. • There is no crossover point for K ≤ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s3 + 3s2 + 3s + K = 0. −jω 3 − 3ω 2 + 3jω + K = 0 which gives (3 − ω 2 )ω = 0
and K − 3ω 2 = 0
The first equation gives two possibilities: 1) ω = 0, which √ is not of interest because it does not lie on the root locus for K < 0, and 2) ω = ± 3. Substituting this into the second equation gives K = 9, which corresponds to the locus for K > 0. • The root locus plot is shown in the following figure.
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Root Locus 2
1.5
1
Imaginary Axis
0.5
0
−0.5
−1
−1.5
−2
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Real Axis
Figure : for Problem 11.9c
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11.9 (d) • Poles: s = 0, s = −5, s = −7 • Zeros: None • Three paths must leave the plot, so there will be three asymptotes. • The locus exists on the real axis in the following intervals: and Re(s) > 0
−7 < Re(s) < −5 • Angles of the asymptotes: θ=
m360◦ = 0◦ , −120◦ , −240◦ 0−3
which are equivalent to θ = 0◦ , −120◦ , 60◦ . • Intersection point of the asymptotes: σ=
$
$
sp − sz 0−5−7−0 = = −4 2−0 3
• The breakaway and breakin points are found as follows. K = −(s3 + 12s2 + 35s) dK = −(3s2 + 24s + 35) = 0 ds The solutions are s = −6.0817 and s = −1.9183, which does not lie on the locus for K < 0. The value of K at the breakaway point is # #
K = − (s3 + 12s2 + 35s)#
s=−6.0817
= −6.0411
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s3 + 12s2 + 35s + K = 0. −jω 3 − 12ω 2 + 35jω + K = 0 which gives (35 − ω 2 )ω = 0
and K − 12ω 2 = 0
The first equation gives two possibilities: 1) ω = 0, which √ is not of interest because it does not lie on the root locus for K < 0, and 2) ω = ± 35. Substituting this into the second equation gives K = 12(35), which corresponds to the locus for K > 0. • The root locus plot is shown in the following figure. 1130 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus 20
15
10
Imaginary Axis
5
0
−5
−10
−15
−20
−25
−20
−15
−10
−5
0
5
10
15
20
25
Real Axis
Figure : for Problem 11.9d
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11.9 (e) • Poles: s = 0, s = −3 • Zero: s = −4 • One path must leave the plot, so there will be one asymptote. • The locus exists on the real axis in the following intervals: −4 < Re(s) < −3 • Angle of the asymptote:
θ=
and Re(s) > 0
m360◦ = 0◦ 1−2
• From the two previous items, it is obvious that there are no breakaway or breakin points for K < 0. • There is no crossover point for K ≤ 0. This can be proved as follows. Substitute s = jω into the characteristic equation s2 + 3s + K = 0. −ω 2 + 3jω + K = 0 which gives ω=0
and
K − ω2 = 0
The first equation gives ω = 0, which is not of interest because it does not lie on the root locus for K < 0. The second equation gives K = 0. • The root locus plot is shown in the following figure.
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Root Locus
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−3
−2
−1
0
1
2
3
4
5
6
Real Axis
Figure : for Problem 11.9e
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11.9 (f) • Poles: s = and s = −6 • Zero: s = 4
• One path must leave the plot, so there will be one asymptote. • The locus exists on the real axis in the following intervals −6 < Re(s) < 0
• Angles of the asymptote:
θ=
and
Re(s) > 4
m360◦ = 0◦ 1−2
• The breakaway and breakin points are found as follows. K=−
s2 + 6s s−4
dK (s − 4)(2s + 6) − (s2 + 6s s2 − 8s − 24 =− = =0 ds (s − 4)2 (s − 4)2 The solutions are s = 10.325 and s = −2.325. The point is at s = 10.325 is a breakin point. The point s = −2.325 is a breakaway point. The value of K at the breakin point is # s2 + 6s ## K=− = −26.65 # s − 4 #s=10.325
The value of K at the breakaway point is
#
s2 + 6s ## K=− = −1.35 # s − 4 #s=−2.325
The breakin and breakaway points are symmetrically located each a distance 6.325 from the zero at s = 4. The locus off the real axis is a circle of radius 6.325 centered at the zero. • The crossover points can be found as follows. Substitute s = jω into the characteristic equation s2 + 6s + K(s − 4) = 0. which gives
−ω 2 + 6jω + K(jω − 4) = 0
(6 + K)ω = 0
and
4K + ω 2 = 0
The first equation gives 1) ω = 0, which is not of interest because it does not lie on the root locus for K √ < 0, and √2) K = −6. Substituting K = −6 into the second equation gives ω = ± 24 = ±2 6.
• The root locus plot is shown in the following figure. 1134
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Root Locus 8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−6
−4
−2
0
2
4
6
8
10
12
Real Axis
Figure : for Problem 11.9f
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11.10 a) The closedloop transfer function is C(s) KP (8 − s) KP (8 − s) = 2 = 2 R(s) s + 2s + 3 + KP (8 − s) s + (2 − KP )s + 3 + 8KP The root locus equation is 1 + KP
s2
s−8 8−s =1+K 2 =0 + 2s + 3 s + 2s + 3
where K = −KP ≤ 0. The poles are s = −1 ± 1.414j, and the zero is s = 8. The root locus plot is shown in the following figure. The breakaway and breakin points are found as follows. K=−
s2 + 2s + 3 s−8
dK (s − 8)(2s + 2) − (s2 + 2s + 3) s2 − 16s − 19 =− = − =0 ds (s − 8)2 (s − 8)2
The solutions are s = 17.11, which is the breakin point, and s = −1.11, which is a breakaway or breakin point for K > 0. From the location of thew breakin point we can tell that the root locus off the real axis is a circle of radius 17.11 − 8 = 9.11 centered on the zero at s = 8. From the RouthHurwitz criterion applied to the characteristic equation s2 + (2 − KP )s + 3 + 8KP = 0 the system is stable if 2 − KP > 0 and 3 + 8KP > 0; that is, if −3/8 < KP < 2. Therefore the crossover points correspond to KP = 2 and can be located by solving the characteristic equation for this value of KP . The result is s2 + 19 = 0 √ or s = ± 19.
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Root Locus 10
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−10 −5
0
5
10
15
20
Real Axis
Figure : for Problem 11.10a
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11.10 b) To achieve τ = 2/3, either 1) the two roots must be real with the dominant root at s = −3/2, or 2) the roots must be complex with the real part equal to −3/2. Since the locus does not exist on the real axis for Re(s) < 0, it is impossible for case (1) to occur. Case (2) is also impossible because the circular locus lies entirely to the right of the vertical line passing through s = −3/2. Thus we see that the smallest time constant possible for this system is 1/1 = 1, which corresponds to KP = 0. c) With KP = 1 the closedloop transfer function becomes C(s) 4−s = 2 R(s) s + s + 3 + 11 The characteristic roots are s = −0.5 ± 3.2787j. The unit step response can be found with the Laplace transform. C(s) =
4−s C1 3.2787 s + 0.5 = + C2 2 + C3 2 (ss2 + s + 3 + 11) s s + s + 3 + 11 s + s + 3 + 11
where C1 + C3 = 0
C1 + 3.2787C2 + 0.5C3 = −1
11C1 = 8
The solution is C1 = 8/11 = 0.7273, C2 = −0.4159, and C3 = −0.7273. Thus c(t) = 0.7273 − 0.4149e−0.5t sin 3.2787t − 0.7273e−0.5t cos 3.2787t The response is plotted in the following figure. The negative sign in the numerator causes the negative initial slope in the response. Step Response 0.7
0.6
0.5
Amplitude
0.4
0.3
0.2
0.1
0
−0.1
0
2
4
6
8
10
12
Time (sec)
Figure : For Problem 11.10c
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11.11 a) The root locus equation is 1 + KP
−2s2 + s + 26 s2 − 0.5s − 13 =1+K =0 s(s + 2)(s + 3) s(s + 2)(s + 3)
where K = −2KP ≤ 0. The poles are s = 0, −2, −3, and the zeros are s = 3.86 and −3.36. The root locus plot is shown in the following figure. The characteristic equation is s(s + 2)(s + 3) + K(s2 − 0.5s − 13) = s3 + (5 + K)s2 + (6 − 0.5K)s − 13K = 0 From the RouthHurwitz criterion, the system is stable if 5+K >0
6 − 0.5K > 0
− 13K > 0
and (5 + K)(6 − 0.5K) > −13K
These reduce to −1.7277 < K < 0, or 0 < KP < 0.8639. At a breakaway or breakin point, ζ = 1. These points are found as follows. K=−
s3 + 5s2 + 6s s2 − 0.5s − 13
dK (s2 − 0.5s − 13)(3s2 + 10s + 6) − (s3 + 5s2 + 6s)(2s − 0.5) =− =0 ds (s2 − 0.5s − 13)2
This is true if s4 − s3 − 4.75s2 − 130s − 78 = 0, which gives s = 8.5, −3.96, −2.68, and −0.8628. The first solution is an unstable point, the second and third solutions correspond to KP < 0, which is unstable, and the fourth is the desired solution. At this point, Ks=−0.8628 =
)
s3 + 5s2 + 6s − 2 s − 0.5s − 13
which gives KP = 0.177/2 = 0.088.
*# # # # #
s=−0.8628
= −0.177
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Root Locus 5
4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−5 −4
−2
0
2
4
6
8
10
Real Axis
Figure : for Problem 11.11a
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11.11 b) The closedloop transfer function is C(s) KP (−2s2 + 1s + 26) = 3 R(s) s + (5 − 2KP )s2 + (KP + 6)s + 26KP The step response with KP = 0.088 is shown in the following figure. The negative sign in the numerator causes the negative initial slope in the response. Step Response 1.2
1
0.8
Amplitude
0.6
0.4
0.2
0
−0.2
0
1
2
3
4
5
6
7
8
Time (sec)
Figure : for Problem 11.11b
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11.12 For ∆ = 0, the closedloop transfer function is C(s) KP + KD s = 2 R(s) s + KD s + KP − 5 Assuming the system is stable, the steadystate response for a unitstep input is css = lim
s→0 s2
KP + KD s 1 KP = &= 1 + KD s + KP − 5 s KP − 5
Thus we cannot achieve the specification that css = 1. For ζ < 1, the time constant expression is τ= Thus KD = 20. For the damping ratio,
2 = 0.1 KD
KD ζ= √ = 0.707 2 KP − 5
which gives KP = 205. Thus the steadystate response will be css = 205/200 = 1.025, which is close to the desired value. With KP = 205 and KD = 20, for ∆ &= 0, the closedloop transfer function is C(s) 205 + 20s = 2 R(s) s + 20s + 200 − ∆ The root locus equation is
1 =0 s2 + 20s + 200 where K = −∆ and −1 ≤ K ≤ 0. The root locus for K ≤ 0 is shown in the following figure. The breakin point is at s = −0.1 when K = −100. It shows that the time constant remains fixed at τ = 0.1 for −100 ≤ K ≤ 0, that is for 0 ≤ ∆ ≤ 100. The damping ratio variation as K varies from 0 to −1 is too small to be seen on the plot, but we can calculate the sensitivity of ζ to changes in K as follows. 1+K
20 ζ= √ 2 200 + K From this we see that ζ varies from ζ = 0.707 to ζ = 0.709 as K varies from 0 to −1. Thus the system is insensitive to K and therefore to ∆.
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Root Locus 10
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−10
−20
−15
−10
−5
0
Real Axis
Figure : for Problem 11.12
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11.13 a) With proportional control, the root locus equation is 1 + KP
6 =0 s(2s + 2)(3s + 24)
or 1+K
1 =0 s(s + 1)(s + 8)
where K = KP . The root locus is similar in shape to that shown in Figure 11.1.20. The breakaway point is found as follows. !
K = −s(s + 1)(s + 8) = − s3 + 9s2 + 8s
"
dK = −(3s2 + 18s + 8) = 0 ds which gives s = −5.517 and s = −0.4833. The first solution corresponds to a breakin point for K < 0. The second solution is the breakaway point for K > 0. For this plot, the farthest to the left the dominant root can lie is at the breakaway point. Thus the smallest possible dominant time constant is 1/0.4833 = 2.07.
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11.13 b) PD action will add a zero that will pull the locus to the left, stabilize the system, and allow a smaller time constant to be obtained, if the zero is placed properly. With PD action, the openloop transfer function becomes !
"
KP TD s + T1D KP (1 + TD s) = s(s + 1)(s + 8) s(s + 1)(s + 8)
The poles are still at s = 0, s = −1, and s = −8, and the zero is at s = −1/TD . The root locus parameter is K = KP TD . We want τ = 0.5. Thus the zero at s = −1/TD must be placed far enough to the left to pull the locus to the left of s = −2. Thus we choose TD = 0.5 to place the zero at s = −2. The resulting root locus is shown in the following figure. Next we choose ζ = 0.707, which along with τ = 0.5, specifies the point s = −2 ± 2j. From the plot we can see that the line corresponding to ζ = 0.707 passes through the locus, so we know that a solution exists. At s = −2 ± 2j, we find that K = 19.7. Thus KP = 19.7/TD = 2(19.7) = 39.4. The third root lies at s = −5 and thus is not dominant. Therefore, one solution is TD = 0.5, KP = 39.4. Other solutions are possible, depending on the choices made for TD and ζ. Root Locus 0.707 4
3
s = −2 + 2j
Imaginary Axis
2
1
0
−1
−2 0.707 −5
−4
−3
−2
−1
0
1
2
3
4
Real Axis
Figure : for Problem 11.13
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11.14 The root locus equation is 1 + KP
s + 10 =0 (s + 2)(s + 3)
The root locus plot is shown in the following figure. The line tangent to the circle gives the smallest ζ. This line has an angle of 49◦ . Thus the smallest ζ is ζ = cos 49◦ = 0.66. The 45◦ line corresponds to ζ = 0.707. This line intersects the circle at two points. The leftmost point gives the smallest time constant. This point, which can be found graphically or analytically, is s = −6.73 + 6.73j, which corresponds to #
(s + 2)(s + 3) ## KP = − = 8.46 # s + 10 s=−6.73+6.73j
The time constant is τ = 1/6.73 = 0.149. Root Locus 8
6
! = 0.66
! = 0.707 4
Imaginary Axis
2
0
−2
−4
−6
−8
−18
−16
−14
−12
−10
−8
−6
−4
−2
0
Real Axis
Figure : for Problem 11.14
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11.15 The characteristic equation is found from 1 + KP Gp (s) = 0 which becomes 1+2
5(s + 4) =0 (s + 3)(s + p)
or s2 + 13s + 40 + p(s + 3) = 0 Let K = p − 7. Then the characteristic equation becomes s2 + 20s + 61 + K(s + 3) = 0 where K ≥ 0. The poles are s = −3.76 and s = −16.2. The zero is s = −3. The root locus plot is sketched in the following figure. As p increases above 7, the dominant root moves from −3.76 to −3, and the other root moves from −16.2 to −∞. The roots are always real, so there will be no oscillations in the free response. The dominant time constant varies from τ = 1/3.76 = 0.266 to τ = 0.333 as p increases above p = 7. Root Locus
8
6
4
Imaginary Axis
2
0
−2
−4
−6
−8
−25
−20
−15
−10
−5
0
Real Axis
Figure : for Problem 11.15
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11.16 Express the controller transfer function as Gc (s) =
KP (s +
KI KP
)
s
=
KP (s + b) s
where b = KI /KP . The root locus equation is 1 + KP
s+b =0 s(s + 1)(s + 2)
The poles are s = 0, s = −1, and s = −2. The zero is at s = −b. The asymptotic angles are θ = ±90◦ . The asymptotes intersect at σ=
b−3 2
Sketches of the root locus plots for three cases are shown in the following figure. These three cases correspond to a) 0 < b < 1, using b = 0.5 b) 1 < b < 2, using b = 1.5, and c) b > 2, using b = 3. Case (a) does not satisfy the specification that the dominant root have a damping ratio of ζ = 0.707. Case (b) allows a smaller dominant time constant than Case (c). Root Locus 4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−6
−5
−4
−3
−2
−1
0
1
2
3
4
Real Axis
Figure : for Problem 11.16, case (a)
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Root Locus 5
4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−5
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 11.16, case (b)
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Root Locus 5
4
3
2
Imaginary Axis
1
0
−1
−2
−3
−4
−5 −8
−6
−4
−2
0
2
4
Real Axis
Figure : for Problem 11.16, case (c) The characteristic equation is s3 + 3s2 + (KP + 2)s + bKP = 0 Because we require that ζ = 0.707, the dominant root pair has the form s = −c ± cj. The characteristic equation can be factored as [(s + c)2 + c2 ](s − s3 ) = s3 + (2c − s3 )s2 + (2c2 − 2cs3 )s − 2c2 s3 = 0 where the third root is s3 and must be real. Comparing coefficients, we see that 2c − s3 = 3
2c2 − 2cs3 = KP + 2
− 2c2 s3 = bKP
Thus the third root is s3 = 2c − 3. In order for s = −c ± cj to be dominant, s3 < −c, and thus we must choose c < 1. To minimize the dominant time constant τ = 1/c, we should choose c as large as possible, subject to the restriction that c < 1. However, the closer c is to 1, the closer the secondary root s3 is to the dominant root. If s3 is too close to the dominant root, the response expected for ζ = 0.707 might not be obtained. Thus we should try a value for c and check the effects of the secondary root by analysis or simulation. For example, one solution is c = 0.9. This gives s3 = −1.2, b = 1.09, and KP = 1.78.
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11.17 The characteristic equation is 1 + Gc (s)Gp (s) = 0 which becomes
&
'
KI 4 1 + KP + + KD s =0 2 s 3s + 3
or
3s3 + 4KD s2 + (3 + 4KP )s + 4KI = 0
(1)
√ To achieve τ = 1 and ζ = 0.5, the desired dominant roots are s = −1 ± j 3. The third root is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will not be the dominant root). The polynomial corresponding to these three roots is [(s + 1)2 + 3](s + b) = 0 or s3 + (b + 2)s2 + (4 + 2b)s + 4b = 0 To compare this with equation (1), we must multiply by 3: 3s3 + 3(b + 2)s2 + 3(4 + 2b)s + 12b = 0
(2)
Comparing the coefficients of (1) and (2), we obtain 9 + 6b 4 KI = 3b 3(b + 2) KD = 4 The gains can be computed once a value for b > 1 has been selected. KP =
11.18 With a PD compensator, the closedloop transfer function is C(s) (KP + KD s)Gp (s) KP + KD s = = 2 R(s) 1 + (KP + KD s)Gp (s) 5s + KD s + KP − 6
The specifications require that
ζ = 0.707 Thus ωn =
+
ωn = 0.5
KP − 6 = 0.5 5
which gives KP = 7.25. Also,
which gives KD = 3.5355.
KD KD ζ= % = √ = 0.707 2 5(KP − 6) 2 6.25 1151
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√ 11.19 a) For K√= 4 the roots are s = −1 ± 3j. For ωn = 4 and ζ = 0.5, the roots must be s = −2 ± 2 3j. A simple gain adjustment will not give the desired roots. The angle deficiency is # # 4 # $ = $ 4 − $ s − $ (s + 2) = −210◦ s(s + 2) #s=−2+2√3j
Thus the compensator must add (210 − 180) = 30◦ to the system. This implies a lead compensator. Place the compensator’s pole and zero to the left of s = −2. Using the same procedure as in Example 11.2.2, with µ = 10, we obtain T = 0.0138 and T = 0.293. Choose T = 0.0138 because the second value would place the zero to the right of s = −2, and thus violate the assumed geometry. The compensator’s pole is s = −1/T = −72.6, and its zero is s = −1/aT = −7.26. The compensated openloop transfer function is s + 7.26 K s + 72.6 s(s + 2) √ The value of K required to place the root at s = −2 ± 2 3j is K = 155.6. Other solutions are possible, depending on the choice for µ. For C = 1µF , R1 R2 T = 0.0138 = 10−6 R1 + R2 Gc (s)G(s) =
µ=
R1 + R2 = 10 R2
These give R1 = 1.38 × 105 and R2 = 1.53 × 104 Ω.
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√ 11.19 b) A ramp error of 0.05 implies that Cv = 20. With K = 4, s = −1 ± j 3 and Cv = 2. To obtain Cv = 20 we need to increase the gain by a factor of 10, but without changing the root locations appreciably. Therefore we use a lag compensator. Because the PM and GM are not specified, we use a root locus design method. Step 1: With no compensation, K = 4 will place the roots at the desired locations. Thus KP 1 = 4. Step 2: To obtain Cv = 20, K must be increased from 4 to 40. Thus KP 2 = 40 and µ = KP 1 KP 2 = 0.1. Step 3: Choose T large. Note that the plant has a pole at s = −2. Select the compensator’s pole and zero to be well to the right of s = −2. So try s = −0.02 and s = −0.2. This gives T = 100. Step 4: The openloop compensated transfer function is 1 2 + 0.2 K 10 s + 0.02 s(s + 2) √ The precise value of the desired roots s = −1 ± j 3 probably will not lie exactly on the new root locus, but if T is chosen large enough, they should be close. The two specifications are ζ = 0.5 and ωn = 2, so we can look for a root that satisfies one specification exactly, and hope that the other is close to being satisfied. For T = 100, the root locus shows that s = −0.8989 + 1.56j gives ζ = 0.5, but ωn = 1.8, when K = 36. If ωn is close enough to 2, we can stop. Otherwise a larger value of T can be chosen. Note that the roots s = −0.8989 ± 1.56j are “dominant” only if we consider the pole at s = −0.2 to be canceled by the zero at s = −0.2. Gc (s)G(s) =
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11.20 Try a PID compensator. The transfer function is C(s) 5(KD s2 + KP s + KI ) = 3 R(s) s + 5KD s2 + 5KP s + 5KI The steadystate error for a step command input will be zero, as required. The characteristic equation is s3 + 5KD s2 + 5KP s + 5KI = 0 (1) To achieve τ = 1 and ζ = 0.45, the desired dominant roots are s = −1 ± j1.985. The third root is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will not be the dominant root). The polynomial corresponding to these three roots is [(s + 1)2 + (1.985)2 ](s + b) = 0 or s3 + (b + 2)s2 + (4.94 + 2b)s + 4.94b = 0
(2)
Comparing the coefficients of (1) and (2), we obtain KP =
4.94 + 2b 5
4.94b 5 b+2 KD = 5 The gains can be computed once a value for b > 1 has been selected. KI =
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11.21 In the first printing of the text, the last line of the problem statement says ”but increases Cv to 50”. It should say ”but increases Cv to 5”. desired valu Because the desired roots are obtainable with a simple gain selection, we need a lag compensator to achieve the desired Cv value. The lag compensator has the form Gc (s) = K Thus Cv =
s+z s+p
Kz 2p
To achieve Cv = 5 with K = 1, we need z/p = 10. Choosing z = 0.05 and p = 0.005 satisfies this requirement, and places the compensator’s pole and zero near the origin, far away from the desired root locations of s = −0.338 ± 0.562j. Using the root locus plot, we can determine the value of K required to place the closedloop roots near the desired location. This value is K = 1.024. Thus one solution is the compensator transfer function Gc (s) = 1.024
s + 0.05 s + 0.005
The resulting closedloop roots are s = −2.326, −0.055, and −0.312 ± 0.55j, which are close to the desired locations. The closedloop zero at s = 0.05 approximately cancels the closedloop pole at s = −0.055. The resulting Cv value is Cv = 5.12, which is close to the desired value.
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11.22 Note: In the first printing of the text, the transfer function was printed incorrectly. The correct transfer function is 1 Gp (s) = 2 s +s Also, the desired value of Cv was incorrectly given to be 5. It should be 10. √ The values ζ = 0.5 and ωn = 2 correspond to a dominant root location of s = −0.5±j 3. These roots can be obtained by setting K = 1. So the gain KP required to achieve the desired transient performance has already been established as KP 1 = K = 1. A lag compensator is indicated, because the steadystate error is too large. The second step is to determine the value of the parameter µ. For this system, the coefficient Cv is Cv = lim s s→0
KP = KP s(s + 1)
and KP 2 is the value of KP that gives Cv = 10. Thus, KP 2 = 10, and the parameter µ = KP /KP 2 = 1/10. The compensator’s pole and zero must be placed close to the imaginary axis, with the ratio of their distances being 1/10. Noting that the plant has a pole at s = −1, we select locations well to the right of this pole, say, at s = −0.01 and s = −0.1 for the pole and zero, respectively. This gives T = 100. The openloop transfer function of the compensated system is thus Gc (s)G(s)H(s) =
0.1Kc (s + 0.1) s(s + 1)(s + 0.01)
The root locus is shown in the following figure. For the desired damping ratio of 0.5, the locus essentially lies on the asymptote that passes through σ=
$
$
sp − sz 0 − 1 − 0.01 + 0.1 = = −0.455 P −Z 3−1
So the real part of the root is approximately −0.455, and the imaginary part will be 0.455 tan 60◦ = 0.788 to obtain ζ = 0.5. Thus the root is approximately s = −0.455 + 0.788j. The value of Kc required to obtain this root is found from # # # s(s + 1)(s + 0.01) # # # s + 0.1
0.1Kc = − ##
= 0.9128 s=−0.455+0.788j
or Kc = 9.128. This is only a tentative estimate because the root does not lie exactly on the locus. Using this value of Kc in the characteristic equation, we find that the actual roots are s = −0.4496 ± 0.788j and s = −0.1109. So our estimate turns out to be very close to the true value. The error coefficient is thus Cv = 9.128 and less than the desired value of 10. Also, the dominant roots differ somewhat from the desired locations at s = −0.5 + j0.866. If these differences are too large, the compensator’s pole and zero can be placed closer to the imaginary axis, say, at s = −0.01 and s = −0.001, respectively, with T = 1000. This will decrease the compensator’s influence on the locus near the desired root locations. 1156 PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Root Locus 0.4
0.3
0.2
Imaginary Axis
0.1
0
−0.1
−0.2
−0.3
−0.4
−1
−0.9
−0.8
−0.7
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
Real Axis
Figure : for Problem 11.22
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11.23 Step 1: Adjust KP to meet Cv = 20. Cv = K/2 = 20. Thus KP = 40. Use KP = 40 in the following steps. Step 2: Check phase margin and gain margin with KP = 40, for G(s) =
20 s(0.5s + 1)
The phase margin is 17◦ and the gain margin is infinite. Thus we must add 40 − 17 = 23◦ to obtain a phase margin of 40◦ . This requires a lead compensator. Following the method of Example 11.3.3, we add a 5◦ safety factor and choose φm to be 23 + 5 = 28◦ . Then 1 + sin 28◦ µ= = 2.7698 1 − sin 28◦ √ Now find the frequency at which the uncompensated gain equals −20 log µ = −4.42 db. √ This occurs at about ω = 8.2 rad/sec. Thus choose ωm = 8.2 and ωm = 1/T µ = 8.2, which gives T = 0.073. Thus the compensator’s parameters are µ = 2.7698 and T = 0.073. The pole and zero are s = −1/T = −13.699 and s = −4.946. The openloop transfer function of the compensated system is Gc (s)G(s) =
s + 4.946 40 s + 13.699 s(s + 2)
As s → 0, the compensator’s gain is seen to introduce a gain factor of 4.946/13/699 = 0.361 = 1/µ. Thus, in order to preserve Cv = 20, we must increase KP by a factor of 1/0.361 = 2.7698. Therefore we set KP = 40(2.7698) = 110.79. The compensated openloop transfer function is Gc (s)G(s) = 110.79
s + 4.946 1 20(0.2022s + 1) = s + 13.699 s(s + 2) s(0.073s + 1)(0.5s + 1)
The frequency response plots show that the phase margin is 42◦ , the gain crossover frequency is ωg = 8, and the gain margin is infinite. Thus the specifications are satisfied. The closedloop transfer function is Gc (s)G(s) 20(0.2022s + 1) = 1 + Gc (s)G(s) 0.0365s3 + 0.573s2 + 5.044s + 20 The roots are s = −7.104 and s = −4.297 ± 7.659j, with a dominant time constant of 0.233, a dominant damping ratio of 0.489, and natural frequency of 8.78.
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11.24 In the first printing of the text, the transfer function was printed incorrectly. The correct transfer function is 2.5KP G(s) = s(s + 2)(0.25s + 2) With no compensation, Cv = 0.625, which is much less than the desired value of Cv = 80. Thus a lag compensator is needed to increase the gain by 80/0.625 = 128. Lead compensation is needed because it is not possible to place the roots at the desired location with a simple gain adjustment. Thus a simple lead or a simple lag compensator will not work. We therefore will design a laglead compensator. The angle deficiency is found from $
G(s)s=−2+2
√
3j
=$
#
# 10 # = 120 − 90 − 30 = −240◦ s(s + 2)(s + 8) #s=−2+2√3j
Thus, the angle deficiency is 180 − 240 = 60◦ . For the lead compensator G1 (s), $
G1 (s)s=−2+2√3j = $
#
1 + µT1 s ## = 60◦ 1 + T1 s #s=−2+2√3j
This is satisfied if the lead compensator has a pole at s = −53 and a zero at s = −3.65. Thus 1/µT1 = 3.65 and 1/T1 = 53. These give T1 = 0.0189 and µ = 14.5. For the lag compensator, the choice of T2 = 100 gives # #s + # # #s +
µ T2 1 T2
# # # # s + 0.145 ## # # # = 0.967 #=# # s + 0.01 #s=−2+2√3j
which we take to be close enough to 1 to indicate adequate polezero cancelation. The laglead compensator is thus given by Gc (s) = 128
s + 3.65 s + 0.145 s + 53 s + 0.01
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11.25 We try a compensator of the form Gc (s) = K
s+a s+b
The characteristic equation is found from 1 + Gc (s)Gp (s) = 0 which becomes 1+K
s+a 10 =0 2 s + b s (0.1s + 1)
or 0.1s4 + (1 + 0.1b)s3 + bs2 + 10Ks + 10Ka = 0
(1)
For τ = 1 and ζ = 0.5, the required dominant root pair is s = −1 ± 1.732j. So the characteristic equation must be factored as 0.1[(s + 1)2 + (1.732)2 ](s2 + cs + d) = 0 or 0.1(s2 +2s+4)(s2 +cs+d) = 0.1s4 +0.1(2+c)s3 +0.1(4+2c+d)s2 +0.1(4c+2d)s+0.4d = 0
(2)
Comparing the coefficients of equations (1) and (2), we obtain 1 + 0.1b = 0.1(2 + c)
(3)
b = 0.1(4 + 2c + d)
(4)
10K = 0.1(4c + 2d)
(5)
10Ka = 0.4d
(6)
We can choose c and solve (3) for b and (4) for d. The choice of c = 14 gives b = 6 and d = 28. So the secondary roots from s2 + cs + d = 0 are s = −2.417 and s = −11.58. Then solve (5) for K to obtain K = 1.12. Finally, solve (6) for a to obtain a = 1. The resulting compensator is s+1 Gc (s) = 1.12 s+6
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11.26 The root locus equation is 1+
K s+a =0 5 s2 (s + b)
The poles are s = 0, s = 0, and s = −b. The zero is s = −a. The asymptotic angles are θ = ±90◦ . The asymptotes intersect at σ=
a−b 2
For stability, the intersection point should be negative; thus, we should select b > a. For a 5% overshoot, the damping ratio of the dominant root must be ζ = 0.69, which corresponds to a complex root. For a time constant of τ = 0.5, the dominant roots must be s = −2 ± 2.096j to achieve ζ = 0.69. The corresponding factor is (s + 2)2 + (2.096)2 = s2 + 4s + 8.393. Thus the characteristic equation can be factored as (s2 + 4s + 8.393)(s − s3 ) = s3 + (4 − s3 )s2 + (8.393 − 4s3 )s − 8.393s3 = 0 The characteristic equation is s3 + bs2 +
K K s+a =0 5 5
Comparing coefficients, we see that s3 = 4 − b
K = 8.393 − 4s3 5
a=−
8.393s3 K/5
The third root must lie to the left of the dominant root, whose real part is −2. Therefore, b must be chosen so that b > 6. The other restriction is that b > a. One solution is b = 10, which gives s3 = −6, K = 161.965, and a = 1.55. If we try to cancel one of the poles at s = 0 by letting a → 0, the root locus equation becomes K 1 1+ =0 5 s(s + b) and the characteristic equation is s2 + bs +
K =0 5
The specifications that ζ = 0.69 and τ = 0.5 require that s = −2 ± 2.096j. This is achieved with b = 4 and K = 41.95.
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11.27 The openloop poles are s = 0 and −2 ± 3j. a) The root locus equation for the lead compensator is 1+
s+ s
1 µT + T1
s(s2
Kc =0 + 4s + 13)
where µ > 1 and the root locus parameter is K = Kc ≥ 0. b) The root locus equation for the lag compensator is 1+
s+ s
1 µT + T1
µKc =0 s(s2 + 4s + 13)
where µ < 1 and the root locus parameter is K = µKc ≥ 0. c) The root locus equation for the reversereaction compensator is 1+
s− s+
1 T1 1 T2
− TT12 Kc
s(s2 + 4s + 13)
=0
where the root locus parameter is K = −T1 Kc /T2 ≤ 0. The rootlocus plots are shown in the following figures. All three compensators can produce an unstable system if the gain Kc is too large. However, only the reversereaction compensator can pull the complexroot paths starting at s = −2±3j to the left. This enables a dominant root to be obtained that is not close to the imaginary axis. The optimum root locations are shown on the plot.
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Figure : For Problem 11.27
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11.28 a) The plot for K = 2 is shown in figure (a). It can be obtained with the following MATLAB file. K = 2; sys = tf(5*K,[1,6,5,0]) margin(sys) The results are that the gain margin is 9.54 dB; the phase crossover frequency is 2.24 rad/sec; the phase margin is 25.4◦ ; and the gain crossover frequency is 1.23 rad/sec. b) The plot for K = 20 is shown in figure (b). It can be obtained in a manner similar to that used for part (a). The results are that the gain margin is −10.5 dB (which means that the system is unstable); the phase crossover frequency is 2.24 rad/sec; the phase margin is −23.7◦ (another indication of instability); and the gain crossover frequency is 3.91 rad/sec. c) For K = 6 both the gain and phase margins are 0. Thus they both are a limiting factor for stability.
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11.29 The openloop transfer function is G(s) = Gc (s)Gp (s) =
2 + 19s 100s2 + s
The Bode plots are shown in following figure. They can be obtained with the MATLAB margin function. The gain margin is infinite because the phase curve is always above the −180◦ line. The phase margin is 66◦ . The system is stable. Bode Diagram Gm = Inf , Pm = 66.3 deg (at 0.212 rad/sec) 150
Magnitude (dB)
100
50
0
Phase (deg)
−50 −90
−120
−150 −4
10
−3
10
−2
10
−1
10
0
10
1
10
Frequency (rad/sec)
Figure : For Problem 11.29
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11.30 The openloop transfer function is G(s) = Gc (s)Gp (s) =
25(7s + 64) 175s + 1600 = 3 5s3 + 6s2 + 5s 5s + 6s2 + 5s
The Bode plots are shown in following figure. They can be obtained with the MATLAB margin function. Both the gain and phase margins are negative, so the system is unstable. Bode Diagram Gm = −47.3 dB (at 1.07 rad/sec) , Pm = −41.5 deg (at 7.46 rad/sec) 150
Magnitude (dB)
100
50
0
−50
−100 −90
Phase (deg)
−135
−180
−225
−270 −2
10
−1
10
0
1
10
10
2
10
3
10
Frequency (rad/sec)
Figure : For Problem 11.30
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11.31 The openloop transfer function is G(s) =
0.1KP (10) KP = (0.2s + 1)(s2 + s + 10) 0.2s3 + 1.2s2 + 3s + 10
The Bode plots, phase margins, and gain margins can be obtained with the MATLAB margin function. The results are as follows: a) KP = 1: Gain margin = 20.9 dB, Phase margin = 88.3◦ b) KP = 10: Gain margin = 0.916 dB, Phase margin = 70.4◦ c) KP = 100: Gain margin = −19.1 dB, Phase margin = −115◦ Cases (a) and (b) are stable, but Case (c) is unstable. This means that any initial roll angle will continue to increase.
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11.32 One solution is K = 0.376. 11.33 The answers are given in the following table.
Case (a) (b) (c)
Type No. 1 0 2
Cp ∞ 20 ∞
Cv 20 0 ∞
Table : for Problem 11.33 Ca Step Error Ramp Error 0 0 1/20 0 1/21 ∞ 7 0 0
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11.34 The openloop transfer function is G(s) = KP e−(D1 +D2 )s
1 1 = KP e−100s 100s + 1 100s + 1
The delay of 100 s lowers the phase curve by 100ω rad, or 100ω(180/π) degrees . A plot of m and φ for KP = 1 shows that instability is caused by the negative gain margin. The phase crossover frequency is approximately 0.02 rad/sec, and m at this frequency is approximately −7 dB. To achieve a positive gain margin, we must therefore increase KP to at least 107/20 = 2.239. For this value of KP both the phase margin and the gain margin are zero. The calculations for KP = 1 can be done in MATLAB as follows. sys = tf(1,[100,1]); w = [0.001:0.001:0.02]’; [mag, phase] = bode(sys,w); m = 20*log10(mag(:)); phasetotal = phase(:)100*w*(180/pi); semilogx(w,m,w,phasetotal),grid
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11.35 10.31 With D = 0, the openloop frequency response plot of 10/(0.1s + 1) has a gain crossover at ω ≈ 100, and a phase margin of 95◦ . For PM = 40◦ , the dead time D can reduce the phase curve at ω = 100 by no more than 95 − 40 = 55◦ . Because $
P (iω)e−iωD = $ P (iω) − ωD
we have 100D ≤ 55
π rad 180
or D ≤ 0.0096.
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11.36 In the first printing of the text, the required value for Cv was incorrectly stated as Cv = 0.2. The required value should be Cv = 5. One solution is the compensator transfer function Gc (s) =
s + 0.1 s + 0.01
This gives a gain margin of 14.3 dB and a phase margin of 41.6◦ .
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11.37 One solution is the compensator transfer function Gc (s) = 42
s + 4.4 s + 18
This gives an infinite gain margin and a phase margin of approximately 50◦ .
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11.38 The problem statement should read “at least 45◦ ”. One solution is the compensator transfer function Gc (s) = 20
6.66s + 1 66.6s + 1
This gives an infinite gain margin and a phase margin of approximately 57◦ .
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11.39 In the first printing of the text, the required value for Cv was incorrectly stated as Cv = 0.05. The required value should be Cv = 20. Also, in the second line after the equation, delete the phrase “the steadystate ramp error will be”. One solution is the compensator transfer function Gc (s) = 20
0.34s + 1 0.07s + 1
This gives an infinite gain margin and a phase margin of approximately 50◦ .
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11.40 In the first printing of the text, the required value for Cv was incorrectly stated as Cv = 0.02. The required value should be Cv = 50. One solution is the compensator transfer function Gc (s) = 50
(s + 1)(s + 0.16) (s + 0.01)(s + 16)
The closedloop transfer function is C(s) 50s2 + 58s + 8 = R(s) 0.2s5 + 4.402s4 + 20.24s3 + 66.2s2 + 58.16s + 8 There is polezero cancelation of the factor s + 1, so the transfer function becomes C(s) s + 0.16 = 250 R(s) (s + 17.185)(s + 0.168)(s2 + 3.656s + 13.846) The roots are s = −17.1837, s = −0.1681, and s = −1.8291 ± 3.2401j. If we take the roots s = −0.1681 to be canceled approximately by the compensator zero at s = −0.16, the root pair s = −1.8291 ± 3.2401j is the dominant root. It has a damping ratio of ζ = cos(tan−1 (3.2401/1.8291)) = 0.4916, which is approximately the desired value of 0.5.
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11.41 One solution is the compensator transfer function Gc (s) = 20
(s + 0.7)(s + 0.15) (s + 7)(s + 0.015)
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c Solutions Manual! to accompany System Dynamics, First Edition by William J. Palm III University of Rhode Island
Solutions to Problems in Chapter Twelve
c Solutions Manual Copyright 2004 by McGrawHill Companies, Inc. Permission required ! for use, reproduction, or display.
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12.1 The equation of motion is m¨ x + kx = ky where y(t) = Y sin 6πt = 4 × 10−3 sin 6πt. Also, ωn2 =
500 k = = 1000 rad/s m 0.5
From (12.1.9), with ζ = 0 and r 2 = (6π)2 /1000, ! !
X = Y !!
!
1 !! = 6.2 × 10−3 m 1 − r2 !
or 6.2 mm
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12.2 For a period of 20 feet and a vehicle speed of v (mph), the frequency ω is ω=
"
5280 20
#"
#
1 (2π) = 0.4608v 3600
rad/sec
Thus ω = 0.4608(20) = 9.216 rad/sec for v = 20 mph, and ω = 0.4608(50) = 23.04 rad/sec for v = 50 mph. For a car weighing lb, the quartercar mass is m = 500/32.2 slugs. Its natural $ 2000 $ frequency is ωn = k/m = 2000/(500/32.2) = 11.35 rad/sec. Its frequency ratio at 20 mph is r = ω/ωn = 9.216/11.35 = 0.812, and at 50 mph it is r = 23.04/11.35 = 2.03. Its damping ratio is 360 ζ= $ = 1.02 2 2000(500/32.2)
Now substitute these values of r and ζ and Y = 0.03 ft into the following expressions, obtained from (12.1.9) and (12.1.12). X=Y
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
Ft = r 2 kX This gives the following table.
v (mph) 20 50
2000 lb Car (ζ = 1.02) r X (ft) Ft (lb) 0.812 0.034 45.12 2.03 0.025 203.4
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12.3 We are given m = 1500 kg, k = 20, 000 N/m, ζ = 0.04, and Y = 0.01 m. At resonance, $
ωr ωn 1 − 2ζ 2 r= = = 0.998 ωn ωn and Ft = r kY 2
%
4ζ 2 r 2 + 1 = 2500 N (1 − r 2 )2 + 4ζ 2 r 2
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12.4 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also, ω = 40 Hz = 215 rad/s r = 2
From (12.1.9) with c = 0,
"
ω ωn
ωn =
%
#2
"
=
k = m
251 57.2
%
#2
66 667 = 57.2 rad/s (200/9.81)
= 19.26
X 1 = = 0.055 Y 1 − r 2 
Thus 5.5% of the airframe motion is transmitted to the module.
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12.5 a) Neglect damping in the isolator, and determine its required stiffness k. From (12.1.9), 1 X = = 0.1 Y 1 − r 2 
which gives r 2 = 11. Thus
ωn2 =
ω2 [3000(2π)/60]2 (314)2 = = r2 11 11
and
2 (314)2 = 556.7lb/ft k = mωn2 = 32.2 11 √ b) r = ω/ωn and ωn = 314/ 11. Thus
From (12.1.9)
r1 =
2500(2π)/60 √ = 2.77 314/ 11
r2 =
3500(2π)/60 √ = 3.87 314/ 11 1 X = Y 1 − r 2 
Thus the highest percentage of motion will be transmitted at the lowest r value, which corresponds to 2500 rpm. For 2500 rpm, X 1 = = 0.15 Y 1 − r12  For 3500 rpm,
1 X = = 0.07 Y 1 − r22 
Thus at most, 15% of the crane motion will be transmitted to the module.
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12.6 We are given m = 5/32.2 slug and r = 2
From (12.1.9) with ζ = 0,
"
ω ωn
#2
=
(30(2π)/60)2 32.2k/5
1 X = = 0.1 Y 1 − r 2 
which gives r 2 = 11. Thus
(30(2π)/60)2 = 11 32.2k/5
Solve for k to obtain k = 0.139 lb/ft. The transmitted force is "
Ft = r
2X
Y
#
kY = 11(0.1)0.139(0.003) = 4.59 × 10−4 lb
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12.7 a) From Newton’s law, m¨ x = c(y˙ − x) ˙ − kx
or
mx˙ + cx˙ + kx = cy˙ T (s) =
X(s) cs = 2 Y (s) ms + cs + k
Thus T (jω) = This gives T (jω) =
cωm mk j
−mω 2
cωj/k cωj = + cωj + k 1 − r 2 + cωj/k
1 − r 2 + cωmj/mk
=
$
2ζωn ωj/ωn2 2ζrj = 2 2 1 − r + 2ζωn ωj/ωn 1 − r 2 + 2ζrj
where we have used the fact that ωn = k/m, r = ω/ωn , and c/m = 2ζωn . The magnitude is 2ζr X=$ Y 2 (1 − r )2 + (2ζr)2 b) Thus
2ζrk Ft = kX = $ Y (1 − r 2 )2 + (2ζr)2
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12.8 The displacement transmissibility with r = 2 is X = Y
%
4ζ 2 r 2 + 1 = (1 − r 2 )2 + 4ζ 2 r 2
%
1 + 16ζ 2 9 + 16ζ 2
This has a minimum of 1/3 when ζ = 0. This is the best choice for ζ if the velocity increases (r > 2). However, if the velocity decreases (so that r → 1, and the system approaches resonance), an nonzero value of ζ would be a better choice to limit the resonant response. With ζ = 0 and a 20% increase in r (r = 2.4), we have X/Y = 0.21. With ζ = 0 and a 20% decrease in r (r = 1.6), we have X/Y = 0.64.
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12.9 The relations are Ft = r 2 kX
X=Y where
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
c ζ= √ 2 mk
r=
ω ωn
At 40 mph, the forcing frequency is ω=
"
#"
5280 20
#
1 2π(4) = 18.432 rad/sec 3600
and r = 18.432/ωn , where ωn =
%
k m
To minimize Ft , choose either 1. r small (with ζ near 1 and ωn large; this means k large and c large), or 2. r > 2 (this means ωn small). For example, with m = 800/32.2, choosing ζ = 1 and r = 0.5, implies that ωn = 2(18.432) = 36.864 k = m(36.864)2 = 33 763 lb/ft √ c = 2 mkζ = 1832 lb sec/ft This gives X =Y
%
and
4ζ 2 r 2 + 1 = 0.0589 ft (1 − r 2 )2 + 4ζ 2 r 2 Ft = r 2 kX = 497 lb
On the other hand, choosing ζ = 1 and r = 3 for example, we obtain k=
mωn2
800 = 32.2
c=2 This gives X =Y
&
%
"
18.432 3
#2
= 937.9 lb/ft
800 937.9 = 305 lb sec/ft 32.2
4ζ 2 r 2 + 1 = 0.0317 ft (1 − r 2 )2 + 4ζ 2 r 2
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and Ft = r 2 kX = 268 lb Note that the plot in Figure 12.1.3 is a plot of Ft /kY . Thus the curves in the plot must be interpreted with the value of k kept in mind. Even though the curve for r = 0.5 is lower that the curve for r = 3, for ζ = 1, the corresponding values of k are different, and thus the case r = 3 gives a lower value of Ft .
1211 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.10 For the lumpedmass equivalent system, we have the following spring constant and equivalent mass (after converting inches to feet): (4.32 × 109 )(0.333)(0.03125)3 Ewh3 = 4L3 4(0.5)3 = 8.78 × 104 lb/ft
k =
We compute the equivalent mass of the beam as follows. Using 15.2 slug/ft3 for the density of steel, and including 23% of the beam’s mass, we obtain me =
20 + 0.23(15.2)(0.333)(0.5)(0.03125) = 0.640 32.17
slug
The unbalanced mass is m = 1/32.17 = 0.0311 slug. The model for the system is ¨ + cx˙ + kx = f (t) = mu %ω 2 sin ωt me x This gives the transfer function 1 X(s) = 2 F (s) me s + cs + k 1/k me 2 c k s + ks + 1
T (s) = = Thus,
T (jω) = where 1/k = 1.139 × 10−5 and ωn = ratio is.
$
1−
'
1/k ω ωn
(2
+
2ζω ωn j
k/me = 370 rad/sec = 3537 rpm. The log magnitude +
ω2 m(ω) = 20 log(1/k) − 10 log 1 − 2 ωn
,2
+
"
2ζω ωn
#2
The motor speed of 1750 rpm gives a forcing frequency of ω = 183 rad/sec. Using ζ = 0.1 and ω/ωn = 183/370, we see that m = 2.36 − 98.9 = −96.5 db, which corresponds to a magnitude ratio of 1.5 × 10−5 . The amplitude of the forcing function is mu %ω 2 = 0.0311(0.01)(183)2 = 10.4 lb. Thus, the steadystate amplitude is 10.4(1.5 × 10−5 ) = 15.6×10−5 ft. On the downward oscillation, the total amplitude as measured from horizontal is the preceding value plus the static deflection, or 15.6 × 10−5 + 20/87800 = 3.84 × 10−4 ft. Because ω is not close to ωn in this problem, the preceding results are not very sensitive to the assumed value of ζ = 0.1. For example, the calculated amplitudes of vibration for ζ = 0.05 and 0.2 are close to the amplitude for ζ = 0.1 (the amplitudes are 15.7 × 10−5 ft and 15.2 × 10−5 ft, respectively). However, if we had used a motor with a speed of 3500 rpm = 366 rad/sec, this choice would put the forcing frequency very close to the 1212 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
natural frequency. In this region, the assumed value of ζ would be critical in the amplitude calculation. In practice, such a design would be avoided. That is, in vibration analysis, the most important quantity to know is the natural frequency ωn . If the damping is slight, the resonant frequency is near ωn . If ωn is designed so that it is not close to the forcing frequency, it is often not necessary to know the precise amount of damping.
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12.11 We are given that the natural frequency is ωn1 = 900 rpm. Thus, at 1750 rpm, r1 =
1750 = 1.94 900
If we decrease the stiffness k by 1/2, then the new natural frequency will be ωn2 =
%
1 900 k/2 = √ ωn1 = √ rpm m 2 2
Therfore, r2 =
1750 √ = 2.75 900/ 2
From the rotating unbalance equation (12.2.6) with ζ = 0, X=
mu % r 2 m 1 − r 2 
(1)
We are given that at r1 , X1 = 8 mm. We need to compute X2 . From (1), X2 = X1
"
r2 r1
#2
−2.7636 1 − r12 = 0.842 =2 2 −6.5625 1 − r2
Thus X2 = 0.842X1 = 6.73 mm
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12.12 We are given that k= Thus ωn1 =
%
500 = 24, 000 lb/ft 0.25/12
k = m
%
24, 000 = 39.31 rad/sec 500/32.2
From the rotating unbalance equation (12.2.6) with ζ = 0, mu % r 2 m 1 − r 2 
X= and thus
X2 = X1
where
"
r2 r1
#2
1 − r12 1 − r22
1750(2π)/60 = 4.66 39.31 After the block is added, the new mass is m2 = 4m1 and the new natural frequency is r1 =
ωn2 = Therefore, r2 = Thus,
%
1 k = ωn1 4m1 2
ω ω =2 = 2r1 ωn2 ωn1
X2 −20.7156 1 − r12 = 0.9651 = (2)2 =4 X1 −85.8624 1 − r22
and given that X1 = 0.1 in, we have
X2 = 0.9651(0.1) = 0.0965 in
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12.13 The transmitted force is given by Ft = mu %ω
2
%
1 + 4ζ 2 r 2 (1 − r 2 )2 + 4ζ 2 r 2
We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. R = 0.1/12 = 0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus r= Thus
ω 104.7 =$ = 5.83 ωn 500/1.55
Ft = (0.00155)(0.0083)(104.7)2
%
1 + 136ζ 2 = 0.141 1088 + 136ζ 2
%
1 + 136ζ 2 1088 + 136ζ 2
a) For ζ = 0.05, Ft = 0.0049 lb. b) For ζ = 0.7, Ft = 0.034 lb.
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12.14 We have Ft = mu %ω 2 Tr where Ft = 15, ω = 200(2π)/60 = 20.9 rad/s, and because c = 0, Tr = We have r= Thus Tr = 0.0826 and
r2
1 −1
ω 20.9 20.9 =$ =$ = 3.62 ωn k/M 2500/75 Ft = mu %(20.9)2 (0.0826) = 15
Solve for mu % to obtain mu % = 0.416 kg m.
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12.15 We want Tr = 0.1. Neglecting damping, we have from (12.2.12) r2 = But r= √ Thus ωn = 314/ 11 = 94.7. But
1 + Tr = 11 Tr
314 √ 3000(2π)/60 = = 11 ωn ωn
ωn =
%
k = m
%
k 3
from which we obtain k = 3ωn2 = 3(94.7)2 = 26 904 N/m.
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12.16 Damping is assumed to be negligible. Thus for the vertical motion, Ft 1 = Tr = 2 F r −1
where r=
1750(2π)/60 ω ω = $ = 10.24 =$ ωn 4k/m 8000/25
Thus Tr = 0.0096. The equation of motion for rotation is
I θ¨ = −(4kRθ)D/2 $
Thus the natural frequency for rotation is ωn = 4kD 2 /4I = rotational motion, 1 Tt = Tr = 2 T r −1 where 1750(2π)/60 ω = 9.16 = r= ωn 20
$
80/0.2 = 20. For the
Thus Tr = 0.012.
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12.17 We are given m = 20/g slug, mu = 1/g slug, % = 0.01 ft, and ω = 3500(2π)/60 = 366.5 rad/sec. Thus the unbalance force amplitude is mu %ω = 2
"
#
1 (0.01)(366.5)2 = 41.719 lb 32.2
The beam stiffness is k=
Ewh3 4.32 × 109 (1/3)(3/96)3 = = 8.79 × 104 lb/ft 4L3 4(1/2)3
The beam mass is m = ρV = 15.2(1/3)(3/96)(1/2) = 0.079 slug. The first design equation for the absorber is r2 = 1, or r2 =
ω 366.5 =$ =1 ωn2 k2 /m2
This implies that k2 /m2 = (366.5)2 . The second design equation is X2 = 0.25/12 = 1/48 ft, where 1 41.719 X2 = F = k2 k2 Thus k2 = 41.719(48) = 2002.5 lb/ft. Substitute this value into the first design equation to obtain k2 2002.5 m2 = = 0.0149 slug 2 (366.5) (366.5)2
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12.18 We are given that X2 ≤ 0.08/12 ft, ω = 6000(2π)/60 = 628.3 rad/sec, and that the unbalance force amplitude is mu %ω 2 = 60 lb. Thus 60 60 = = 0.0955 2 ω 628.3
mu % =
The first design equation for the absorber is k2 = The second design equation is
F 60 = = 9000 lb/ft X2 0.08/12 %
Thus m2 =
k2 = ω = 628.3 m2
k2 = 0.023 slug (628.3)2
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12.19 We are given that ω = 200(2π)/60 = 20.94 rad/sec, the amplitude of the unbalance force is mu %ω 2 = 4 lb, and that X2 ≤ 1/12 ft. Assume that the table legs are rigid. The first design equation for the absorber is %
k2 = ω = 20.94 m2
The second design equation is k2 =
F 4 = = 48 lb/ft X2 1/12
m2 =
48 = 0.109 slug (20.94)2
Thus
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12.20 The machine mass is m1 = 8 kg. a) We are given that ωn = 2π(6) = 12π rad/sec, mu %ω 2 = 50 N, ω = 4(2π) = 8π rad/sec, and X2 ≤ 0.1 m. The design equation for the absorber is % k2 = ω = 8π m2 Thus k2 = 50/0.1 = 500 N/m, and m2 =
k2 500 = = 0.792 kg 2 (8π) (8π)2
b) We have that k1 = m1 ωn2 = (12π)2 m1 = 1421m1 = 11368 N/m. From (12.3.7), 1 X1 (jω) = T1 (jω) = F (jω) 11368 where
! ! ! ! 1 − r24 ! ! 0 !/ 2 4 ! ! b r2 − [1 + (1 + µ)b2 ] r22 + 1 !
2 ωn2 8π = = ωn1 12π 3 0.792 m2 = µ= = 6.336 m1 8 13 3.168 + = 1.84 1 + (1 + µ)b2 = 9 8 b=
The amplitude of F (jω) is mu %ω 2 , where r2 = ωωn2 = ω/8π. Thus 1 X1 = mu % 11368 or
! ! X1 1 !! 3 = mu % 11368 !! 4 9
The plot is shown in the following figure.
! ! ! ω 2 11 − r 2 2 ! ! ! 2 !4 4 ! ! r2 − 1.84r22 + 1 ! 9
'
ω2 1 −
ω4 4096π 4
ω2 64π 2
(
2
ω − 1.84 64π 2
! ! ! 4 !! +1 !
1223 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
1
0.9
0.8
0.6
0.5
1
u
X /m " (slug−1)
0.7
0.4
0.3
0.2
0.1
0
0
10
20
30
60 50 40 Forcing frequency ! (rad/s)
70
80
90
100
Figure : for Problem 12.20.
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12.21 The first printing of the text had incorrect values for some of the frequencies. The correct values are as follows. The operating speed range is from 1500 to 3000 rpm. Without an absorber, excessive vibration was observed at 2100 rpm. After the absorber was attached, resonance was observed at 1545 and 2850 rpm. An absorber tuned to 2100 rpm requires that %
k2 = m2
%
2100(2π) k1 = 219.9 rad/sec = m1 60
Because we are given that m2 = 5/32.2 = 0.155, we have that k2 = (219.9)2 m2 = 9747.8. Also, k1 = (219.9)2 m1 (1) The characteristic equation of the combined system is given by the denominator of (12.3.1), and is m1 m2 s4 + (m2 k1 + m2 k2 + m1 k2 )s2 + k1 k2 = 0 (1) One of the observed resonances of the combined system is 2850 rpm, or 2850(2π)/60 = 298.45 rad/sec. Thus substituting s = 298.45j and relation (1) into (2), along with the values m2 = 0.155 and k2 = 9747, we can determine the value of m1 , which is m1 = 0.4 slug. Thus gives k1 = 19, 344 lb/ft. With m1 and k1 determined, we now calculate the required values of m2 and k2 . Suppose we choose to put the resonances just outside the operating range, say at 1400 and 3100 rpm (146.6 and 324.6 rad/sec). Let λ = s2 , and let λ1 and λ2 denote the desired values of s2 . We can factor the characteristic equation (2) as follows: m1 m2 (λ − λ1 )(λ − λ2 ) = m1 m2 λ2 − m1 m2 (λ1 + λ2 )λ + m1 m2 λ1 λ2 = 0
(3)
Comparing the coefficients of (3) with (2), we see that
and
(m2 k1 + m2 k2 + m1 k2 = −m1 m2 (λ1 + λ2 ) m1 m2 λ1 λ2 = k1 k2
(4)
(5)
We can solve (5) for k2 as follows. k2 =
m 1 m 2 λ1 λ2 = Dm2 k1
where D=
(6)
m1 λ1 λ2 k1
Substitute (6) into (4) and solve for m2 . m2 = −
m1 (λ1 + λ2 ) + k1 + m1 D D
The desired values are λ1 = −(146.6)2 and λ2 = −(324.6)2 . These give the absorber values m2 = 0.2706 slug and k2 = 12, 672 lb/ft. 1225 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.22 a) From Newton’s law, m1 x ¨1 = f − kx1 + c(x˙ 2 − x˙ 1 ) m2 x ¨2 = −c(x˙ 2 − x˙ 1 )
Transform these equations with zero initial conditions to obtain (m1 s2 + cs + k)X1 (s) − csX2 (s) = F (s) −csX1 (s) + (m2 s2 + cs)X2 (s) = 0
The solutions obtained with Cramer’s rule are X1 (s) =
m2 s2 + cs F (s) D(s)
X2 (s) =
cs F (s) D(s)
where Cramer’s determinant is D(s) = s(m1 m2 s3 + c(m1 + m2 )s2 + km2 s + ck) Define the following parameters: µ=
m2 m1
ω12 =
c ζ= √ 2 m 1 k1
k1 m1
r=
ω ω1
Then D(jω) can be written as
or
! !
! !
D(jω) = !ωj[−m1 m2 ω 3 j − c(m1 + m2 )ω 2 + km2 ωj + kc]! 5
D(jω) = m21 ω14 r (2ζr)2 [1 − (1 + µ)r 2 ]2 + µ2 r 2 (1 − r 2 )2 $
The numerator of kX1 (jω)/F (jω) is  − m2 ω 2 + cωj = m1 rω12 4ζ 2 + µ2 r 2 . The numerator of X2 (jω)/F (jω) is cωj. Thus, $
and
kX(jω) 4ζ 2 + µ2 r 2 =$ F (jω) (2ζ)2 [1 − (1 + µ)r 2 ]2 + µ2 r 2 (1 − r 2 )2 1 X2 (jω) 2ζ = 2$ 2 F (jω) ω1 (2ζ) [1 − (1 + µ)r 2 ]2 + µ2 r 2 (1 − r 2 )2
1226 ______________________________________________________________________________________________ PROPRIETARY MATERIAL. © 2005 The McGrawHill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
12.23 a) From Newton’s law, m1 x ¨1 = f − kx1 + k2 (x2 − x1 ) + c(x˙ 2 − x˙ 1 ) m2 x ¨2 = −k2 (x2 − x1 ) − c(x˙ 2 − x˙ 1 )
Transform these equations with zero initial conditions to obtain (m1 s2 + cs + k1 + k2 )X1 (s) − (cs + k2 )X2 (s) = F (s) −(cs + k2 )X1 (s) + (m2 s2 + cs + k2 )X2 (s) = 0
The solutions obtained with Cramer’s rule are X1 (s) =
m2 s2 + cs + k2 F (s) D(s)
X2 (s) =
cs + k2 F (s) D(s)
where Cramer’s determinant is D(s) = m1 m2 s4 + c(m1 + m2 )s3 + (m2 k1 + m2 k2 + m1 k2 )s2 + ck1 s + k1 k2 Define the following parameters: µ=
m2 m1
ω12 =
k1 m1
ω22 =
k2 m2
ω1 ω r= ω2 ω1 c k2 ζ= √ λ= k1 2 m 1 k1 α=
Then D(jω) can be written as
or
! !
! !
D(jω) = !m1 m2 ω 4 − (m2 k1 + m2 k2 + m1 k2 )ω 2 + k1 k2 + [ck1 ω − (m1 + m2 )cω 3 ]j ! ! !
! !
D(jω) = !m21 ω 4 [µr 4 − (1 + λ + µλ)r 2 + µα2 ] + [2ζr − 2(1 + µ)r 3 ]j ! !
!
The numerator of k1 X1 (jω)/F (jω) is k2 − m2 ω 2 + cωj = k1 ![λ − µr 2 + 2ζri]!. The numerator of k1 X2 (jω)/F (jω) is cωj + k2  = k1 (λ + 2ζrj). Thus, $
and
k1 X(jω) (λ − µr 2 )2 + (2ζr)2 =5 F (jω) [µr 4 − (1 + λ + µλ)r 2 + µα2 ]2 + (2ζr)2 [1 − (1 + µ)r 2 ]2 $
k1 X2 (jω) λ2 + (2ζr)2 =5 F (jω) [µr 4 − (1 + λ + µλ)r 2 + µα2 ]2 + (2ζr)2 [1 − (1 + µ)r 2 ]2
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12.24 The equations of motion are m1 x ¨1 = −k1 x1 − k2 (x1 − x2 ) m2 x ¨2 = k2 (x1 − x2 )
From these equations we can write the modal amplitude equations by substituting x1 = A1 est and x2 = A2 est , and using the given parameter values. The result is (10s2 + 30, 000)A1 − 20, 000A2 = 0 These give the solution
−20, 000A1 + (30, 000s2 + 20, 000)A2 = 0
s2 + 3000 A1 (1) 2000 The roots are found from Cramer’s determinant of the modal equations, which is A2 =
3s4 + 11, 000s2 + 2 × 106 = 0 The roots are s2 = −3475 and s2 = −192. Substitute s2 = −3475 into (1) to obtain A2 = −0.2375A1 . Substitute s2 = −192 into (1) to obtain A2 = 1.404A1 . √In the first mode, the masses oscillate in opposite directions with a radian frequency of 3475. The displacement amplitude of mass 2 is 0.2375 times that of mass 1. In √ the second mode, the masses oscillate in the same direction with a radian frequency of 192. The displacement amplitude of mass 2 is 1.404 times that of mass 1.
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12.25 The equations of motion are m1 L2 θ¨1 = −mg1 Lθ1 − kd(dθ1 − dθ2 ) m2 L2 θ¨2 = −m2 gLθ2 + kd(dθ1 − dθ2 )
From these equations we can write the modal amplitude equations by substituting θ1 = A1 est and θ2 = A2 est . The result is (m1 L2 s2 + m1 gL + kd2 )A1 − kd2 A2 = 0 −kd2 A1 + (m2 L2 s2 + m2 gL + kd2 )A2 = 0
Using the given parameter values and g = 9.81 m/s2 , these equations give the solution A2 = (3.13s2 + 7.13)A1
(1)
The roots are found from Cramer’s determinant of the modal equations, which is 2500s4 + 10810s2 + 11 586 = 0 The roots are s2 = −1.96 and s2 = −2.36. Substitute s2 = −1.96 into (1) to obtain A2 = 0.995A1 . Substitute s2 = −2.36 into (1) to obtain A2 = −0.257A1 . √ In the first mode, the masses oscillate in the same direction with a radian frequency of 1.96. The displacement amplitude of mass 2 is 0.995 times that of mass 1. In√the second mode, the masses oscillate in the opposite direction with a radian frequency of 2.36. The displacement amplitude of mass 2 is 0.257 times that of mass 1.
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12.26 The equations of motion are I1 θ¨1 = k2 (θ2 − θ1 ) − k1 θ1 I2 θ¨2 = −k2 (θ2 − θ1 )
From these equations we can write the modal amplitude equations by substituting θ1 = A1 est and θ2 = A2 est . The result is (I1 s2 + k1 + k2 )A1 − k2 A2 = 0 −k2 A1 + (I2 s2 + k2 )A2 = 0
Using the given parameter values, these equations give A2 =
I1 s2 + k1 + k2 s2 + 4 A1 θ1 = k2 3
(1)
The roots are found from Cramer’s determinant of the modal equations, which is 5s4 + 23s2 + 3 = 0 The roots are s2 = −0.134 and s2 = −4.47. Substitute s2 = −0.134 into (1) to obtain A2 = 1.29A1 . Substitute s2 = −4.47 into (1) to obtain A2 = −0.157A1 . √ In the first mode, the masses oscillate in the same direction with a radian frequency of 0.134. The displacement amplitude of mass 2 is 1.29 times that of mass 1. In√the second mode, the masses oscillate in the opposite direction with a radian frequency of 4.47. The displacement amplitude of mass 2 is 0.157 times that of mass 1.
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12.27 The equations of motion are m¨ x = −2k sin 45◦ x = −1.41kx m¨ y = −ky − 2k sin 45◦ x = −2.41kx
$
In the first mode, the mass oscillates in the x direction with a radian frequency of 1.41k/m. In $ the second mode, the mass oscillates in the y direction with a radian frequency of 2.41k/m.
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12.28 From (12.4.8): (750)(1350)s4
+
6
7
730[(1.95 × 104 )(1.5)2 + 2.3 × 104 (1.1)2 ] + 1350(4.25 × 104 ) s2
+ 1.95(2.3) × 108 (2.6) = 0 or
s4 + 111.329s2 + 1.166 × 104 = 0
This gives or
s2 = −99.43
and
s2 = −11.9
s = ±9.971j
and
s = ±3.45j
These correspond to frequencies of 1.587 Hz and 0.549 Hz.
x 3.95 × 103 5.411 k1 L1 − k2 L2 A1 = = = = 2 2 2 4 A2 θ ms + k1 + k2 730s + 4.25 × 10 s + 58.2192 For mode 1 (s2 = −99.43), x = −0.131 m ahead of the mass center θ For mode 2 (s2 = −11.9), x = 0.1168 m behind the mass center θ
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12.29 Referring to Figure 12.4.4, we are given that m1 g = 1000 lb, c1 = 0, and k2 = 1300 lb/in. The ride rate should be ke =
m1 g 1000 = = 102 ∆ 9.8
lb/in
Thus the suspension stiffness should be k1 =
ke k2 102(1300) = = 111 lb/in k2 − ke 1300 − 102
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12.30 Define the following (refer to Figure 12.4.2): k1 = rear quartercar suspension stiffness. k2 = front quartercar suspension stiffness. kr = total rear suspension stiffness = 2k1 . kf = total front suspension stiffness = 2k2 . ker = total rear ride rate (including suspension and tire stiffness). kef = total front ride rate (including suspension and tire stiffness). ke1 =quartercar rear ride rate. ke2 =quartercar front ride rate. kt = individual tire stiffness. Referring to the guidelines on page 851, and using equations (4) and (5) of Example 12.4.3 as approximations for the bounce and pitch dynamics, we have kef = 0.7ker 1 ωbounce = 2π 2π ωpitch 1 = 2π 2π
%
%
(1)
kf + kr ≤ 1.3 Hz m
(2)
kf L21 + kr L22 ≤ 1.3 Hz IG
ke1 =
(m/4)g ∆1
(4a)
ke2 =
(m/4)g ∆2
(4b)
(3)
From the above definitions, kef = 2ke2
ker = 2ke1
(5)
Because the tire stiffness is in parallel with the suspension stiffness, ke1 =
k1 kt k1 + kt
ke2 =
k2 kt k2 + kt
(6)
These relations must be satisfied by k1 , k2 , and kt . Using the given values m = 4800/32.2, IG = 1800, L1 = 3.5, and L2 = 2.5, these equations can be rearranged as follows: $ 0.02019 k1 + k2 ≤ 1.3 (7) $
0.0375 24.5k1 + 12.5k2 ≤ 1.3 k1 =
k2 =
1200 ∆ 1 kt kt − 1200 ∆1
1200 ∆ 2 kt kt − 1200 ∆2
(8)
(9) (10)
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∆1 = 0.7∆2
(11)
The procedure is to select suitable values for kt and ∆2 , solve (9) and (10) for k1 and k2 , and see if (7) and (8) are satisfied. Trying ∆2 = 9.8/12 ft, as suggested on page 851, and kt = 1200(12) = 14400 lb/ft, we obtain from (9) and (10) k1 = 2457 lb/ft and k2 = 1636 lb/ft. With these values, the lefthand sides of (7) and (8) are 1.29 and 1.07. Thus the requirements have been met.
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