CHAPTER 1 1.1 to 1.41  part of text
1.42
(a) Periodic: Fundamental period = 0.5s (b) Nonperiodic (c) Periodic Fundame...
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CHAPTER 1 1.1 to 1.41  part of text
1.42
(a) Periodic: Fundamental period = 0.5s (b) Nonperiodic (c) Periodic Fundamental period = 3s (d) Periodic Fundamental period = 2 samples (e) Nonperiodic (f) Periodic: Fundamental period = 10 samples (g) Nonperiodic (h) Nonperiodic (i) Periodic: Fundamental period = 1 sample
l.43
π 2 y ( t ) = 3 cos 200t +  6 2 π = 9 cos 200t +  6
9 π =  cos 400t +  1 2 3 9 (a) DC component = 2 9 π (b) Sinusoidal component =  cos 400t +  2 3 9 Amplitude = 2
1
200 Fundamental frequency =  Hz π
1.44
The RMS value of sinusoidal x(t) is A ⁄ 2 . Hence, the average power of x(t) in a 1ohm 2
resistor is ( A ⁄ 2 ) = A2/2.
1.45
Let N denote the fundamental period of x[N]. which is defined by 2π N = Ω The average power of x[n] is therefore N 1
1 2 P =  ∑ x [ n ] N 1 = N
n=0 N 1
∑A
n=0 2 N 1
A = N
1.46
2
2 2πn cos  + φ N
∑ cos
n=0
2
2πn  + φ N
The energy of the raised cosine pulse is E =
π⁄ω
1
∫–π ⁄ ω 4 ( cos ( ωt ) + 1 )
2
dt
1 π⁄ω 2 =  ∫ ( cos ( ωt ) + 2 cos ( ωt ) + 1 ) dt 2 0 1 π ⁄ ω 1 1  cos ( 2ωt ) +  + 2 cos ( ωt ) + 1 dt =  ∫ 2 0 2 2 1 3 π =    = 3π ⁄ 4ω 2 2 ω
1.47
The signal x(t) is even; its total energy is therefore 5 2
E = 2 ∫ x ( t ) dt 0
2
4
5
= 2 ∫ ( 1 ) dt + 2 ∫ ( 5 – t ) dt 2
0
2
4
5
1 4 3 = 2 [ t ] t=0 + 2 –  ( 5 – t ) 3
t=4
2 26 = 8 +  = 3 3
1.48
(a) The differentiator output is 1 y ( t ) = –1 0
for – 5 < t < – 4 for 4 < t < 5 otherwise
(b) The energy of y(t) is E =
–4
∫–5
5
( 1 ) dt + ∫ ( – 1 ) dt 2
2
4
= 1+1 = 2
1.49
The output of the integrator is t
y ( t ) = A ∫ τ dτ = At
for 0 ≤ t ≤ T
0
Hence the energy of y(t) is E =
1.50
T
∫0
2 3
A T A t dt = 3 2 2
(a) x(5t) 1.0
1
0.8
(b)
0
0.8
1
t
25
t
x(0.2t) 1.0
25
20
0
20
3
1.51 x(10t  5)
1.0
0
1.52
0.1
0.5
0.9
1.0
t
(a) x(t) 1 1
1
2
t
3
1 y(t  1)
1
1
2
t
3
1 x(t)y(t  1) 1 1 1
2
t
3
1
4
1.52
(b) x(t + 1) x(t  1) 1
1 1
1
2
3
4
t
1 y(t)y(t) 1 2
1
1
2
3
4
3
4
t
1 x(t  1)y(t) 1 t 2
1
1
2
1
1.52
(c)
2 1
1
2
3
3
4
3
4
t
1
2
1 1
2
t
x(t + 1)y(t  2)
2
1
1
2
5
t
1.52
(d) x(t) 1
3
2
1
1
2
t
3
1
y(1/2t + 1) 6
5
4 3
2
1 1
2
4
6
t
1.0
x(t  1)y(t) 1 t 3
1.52
2
1 1
1
2
3
(e) x(t) 1 4
3
2
1 1
2
3
t
1
y(2  t)
1 4
3
2
2
3
t
1
x(t)y(2  t) 1 1
2
3
1
6
t
1.52
(f) x(t) 1
2
1
1
t
2
1
y(t/2 + 1) 1.0
5
3
2
1
6
1
1
2
t
3
1.0
x(2t)y(1/2t + 1) +1 0.5 1
1
2
t
1
1.52
(g) x(4  t) 1 7
6
5 4
3
t
2 1
y(t)
2
1
1
2
4
t
x(4  t)y(t) = 0
3
2
1
1
2
3
7
t
1.53
We may represent x(t) as the superposition of 4 rectangular pulses as follows: g1(t) 1 1
2
11
2
3
4
t
g2(t) 1 3
4
t
g3(t) 1 1
2
3
4
t
g4(t) 1 1
2
3
4
t
0
To generate g1(t) from the prescribed g(t), we let g 1 ( t ) = g ( at – b ) where a and b are to be determined. The width of pulse g(t) is 2, whereas the width of pulse g1(t) is 4. We therefore need to expand g(t) by a factor of 2, which, in turn, requires that we choose 1 a = 2 The midpoint of g(t) is at t = 0, whereas the midpoint of g1(t) is at t = 2. Hence, we must choose b to satisfy the condition a(2) – b = 0 or 1 b = 2a = 2  = 1 2 1 Hence, g 1 ( t ) = g  t – 1 2 Proceeding in a similar manner, we find that 2 5 g 2 ( t ) = g  t –  3 3 g3 ( t ) = g ( t – 3 ) g 4 ( t ) = g ( 2t – 7 ) Accordingly, we may express the staircase signal x(t) in terms of the rectangular pulse g(t) as follows:
8
2 5 1 x ( t ) = g  t – 1 + g  t –  + g ( t – 3 ) + g ( 2t – 7 ) 3 3 2
1.54
(a) x(t) = u(t)  u(t  2)
0
1
t
2
(b) x(t) = u(t + 1)  2u(t) + u(t  1) 2
0
1
2
1
t
3 1
(c) x(t) = u(t + 3) + 2u(t +1) 2u(t  1) + u(t  3) 1
3
2
3
t
0 1
(d) x(t) = r(t + 1)  r(t) + r(t  2) 1 2
1
0
1
2
t
3
(e) x(t) = r(t + 2)  r(t + 1)  r(t  1)+ r(t  2) 1
3
2
1
0
1
t
2
9
1.55
We may generate x(t) as the superposition of 3 rectangular pulses as follows: g1(t) 1 4
2
0
2
4
2
4
t
g2(t) 1 4
2
0
t
g3(t) 1 4
2
0
2
4
t
All three pulses, g1(t), g2(t), and g3(t), are symmetrically positioned around the origin: 1. g1(t) is exactly the same as g(t). 2. g2(t) is an expanded version of g(t) by a factor of 3. 3. g3(t) is an expanded version of g(t) by a factor of 4. Hence, it follows that g1 ( t ) = g ( t ) 1 g 2 ( t ) = g  t 3 1 g 3 ( t ) = g  t 4 That is, 1 1 x ( t ) = g ( t ) + g  t + g  t 3 4 1.56
(a) x[2n]
o 2
o o 0
1
1
n
(b) x[3n  1] o
2 o1 o 1
0
1
10
n
1.56
(c) y[1  n] o1
o
o
o
o
4
3
2
1 0 1
o 1
2
3
4
5
o
o
o
o
2
3
4
5
o
o
o
o
n
(d) y[2  2n] o
o
3
2
o
o1 o 1
1 1
n
(e) x[n  2] + y[n + 2] o
o
4 o
o3 o
2 o o
1 7 o
6 o
5 o
4 o
3
o 2
1
0
1
2
3
4
o
o
o
5
6
7
8
o
o
o
5
6
7
n
o
(f) x[2n] + y[n  4] o 5
4
3
1 o
2
2
3
1 o
o
o
o
o 1
11
o
o
o 4
n
1.56
(g) x[n + 2]y[n  2]
o
5
4
3
2 o
1
1
o
n
1
o
o
o
o
o2 3 o
o
(h) x[3  n]y[n] 3
o o
2
o 2
o
o
1 o 3
o
o 1
1
2
o 7
o 8
n
3
4
5
6
3
4 o
5 o
6 o
n
4 o
5 o
6 o
n
(i) x[n] y[n] o
3 o
2 o
o 5
o 4
3
2
1
1 1
o
2
1 o 2
o
3
o
(j) x[n]y[2n] o
3 2 1
o 6
o 5
o 4
2 o
1
1
o
2
3
3 o
1 o 2 3
12
o o
1.56
(k) x[n + 2]y[6n] 3
o
2 o 1 8 o
7 o
6 o
5
4
3 o
o o
1.57
o 2
1
1
o 2
o 3
o 4
o 5
o 6
n
1 2 3
(a) Periodic Fundamental period = 15 samples (b) Periodic Fundamental period = 30 samples (c) Nonperiodic (d) Periodic Fundamental period = 2 samples (e) Nonperiodic (f) Nonperiodic (g) Periodic Fundamental period = 2π seconds (h) Nonperiodic (i) Periodic Fundamental period = 15 samples
1.58
The fundamental period of the sinusoidal signal x[n] is N = 10. Hence the angular frequency of x[n] is 2πm m: integer Ω = N The smallest value of Ω is attained with m = 1. Hence, 2π π Ω =  =  radians/cycle 10 5
13
1.59
The amplitude of complex signal x(t) is defined by 2
2
xR( t ) + xI ( t ) =
2
2
2
2
A cos ( ωt + φ ) + A sin ( ωt + φ ) 2
2
= A cos ( ωt + φ ) + sin ( ωt + φ ) = A 1.60
Real part of x(t) is αt
Re { x ( t ) } = Ae cos ( ωt ) Imaginary part of x(t) is αt
Im { x ( t ) } = Ae sin ( ωt )
1.61
We are given x(t ) =
t ∆ ∆  for –  ≤ t ≤ 2 2 ∆ ∆ 1 for t ≥ 2 ∆ 2 for t < – 2
The waveform of x(t) is as follows x(t)
1 1 2 ∆/2 ∆/2
0
 12
14
t
The output of a differentiator in response to x(t) has the corresponding waveform: y(t) 1 δ(t  1 ) 2 2
1/∆
t
∆/2
0
∆/2
1 δ(t + ∆ ) 2 2
y(t) consists of the following components: 1. Rectangular pulse of duration ∆ and amplitude 1/∆ centred on the origin; the area under this pulse is unity. 2. An impulse of strength 1/2 at t = ∆/2. 3. An impulse of strength 1/2 at t = ∆/2. As the duration ∆ is permitted to approach zero, the impulses (1/2)δ(t∆/2) and (1/2)δ(t+∆/2) coincide and therefore cancel each other. At the same time, the rectangular pulse of unit area (i.e., component 1) approaches a unit impulse at t = 0. We may thus state that in the limit: d lim y ( t ) = lim  x ( t ) ∆→0 ∆ → 0 dt = δ(t )
1.62
We are given a triangular pulse of total duration ∆ and unit area, which is symmetrical about the origin: x(t) 2/∆ slope = 4/∆2
slope = 4/∆2
area = 1
∆/2
∆/2
0
15
t
(a) Applying x(t) to a differentiator, we get an output y(t) depicted as follows: y(t) 4/∆2
area = 2/∆
∆/2
t
∆/2 area = 2/∆ 4/∆2
(b) As the triangular pulse duration ∆ approaches zero, the differentiator output approaches the combination of two impulse functions described as follows: • An impulse of positive infinite strength at t = 0. • An impulse of negative infinite strength at t = 0+. (c) The total area under the differentiator output y(t) is equal to (2/∆) + (2/∆) = 0. In light of the results presented in parts (a), (b), and (c) of this problem, we may now make the following statement: When the unit impulse δ(t) is differentiated with respect to time t, the resulting output consists of a pair of impulses located at t = 0 and t = 0+, whose respective strengths are +∞ and ∞. 1.63
From Fig. P.1.63 we observe the following: x1 ( t ) = x2 ( t ) = x3 ( t ) = x ( t ) x4 ( t ) = y3 ( t ) Hence, we may write y 1 ( t ) = x ( t )x ( t – 1 )
(1)
y2 ( t ) = x ( t )
(2)
y 4 ( t ) = cos ( y 3 ( t ) ) = cos ( 1 + 2x ( t ) )
(3)
The overall system output is y ( t ) = y1 ( t ) + y2 ( t ) – y4 ( t )
(4)
Substituting Eqs. (1) to (3) into (4): y ( t ) = x ( t )x ( t – 1 ) + x ( t ) – cos ( 1 + 2x ( t ) )
(5)
Equation (5) describes the operator H that defines the output y(t) in terms of the input x(t).
16
1.64 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)
1.65
Memoryless ✓ ✓ ✓ x x x ✓ x x ✓ ✓ ✓
Stable ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓
Causal ✓ ✓ ✓ ✓ x ✓ x ✓ x ✓ ✓ ✓
Linear x ✓ x ✓ ✓ ✓ x ✓ ✓ ✓ ✓ x
Timeinvariant ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓
We are given y [ n ] = a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] Let k
S { x(n)} = x(n – k ) We may then rewrite Eq. (1) in the equivalent form 1
2
3
y [ n ] = a0 x [ n ] + a1 S { x [ n ] } + a2 S { x [ n ] } + a3 S { x [ n ] } 1
2
3
= ( a0 + a1 S + a2 S + a3 S ) { x [ n ] } = H { x[n]} where 1
2
H = a0 + a1 S + a2 S + a3 S
.
3
.
(a) Cascade implementation of operator H: x[n]
a0
S
S
.
S
a2
a1
a3
Σ y[n]
17
(1)
(b) Parallel implementation of operator H:
x[n]
1.66
a0
.. .
S1
a1 Σ
S2
a2
S3
a3
y[n]
Using the given inputoutput relation: y [ n ] = a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] we may write y [ n ] = a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] ≤ a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] ≤ a0 M x + a1 M x + a2 M x + a3 M x = ( a 0 + a 1 + a 2 + a 3 )M x where M x = x ( n ) . Hence, provided that Mx is finite, the absolute value of the output will always be finite. This assumes that the coefficients a0, a1, a2, a3 have finite values of their own. It follows therefore that the system described by the operator H of Problem 1.65 is stable.
1.67
The memory of the discretetime described in Problem 1.65 extends 3 time units into the past.
1.68
It is indeed possible for a noncausal system to possess memory. Consider, for example, the system illustrated below:
.
x(n + k)
ak
.
x[n] Sk
x(n  l) Sl
a0
al
Σ y[n] l{x[n]}
= x[n  l], we have the inputoutput relation That is, with S y [ n ] = a0 x [ n ] + ak x [ n + k ] + al x [ n – l ] This system is noncausal by virtue of the term akx[n + k]. The system has memory by virtue of the term alx[n  l].
18
1.69
(a) The operator H relating the output y[n] to the input x[n] is 1
H = 1+S +S where
2
k
S { x[n]} = x[n – k ]
for integer k
(b) The inverse operator Hinvis correspondingly defined by inv 1 H = 1 2 1+S +S Cascade implementation of the operator H is described in Fig. 1. Correspondingly, feedback implementation of the inverse operator Hinvis described in Fig. 2 x[n]
.
.
S
S
Σ
Fig. 1 Operator H
y[n]
y[n] +
Σ
.
S
.
x[n]
S
Fig. 2 Inverse Operator Hinv Figure 2 follows directly from the relation: x[n] = y[n] – x[n – 1] – x[n – 2] 1.70
For the discretetime system (i.e., the operator H) described in Problem 1.65 to be timeinvariant, the following relation must hold n
0 S H = HS where
n0
for integer n0
(1)
n
S 0 { x [ n ] } = x [ n – n0 ] and 1
2
H = 1+S +S We first note that n
n
1
n
n +1
2
S 0H = S 0(1 + S + S ) = S 0+S 0 Next we note that HS
n0
1
+S
2
= (1 + S + S )S
n0 + 2
(2)
n0
19
n
1+n
2+n
0 0 = S 0+S +S (3) From Eqs. (2) and (3) we immediately see that Eq. (1) is indeed satisfied. Hence, the system described in Problem 1.65 is timeinvariant.
1.71
(a) It is indeed possible for a timevariant system to be linear. (b) Consider, for example, the resistancecapacitance circuit where the resistive component is time variant, as described here: R(t)
i(t) v1(t)
+
C
. .
o+
v2(t) o
This circuit, consisting of the series combination of the resistor R(t) and capacitor C, is time variant because of R(t). The input of the circuit, v1(t), is defined in terms of the output v2(t) by dv 2 ( t ) v 1 ( t ) = R ( t )C  + v 2 ( t ) dt Doubling the input v1(t) results in doubling the output v2(t). Hence, the property of homogeneity is satisfied. Moreover, if N
v1 ( t ) =
∑ v1, k ( t ) k=1
then N
v2 ( t ) =
∑ v2, k ( t ) k=1
where dv 2, k ( t ) v 1, k ( t ) = R ( t )C  + v 2, k ( t ) , dt
k = 1,2,...,N
Hence, the property of superposition is also satisfied. We therefore conclude that the timevarying circuit of Fig. P1.71 is indeed linear.
1.72
We are given the pth power law device: p
y(t ) = x (t )
(1)
20
Let y1(t) and y2(t) be the outputs of this system produced by the inputs x1(t) and x2(t), respectively. Let x(t) = x1(t) + x2(t), and let y(t) be the corresponding output. We then note that p for p ≠ 0, 1 y ( t ) = ( x1 ( t ) + x2 ( t ) ) ≠ y1 ( t ) + y2 ( t ) Hence the system described by Eq. (1) is nonlinear.
1.73
Consider a discretetime system described by the operator H1: H 1 : y [ n ] = a0 x [ n ] + ak x [ n – k ] This system is both linear and time invariant. Consider another discretetime system described by the operator H2: H 2 : y [ n ] = b0 x [ n ] + bk x [ n + k ] which is also both linear and time invariant. The system H1 is causal, but the second system H2 is noncausal.
1.74
The system configuration shown in Fig. 1.56(a) is simpler than the system configuration shown in Fig. 1.56(b). They both involve the same number of multipliers and summer. however, Fig. 1.56(b) requires N replicas of the operator H, whereas Fig. 1.56(a) requires a single operator H for its implementation.
1.75
(a) All three systems • have memory because of an integrating action performed on the input, • are causal because (in each case) the output does not appear before the input, and • are timeinvariant. (b) H1 is noncausal because the output appears before the input. The inputoutput relation of H1 is representative of a differentiating action, which by itself is memoryless. However, the duration of the output is twice as long as that of the input. This suggests that H1 may consist of a differentiator in parallel with a storage device, followed by a combiner. On this basis, H1 may be viewed as a timeinvariant system with memory. System H2 is causal because the output does not appear before the input. The duration of the output is longer than that of the input. This suggests that H2 must have memory. It is timeinvariant. System H3 is noncausal because the output appears before the input. Part of the output, extending from t = 1 to t = +1, is due to a differentiating action performed on the input; this action is memoryless. The rectangular pulse, appearing in the output from t = +1 to t = +3, may be due to a pulse generator that is triggered by the termination of the input. On this basis, H3 would have to be viewed as timevarying.
21
Finally, the output of H4 is exactly the same as the input, except for an attenuation by a factor of 1/2. Hence, H4 is a causal, memoryless, and timeinvariant system.
1.76
H1 is representative of an integrator, and therefore has memory. It is causal because the output does not appear before the input. It is timeinvariant. H2 is noncausal because the output appears at t = 0, one time unit before the delayed input at t = +1. It has memory because of the integrating action performed on the input. But, how do we explain the constant level of +1 at the front end of the output, extending from t = 0 to t = +1? Since the system is noncausal, and therefore operating in a non realtime fashion, this constant level of duration 1 time unit may be inserted into the output by artificial means. On this basis, H2 may be viewed as timevarying. H3 is causal because the output does not appear before the input. It has memory because of the integrating action performed on the input from t = 1 to t = 2. The constant level appearing at the back end of the output, from t = 2 to t = 3, may be explained by the presence of a strong device connected in parallel with the integrator. On this basis, H3 is timeinvariant. Consider next the input x(t) depicted in Fig. P1.76b. This input may be decomposed into the sum of two rectangular pulses, as shown here:
xA(t)
x(t) 2
2
1
1 0
1
2
t
xB(t)
+ 0
1
2
2 1
t
0
1
2
t
2
t
Response of H1 to x(t): y1,A(t)
y1,B(t)
2
+
1 0
1
2
t
y1(t)
2
2
1
1 0
1
2
22
t
0
1
Response of H2 to x(t): y2,A(t) 2
2
+
1 1 1
y2(t)
y2,B(t)
1 2
t
0 1
2
0 1
2
t
1
1 0
1
t
1
2
2
The rectangular pulse of unit amplitude and unit duration at the front end of y2(t) is inserted in an offline manner by artificial means Response of H3 to x(t): y3,A(t) 2
+
1 0
1.77
1
y3(t)
y3,B(t)
2
3
2
2
1
1
t
0
1
2
3
t
0
1
2
3
t
(a) The response of the LTI discretetime system to the input δ[n1] is as follows: y[n]
o
2
o
1
o
4
2
o
1
1 1
5
o
o
n
3
o
(b) The response of the system to the input 2δ[n]  δ[n  2] is as follows y[n] 4o 3 2
o
1
o 2
1
o
1 0 1 2
4
o 2
3
o
o
5
6
o o
23
n
(c) The input given in Fig. P1.77b may be decomposed into the sum of 3 impulse functions: δ[n + 1], δ[n], and 2δ[n  1]. The response of the system to these three components is given in the following table: Time n
δ[n + 1]
1 0 1 2 3
+2 1 +1
δ[n]
2δ[n  1]
2 +1 1
+4 2 +2
Total response +1 3 +6 3 2
Thus, the total response y[n] of the system is as shown here: y[n]
o
6 5 4 3
o
2
o 1 o
3
2
o
2
0
1
1
4 3
5
6
o
o
o
1 2 3
o
o
Advanced Problems 1.78
(a) The energy of the signal x(t) is defined by E =
∞
∫–∞ x
2
( t ) dt
Substituting x ( t ) = xe ( t ) + xo ( t ) into this formula yields E =
∞
∫–∞ [ xe ( t ) + xo ( t ) ] ∞
2
dt 2
=
∫–∞ [ xe ( t ) + xo ( t ) + 2xe ( t )xo ( t ) ]
=
∫–∞ xe ( t ) dt + ∫–∞ xo ( t ) dt + 2 ∫–∞ xe ( t )xo ( t ) dt
∞
2
2
2
∞
2
∞
24
dt (1)
With xe(t) even and xo(t) odd, it follows that the product xe(t)xo(t) is odd, as shown by x e ( – t )x o ( – t ) = x e ( t ) [ – x o ( t ) ] = – x e ( t )x o ( t ) Hence, ∞
∫–∞
( x e ( t )x o ( t ) ) dt =
0
∫–∞
∞
x e ( t )x o ( t ) dt + ∫ x e ( t )x o ( t ) dt 0
∞
0
∫–∞ ( – xe ( t )xo ( t ) ) ( –dt ) + ∫0 ( xe ( t )xo ( t ) ) dt
=
∞
∞
0
0
= – ∫ x e ( t )x o ( t ) dt + ∫ x e ( t )x o ( t ) dt = 0 Accordingly, Eq. (1) reduces to E =
∞
∫–∞
∞
x e ( t ) dt + ∫ 2
2
–∞
x o ( t ) dt
(b) For a discretetime signal x[n], – ∞ ≤ n ≤ ∞ , we may similarly write ∞
E =
∑x
2
[n]
n=∞ ∞
=
∑ [ xe [ n ] + xo [ n ] ]
n=∞ ∞
=
∑
∞
2 xe [ n ]
∑
+
n=∞
2
2 2 xo [ n ]
∞
+2
n=∞
∑ xe [ n ]xo [ n ] n=∞
With x e [ – n ]x o [ – n ] = – x e [ n ]x o [ n ] it follows that ∞
∑ xe [ n ]xo [ n ]
=
n=∞
=
0
∞
n=∞ 0
n=0
∑ xe [ n ]xo [ n ] + ∑ xe [ n ]xo [ n ] ∞
∑ xe [ –n ]xo [ –n ] + ∑ xe [ n ]xo [ n ]
n=∞ 0
= – ∑ x e [ n ]x o [ n ] + n=∞
n=0 ∞
∑ xe [ n ]xo [ n ] n=0
= 0
25
(2)
Accordingly, Eq. (2) reduces to ∞
E =
∑
∞
2 xe [ n ]
+
n=∞
1.79
∑ xo [ n ] 2
n=∞
(a) From Fig. P1.79, i ( t ) = i1 ( t ) + i2 ( t )
(1)
di 1 ( t ) 1 t L  + Ri 1 ( t ) =  ∫ i 2 ( τ ) dτ dt C –∞
(2)
Differentiating Eq. (2) with respect to time t: 2
di 1 ( t ) d i1 ( t ) 1  =  i 2 ( t ) + R L 2 dt C dt
(3)
Eliminating i2(t) between Eqs. (1) and (2): 2
d i1 ( t ) di 1 ( t ) 1  =  [ i ( t ) – i 1 ( t ) ] L + R 2 dt C dt Rearranging terms: 2
d i 1 ( t ) R di 1 ( t ) 1 1  +   + i 1 ( t ) = i ( t ) 2 dt LC L LC dt
(4)
(b) Comparing Eqs. (4) with Eq. (1.108) for the MEMS as presented in the text, we may derive the following analogy: MEMS of Fig. 1.64
LRC circuit of Fig. P1.79
y(t)
i1(t)
ωn
1 ⁄ LC
Q
ωn L 1 L  =  R C R
x(t)
1 i ( t ) LC
26
1.80
(a) As the pulse duration ∆ approaches zero, the area under the pulse x∆(t) remains equal to unity, and the amplitude of the pulse approaches infinity. (b) The limiting form of the pulse x∆(t) violates the evenfunction property of the unit impulse: δ ( –t ) = δ ( t )
1.81
The output y(t) is related to the input x(t) as y(t ) = H { x(t )}
(1)
Let T0 denote the fundamental period of x(t), assumed to be periodic. Then, by definition, x(t ) = x(t + T 0)
(2)
Substituting t +T0 for t into Eq. (1) and then using Eq. (2), we may write y(t + T 0) = H { x(t + T 0)} = H { x(t )} = y(t )
(3)
Hence, the output y(t) is also periodic with the same period T0.
1.82
(a) For 0 ≤ t < ∞ , we have 1 –t ⁄ τ x ∆ ( t ) = e ∆ At t = ∆/2, we have A = x∆ ( ∆ ⁄ 2 ) 1 –∆ ⁄ ( 2τ ) = e ∆ Since x∆(t) is even, then 1 –∆ ⁄ ( 2τ ) A = x ∆ ( ∆ ⁄ 2 ) = x ∆ ( – ∆ ⁄ 2 ) = e ∆ (b) The area under the pulse x∆(t) must equal unity for δ ( t ) = lim x ∆ ( t ) ∆→0
The area under x∆(t) is
27
∞
∫–∞
∞
x ∆ ( t ) dt = 2 ∫ x ∆ ( t ) dt 0
∞ 1 –t ⁄ τ = 2 ∫ e dt 0 ∆ 2 –t ⁄ τ ∞ =  ( – τ )e 0 ∆ 2τ = ∆
For this area to equal unity, we require ∆ τ = 2 (c) 8
∆=1 ∆ = 0.5 ∆ = 0.25 ∆ = 0.125
7
6
Amplitude
5
4
3
2
1
0 −2
1.83
−1.5
−1
−0.5
0 Time
0.5
1
1.5
2
(a) Let the integral of a continuoustime signal x(t), – ∞ < t < ∞ , be defined by y(t ) = =
t
∫–∞ x ( τ ) dτ 0
t
∫–∞ x ( t ) dt + ∫0 x ( τ ) dτ
28
The definite integral
0
∫–∞ x ( t ) dt , representing the initial condition, is a constant.
With differentiation as the operation of interest, we may also write dy ( t ) x ( t ) = dt Clearly, the value of x(t) is unaffected by the value assumed by the initial condition 0
∫–∞ x ( t ) dt It would therefore be wrong to say that differentiation and integration are the inverse of each other. To illustrate the meaning of this statement, consider the two following two waveforms that differ from each other by a constant value for – ∞ < t < ∞ : x2(t)
x1(t) slope = a
0
slope = a
t
0
d x1 ( t ) d x2 ( t ) = Yet, y ( t ) = , as illustrated below: dt dt y(t)
a
0
t
(b) For Fig. P1.83(a): R t y ( t ) +  ∫ y ( τ ) dτ = x ( t ) L –∞ For R/L large, we approximately have R t  ∫ y ( τ ) dτ ≈ x ( t ) L –∞ Equivalently, we have a differentiator described by L dx ( t ) R  large y ( t ) ≈   , R dt L
29
t
For Fig. P1.83(b): L dy ( t ) y ( t ) +   = x ( t ) R dt For R/L small, we approximately have L dy ( t )   ≈ x ( t ) R dt Equivalently, we have an integrator described by R R t  small y ( t ) ≈  ∫ x ( τ ) dτ L L –∞ (c) Consider the following two scenarios describing the LR circuits of Fig. P1.83 • The input x(t) consists of a voltage source with an average value equal to zero. • The input x(t) includes a dc component E (exemplified by a battery). These are two different input conditions. Yet for large R/L, the differentiator of Fig. P1.83(a) produces the same output. On the other hand, for small R/L the integrator of Fig. P1.83(b) produces different outputs. Clearly, on this basis it would be wrong to say that these two LR circuits are the inverse of each other.
1.84
(a) The output y(t) is defined by y ( t ) = A 0 cos ( ω 0 t + φ )x ( t )
(1)
This inputoutput relation satisfies the following two conditions: • •
Homogeneity: If the input x(t) is scaled by an arbitrary factor a, the output y(t) will be scaled by the same factor. Superposition: If the input x(t) consists of two additive components x1(t) and x2(t), then y ( t ) = y1 ( t ) + y2 ( t ) where k = 1,2 y ( t ) = A 0 cos ( ω 0 t + φ )x k ( t ) ,
Hence, the system of Fig. P1.84 is linear. (b) For the impulse input x(t ) = δ(t ) , Eq. (1) yields the corresponding output y′ ( t ) = A 0 cos ( ω 0 t + φ )δ ( t )
30
A cos φδ ( 0 ), = 0 0,
t=0 otherwise
For x ( t ) = δ ( t – t 0 ) , Eq. (1) yields y″ ( t ) = A 0 cos ( ω 0 t + φ )δ ( t – t 0 ) A cos ( ω 0 t 0 + φ )δ ( 0 ), = 0 0,
t = t0 otherwise
Recognizing that y′ ( t ) ≠ y″ ( t ) , the system of Fig. P1.84 is timevariant.
1.85
(a) The output y(t) is related to the input x(t) as t y ( t ) = cos 2π f c t + k ∫ x ( τ ) dτ –∞
(1)
The output is nonlinear as the system violates both the homogeneity and superposition properties: •
Let x(t) be scaled by the factor a. The corresponding value of the output is t y a ( t ) = cos 2π f c t + ka ∫ x ( τ ) dτ –∞
For a ≠ 1 , we clearly see that y a ( t ) ≠ y ( t ) . •
Let x ( t ) = x1 ( t ) + x2 ( t ) Then t t y ( t ) = cos 2π f c t + k ∫ x 1 ( τ ) dτ + k ∫ x 2 ( τ ) dτ –∞ –∞
≠ y1 ( t ) + y2 ( t ) where y1(t) and y2(t) are the values of y(t) corresponding to x1(t) and x2(t), respectively. (b) For the impulse input x(t ) = δ(t ) , Eq. (1) yields t y′ ( t ) = cos 2π f c t + k ∫ δ ( τ ) dτ –∞
31
= cos k, t = 0
+
For the delayed impulse input x ( t ) = δ ( t – t 0 ) , Eq. (1) yields t y″ ( t ) = cos 2π f c t + k ∫ δ ( τ – t 0 ) dτ –∞ +
= cos ( 2π f c t 0 + k ) ,
t = t0
Recognizing that y′ ( t ) ≠ y″ ( t ) , it follows that the system is timevariant. 1.86
For the squarelaw device 2
y(t ) = x (t ) , the input x ( t ) = A 1 cos ( ω 1 t + φ 1 ) + A 2 cos ( ω 2 t + φ 2 ) yields the output 2
y(t ) = x (t ) = [ A 1 cos ( ω 1 t + φ 1 ) + A 2 cos ( ω 2 t + φ 2 ) ] 2
2
2
= A 1 cos 2 ( ω 1 t + φ 1 ) + A 2 cos 2 ( ω 2 t + φ 2 ) + 2 A 1 A 2 cos ( ω 1 t + φ 1 ) cos ( ω 2 t + φ 2 ) 2
A1 =  [ 1 + cos ( 2ω 1 t + 2φ 1 ) ] 2 2
A2 +  [ 1 + cos ( 2ω 2 t + 2φ 2 ) ] 2 + A 1 A 2 [ cos ( ( ω 1 + ω 2 )t + ( φ 1 + φ 2 ) ) + cos ( ( ω 1 – ω 2 )t + ( φ 1 – φ 2 ) ) ] The output y(t) contains the following components:
1.87
2
2
•
DC component of amplitude ( A 1 + A 2 ) ⁄ 2
•
Sinusoidal component of frequency 2ω1, amplitude A 1 ⁄ 2 , and phase 2φ1
• • •
Sinusoidal component of frequency 2ω2, amplitude A 2 ⁄ 2 , and phase 2φ2 Sinusoidal component of frequency (ω1  ω2), amplitude A1A2, and phase (φ1  φ2) Sinusoidal component of frequency (ω1 + ω2), amplitude A1A2, and phase (φ1 + φ2)
2 2
The cubiclaw device 3
y(t ) = x (t ) , in response to the input,
32
x ( t ) = A cos ( ωt + φ ) , produces the output 3
3
y ( t ) = A cos ( ωt + φ ) 1 3 3 = A cos ( ωt + φ ) ⋅  ( cos ( ( 2ωt + 2φ ) + 1 ) ) 2 3
A =  [ cos ( ( 2ωt + 2φ ) + ( ωt + φ ) ) 2 3
A + cos ( ( 2ωt + 2φ ) – ( ωt + φ ) ) ] +  cos ( ωt + φ ) 2 3
3
A A =  [ cos ( 3ωt + 3φ ) + cos ( ωt + φ ) ] +  cos ( ωt + φ ) 2 2 3
A 3 =  cos ( 3ωt + 3φ ) + A cos ( ωt + φ ) 2 The output y(t) consists of two components: • •
Sinusoidal component of frequency ω, amplitude A3 and phase φ Sinusoidal component of frequency 3ω, amplitude A3/2, and phase 3φ
To extract the component with frequency 3ω (i.e., the third harmonic), we need to use a bandpass filter centered on 3ω and a passband narrow enough to suppress the fundamental component of frequency ω. From the analysis presented here we infer that, in order to generate the pth harmonic in response to a sinusoidal component of frequency ω, we require the use of two subsystems: •
Nonlinear device defined by
•
p = 2,3,4,... y(t ) = x (t ) , Narrowband filter centered on the frequency pω.
p
1.88
(a) Following the solution to Example 1.21, we start with the pair of inputs: 1 ∆ x 1 ( t ) = u t +  ∆ 2 1 ∆ x 2 ( t ) = u t –  ∆ 2 The corresponding outputs are respectively given by
33
1 y 1 ( t ) =  1 – e ∆ 1 y 2 ( t ) =  1 – e ∆
∆ – α t +  2
∆ – α t –  2
∆ ∆ cos ω n t +  u t +  2 2 ∆ ∆ cos ω n t –  u t –  2 2
The response to the input x∆ ( t ) = x1 ( t ) – x2 ( t ) is given by 1 ∆ ∆ y ∆ ( t ) =  u t +  – u t –  ∆ 2 2 –α t + ∆ ω n ∆ ∆ 2 1 –  e cos ω n t +  u t +  ∆ 2 2
–e
∆ – α t –  2
ω n ∆ ∆ cos ω n t –  u t –  2 2
As ∆ → 0, x ∆ ( t ) → δ ( t ) . We also note that 1 d ∆ ∆  z ( t ) = lim  z t +  – z t –  dt 2 2 ∆ → 0 ∆ Hence, with z ( t ) = e system is y ( t ) = lim y ∆ ( t )
– αt
cos ( ω n t )u ( t ) , we find that the impulse response of the
∆→0
d –αt = δ ( t ) –  [ e cos ( ω n t )u ( t ) ] dt d d –αt – αt = δ ( t ) –  [ e cos ( ω n t ) ] ⋅ u ( t ) – [ e cos ( ω n t ) ]  u ( t ) dt dt = δ ( t ) – [ – αe –e
– αt
– αt
cos ( ω n t ) – ω n e
cos ( ω n t )δ ( t )
– αt
sin ( ω n t ) ]u ( t ) (1)
34
Since e
– αt
cos ( ω n t ) = 1 at t = 0, Eq. (1) reduces to
y ( t ) = [ αe
– αt
cos ( ω n t ) + ω n e
– αt
sin ( ω n t ) ]u ( t )
(b) ω n = jα n where α n < α Using Euler’s formula, we can write – jω t
jω t
–α t
α t
n +e n e n +e n e cos ( ω n t ) =  = 2 2
The step response can therefore be rewritten as – ( α – α n )t 1 –( α + αn )t y ( t ) = 1 –  ( e +e ) u(t ) 2
Again, the impulse response in this case can be obtained as dy ( t ) h ( t ) =  = dt
– ( α – α n )t 1 –( α + αn )t 1 –  ( e +e ) δ(t ) 2
=0 α + α n –( α + αn )t α – α n –( α – αn )t +  e +  e u(t ) 2 2 α 2 –α t α 1 –α t =  e 2 +  e 1 u ( t ) 2 2 where α1 = α  αn and α2 = α + αn.
1.89
Building on the solution described in Fig. 1.69, we may relabel Fig. P1.89 as follows x[n] + +
Σ
.
y′[n]
0.5
S
.
+
Σ
y[n]
+ 0.5
where (see Eq. (1.117)) ∞
y′ [ n ] = x [ n ] +
∑ 0.5
k
x[n – k ]
k=1
and y [ n ] = y′ [ n ] + 0.5y′ [ n – 1 ]
35
(2)
∞
∑ 0.5
= x[n] +
∞
k
x [ n – k ] + 0.5x [ n – 1 ] +
k=1 ∞
=
∑ 0.5
k
x[n – 1 – k ]
k=1 ∞
k
x [ n – k ] + 0.5 ∑ 0.5 x [ n – 1 – k ]
k=0
1.90
∑ 0.5
k
k=0
According to Eq. (1.108) the MEMS accelerometer is described by the secondorder equation 2 2 d y ( t ) ω n dy ( t )  +   + ω n y ( t ) = x ( t ) (1) 2 Q dt dt Next, we use the approximation (assuming that Ts is sufficiently small) T T 1 d  y ( t ) ≈  y t + s – y t – s Ts dt 2 2
(2)
Applying this approximation a second time: 2 d y ( t ) 1 d T s t – T y t ≈ + – y s 2 T s dt 2 2 dt T T 1 d 1 d =   y t + s –   y t – s T s dt 2 T s dt 2 11 ≈   [ y ( t + T s ) – y ( t ) ] T sT s 11 –   [ y ( t ) – y ( t – T s ) ] T sT s 1 = 2 [ y ( t + T s ) – 2y ( t ) + y ( t – T s ) ] Ts
(3)
Substituting Eqs. (2) and (3) into (1): 2
ω n T s T 1 2 2 [ y ( t + T s ) – 2y ( t ) + y ( t – T s ) ] +  y t +  – y t – s + ω n y ( t ) = x ( t ) QT 2 2 s Ts Define 2
ωn T s  = a1 , Q 2 2
ωn T s – 2 = a2 ,
36
(4)
1 2 = b 0 , Ts t = nT s ⁄ 2 , y [ n ] = y ( nT s ⁄ 2 ) , where, in effect, continuous time is normalized with respect to Ts/2 to get n. We may then rewrite Eq. (4) in the form of a noncausal difference equation: y [ n + 2 ] + a1 y [ n + 1 ] + a2 y [ n ] – a1 y [ n – 1 ] + y [ n – 2 ] = b0 x [ n ]
(5)
Note: The difference equation (5) is of order 4, providing an approximate description of a secondorder continuoustime system. This doubling in order is traced to Eq. (2) as the approximation for a derivative of order 1. We may avoid the need for this order doubling by adopting the alternative approximation: d 1  y ( t ) ≈  [ y ( t + T s ) – y ( t ) ] dt Ts However, in general, for a given sampling period Ts, this approximation may not be as accurate as that defined in Eq. (2).
1.91
Integration is preferred over differentiation for two reasons: (i) Integration tends to attenuate high frequencies. Recognizing that noise contains a broad band of frequencies, integration has a smoothing effect on receiver noise. (ii) Differentiation tends to accentuate high frequencies. Correspondingly, differentiation has the opposite effect to integration on receiver noise.
1.92
From Fig. P1.92, we have i ( t ) = i1 ( t ) + i2 ( t ) 1 t v ( t ) = Ri 2 ( t ) =  ∫ i 1 ( τ ) dτ C –∞ This pair of equations may be rewritten in the equivalent form: 1 i 1 ( t ) = i ( t ) – v ( t ) R 1 t v ( t ) =  ∫ i 1 ( τ ) dτ C –∞
37
Correspondingly, we may represent the parallel RC circuit of Fig. P1.92 by the block diagram i(t) +
Σ
i1(t)
1 t  ∫ dτ C –∞

.
v(t)
1 R
The system described herein is a feedback system with the capacitance C providing the forward path and the conductance 1/R providing the feedback path.
%Solution to Matlab Experiment 1.93 f = 20; k = 0:0.0001:5/20; amp = 5; duty = 60; %Make Square Wave y1 = amp * square(2*pi*f*k,duty); %Make Sawtooth Wave y2 = amp * sawtooth(2*pi*f*k); %Plot Results figure(1); clf; subplot(2,1,1) plot(k,y1) xlabel(’time (sec)’) ylabel(’Voltage’) title(’Square Wave’) axis([0 5/20 6 6]) subplot(2,1,2) plot(k,y2) xlabel(’time (sec)’) ylabel(’Voltage’) title(’Sawtooth Wave’)
38
axis([0 5/20 6 6]) Square Wave 6 4
Voltage
2 0 −2 −4 −6
0
0.05
0.1
0.15
0.2
0.25
0.2
0.25
time (sec) Sawtooth Wave 6 4
Voltage
2 0 −2 −4 −6
0
0.05
0.1
0.15 time (sec)
% Solution to Matlab Experiment 1.94 t = 0:0.01:5; x1 = 10*exp(t)  5*exp(0.5*t); x2 = 10*exp(t) + 5*exp(0.5*t); %Plot Figures figure(1); clf; subplot(2,1,1) plot(t,x1) xlabel(’time (sec)’) ylabel(’Amplitude’) title(’x(t) = e^{t}  e^{0.5t}’) subplot(2,1,2); plot(t,x2) xlabel(’time (sec)’) ylabel(’Amplitude’) title(’x(t) = e^{t} + e^{0.5t}’)
39
x(t) = e−t − e−0.5t 5
Amplitude
4 3 2 1 0 −1
0
0.5
1
1.5
2
2.5 time (sec)
3
3.5
4
4.5
5
3
3.5
4
4.5
5
x(t) = e−t + e−0.5t
Amplitude
15
10
5
0
0
0.5
1
1.5
2
2.5 time (sec)
% Solution to Matlab Experiment 1.95 t = (2:0.01:2)/1000; a1 = 500; x1 = 20 * sin(2*pi*1000*t  pi/3) .* exp(a1*t); a2 = 750; x2 = 20 * sin(2*pi*1000*t  pi/3) .* exp(a2*t); a3 = 1000; x3 = 20 * sin(2*pi*1000*t  pi/3) .* exp(a3*t); %Plot Resutls figure(1); clf; plot(t,x1,’b’); hold on plot(t,x2,’k:’); plot(t,x3,’r’); hold off xlabel(’time (sec)’)
40
ylabel(’Amplitude’) title(’Exponentially Damped Sinusoid’) axis([2/1000 2/1000 120 120]) legend(’a = 500’, ’a = 750’, ’a = 1000’)
Exponentially Damped Sinusoid a = 500 a = 750 a = 1000
100
Amplitude
50
0
−50
−100
−2
−1.5
−1
−0.5
0 time (sec)
% Solution to Matlab Experiment 1.96 F = 0.1; n = 1/(2*F):0.001:1/(2*F); w = cos(2*pi*F*n); %Plot results figure(1); clf; plot(n,w) xlabel(’Time (sec)’) ylabel(’Amplitude’) title(’Raised Cosine Filter’)
41
0.5
1
1.5
2 −3
x 10
Raised Cosine Filter 1
0.8
0.6
0.4
Amplitude
0.2
0
−0.2
−0.4
−0.6
−0.8
−1 −5
−4
−3
−2
−1
0 Time (sec)
% Solution to Matlab Experiment 1.97 t = 2:0.001:10; %Generate first step function x1 = zeros(size(t)); x1(t>0)=10; x1(t>5)=0; %Generate shifted function delay = 1.5; x2 = zeros(size(t)); x2(t>(0+delay))=10; x2(t>(5+delay))=0;
42
1
2
3
4
5
%Plot data figure(1); clf; plot(t,x1,’b’) hold on plot(t,x2,’r:’) xlabel(’Time (sec)’) ylabel(’Amplitude’) title(’Rectangular Pulse’) axis([2 10 1 11]) legend(’Zero Delay’, ’Delay = 1.5’); Rectangular Pulse Zero Delay Delay = 1.5
10
8
Amplitude
6
4
2
0
−2
0
2
4 Time (sec)
43
6
8
10
Solutions to Additional Problems 2.32. A discretetime LTI system has the impulse response h[n] depicted in Fig. P2.32 (a). Use linearity and time invariance to determine the system output y[n] if the input x[n] is Use the fact that:
δ[n − k] ∗ h[n] = h[n − k] (ax1 [n] + bx2 [n]) ∗ h[n] = ax1 [n] ∗ h[n] + bx2 [n] ∗ h[n]
(a) x[n] = 3δ[n] − 2δ[n − 1]
y[n]
=
3h[n] − 2h[n − 1]
=
3δ[n + 1] + 7δ[n] − 7δ[n − 2] + 5δ[n − 3] − 2δ[n − 4]
(b) x[n] = u[n + 1] − u[n − 3]
x[n] = δ[n] + δ[n − 1] + δ[n − 2] y[n] = h[n] + h[n − 1] + h[n − 2] = δ[n + 1] + 4δ[n] + 6δ[n − 1] + 4δ[n − 2] + 2δ[n − 3] + δ[n − 5]
(c) x[n] as given in Fig. P2.32 (b) . x[n]
=
2δ[n − 3] + 2δ[n] − δ[n + 2]
y[n]
=
2h[n − 3] + 2h[n] − h[n + 2]
= −δ[n + 3] − 3δ[n + 2] + 7δ[n] + 3δ[n − 1] + 8δ[n − 3] + 4δ[n − 4] − 2δ[n − 5] + 2δ[n − 6]
2.33. Evaluate the discretetime convolution sums given below. (a) y[n] = u[n + 3] ∗ u[n − 3] Let u[n + 3] = x[n] and u[n − 3] = h[n]
1
x[k]
h[n−k]
...
... k
−3 −2 −1
n−3
k
Figure P2.33. (a) Graph of x[k] and h[n − k]
for n − 3 < −3
n 0
k=−J
we also have As J increases: ,
1 T
T
x(t)2 dt > 0 0
J
2
X[k] increases or remains constant and min M SEJ decreases.
k=−J
3.100. Generalized Fourier Series. The concept of the Fourier Series may be generalized to sums of signals other than complex sinusoids. That is, given a signal x(t) we may approximate x(t) on an interval [t1 , t2 ] as a weighted sum of N functions φ0 (t), φ2 (t), . . . , φN −1 (t) x(t) ≈
N −1
ck φk (t)
k=0
We shall assume that these N functions are mutually orthogonal on [t1 , t2 ]. This means that
t2 0, k = l, ∗ φk (t)φl (t)dt = fk , k = l t1 The M SE in using this approximation is 1 M SE = t 2 − t1
t2
t1
)2 ) N ) ) ) ) ck φk (t)) dt )x(t) − ) ) k=1
(a) Show that the M SE is minimized by choosing ck = lined in Problem 3.99 (a)  (d) to this problem. )2 ) N ) ) ) ) ck φk (t)) )x(t) − ) )
= x(t)2 − x(t)
k=1
1 fk
N −1
t2 t1
x(t)φ∗k (t)dt. Hint: Generalize steps out
c∗k φ∗k (t) − x∗ (t)
k=0
+
−1 N −1 N
N
ck φk (t)
k=0
ck c∗m φk (t)φ∗m (t)
m=0 k=0
Thus:
t2 1 ∗ x(t) dt − x(t)φk (t)dt M SE = t2 − t1 t1 t1 k=0 N t2 t2 N −1 −1 N −1 1 1 ∗ ∗ ∗ − ck x (t)φk (t)dt + ck cm φk (t)φm (t)dt t2 − t1 t1 t2 − t1 t1 m=0 k=0 k=0 t2 1 Let A[k] = x(t)φ∗k (t)dt and use orthogonality fk t1 1 t2 − t1
t2
73
2
N −1
c∗k
M SE
1 t2 − t1
=
t2 t1
(b) Show that the M SE is zero if
N −1 fk ∗ ck A[k] t 2 − t1 k=0
N −1
N −1 fk fk ck A∗ [k] + ck 2 t2 − t 1 t 2 − t1 k=0 k=0 t2 N −1 N −1 1 fk fk 2 2 = x(t)2 dt + ck − A[k] − A[k] t2 − t1 t1 t 2 − t1 t2 − t1 k=0 k=0 t2 1 = A[k] = x(t)φ∗k (t)dt fk t1
−
Analogous to Problem 3.99(d), ck
x(t)2 dt −
t2
N −1
x(t)2 dt =
t1
fk ck 2
k=0
If this relationship holds for all x(t) in a given class of functions, then the basis functions φ0 (t), φ2 (t), . . . , φN −1 (t) is said to be “complete” for that class.
min M SE
min M SE t2
x(t)2 dt
=
1 t2 − t1
=
1 t2 − t1
= =
t1
N −1 fk 2 ck  t − t 2 1 t1 k=0 N −1 t2 2 2 x(t) dt − fk ck  t2
x(t)2 dt −
t1
k=0
0 when: N −1 fk ck 2 k=0
(c) The Walsh functions are one set of orthogonal functions that are used for signal representation on [0, 1]. Determine the ck and M SE obtained by approximating the following signals with the ﬁrst six Walsh functions
depicted in Fig. P3.100. Sketch the signal and the Walsh function approximation. 2, 12 < t < 34 (i) x(t) = 0, 0 < t < 12 , 34 < t < 1 We can see that the orthogonality relation for the Walsh function is as follows: 0
1
φk (t)φ∗l (t)
=
c0
1 0
k=l k = l, so fk = 1
0.75
=
2dt = 0.5
c1
1 2
0.75
= −
2dt = −
1 2
2dt = −
1 2
0.5 0.75
c2
= − 0.5
c3
=
c4
=
1 1 2 = 4 2 2(0) = 0 74
c5
=
2(0) = 0
Let x ˆ(t)
=
5
ck φk (t)
k=0
x ˆ(t)
=
1 (φ0 (t) − φ1 (t) − φ2 (t) + φ3 (t)) 2
Approximation of x(t) by xhat(t) 2.5 2
x(t)
1.5 1 0.5 0
0
0.1
0.2
0.3
0.4
0.5 t
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5 t
0.6
0.7
0.8
0.9
1
2.5
xhat(t)
2 1.5 1 0.5 0
Figure P3.100. Plot 1 of 2 From the sketch, we can see that x(t) = x ˆ(t), so M SE = 0. (ii) x(t) = sin(2πt) c0
1
=
sin(2πt)dt = 0 0
c1
1 2
=
sin(2πt)dt −
0
1
sin(2πt)dt = 1 2
1
c2
=
φ2 (t) sin(2πt)dt =
c3
=
0, by inspection.
c4
=
0, by inspection.
0
c5
=
2 0
=
1 8
sin(2πt)dt −
2 π
1 π
3 8
sin(2πt)dt + 1 8
√ 2 1 (1 − 2) ≈ − (0.83) π π
75
1 2
sin(2πt)dt 3 8
Approximation of x(t) by xhat(t) 1
x(t)
0.5
0
−0.5
−1
0
0.1
0.2
0.3
0.4
0.5 t
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5 t
0.6
0.7
0.8
0.9
1
1.5 1
x(t)
0.5 0 −0.5 −1 −1.5
Figure P3.100. Plot 2 of 2 M SE
1
2
sin(2πt) − x ˆ(t) dt
= 0
) )2 28 ) 2.17 )) ) = 2 )sin(2πt) − π ) + 2 1 0 8 )2 12 ) ) ) 0.17 ) dt )sin(2πt) − +2 ) ) 3 π 8
1 8
) )2 38 ) ) )sin(2πt) − 3.83 ) dt + 2 ) 2 π ) 8
) )2 ) ) )sin(2πt) − 1.83 ) dt ) π )
≈ 0.1265
(d) The Legendre polynomials are another set of orthogonal functions on the interval [−1, 1]. They are obtained from the diﬀerence equation 2k − 1 k−1 tφk−1 (t) − φk−2 (t) k k using the initial functions φ0 (t) = 1 and φ1 (t) = t. Determine the ck and M SE obtained by approximating the following signals with the ﬁrst six Legendre polynomials. φk (t) =
φ0 (t)
=
φ1 (t)
= t 1 1 3 t(t) − (1) = (3t2 − 1) = 2 2 2 5 1 2 2 1 = t (3t − 1) − t = (5t3 − 3t) 3 2 3 2 7 1 3 31 2 1 = t (5t − 3t) − (3t − 1) = (35t4 − 30t2 + 3) 4 2 42 8 9 1 41 3 1 = t (35t4 − 30t2 + 3) − (5t − 3t) = (315t5 − 350t3 + 75t) 5 8 52 40
φ2 (t) φ3 (t) φ4 (t) φ5 (t)
1
76
The orthogonality relation for Legendre polynomials is: 1 2 φ (t)φ∗l (t)dt = δk,l 2k+1 −1 k 2 so: fk = 2k+1
2, 0 < t < 12 (i) x(t) = 0, −1 < t < 0, 12 < t < 1 c0 c1 c2 c3 c4 c5
1 2
=
2(1)dt = 0
3 2
=
0
1 2
1 2
3 8
(3t2 − 1)dt = −
0
7 2
=
1 2
2tdt =
5 2
=
1 2
1 2
15 16
(5t3 − 3t)dt = −1.039
0
=
9 2
=
11 2
1 2
0
0
x ˆ(t)
=
1 (35t4 − 30t2 + 3))dt = 0.527 4 1 2
1 (315t5 − 350t3 + 75t))dt = 1.300 20
5
ck φk (t)
k=0
M SE
= =
1 2
1 2
4 dt −
1
5 k=0
2 ck 2 2k + 1
0.1886
(ii) x(t) = sin(πt)
c0
=
c1
=
c2
=
c3 c4 c5
1 2 3 2 5 2
1
sin(πt) dt = 0 −1 1
t sin(πt) dt = 0.955 −1 1
(3t2 − 1) sin(πt) dt = 0 0
=
7 2
=
9 2
=
11 2
1
(5t3 − 3t) sin(πt) dt = −1.158 0
1
0
0
1 (35t4 − 30t2 + 3)) sin(πt) dt = 0 4 1
1 (315t5 − 350t3 + 75t)) sin(πt) dt = 0.213 20
77
x ˆ(t)
5
=
ck φk (t)
k=0
M SE
1 2
=
1
−1
 sin(πt)2 dt −
5 k=0
2 ck 2 2k + 1
= 3 ∗ 10−4
3.101. We may derive the FT from the FS by describing an nonperiodic signal as the limiting form of a periodic signal whose period, T , approaches inﬁnity. In order to take this approach, we assume that the FS of the periodic version of the signal exists, that the nonperiodic signal is zero for t > T2 , and that the limit as T approaches inﬁnity is taken in a symmetric manner. Deﬁne the ﬁnite duration nonperiodic signal x(t) as one period of the T periodic signal x ˜(t)
x ˜(t), − T2 < t < T2 x(t) = 0, t > T2 (a) Graph an example of x(t) and x ˜(t) to demonstrate that as T increases, the periodic replicates of x(t) in x ˜(t) are moved farther and farther away from the origin. Eventually, as T approaches inﬁnity, these replicates are removed to inﬁnity. Thus, we write x(t) = lim x ˜(t) T →∞
Notice how as T → ∞, x(t) → x ˜(t) Plot of xhat(t), with T = 15, x(t) = T/3 − t 5
xhat(t)
4 3 2 1 0 −20
−15
−10
−5
0 time
5
10
15
20
20
30
40
Plot of xhat(t), with T = 30, x(t) = T/3 − t 5
xhat(t)
4 3 2 1 0 −40
−30
−20
−10
0 time
10
Figure P3.101. Plot 1 of 1 (b) The FS representation for the periodic signal x ˜(t) is x ˜(t) =
∞
X[k]ejkωo t
k=−∞
X[k] =
1 T
T 2
x ˜(t)e−jkωo t dt
− T2
78
Show that X[k] =
1 T
X(jkωo ) where
∞
X(jω) =
x(t)ejωt dt
−∞
X[k]
X[k]
=
1 T
T 2
x ˜(t)e−jkωo t dt
− T2
T T Since x ˜(t) = x(t) for − < t < 2 2 T2 1 x(t)e−jkωo t dt = T − T2 1 ∞ x(t)e−jkωo t dt = T −∞ Since x(t) = 0 for t >
X[k]
T 2
Compare with the equation of the FT, and one can see, 1 X(jkωo ) = T
(c) Substitute this deﬁnition for X[k] into the expression for x ˜(t) in (b) and show that x ˜(t) =
∞ 1 X(jkωo )ejkωo t ωo 2π k=−∞
x ˜(t)
=
∞ 1 X(jkωo )ejkωo t T
k=−∞
since T = =
2π ωo
∞ 1 X(jkωo )ejkωo t ωo 2π k=−∞
(d) Use the limiting expression for x(t) in (a) and deﬁne ω ≈ kωo to express the limiting form of the sum in (c) as the integral ∞ 1 X(jω)ejωt dω x(t) = 2π −∞ x(t)
= =
lim x ˜(t)
T →∞
∞ 1 X(jkωo )ejkωo t ωo T →∞ 2π
lim
k=−∞
let ω = kωo as T → ∞, ωo → 0, k → ∞, ω → ∞ implies: x(t)
∞ 1 X(jω)ejωt ωo ωo →0 2π ω=−∞ ∞ 1 = X(jω)ejωt dω 2π −∞
=
lim
79
Solutions to Computer Experiments
0.5
1 x1[n]
B[1]cos(omega*n)
3.102. Use MATLAB to repeat Example 3.7 for N = 50 and (a) M = 12. (b) M = 5. (c) M = 20.
0
0.5
−0.5
0 −10
0 n
10
20
0.2
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
1
0.1 0
0.5
−0.1 0
−0.2 −20
−10
0 n
10
20 1
0.1 0.05 x5[n]
B[5]cos(omega*n)
−20
x3[n]
B[3]cos(omega*n)
−20
0
0.5
−0.05 −0.1
0 −20
−10
0 n
10
20
Figure P3.102. Plot 1 of 6 1 0.8
0.02
0.6
x23[n]
B[23]cos(omega*n)
0.04
0
0.4
−0.02
−0.04
0.2 0 −20
−10
0 n
10
20
1 0.8
0.01
0.6
x25[n]
B[25]cos(omega*n)
0.02
0
0.4
−0.01
−0.02
0.2
−20
−10
0 n
10
20
0
Figure P3.102. Plot 2 of 6
80
0.4
0
0.2
−0.2 −0.4
B[3]cos(omega*n)
x1[n]
0.6
0.2
0 −20
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
1 0.1 x3[n]
B[1]cos(omega*n)
0.4
0
0.5
−0.1 0 −20
−10
0 n
10
20 1
0.02 x5[n]
B[5]cos(omega*n)
0.04
0
0.5
−0.02 0 −20
−10
0 n
10
20
Figure P3.102. Plot 3 of 6 −3
x 10
1 0.8
4 2
0.6
x23[n]
B[23]cos(omega*n)
6
0
0.4
−2 −4
0.2
−6 0 −20
−10
0 n
10
20
1 0.8
0.01
0.6
x25[n]
B[25]cos(omega*n)
0.02
0
0.4
−0.01
−0.02
0.2
−20
−10
0 n
10
20
0
Figure P3.102. Plot 4 of 6
81
x1[n]
B[1]cos(omega*n)
1
0.2 0
0.5
−0.2 −10
0 n
10
0
20
0.2
−10
0 n
10
20
−20
−10
0 n
10
20
−20
−10
0 n
10
20
1
0.1 0
0.5
−0.1 −0.2
0 −20
−10
0 n
10
20 1
0.02 x5[n]
B[5]cos(omega*n)
−20
x3[n]
B[3]cos(omega*n)
−20
0.5
0 −0.02
0 −20
−10
0 n
10
20
0.015
1
0.01
0.8
0.005
0.6
x23[n]
B[23]cos(omega*n)
Figure P3.102. Plot 5 of 6
0
0.4
−0.005
0.2
−0.01 −0.015
0 −20
−10
0 n
10
20
−10
0 n
10
20
−20
−10
0 n
10
20
1 0.8
0.01
0.6
x25[n]
B[25]cos(omega*n)
0.02
−20
0
0.4
−0.01
−0.02
0.2
−20
−10
0 n
10
20
0
Figure P3.102. Plot 6 of 6
82
3.103. Use MATLAB’s ﬀt command to repeat Problem 3.48.  Xa[k] 
< Xa[k] deg
0.4 0.3 0.2
0
−100
0.1 0
100
0
5
10
15
0
5
10
15
1
 Xb[k] 
< Xb[k] deg
100 0.5
0
−100 0
0
5
10
15
0
10
15
150
 Xc[k] 
< Xc[k] deg
0.4 0.3
100
0.2 0.1 0
5
0
2
4
6
8
50 0
10
0
2
4
6
8
10
Figure P3.103. Plot 1 of 2 0.25 0.2 < Xd[k] deg
50
 Xd[k] 
0.15 0.1
−50
0.05 0
0
2
4
6
0
0.25
4
6
100 < Xe[k] deg
0.15
 Xe[k] 
2
150
0.2
0.1
0.05 0
0
50 0 −50
−100 −150 0
2
4
6
8
0
2
4
Figure P3.103. Plot 2 of 2
83
6
8
3.104. Use MATLAB’s iﬀt command to repeat Problem 3.2. < xa[n] deg
 xa[n] 
10
5
100 0
−100 0
0
5
10
15
20
0
< xb[n] deg
8  xb[n] 
6 4
10
15
20
100 0
−100
2 0
5
0
5
10
15
0
5
10
15
8 150
 xc[n] 
< xc[n] deg
6
100
4 2 0
0
2
4
6
8
50 0
10
0
2
4
6
8
10
Figure P3.104. Plot 1 of 2 150 < xd[n] deg
 xd[n] 
1.5
100
1
0.5 0
0
2
4
< xe[n] deg
 xe[n] 
0
2
4
6
0
2
4
6
150
1.5
100
1
0.5 0
0
6
2
50
0
2
4
50 0
6
< xf[n] deg
8  xf[n] 
6 4
0
−50
2 0
50
0
5
10
0
5
10
Figure P3.104. Plot 2 of 2
3.105. Use MATLAB’s ﬀt command to repeat Example 3.8. Simply take the ﬀt of the samples of one period of the time waveform and scale by the number of samples used. 84
1.5
x[n]:normal
1
0.5
0
−0.5
0
50
100
150 200 Time index(n)
250
300
350
2.5
y[n]:vent tach
2 1.5 1 0.5 0 −0.5 −1
0
50
100
150
200 250 Time index(n)
300
350
400
450
Figure P3.105. Plot 1 of 2 0.25
X[k]:normal
0.2 0.15 0.1 0.05 0
0
10
20
30 Frequency index(k)
40
50
60
0
10
20
30 Frequency index(k)
40
50
60
0.25
Y[k]:vent tach
0.2 0.15 0.1 0.05 0
Figure P3.105. Plot 2 of 2
3.106. Use MATLAB to repeat Example 3.14. Evaluate the peak overshoot for J = 29, 59, and 99. The peak overshoot for J = 29 is 0.0891 above 1.
85
B[29]*cos(29*wo.t)
0.02 0.01 0 −0.01 −0.02 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
1 0.8 x29(t)
0.6 0.4 0.2 0
Figure P3.106. Plot 1 of 3 The peak overshoot for J = 59 is 0.0894 above 1.
B[59]*cos(59*wo.t)
0.01
0.005
0
−0.005
−0.01 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
1 0.8 x59(t)
0.6 0.4 0.2 0
Figure P3.106. Plot 2 of 3 The peak overshoot for J = 99 is 0.0895 above 1.
86
−3
x 10 6 B[99]*cos(99*wo.t)
4 2 0 −2 −4 −6 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
1 0.8 x99(t)
0.6 0.4 0.2 0
Figure P3.106. Plot 3 of 3
3.107. Let x(t) be the triangular wave depicted in Fig. P3.107. (a) Find the FS coeﬃcients, X[k].
X[k] =
1 T
T
x(t)e−j T
2π
kt
dt, T = 1
0
From Example 3.39 with T = 1, x(t) = 2y(t + 14 ) − 1
Y [k] x(t − t0 ) X[k]
=
1 2, 2 sin( kπ 2 ) jk2 π 2 ,
k = 0, k = 0
F S; ωo
←−−−→ e−jkωo t X[k] =
1
2ejk2π 4 Y [k] − δ[k] implies:
X[k]
− 12 , π 4 sin( kπ 2 ) j 2 (k−1) , k2 π 2 e
=
k = 0, k = 0
(b) Show that the FS representation for x(t) can be expressed in the form x(t) =
∞
B[k] cos(kωo t)
k=0
since x(t) is real and even, X[k] is also real and even with X[k] = X[−k] 87
x(t)
∞
=
k=1 ∞
=
−1
X[k]ejωo kt + X[k]ejωo kt +
k=1
X[k]ejωo kt + X[0]
k=−∞ ∞
X[−k]e−jωo kt + X[0]
k=1
since X[k] = X[−k] ∞ X[k] ejωo kt + e−jωo kt + X[0] = k=1 ∞
=
B[k] cos(kωo t)
k=0
B[k]
where
X[0], = 2X[k],
k = 0, k = 0
(c) Deﬁne the J term partial sum approximation to x(t) as
x ˆJ (t) =
J
B[k] cos(kωo t)
k=0
Use MATLAB to evaluate and plot one period of the J th term in this sum and x ˆJ (t) for J = 1, 3, 7, 29, and 99.
B[1]*cos(1*wo.t)
0.4
0.2
0
−0.2
−0.4 −0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
0.1
0.2
0.3
0.4
0.5
t
−0.2
x1(t)
−0.4
−0.6
−0.8 −0.4
−0.3
−0.2
−0.1
0 t
Figure P3.107. Plot 1 of 5
88
B[3]*cos(3*wo.t)
0.04 0.02 0 −0.02 −0.04 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.2
x3(t)
−0.4 −0.6 −0.8
Figure P3.107. Plot 2 of 5 −3
x 10
B[7]*cos(7*wo.t)
5
0
−5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.2
x7(t)
−0.4 −0.6 −0.8
Figure P3.107. Plot 3 of 5
89
−4
x 10
B[29]*cos(29*wo.t)
4 2 0 −2 −4 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.2
x29(t)
−0.4 −0.6 −0.8
Figure P3.107. Plot 4 of 5 −5
x 10
B[99]*cos(99*wo.t)
4
2
0
−2
−4 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.2
x99(t)
−0.4 −0.6 −0.8
Figure P3.107. Plot 5 of 5
3.108. Repeat Problem 3.107 for the impulse train given by x(t) =
∞ n=−∞
90
δ(t − n)
(a) X[k]
1 1
=
1 2
δ(t)e−j2πkt dt = 1 for all k
− 12
(b)
B[k] =
1, 2,
k = 0, k = 0
B[1]*cos(1*wo.t)
2
1
0
−1
−2
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
0.1
0.2
0.3
0.4
0.5
t
3
x1(t)
2
1
0
−1
−0.4
−0.3
−0.2
−0.1
0 t
Figure P3.108. Plot 1 of 5
B[3]*cos(3*wo.t)
2
1
0
−1
−2
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
7 6 5 x3(t)
4 3 2 1 0 −1
91
Figure P3.108. Plot 2 of 5
B[7]*cos(7*wo.t)
2
1
0
−1
−2
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
15
x7(t)
10
5
0
Figure P3.108. Plot 3 of 5
B[29]*cos(29*wo.t)
2
1
0
−1
−2
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
50
x29(t)
40 30 20 10 0 −10
Figure P3.108. Plot 4 of 5
92
B[99]*cos(99*wo.t)
2
1
0
−1
−2
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
150
x99(t)
100 50 0
Figure P3.108. Plot 5 of 5
3.109. Use MATLAB to repeat Example 3.15 using the following values for the time constant. (a) RC = .01 s. (b) RC = 0.1 s. (c) RC = 1 s.
Y [k] = y(t)
=
1 RC
j2πk + 100 k=−100
93
1 RC
sin( kπ 2 ) kπ
Y [k]ej2πkt
1
y(t)
0.8 0.6 0.4 0.2 −0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
y(t)
0.8 0.6 0.4 0.2
0.6
y(t)
0.55 0.5 0.45 0.4
Figure P3.109. Plot 1 of 1
3.110. This experiment builds on Problem 3.71. (a) Graph the magnitude response of the circuit depicted in Fig. P3.71 assuming the voltage across the inductor is the output. Use logarithmically spaced frequencies from 0.1 rads/S to 1000 rads/S. You may generate N logarithmically spaced values between 10d1 and 10d2 using the MATLAB command logspace(d1,d2,N).
H(jω) =
(ωL)2 1 + (ωL)2
(b) Determine and plot the voltage across the inductor if the input is the square wave depicted in Fig. 3.21 with T = 1 and To = 1/4. Use at least 99 harmonics in a truncated FS expansion for the output.
Y [k]
=
y(t)
=
2j sin(k π2 ) 1 L + j2kπ 99
Y [k]ejk2πt
k=−99
(c) Graph the magnitude response of the circuit depicted in Fig. P3.71 assuming the voltage across the resistor is the output. Use logarithmically spaced frequencies from 0.1 rad/s to 1000 rad/s.
H(jω) =
94
1 1 + (ωL)2
(d) Determine and plot the voltage across the resistor if the input is the square wave depicted in Fig. 3.21 with T = 1 and To = 1/4. Use at least 99 harmonics in a truncated FS expansion for the output.
Y [k]
=
y(t)
=
2j sin(k π2 ) kπ(1 + j2kπL) 99
Y [k]ejk2πt
k=−99
(a) Magnitude response across the inductor 1
H(jw)
0.8 0.6 0.4 0.2 0
0
1000
2000
3000
4000
5000
6000
7000
w(rad/s) (b) One period of the output for yL(t) using 200 harmonics 1
yL(t)
0.5
0
−0.5
−1 −0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
0.4
0.5
Figure P3.110. Plot 1 of 2 (c) Magnitude response across the resistor 1
H(jw)
0.8 0.6 0.4 0.2 0
0
1000
2000
3000
4000
5000
6000
7000
w(rad/s) (d) One period of the output for yR(t) using 200 harmonics 1 0.8
yR(t)
0.6 0.4 0.2 0 −0.5
−0.4
−0.3
−0.2
−0.1
0 t
0.1
0.2
0.3
Figure P3.110. Plot 2 of 2 95
0.4
0.5
3.111. This experiment builds on Problem 3.72. (a) Graph the magnitude response of the circuit depicted in Fig. P3.71 assuming the voltage across the inductor is the output. Use 401 logarithmically spaced frequencies from 100 rads/S to 104 rad/S. You may generate N logarithmically spaced values between 10d1 and 10d2 using the MATLAB command logspace(d1,d2,N). (i) Assume L = 10mH. (a) L = 10mH Magnitude response in dB 0
−20
−40
20log10(H(jw))
−60
−80
−100
−120
−140
−160
0
1000
2000
3000
4000
5000
6000
7000
w(rad/s)
Figure P3.111. Plot 1 of 2 (b) Determine and plot the output if the input is the square wave depicted in Fig. 3.21 with T = 2π ∗10−3 and To = π2 ∗ 10−3 . Use at least 99 harmonics in a truncated FS expansion for the output. (i) Assume L = 10mH.
Y [k] y(t)
= =
j1000 sin( πk 2 ) π(1/C + j1000k + L(j1000k)2 ) 99
Y [k]ejk1000t
k=−99
96
(b) L = 10mH, One period of the output for y(t) using 99 harmonics 0.8
0.6
0.4
y(t)
0.2
0
−0.2
−0.4
−0.6
−0.8 −4
−3
−2
−1
0 t
1
2
3
4 −3
x 10
Figure P3.111. Plot 2 of 2
3.112. Evaluate the frequency response of the truncated ﬁlter in Example 3.46. You may do this in MATLAB by writing an mﬁle to evaluate Ht (ejΩ ) =
M
h[n]e−jΩn
n=−M
for a large number (> 1000) values of Ω in the interval −π < Ω ≤ π. Plot the frequency response magnitude in dB (20 log10 Ht (ejΩ )) for the following values of M . (a) M = 4 (b) M = 10 (c) M = 25 (d) M = 50 Discuss the eﬀect of increasing M on the accuracy with which Ht (ejΩ ) approximates H(ejΩ ). It can be seen that as M increases: (1) The ripple decreases (2) The transition rolls oﬀ faster 1 sin( πn Observe that h[n] = πn 2 ) is symmetric, hence:
Ht (ejΩ )
=
M
h[n]e−jΩn +
n=1
=
M
−1
h[n]e−jΩn + h[0]
n=−M
2h[n] cos(Ωn) + h[0]
n=1
This is the same basic form as a Fourier Series approximation using M terms. The approximation is 97
more accurate as M increases. Ht(Omega) M=4
1 0.8 0.6 0.4 0.2
−3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
Ht(Omega) M=10
1 0.8 0.6 0.4 0.2
Figure P3.112. Plot 1 of 2 Ht(Omega) M=25
1 0.8 0.6 0.4 0.2
−3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
Ht(Omega) M=50
1 0.8 0.6 0.4 0.2
Figure P3.112. Plot 2 of 2
3.113. Use the MATLAB command freqs or freqz to plot the magnitude response of the following systems. Determine whether the system has a lowpass, highpass, or bandpass characteristic. 8 (a) H(jω) = (jω)3 +4(jω) 2 +8jω+8 Low pass ﬁlter 3
(jω) (b) H(jω) = (jω)3 +2(jω) 2 +2jω+1 High pass ﬁlter
98
(c) H(ejΩ ) = 1+3e Low pass ﬁlter
−jΩ
+3e−j2Ω +e−j3Ω 6+2e−j2Ω
−jΩ 4
0.02426(1−e ) (d) H(ejΩ ) = (1+1.10416e−jΩ +0.4019e −j2Ω )(1+0.56616e−jΩ +0.7657e−j2Ω ) High pass ﬁlter
(a)
(b)
0
0 −20 Magnitude in dB
Magnitude in dB
−20 −40 −60
−40 −60 −80 −100
−80 −120 −100 −2 10
0
−140 −2 10
2
10 omega
10
0
2
10 omega
(c)
10
(d)
0
0
Magnitude in dB
Magnitude in dB
−50 −50
−100
−150
−100 −150 −200 −250
−200 −4
−2
0 Omega
2
4
−2
0 Omega
2
Figure P3.113. Plot 1 of 1
3.114. Use MATLAB to verify that the timebandwidth product for a discretetime square wave is approximately independent of the number of nonzero values in each period when duration is deﬁned as the number of nonzero values in the square wave and bandwidth is deﬁned as the mainlobe width. Deﬁne one period of the square wave as
x[n] =
1, 0 ≤ n < M 0, M ≤ n ≤ 999
Evaluate the bandwidth by ﬁrst using ﬀt and abs to obtain the magnitude spectrum and then count the number of DTFS coeﬃcients in the mainlobe for M = 10, 20, 40, 50, 100, and 200. 99
P3.114−1
M=10
10
5
0 300
350
400
450
500
550
600
650
700
20
M=20
15 10 5 0 400
420
440
460
480
500
520
540
560
580
600
460
470
480
490
500
510
520
530
540
550
40
M=40
30 20 10 0 450
Figure P3.114. Plot 1 of 2 P3.114−2 50
M=50
40 30 20 10 0 450
460
470
480
490
500
510
520
530
540
550
M=100
100
50
0 480
485
490
495
500
505
510
515
520
200
M=200
150 100 50 0 490
492
494
496
498
500
502
504
Figure P3.114. Plot 2 of 2
100
506
508
510
M 10 20 40 50 100 200
Bw 201 101 51 41 21 11
3.115. Use the MATLAB function TdBw introduced in Section 3.18 to evaluate and plot the timebandwidth product as a function of duration for the following classes of signals: (a) Rectangular pulse trains. Let the pulse in a single period be of length M and vary M from 51 to 701 in steps of 50. (b) Raised cosine pulse trains. Let the pulse in a single period be of length M and vary M from 51 to 701 in steps of 50. 2 (c) Gaussian pulse trains. Let x[n] = e−an , −500 ≤ n ≤ 500 represent the Gaussian pulse in a single period. Vary the pulse duration by letting a take the following values: 0.00005, 0.0001, 0.0002, 0.0005, 0.001, 0.002, and 0.005. TdBw
2000 1500 1000
100
200
100
200
300 400 500 Duration M : rectangular pulse
600
700
600
700
TdBw
600 500 400 300 300 400 Duration M : raised cosine
500
TdBw
82
81.5
81
0
0.5
1
1.5
2 2.5 3 3.5 Pulse duration a : Gaussian pulse
4
4.5
5 −3
x 10
Figure P3.115. Plot 1 of 1
3.116. Use MATLAB to evaluate and plot the solution to Problem 3.96 on the interval 0 ≤ l ≤ 1 at t = 0, 1, 2, . . . , 20 assuming c = 1. Use at least 99 harmonics in the sum. 101
t=0
t=0.25
0.1
0.1
0.05
0.05
0
0
−0.05
−0.05
−0.1
0
0.2
0.4
0.6
0.8
1
−0.1
0
0.2
t=0.5 0.1
0.05
0.05
0
0
−0.05
−0.05
0
0.2
0.4
0.6
0.8
1
0.8
1
0.8
1
0.8
1
t=0.75
0.1
−0.1
0.4
0.6
0.8
1
−0.1
0
0.2
0.4
0.6
Figure P3.116. Plot 1 of 2 t=1.0
t=1.25
0.1
0.1
0.05
0.05
0
0
−0.05
−0.05
−0.1
0
0.2
0.4
0.6
0.8
1
−0.1
0
0.2
t=1.5 0.1
0.05
0.05
0
0
−0.05
−0.05
0
0.2
0.4
0.6
t=1.75
0.1
−0.1
0.4
0.6
0.8
1
−0.1
0
0.2
0.4
0.6
Figure P3.116. Plot 2 of 2
3.117. The frequency response of either a continuous or discretetime system described in statevariable form (see problem 3.87) may be computed using freqresp. The syntax is h = freqresp(sys,w) 102
where sys is the object containing the statevariable description (see section 2.13) and w is the vector containing the frequencies at which to evaluate the frequency response. freqresp applies in general to multipleinput, multipleoutput systems so the output h is a multidimensional array. For the class of singleinput, singleoutput systems considered in this text and N frequency points in w,h is a dimension of 1by1byN. The command squeeze(h) converts h to an Ndimensional vector that may be displayed with the plot command. Thus we obtain the frequency response using the commands: h = freqresp(sys,w); hmag = abs(squeeze(h)); plot(w,hmag); title(‘System Magnitude Response’); xlabel(‘Frequency(rad/s)’); ylabel(’Magnitude’); Use MATLAB to plot the magnitude and phase response for the systems with statevariable descriptions given in Problems 3.88 and 3.89. 3.88(a) System Magnitude Response 2
Magnitude
1.5
1
0.5
0
0
1
2
3
4
5 6 Frequency (rad/s)
7
8
9
10
7
8
9
10
3.88(a) System Phase Response 0
Phase
−0.5
−1
−1.5
0
1
2
3
4
5 6 Frequency (rad/s)
Figure P3.117. Plot 1 of 4
103
3.88(b) System Magnitude Response 2.5
Magnitude
2 1.5 1 0.5 0
0
1
2
3
4
5 6 Frequency (rad/s)
7
8
9
10
7
8
9
10
3.88(b) System Phase Response 4 3
Phase
2 1 0 −1 −2
0
1
2
3
4
5 6 Frequency (rad/s)
Figure P3.117. Plot 2 of 4 3.89(a) System Magnitude Response 3
Magnitude
2.5 2 1.5 1 0.5 0 −4
−3
−2
−1
0 Frequency (rad/s)
1
2
3
4
2
3
4
3.89(a) System Phase Response 1.5 1
Phase
0.5 0 −0.5 −1 −1.5 −4
−3
−2
−1
0 Frequency (rad/s)
1
Figure P3.117. Plot 3 of 4
104
3.89(b) System Magnitude Response 1.5 1.4 Magnitude
1.3 1.2 1.1 1 0.9 0.8 −4
−3
−2
−1
0 Frequency (rad/s)
1
2
3
4
2
3
4
3.89(b) System Phase Response 4
Phase
2
0
−2
−4 −4
−3
−2
−1
0 Frequency (rad/s)
1
Figure P3.117. Plot 4 of 4
105
Solutions to Additional Problems 4.16. Find the FT representations for the following periodic signals: Sketch the magnitude and phase spectra. (a) x(t) = 2 cos(πt) + sin(2πt)
x(t)
= ejπt + e−jπt +
ωo
= lcm(π, 2π) = π
1 j2πt 1 e − e−j2πt 2j 2j
x[1] = x[−1] = 1 x[2] = −x[−2] = X(jω)
=
∞
2π
1 2j
X[k]δ(ω − kωo )
k=−∞
X(jω)
2 [πδ(ω − π) + πδ(ω + π)] +
=
1 [πδ(ω − 2π) − πδ(ω + 2π)] j
(a) Magnitude and Phase plot 7 6
X(jω)
5 4 3 2 1 0 −10
−5
0
5
10
15
20
2
arg(X(jω))
1
0
−1
−2 −8
−6
−4
−2
0 ω
2
4
6
Figure P4.16. (a) Magnitude and Phase plot (b) x(t) =
4
(−1)k k=0 k+1
cos((2k + 1)πt)
x(t)
=
4 1 (−1)k j(2k+1)πt e + e−j(2k+1)πt 2 k+1 k=0
1
8
X(jω)
= π
4 (−1)k k=0
k+1
[δ(ω − (2k + 1)π) + δ(ω + (2k + 1)π)]
(b) Magnitude and Phase plot 3.5 3
X(jω)
2.5 2 1.5 1 0.5 0 −30
−20
−10
0
10
20
30
−20
−10
0 ω
10
20
30
3.5 3
arg(X(jω))
2.5 2 1.5 1 0.5 0 −30
Figure P4.16. (b) Magnitude and Phase plot (c) x(t) as depicted in Fig. P4.16 (a).
x(t)
=
X(jω)
=
t ≤ 1 2 t ≤ 12 + 0 otherwise otherwise ∞ 2 sin(k 2π ) 4 sin(k π ) 2π 3 3 + δ(ω − k ) k k 3 1 0
k=−∞
2
(c) Magnitude and Phase plot 10
X(jω)
8 6 4 2 0
0
5
10
0
5
10
15
20
25
15
20
25
3.5 3
arg(X(jω))
2.5 2 1.5 1 0.5 0
ω
Figure P4.16. (c) Magnitude and Phase plot (d) x(t) as depicted in Fig. P4.16 (b).
π ωo = 2 π 1 2 2te−j 2 kt dt X[k] = 4 −2 0 k=0 = 2j cos(πk) k = 0 πk
T =4
X(jω)
=
2π
∞ k=−∞
3
X[k]δ(ω −
π k) 2
(d) Magnitude and Phase plot 0.7 0.6
X(jω)
0.5 0.4 0.3 0.2 0.1 0
0
2
4
6
8
10
12
14
16
0
2
4
6
8 ω
10
12
14
16
2
arg(X(jω))
1
0
−1
−2
Figure P4.16. (d) Magnitude and Phase plot
4.17. Find the DTFT representations for the following periodic signals: Sketch the magnitude and phase spectra. (a) x[n] = cos( π8 n) + sin( π5 n)
x[n]
=
Ωo
=
X[5]
=
X[8]
=
X(ejΩ )
=
X(ejΩ )
π π 1 jπn 1 jπn e 8 + e−j 8 n + e 5 − e−j 5 n 2 2j π π π lcm( , ) = 8 5 40 1 X[−5] = 2 1 −X[−8] = 2j ∞ 2π X[k]δ(Ω − kΩo )
k=−∞
= π δ(Ω −
π π π π π ) + δ(Ω + ) + δ(Ω − ) − δ(Ω + ) 8 8 j 5 5
4
(a) Magnitude and Phase plot 2
X(ejΩ)
1.5
1
0.5
0 −0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
−0.6
−0.4
−0.2
0 Ω
0.2
0.4
0.6
0.8
2
arg(X(ejΩ)
1
0
−1
−2 −0.8
Figure P4.17. (a) Magnitude and Phase response (b) x[n] = 1 +
∞ m=−∞
N =8
cos( π4 m)δ[n − m]
Ωo =
π 4 ∞
π cos( m)δ[n − m] 4 m=−∞ π = 1 + cos( n) 4 3 π 1 X[k] = x[n]e−jk 4 n 8 n=−4 x[n]
=
1+
For one period of X[k], k ∈ [−4, 3] X[−4]
=
X[−3]
=
X[−2]
=
X[−1]
=
X[0]
=
X[1]
=
X[2]
=
0 1 − 2−0.5 jk 3π e 4 8 1 jk 2π e 4 8 1 + 2−0.5 jk π e 4 8 2 8 1 + 2−0.5 −jk π 4 e 8 1 −jk 2π 4 e 8 5
X[3]
=
X(ejΩ )
=
1 − 2−0.5 −jk 3π 4 e 8 ∞ 2π X[k]δ(Ω − kΩo ) k=−∞
(1 − 2−0.5 ) δ(Ω + 4 (1 + 2−0.5 ) π δ(Ω − 4
= π
3π 1 π (1 + 2−0.5 ) π 1 ) + δ(Ω + ) + δ(Ω + ) + δ(2Ω) + 4 4 2 4 4 4 −0.5 π 1 π (1 − 2 3π ) ) + δ(Ω − ) + δ(Ω − ) 4 4 2 4 4
(b) Magnitude and Phase plot 2
X(ejΩ)
1.5
1
0.5
0 −2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
−2
−1.5
−1
−0.5
0 Ω
0.5
1
1.5
2
2.5
1
arg(X(ejΩ))
0.5
0
−0.5
−1 −2.5
Figure P4.17. (b) Magnitude and Phase response (c) x[n] as depicted in Fig. P4.17 (a).
Ωo =
N =8
π 4
sin(k 5π 8 ) π 8 sin( 8 k) 5 X[0] = 8 ∞ π jΩ X(e ) = 2π X[k]δ(Ω − k ) 4 X[k]
=
k=−∞
6
(c) Magnitude and Phase plot 4
X(ejΩ)
3
2
1
0
0
2
4
6
8
10
12
0
2
4
6 Ω
8
10
12
3.5 3 arg(X(ejΩ))
2.5 2 1.5 1 0.5 0
Figure P4.17. (c) Magnitude and Phase response (d) x[n] as depicted in Fig. P4.17 (b).
2π Ωo = 7
2π 2π 1 X[k] = 1 − ejk 7 − e−jk 7 7 1 2π = 1 − 2 cos(k ) 7 7 ∞ 2π X(ejΩ ) = 2π X[k]δ(Ω − k ) 7 N =7
k=−∞
7
(d) Magnitude and Phase plot 0.35 0.3
X(ejΩ)
0.25 0.2 0.15 0.1 0.05 0
0
2
4
6
8
10
12
0
2
4
6 Ω
8
10
12
1
arg(X(ejΩ))
0.5
0
−0.5
−1
Figure P4.17. (d) Magnitude and Phase response (e) x[n] as depicted in Fig. P4.17 (c).
π Ωo = 2
π 3π 1 X[k] = 1 + e−jk 2 − e−jkπ − e−jk 2 4 π 1 j (1 − (−1)k ) − sin(k ) = 4 2 2 ∞ π X(ejΩ ) = 2π X[k]δ(Ω − k ) 2 N =4
k=−∞
8
(e) Magnitude and Phase plot 5
X(ejΩ)
4 3 2 1 0
0
2
4
6
8
10
12
0
2
4
6 Ω
8
10
12
1
arg(X(ejΩ))
0.5 0 −0.5 −1 −1.5 −2
Figure P4.17. (e) Magnitude and Phase response
4.18. An LTI system has the impulse response sin(2πt) cos(7πt) πt Use the FT to determine the system output for the following inputs, x(t). h(t) = 2
sin(2πt) Let a(t) = πt
FT
←−−−→ A(jω) =
1 0
ω < 2π otherwise
FT
h(t) = 2a(t) cos(7πt) ←−−−→ H(jω) = A(j(ω − 7π)) + A(j(ω + 7π))
(a) x(t) = cos(2πt) + sin(6πt)
X(jω)
=
2π
∞
X[k]δ(ω − kωo )
k=−∞
X(jω)
= πδ(ω − 2π) + πδ(ω + 2π) +
Y (jω)
= X(jω)H(jω) π π = δ(ω − 6π) − δ(ω + 6π) j j = sin(6πt)
y(t)
9
π π πδ(ω − 6π) − πδ(ω + 6π) j j
(b) x(t) =
∞
m m=−∞ (−1) δ(t
− m) 1 1 (1 − e−jkπ ) = (1 − (−1)k ) 2 2 0 n even = 1 n odd
X[k] =
X(jω)
=
2π
∞
δ(ω − l2π)
l=−∞
Y (jω)
= X(jω)H(jω) =
2π
4
[δ(ω − l2π) + δ(ω + l2π)]
l=3
y(t)
=
2 cos(6πt) + 2 cos(8πt)
(c) x(t) as depicted in Fig. P4.18 (a).
T =1 X(jω)
ωo = 2π ∞ sin(k π4 ) −jkπ (1 − e ) δ(ω − k2π) = 2π kπ k=−∞
Y (jω)
= X(jω)H(jω) sin(−3 π4 ) sin(3 π4 ) −j3π j3π )δ(ω − 6π) + = 2π (1 − e (1 − e )δ(ω + 6π) 3π −3π 4 sin(3 π4 ) 4 sin(3 π4 ) = δ(ω − 6π) + δ(ω + 6π) 3 3 3π 4 sin( 4 ) y(t) = cos(6πt) 3π
(d) x(t) as depicted in Fig. P4.18 (b). X[k]
=
1 8
4
16te−jk8πt dt
− 18
= X(jω)
=
2j cos(πk) πk ∞ 2π X[k]δ(ω − k8π) k=−∞
= −4jδ(ω − 8π) + 4jδ(ω + 8π) 4 y[n] = − sin(8πn) π
Y (jω)
(e) x(t) as depicted in Fig. P4.18 (c).
T =1
ωo = 2π 10
X[k]
1
=
e−t e−jk2πt dt
0
1 − e−(1+jk2π) = 1 + jk2π 1 − e−1 = 1 + jk2π ∞ X(jω) = 2π X[k]δ(ω − k2π) k=−∞
Y (jω)
= X(jω)H(jω) 1 − e−1 1 − e−1 1 − e−1 1 − e−1 δ(ω − 6π) + δ(ω − 8π) + δ(ω + 6π) + δ(ω + 8π) = 2π 1 + j6π 1 + j8π 1 − j6π 1 − j8π j6πt j8πt −j6πt −j8πt e e e e Y (jω) = (1 − e−1 ) + + + 1 + j6π 1 + j8π 1 − j6π 1 − j8π j6πt ej8πt e −1 y(t) = 2(1 − e ) Re + 1 + j6π 1 + j8π y(t)
=
Where
2(1 − e−1 ) [r1 cos(6πt + φ1 ) + r2 cos(8πt + φ2 )]
1 + (6π)2
r1
=
φ1 r2
= − tan−1 (6π) = 1 + (8π)2
φ1
= − tan−1 (8π)
4.19. We may design a dc power supply by cascading a fullwave rectiﬁer and an RC circuit as depicted in Fig. P4.19. The full wave rectiﬁer output is given by z(t) = x(t) Let H(jω) =
Y (jω) Z(jω)
be the frequency response of the RC circuit as shown by H(jω) =
1 jωRC + 1
Suppose the input is x(t) = cos(120πt). (a) Find the FT representation for z(t).
ωo = 240π Z[k]
1 T = 120 1 240 1 j120πt e + e−j120πt e−jk240πt dt = 120 1 2 − 240 =
Z(jω)
=
2(−1)k π(1 − 4k 2 ) ∞ (−1)k δ(ω − k240π) 4 π(1 − 4k 2 ) k=−∞
11
(b) Find the FT representation for y(t).
H(jω)
Y (jω) Z(jω)
=
In the time domain: FT d z(t) − y(t) = RC y(t) ←−−−→ Z(jω) = (1 + jωRC)Y (jω) dt 1 H(jω) = 1 + jωRC Y (jω) = Z(jω)H(jω) ∞ (−1)k 1 = 4 δ(ω − k240π) π(1 − 4k 2 ) 1 + jk240πRC k=−∞
(c) Find the range for the time constant RC such that the ﬁrst harmonic of the ripple in y(t) is less than 1% of the average value. The ripple results from the exponential terms. Let τ = RC.
Use ﬁrst harmonic only: Y (jω) ≈ y(t)
=
ripple =
1 δ(ω − 240π) δ(ω + 240π) 4 δ(ω) + + π 3 1 + j240πRC 1 + j240πRC 2 e−j240πt ej240πt 2 + + 2 π2 3π 1 + j240πRC 1 − j240πRC 2 2 2 < 0.01 2 2 3π 2 π 1 + (240πτ )
240πτ
>
66.659
τ
>
0.0884s
4.20. Consider the system depicted in Fig. P4.20 (a). The FT of the input signal is depicted in Fig. 4.20 FT
FT
(b). Let z(t) ←−−−→ Z(jω) and y(t) ←−−−→ Y (jω). Sketch Z(jω) and Y (jω) for the following cases. (a) w(t) = cos(5πt) and h(t) = sin(6πt) πt 1 X(jω) ∗ W (jω) 2π W (jω) = π (δ(ω − 5π) + δ(ω + 5π)) 1 (X(j(ω − 5π)) + X(j(ω + 5π))) Z(jω) = 2 1 ω < 6π H(jω) = 0 otherwise Z(jω) =
C(jω) = H(jω)Z(jω) = Z(jω) 1 C(jω) ∗ [π (δ(ω − 5π) + δ(ω + 5π))] Y (jω) = 2π 1 [Z(j(ω − 5π)) + Z(j(ω + 5π))] Y (jω) = 2 12
1 [X(j(ω − 10π)) + 2X(jω) + X(j(ω + 10π))] 4
=
Z(jw) 0.5
w
5 Figure P4.20. (a) Sketch of Z(jω)
Y(jw) 0.5 0.25
w
10 Figure P4.20. (a) Sketch of Y (jω) (b) w(t) = cos(5πt) and h(t) =
sin(5πt) πt
Z(jω) H(jω)
1 [X(j(ω − 5π)) + X(j(ω + 5π))] 2 1 ω < 5π = 0 otherwise
=
C(jω)
= H(jω)Z(jω) 1 Y (jω) = C(jω) ∗ [π (δ(ω − 5π) + δ(ω + 5π))] 2π 1 Y (jω) = [C(j(ω − 5π)) + C(j(ω + 5π))] 2
Z(jw) 0.5
w
5 Figure P4.20. (b) Sketch of Z(jω)
13
Y(jw) 0.25
w
10
Figure P4.20. (b) Sketch of Y (jω) (c) w(t) depicted in Fig. P4.20 (c) and h(t) = T = 2, ωo = π, To = 12
W (jω)
=
sin(2πt) πt
cos(5πt)
∞ 2 sin(k π2 ) δ(ω − kπ) k
k=−∞
Z(jω)
= =
1 X(jω) ∗ W (jω) 2π ∞ sin(k π2 ) X(j(ω − kπ)) kπ
k=−∞
C(jω)
= H(jω)Z(jω) =
7 sin(k π ) 2
k=3
Y (jω)
=
kπ
X(j(ω − kπ)) +
−7 sin(k π2 ) X(j(ω − kπ)) kπ
k=−3
1 [C(j(ω − 5π)) + C(j(ω + 5π))] 2
Z(jw) 0.5
. . .
−7
−3
7
3
. . . w
Figure P4.20. (c) Sketch of Z(jω)
14
Y(jw) 2 1 8 −5 3
12 10
−50 21 Figure P4.20. (c) Sketch of Y (jω)
4.21. Consider the system depicted in Fig. P4.21. The impulse response h(t) is given by h(t) = and we have x(t) =
sin(11πt) πt
∞ 1 cos(k5πt) k2
k=1
g(t) =
10
cos(k8πt)
k=1
Use the FT to determine y(t).
y(t)
h(t) = x(t) =
sin(11πt) πt
=
[(x(t) ∗ h(t))(g(t) ∗ h(t))] ∗ h(t)
=
[xh (t)gh (t)] ∗ h(t)
=
m(t) ∗ h(t)
FT
←−−−→ H(jω) =
g(t) =
10
k=1
cos(k8πt)
k=1
=
π
10 k=1
= X(jω)H(jω) = π
ω ≤ 11π otherwise
∞ ∞ FT 1 1 cos(k5πt) ←− − −→ X(jω) = π [δ(ω − 5kπ) + δ(ω + 5kπ)] 2 k k2
k=1
Xh (jω)
1 0
2 1 [δ(ω − 5kπ) + δ(ω + 5kπ)] k2
k=1
15
[δ(ω − 8kπ) + δ(ω + 8kπ)]
−5 7
w
Gh (jω)
= G(jω)H(jω)
= πδ(ω − 8π) + πδ(ω − 8π) 1 Xh (jω) ∗ Gh (jω) M (jω) = 2π 1 [Xh (j(ω − 8π)) + Xh (j(ω + 8π))] = 2 2 1 [(δ(ω − 8π − 5kπ) + δ(ω − 8π + 5kπ)) + (δ(ω + 8π − 5kπ) + δ(ω + 8π + 5kπ))] = π k2 k=1
Y (jω)
= M (jω)H(jω) π π = [δ(ω + 3π) + δ(ω − 3π)] + [δ(ω − 2π) + δ(ω + 2π)] 2 8 1 1 cos(3πt) + cos(2πt) y(t) = 2 8
4.22. The input to a discretetime system is given by π 3π x[n] = cos( n) + sin( n) 8 4 Use the DTFT to ﬁnd the output of the system, y[n], for the following impulse responses h[n], ﬁrst note that
jΩ
X(e )
(a) h[n] =
π π 3π 3π π δ(Ω − ) − δ(Ω + ) = π δ(Ω − ) + δ(Ω + ) + 8 8 j 4 4
sin( π 4 n) πn
jΩ
H(e )
=
1 Ω ≤ π4 0 π4 ≤ Ω < π, 2π periodic.
Y (ejΩ )
= H(ejΩ )X(ejΩ ) π π = π δ(Ω − ) + δ(Ω + ) 8 8 π y[n] = cos( n) 8
(b) h[n] = (−1)n
sin( π 2 n) πn
sin( π2 n) h[n] = ejπn πn 1 Ω − π ≤ π4 H(ejΩ ) = 0 π4 ≤ Ω − π < π, 2π periodic. Y (ejΩ )
= H(ejΩ )X(ejΩ ) 16
= y[n] =
(c) h[n] = cos( π2 n)
3π 3π π δ(Ω − ) + δ(Ω + ) j 4 4 3π sin( n) 4
sin( π 5 n) πn
h[n] H(ejΩ ) Y (ejΩ )
π sin( π5 n) = cos( n) πn 2 1 1 Ω − π2  ≤ π5 2 2 + = π π 0 5 ≤ Ω − 2  < π 0
Ω + π2  ≤ π5 π π 5 ≤ Ω + 2  < π, 2π periodic.
= H(ejΩ )X(ejΩ ) = 0
y[n]
=
0 sin( π n)
4.23. Consider the discretetime system depicted in Fig. P4.23. Assume h[n] = πn2 . Use the DTFT to determine the output, y[n] for the following cases: Also sketch G(ejΩ ), the DTFT of g[n].
y[n]
h[n] =
sin( π2 n) πn
=
(x[n]w[n]) ∗ h[n]
=
g[n] ∗ h[n]
FT
←−−−→ H(e ) = jΩ
1 0
Ω ≤ π2 π 2 ≤ Ω < π
H(ejΩ ) is 2π periodic.
(a) x[n] =
sin( π 4 n) πn ,
w[n] = (−1)n
sin( π4 n) x[n] = πn w[n] = ejπn G(ejΩ )
DT F T
←−− −−→ X(ejΩ ) =
1 0
DT F T
←−− −−→ W (ejΩ ) = 2πδ(Ω − π) 1 X(ejΩ ) ∗ W (ejΩ ) = 2π 1 Ω − π ≤ π4 = 0 π4 ≤ Ω − π < π
g[n]
=
Y (ejΩ )
=
sin( π4 n) πn G(ejΩ )H(ejΩ )
=
0
=
0
y[n]
Ω ≤ π4 π 4 ≤ Ω < π
ejπn
17
G(e j
)
1 5 4
3 4 Figure P4.23. (a) The DTFT of g[n] (b) x[n] = δ[n] −
sin( π 4 n) πn ,
w[n] = (−1)n
sin( π4 n) x[n] = δ[n] − πn w[n] = ejπn G(ejΩ )
DT F T
←−− −−→ X(e ) = jΩ
DT F T
=
Y (ejΩ )
=
sin( 3π 4 n) πn G(ejΩ )H(ejΩ ) H(ejΩ ) sin( π2 n) πn
=
G(e j
=
)
1 3 4 Figure P4.23. (b) The DTFT of g[n] (c) x[n] =
sin( π 2 n) πn ,
Ω ≤ π4 π 4 ≤ Ω < π
←−− −−→ W (ejΩ ) = 2πδ(Ω − π) 1 = X(ejΩ ) ∗ W (ejΩ ) 2π 0 Ω − π ≤ 3π 4 = 1 3π ≤ Ω − π ωm and is sampled with a sampling interval of Ts . Determine the constraints on the magnitude response of the antiimaging ﬁlter so that the overall magnitude response of this reconstruction system is between 0.99 and 1.01 in the signal passband and less than 10−4 to the images of the signal spectrum for the following values:
x [n] s
Ho(jw) zero order hold
Hc(jw)
x (t) r
xs[n] = x(nTs ) Figure P4.32. Reconstruction system. (1) 0.99 < Ho (jω)Hc (jω) < 1.01, −ωm ≤ ω ≤ ωm Thus: (i) Hc (jω) > 32
0.99ω 2 sin(ω T2s )
(ii) Hc (jω)
ωm (c) Determine the constraints on Hc (jω) so that the overall magnitude response of this reconstruction system is between 0.99 and 1.01 in the signal passband and less than 10−4 to the images of the signal spectrum for the following values. Assume x(t) is bandlimited to 12π, that is, X(jω) = 0 for ω > 12π. Constraints: (1) In the pass band:
0.99 < H1 (jω)Hc (jω) < 1.01
0.99ω 2 1.01ω 2 < H (jω) < c 4 sin2 (ω T2s ) 4 sin2 (ω T2s ) (2) In the image region: ω =
2π Ts
− ωm
H1 (jω)Hc (jω) < 10−4 10−4 ω 2 Hc (jω) < 4 sin2 (ω T2s ) (i) Ts = .05 2π − ωm Ts 2π = − 12π = 28π 0.05 10−4 (28π)2 Hc (jω) < 4 sin2 (28π( 0.05 2 )) ω
=
≈ 0.2956
1.01ω 2 0.99ω 2 (jω) < < H c 4 sin2 (ω T2s ) 4 sin2 (ω T2s ) 2926.01 < Hc (jω) < 2985.12
35
(ii) Ts = .02 2π − 12π = 88π 0.02 10−4 (88π)2 Hc (jω) < 4 sin2 (88π( 0.02 )) 2 ω
=
≈ 14.1
1.01ω 2 0.99ω 2 < Hc (jω) < 2 Ts 4 sin (ω 2 ) 4 sin2 (ω T2s ) 139589 < Hc (jω) < 142409
4.34. Determine the maximum factor q by which x[n] with DTFT X(ejΩ ) depicted in Fig. P4.34 can be decimated without aliasing. Sketch the DTFT of the sequence that results when x[n] is decimated by this amount. Looking at the following equation:
Y (ejΩ )
=
q−1 1 j q1 (Ω−m2π) X e q m=0
For the bandlimited signal, overlap starts when:
2qW
>
2π
=
π =3 W
Thus: qmax After decimation:
2 1 j 1 (Ω−m2π) X e 3 3 m=0
Y (ejΩ ) =
Y(e . . .
j
)
1 3
. . . 2
4 36
Figure P4.34. Sketch of the DTFT
4.35. A discretetime system for processing continuoustime signals is shown in Fig. P4.35. Sketch the magnitude of the frequency response of an equivalent continuoustime system for the following cases: Ts 1 jωTs 2 sin(ω 2 ) ) HT (jω) = Ha (jω) H(e Hc (jω) Ts ω
(a) Ω1 =
π 4,
Wc = 20π
ωmax
= min(10π,
π (20), 20π) = 5π 4
H (jw) T
1
5
w
Figure P4.35. (a) Magnitude of the frequency response (b) Ω1 =
3π 4 ,
Wc = 20π
ωmax
= min(10π,
3π (20), 20π) = 10π 4
H (jw) T
1
10
w
Figure P4.35. (b) Magnitude of the frequency response (c) Ω1 =
π 4,
Wc = 2π 37
ωmax
= min(10π,
π (20), 2π) = 2π 4
H (jw) T
1
w
2
Figure P4.35. (c) Magnitude of the frequency response
DT F S; Ω
11Ω
o sin( ) ˜ = X(ejkΩo ). Find and sketch x ˜[n] where x ˜[n] ←−− −−→ 4.36. Let X(ejΩ ) = sin( 2Ω and deﬁne X[k] 2 ˜ X[k] for the following values of Ωo :
˜ X[k] o sin( 11kΩ 2 ) ˜ X[k] = o sin( kΩ 2 )
(a) Ωo =
2π 15 ,
DT F S; Ωo
←−− −−→ x ˜[n] =
x ˜[n]
π 10 ,
N 0
n ≤ 5 5 < n
120π rad/s. Determine the frequency range −ωa < ω < ωa over which the DTFS oﬀers a reasonable approximation to 40
X(jω), the eﬀective resolution of this approximation, ωr , and the frequency interval between each DTFS coeﬃcient, ∆ω. x(t)
Ts = 0.01
100 samples
M = 100
Use N = 200 DTFS to approximate X(jω), X(jω) ≈ 0, ω > 120π, ωm = 120π
Ts
2π ωs > ∆ω ωs > N 2π = N Ts
Therefore: ∆ω
>
π
4.39. A signal x(t) is sampled at intervals of Ts = 0.1 s. Assume X(jω) ≈ 0 for ω > 12π rad/s. Determine the frequency range −ωa < ω < ωa over which the DTFS oﬀers a reasonable approximation to X(jω), the minimum number of samples required to obtain an eﬀective resolution ωr = 0.01π rad/s, and the length of the DTFS required so the frequency interval between DTFS coeﬃcients is ∆ω = 0.001π rad/s.
Ts
ω2 . Assume ω1 > ω2 − ω1 . In this case we can sample x(t) at a rate less than that indicated by the sampling interval and still perform perfect reconstruction by using a bandpass reconstruction ﬁlter Hr (jω). Let x[n] = x(nTs ). Determine the maximum sampling interval Ts such that x(t) can be perfectly reconstructed from x[n]. Sketch the frequency response of the reconstruction ﬁlter required for this case. We can tolerate aliasing as long as there is no overlap on ω1 ≤ ω ≤ ω2 We require: ωs − ω 2 ωs
≥
−ω1
≥ ω2 − ω 1
Implies: Ts Hr (jω)
2π ω − ω1 2 Ts ω1 ≤ ω ≤ ω2 = 0 otherwise
≤
4.44. Suppose a periodic signal x(t) has FS coeﬃcients ( 34 )k , k ≤ 4 X[k] = 0, otherwise The period of this signal is T = 1. (a) Determine the minimum sampling interval for this signal that will prevent aliasing.
X(jω)
=
2π
ωm 2π Ts
=
8π
min Ts
=
4 3 ( )k δ(ω − k2π) 4
k=−4
> 2(8π) 1 8
(b) The constraints of the sampling theorem can be relaxed somewhat in the case of periodic signals if we allow the reconstructed signal to be a timescaled version of the original. Suppose we choose a sampling interval Ts = 20 19 and use a reconstruction ﬁlter 1, ω < π Hr (jω) = 0, otherwise Show that the reconstructed signal is a timescaled version of x(t) and identify the scaling factor. Ts
=
Xδ (jω) =
20 19 ∞ 19 X(j(ω − l1.9π)) 20 l=−∞
47
Aliasing produces a “ frequency scaled” replica of X(jω) centered at zero. The scaling is by a factor of t t ), and xreconstructed (t) = 19 20 from ωo = 2π to ωo = 0.1π. Applying the LPF, ω < π gives x( 20 20 x( 20 ) (c) Find the constraints on the sampling interval Ts so that use of Hr (jω) in (b) results in the reconstruction ﬁlter being a timescaled version of x(t) and determine the relationship between the scaling factor and Ts . The choice of Ts is so that no aliasing occurs. (1)
(2)
2π Ts T s 2π 4 2π − Ts
1 period of the original signal. 1 2π 2 Ts 9 8 9 < 8
0
=
6.28. Determine the unilateral Laplace transform of the following signals using the deﬁning equation: (a) x(t) = u(t − 2) X(s)
∞
= − 0 ∞
= − 0 ∞
=
x(t)e−st dt u(t − 2)e−st dt e−st dt
2
=
e−2s s
(b) x(t) = u(t + 2) X(s)
∞
=
u(t + 2)e−st dt
0−
∞
= 0−
=
1 s
(c) x(t) = e−2t u(t + 1) 4
e−st dt
X(s)
∞
=
e−2t u(t + 1)e−st dt
0−
∞
=
e−t(s+2) dt
0−
1 s+2
=
(d) x(t) = e2t u(−t + 2) X(s)
∞
=
e2t u(−t + 2)e−st dt
0− 2
et(2−s) dt
=
0− 2(2−s)
e
=
−1 2−s
(e) x(t) = sin(ωo t) X(s)
= = = =
∞
1 jωo t − e−jωo t e−st dt e 2j 0− ∞ ∞ 1 t(jωo −s) −t(jωo +s) e dt − e dt 2j 0− − 0 1 −1 1 − 2j jωo − s jωo + s ωo s2 + ωo2
(f) x(t) = u(t) − u(t − 2) X(s)
2
=
e−st dt
0−
1 − e−2s s
=
(g) x(t) =
sin(πt), 0 < t < 1 0, otherwise X(s) =
1 jπt e − e−jπt e−st dt 2j π(1 + e−s ) s2 + π 2 1
0−
=
5
6.29. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and D.2 to determine the unilateral Laplace transform of the following signals: (a) x(t) =
d dt
{te−t u(t)} Lu
1 (s + 1)2 Lu d s x(t) = a(t) ←−−−→ X(s) = dt (s + 1)2
a(t) = te−t u(t) ←−−−→ A(s) =
(b) x(t) = tu(t) ∗ cos(2πt)u(t)
a(t) = tu(t)
Lu
←−−−→ A(s) = Lu
b(t) = cos(2πt)u(t) ←−−−→
s2
1 s2
s + 4π 2
Lu
x(t) = a(t) ∗ b(t) ←−−−→ X(s) = A(s)B(s) 1 X(s) = 2 2 s (s + 4π 2 )
(c) x(t) = t3 u(t)
a(t) = tu(t) b(t) = −ta(t) x(t) = −tb(t)
Lu
1 s2 Lu d −2 ←−−−→ B(s) = A(s) = 3 ds s Lu 6 d B(s) = 4 ←−−−→ X(s) = ds s ←−−−→ A(s) =
(d) x(t) = u(t − 1) ∗ e−2t u(t − 1) Lu
1 s Lu e−s b(t) = a(t − 1) ←−−−→ B(s) = s Lu 1 −2t c(t) = e u(t) ←−−−→ C(s) = s+2 Lu e−(s+2) d(t) = e−2 c(t − 1) ←−−−→ D(s) = s+2 a(t) = u(t) ←−−−→ A(s) =
Lu
x(t) = b(t) ∗ d(t) ←−−−→ X(s) = B(s)D(s) X(s)
=
6
e−2(s+1) s(s + 2)
(e) x(t) =
t 0
e−3τ cos(2τ )dτ a(t) = e−3t cos(2t)u(t)
t
a(τ )dτ −∞
X(s)
Lu
s+3 (s + 3)2 + 4
←−−−→ A(s) = Lu
←−−−→ =
1 s
0−
a(τ )dτ + −∞
A(s) s
s+3 s((s + 3)2 + 4)
d (e−t cos(t)u(t)) (f) x(t) = t dt Lu
s+1 (s + 1)2 + 1 Lu d s(s + 1) b(t) = a(t) ←−−−→ B(s) = dt (s + 1)2 + 1 Lu d x(t) = tb(t) ←−−−→ X(s) = − B(s) ds −s2 − 4s − 2 X(s) = (s2 + 2s + 2)2
a(t) = e−t cos(t)u(t) ←−−−→ A(s) =
6.30. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and D.2 to determine
the time
signals corresponding to the following unilateral Laplace transforms: 1 1 (a) X(s) = s+2 s+3 X(s) x(t) d (b) X(s) = e−2s ds
1 (s+1)2
1 (s + 1)2 d A(s) B(s) = ds
A(s) =
X(s) = e−2s B(s)
(c) X(s) =
−1 1 + s+2 s+3 = e−2t − e−3t u(t) =
Lu
←−−−→ a(t) = te−t u(t) Lu
←−−−→ b(t) = −ta(t) = −t2 e−t u(t) Lu
←−−−→ x(t) = b(t − 2) = −(t − 2)2 e−(t−2) u(t − 2)
1 (2s+1)2 +4 Lu s B( ) ←−−−→ ab(at) a Lu 1 1 −t e sin(2t)u(t) ←−−−→ 2 (s + 1) + 4 2 1 −0.5t e x(t) = sin(t)u(t) 4
7
2
d (d) X(s) = s ds 2
1 s2 +9
+
1 s+3
s2
1 +9
Lu
1 sin(3t)u(t) 3 Lu t d A(s) ←−−−→ b(t) = −ta(t) = − sin(3t)u(t) B(s) = ds 3 Lu d t2 C(s) = B(s) ←−−−→ c(t) = −tb(t) = sin(3t)u(t) ds 3 Lu d 2t D(s) = sC(s) ←−−−→ d(t) = c(t) − c(0− ) = sin(3t)u(t) + t2 cos(3t)u(t) dt 3 Lu 1 E(s) = ←−−−→ e(t) = e−3t u(t) s+3 2t x(t) = e(t) + d(t) = e−3t + sin(3t) + t2 cos(3t) u(t) 3 A(s) =
←−−−→ a(t) =
Lu
6.31. Given the transform pair cos(2t)u(t) ←−−−→ X(s), determine the time signals corresponding to the following Laplace transforms: (a) (s + 1)X(s)
sX(s) + X(s)
Lu
d x(t) + x(t) dt [−2 sin(2t) + cos(2t)] u(t)
←−−−→ =
(b) X(3s) s X( ) a x(t)
Lu
←−−−→ ax(at) 2 1 cos( t)u(t) 3 3
=
(c) X(s + 2) Lu
X(s + 2) ←−−−→ e−2t x(t) x(t)
e−2t cos(2t)u(t)
=
(d) s−2 X(s)
B(s) =
1 X(s) s
Lu
←−−−→ Lu
←−−−→ Lu
x(τ )dτ −∞ t
cos(2τ )u(τ )dτ −∞ t
←−−−→
cos(2τ )dτ 0
8
t
Lu
1 sin(2t) 2 t Lu 1 1 B(s) ←−−−→ sin(2τ )dτ s 0 2 Lu 1 − cos(2t) ←−−−→ u(t) 4 B(s) ←−−−→
(e)
d ds
e−3s X(s) Lu
A(s) = e−3s X(s) ←−−−→ a(t) = x(t − 3) = cos(2(t − 3))u(t − 3) Lu d B(s) = A(s) ←−−−→ b(t) = −ta(t) = −t cos(2(t − 3))u(t − 3) ds Lu
6.32. Given the transform pair x(t) ←−−−→ transform of the following time signals:
2s s2 +2 ,
where x(t) = 0 for t < 0, determine the Laplace
(a) x(3t) Lu
x(3t)
←−−−→
X(s)
= =
s 1 X( ) 3 3 2 3s ( 3s )2 + 2 6s 2 s + 18
(b) x(t − 2) Lu
x(t − 2) ←−−−→ e−2s X(s) = e−2s
(c) x(t) ∗
2s +2
s2
d dt x(t)
b(t) =
Lu d x(t) ←−−−→ B(s) = sX(s) dt
y(t) = x(t) ∗ b(t) Y (s)
Lu
←−−−→ Y (s) = B(s)X(s) = s[X(s)]2 2 2s = s 2 s +2
(d) e−t x(t) Lu
e−t x(t) ←−−−→ X(s + 1) =
9
2(s + 1) (s + 1)2 + 2
(e) 2tx(t) Lu
←−−−→
2tx(t) (f)
t 0
−2
d 4s2 − 8 X(s) = 2 ds (s + 2)2
x(3τ )dτ
t
Lu
←−−−→ Y (s) =
x(3τ )dτ 0
Y (s)
=
s2
X( 3s ) 3s
2 + 18 Lu
6.33. Use the sdomain shift property and the transform pair e−at u(t) ←−−−→ unilateral Laplace transform of x(t) = e−at cos(ω1 t)u(t). Lu
1 s+a e−at cos(ω1 t)u(t) 1 −at jω1 t e + e−jω1 t u(t) e 2 Using the sdomain shift property: 1 1 1 + 2 (s − jω1 ) + a (s + jω1 ) + a 1 2(s + a) 2 (s + a)2 + ω12 (s + a) (s + a)2 + ω12
e−at u(t) ←−−−→ x(t)
= =
X(s)
= = =
6.34. Prove the following properties of the unilateral Laplace transform: (a) Linearity
z(t)
= ax(t) + by(t) ∞ z(t)e−st dt Z(s) = 0 ∞ = (ax(t) + by(t)) e−st dt 0 ∞ ∞ ax(t)e−st dt + by(t)e−st dt = 0 0 ∞ ∞ −st x(t)e dt + b y(t)e−st dt = a 0
0
= aX(s) + bY (s)
(b) Scaling
z(t)
= x(at) 10
1 s+a
to derive the
Z(s)
∞
=
x(at)e−st dt
0
= =
s 1 ∞ x(τ )e− a τ dτ a 0 s 1 X( ) a a
(c) Time shift
= x(t − τ ) ∞ Z(s) = x(t − τ )e−st dt z(t)
0
Z(s)
Let m = t − τ ∞ = x(m)e−s(m+τ ) dm
Z(s)
If x(t − τ )u(t) = x(t − τ )u(t − τ ) ∞ x(m)e−sm e−sτ dm =
−τ
0 −sτ
= e
X(s)
(d) sdomain shift
z(t) Z(s)
= eso t x(t) ∞ = eso t x(t)e−st dt 0 ∞ x(t)e−(so −s)t dt = 0
= X(s − so )
(e) Convolution
= x(t) ∗ y(t) ∞ = x(τ )y(t − τ ) dτ ; x(t), y(t) causal 0 ∞ ∞ Z(s) = x(τ )y(t − τ ) dτ e−st dt 0 0 ∞ ∞ x(τ )y(m) dτ e−sm e−sτ dm = 0 0 ∞ ∞ = x(τ )e−sτ dτ y(m)e−sm dm z(t)
0
0
= X(s)Y (s)
11
(f) Diﬀerentiation in the sdomain = −tx(t) ∞ Z(s) = −tx(t)e−st dt 0 ∞ d x(t) (e−st ) dt = ds 0 ∞ d = x(t)e−st dt ds 0 ∞ d (.) dt and Assume: (.) are interchangeable. ds 0 ∞ d Z(s) = x(t)e−st dt ds 0 d X(s) Z(s) = ds z(t)
6.35. Determine the initial value x(0+ ) given the following Laplace transforms X(s): 1 (a) X(s) = s2 +5s−2
(b) X(s) =
x(0+ )
=
x(0+ )
=
lim sX(s) =
s→∞
s2
s =0 + 5s − 2
s+2 s2 +2s−3
lim sX(s) =
s→∞
s2 + 2s =1 + 2s − 3
s2
2
+s (c) X(s) = e−2s s26s +2s−2
x(0+ )
=
lim sX(s) = e−2s
s→∞
6s3 + s2 =0 + 2s − 2
s2
6.36. Determine the ﬁnal value x(∞) given the following Laplace transforms X(s): 2 +3 (a) X(s) = s22s +5s+1
(b) X(s) =
x(∞)
=
x(∞)
=
lim sX(s) =
2s3 + 3s =0 s2 + 5s + 1
lim sX(s) =
s+2 =2 s2 + 2s + 1
s→0
s+2 s3 +2s2 +s
s→0
12
2
2s +1 (c) X(s) = e−3s s(s+2) 2
x(∞)
=
lim sX(s) = e−3s
s→0
2s2 + 1 1 = 2 (s + 2) 4
6.37. Use the method of partial fractions to ﬁnd the time signals corresponding to the following unilateral Laplace transforms: s+3 (a) X(s) = s2 +3s+2
1
s+3 A B = + s2 + 3s + 2 s+1 s+2 = A+B
3
=
X(s) =
2A + B −1 2 + X(s) = s+1 s+2 x(t) = 2e−t − e−2t u(t)
(b) X(s) =
2s2 +10s+11 s2 +5s+6
1 2s2 + 10s + 11 =2− 2 s + 5s + 6 (s + 2)(s + 3) 1 A B = + (s + 2)(s + 3) s+2 s+3 0 = A+B X(s)
=
1
=
3A + 2B 1 1 X(s) = 2 − + s+2 s+3 x(t) = 2δ(t) + e−3t − e−2t u(t)
(c) X(s) =
2s−1 s2 +2s+1
2s − 1 A B = + s2 + 2s + 1 s + 1 (s + 1)2 = A
X(s) = 2 −1
= A+B x(t) = 2e−t − 3te−t u(t)
(d) X(s) =
5s+4 s3 +3s2 +2s
X(s)
=
A B C 5s + 4 = + + s3 + 3s2 + 2s s s+2 s+1 13
0
= A+B+C
5
=
4
=
3A + B + 2C
2A 2 −3 1 X(s) = + + s s+2 s+1 x(t) = 2 − 3e−2t + e−t u(t)
(e) X(s) =
s2 −3 (s+2)(s2 +2s+1)
1
A B C s2 − 3 = + + (s + 2)(s2 + 2s + 1) s + 2 s + 1 (s + 1)2 = A+B
0
=
X(s)
=
2A + 3B + C
−3
= A + 2B + 2C −2 1 + X(s) = s + 2 (s + 1)2 x(t) = e−2t − 2te−t u(t)
(f) X(s) =
3s+2 s2 +2s+10
X(s) x(t)
(g) X(s) =
3(s + 1) − 1 (s + 1)2 + 32 1 −t e sin(3t) u(t) 3
4s2 +8s+10 (s+2)(s2 +2s+5)
2 −2 2(s + 1) + + s + 2 (s + 1)2 + 22 (s + 1)2 + 22 = 2e−2t + 2e−t cos(2t) − e−t sin(2t) u(t)
X(s)
=
x(t)
(h) X(s) =
3s + 2 = s2 + 2s + 10 = 3e−t cos(3t) −
=
3s2 +10s+10 (s+2)(s2 +6s+10)
X(s) 3 10
A Bs + C 3s2 + 10s + 10 = + (s + 2)(s2 + 6s + 10) s + 2 s2 + 6s + 10 = A+B =
= 6A + 2B + C
10 = 10A + 2C X(s) x(t)
1 2(s + 3) 6 + − 2 s + 2 (s + 3) + 1 (s + 3)2 + 1 −2t + 2e−3t cos(t) − 6e−3t sin(t) u(t) = e
=
14
(i) X(s) =
2s2 +11s+16+e−2s s2 +5s+6
s+4 e−2s 2s2 + 11s + 16 + e−2s = 2 + + s2 + 5s + 6 s2 + 5s + 6 s2 + 5s + 6 −1 2 1 1 X(s) = 2 + + − e−2s + e−2s s+3 s+2 s+3 s+2 x(t) = 2δ(t) + 2e−2t − e−t u(t) + e−2(t−2) − e−3(t−2) u(t − 2)
X(s)
=
6.38. Determine the forced and natural responses for the LTI systems described by the following differential equations with the speciﬁed input and initial conditions: d y(t) + 10y(t) = 10x(t), y(0− ) = 1, x(t) = u(t) (a) dt X(s)
=
Y (s)(s + 10)
=
Y f (s)
= = =
y f (t)
=
y(0− ) s + 10 = e−10t u(t)
Y n (s)
=
y n (t)
(b)
d2 dt2 y(t)
1 s 10X(s) + y(0− ) 10X(s) s + 10 10 s(s + 10) 1 −1 + s s + 10 1 − e−10t u(t)
d d + 5 dt y(t) + 6y(t) = −4x(t) − 3 dt x(t), y(0− ) = −1,
Y (s)(s2 + 5s + 6) − 5 + s + 5
=
Y (s)
=
d dt y(t) t=0−
= 5, x(t) = e−t u(t)
1 s+1 −1 s + (s + 1)(s + 2)(s + 3) (s + 2)(s + 3)
(−4 − 3s)
= Y f (s) + Y n (s) −0.5 −2 2.5 Y f (s) = + + s+1 s+2 s+3 y f (t) = −0.5e−t − 2e−2t + 2.5e−3t u(t) −2 3 Y n (s) = + s+2 s+3 y n (t) = −2e−2t + 3e−3t u(t)
(c)
d2 dt2 y(t)
+ y(t) = 8x(t), y(0− ) = 0,
d dt y(t) t=0−
Y (s)(s2 + 1)
=
= 2, x(t) = e−t u(t)
8X(s) + 2 15
4 4s 4 + 2 − 2 s+1 s +1 s +1 y f (t) = 4 e−t + sin(t) − cos(t) u(t) 2 Y n (s) = 2 s +1 y n (t) = 2 sin(t)u(t) Y f (s)
(d)
d2 dt2 y(t)
d + 2 dt y(t) + 5y(t) =
d dt x(t),
=
y(0− ) = 2,
d dt y(t) t=0−
= 0, x(t) = u(t)
= sX(s) + sy(0− ) + 2y(0− ) 1 Y f (s) = (s + 1)2 + 22 1 −t e sin(2t)u(t) y f (t) = 2 2(s + 2) Y n (s) = (s + 2)2 + 1 y n (t) = 2e−t cos(t)u(t)
Y (s)(s2 + 2s + 5)
6.39. Use Laplace transform circuit models to determine the current y(t) in the circuit of Fig. P6.39 assuming normalized values R = 1Ω and L = 12 H for the speciﬁed inputs. The current through the inductor at time t = 0− is 2 A. X(s) + LiL (0− )
=
Y (s)
= =
(R + Ls)Y (s) 1 X(s) iL (0− ) + L s + R/L s + R/L 2 2X(s) + s+2 s+2
(a) x(t) = e−t u(t)
Y (s)
= = =
y(t)
=
2 2 + (s + 1)(s + 2) s + 2 2 2 2 − + s+1 s+2 s+2 2 s+1 2e−t u(t)
(b) x(t) = cos(t)u(t)
Y (s)
=
(s2
2 2s + + 1)(s + 2) s + 2 16
2 2 2 2s + 1 − + 2 5 s +1 s+2 s+2 4 s 2 1 1 + + 5 s2 + 1 5 s2 + 1 5 s+2 1 4 cos(t) + 2 sin(t) + e−2t u(t) 5
= = y(t)
=
6.40. The circuit in Fig. P6.40 represents a system with input x(t) and output y(t). Determine the forced response and the natural response of this system under the speciﬁed conditions.
Forced Response: = RI(s) + Y (f ) (s) +
X(s)
1 I(s) SC
but Y (f ) (s) sL 1 Y (f ) (s) Y (f ) (s) + Y (f ) (s) + X(s) = R sL SC sL 2 (f ) 2 (f ) s CLX(s) = sRCY (s) + s CLY (s) + Y (f ) (s) X(s)s2 CL Y (f ) (s) = s2 CL + sRC + 1 I(s)
=
Natural Response: 0 iL (0− ) s
=
Ic (s) = I(s) + Cvc (0− )
=
I(s) = IL (s) +
So: sCRY
(n)
(s) sL (n)
RY
+
RiL (0− ) s
= RI(s) + Y (n) (s) +
Y (n) (s) iL (0− ) + sL s Y (n) (s) iL (0− ) + + Cvc (0− ) sL s
− Y (n) (s) 1 + iL (0s ) SC sL (n) (n)
+ Y (n) (s) +
(s) + sRCLiL (0− ) + s2 CLY
Y (n) (s)
(s) + Y
=
1 Ic (s) SC
+ Cvc (0− ) = 0
(s) + LiL (0− ) + sLCvc (0− ) = 0
−L(sRC + 1)iL (0− ) − sLCvc (0− ) 1 + sCR + s2 CL
(a) R = 3Ω, L = 1 H, C = 12 F, x(t) = u(t), current through the inductor at t = 0− is 2 A, and the voltage across the capacitor at t = 0− is 1 V.
Y (f ) (s)
=
1 2 2s
1 2s + 32 s
17
+2
2 −1 + s+1 s+2 = 2e−2t − e−t u(t)
= y (f ) (t)
−( 32 s + 1)2 − 12 s 1 + 32 s + 12 s2 −(3s + 2)2 − s = s2 + 3s + 2 10 3 − = s+1 s+2 y (n) (t) = 3e−t − 10e−2t u(t)
Y (n) (s)
=
(b) R = 2Ω, L = 1 H, C = 15 F, x(t) = u(t), current through the inductor at t = 0− is 2 A, and the voltage across the capacitor at t = 0− is 1 V.
Y (f ) (s)
=
1 2 5s
1 5s + 25 s
+1
s+1 2 −1 + (s + 1)2 + 22 2 (s + 1)2 + 22 1 −t −t = e cos(2t) − e sin(2t) u(t) 2
= y (f ) (t)
−( 25 s + 1)2 − 15 s 1 + 25 s + 15 s2 2 s+1 5 = −5 − 2 2 (s + 1) + 2 2 (s + 1)2 + 22 5 −t (n) −t y (t) = −5e cos(2t) − e sin(2t) u(t) 2
Y (n) (s)
=
6.41. Determine the bilateral Laplace transform and the corresponding ROC for the following signals: (a) x(t) = e−t/2 u(t) + e−t u(t) + et u(−t) X(s)
∞
x(t)e−st dt
= −∞ ∞
=
e−t/2 dt +
0
=
∞
e−t dt +
0
0
et dt −∞
1 1 1 + − s + 0.5 s + 1 s − 1 ROC: 0.5 < Re(s) < 1
(b) x(t) = et cos(2t)u(−t) + e−t u(t) + et/2 u(t)
X(s)
= −
1 1 s−1 + + (s − 1)2 + 4 s + 1 s − 18
1 2
ROC: 0.5 < Re(s) < 1
(c) x(t) = e3t+6 u(t + 3) e−3 e3(t+3) u(t + 3) L 1 a(t) = e3t u(t) ←−−−→ A(s) = s−3 x(t)
=
L
b(t) = a(t + 3) ←−−−→ B(s) = e3s A(s) = e3s X(s)
1 s−3
e3(s−1) s−3 ROC: Re(s) > 3
=
(d) x(t) = cos(3t)u(−t) ∗ e−t u(t) L
a(t) ∗ b(t) ←−−−→ A(s)B(s) s 1 X(s) = − 2 s +9 s+1 ROC: 1 < Re(s) < 0
(e) x(t) = et sin(2t + 4)u(t + 2)
x(t)
=
e−2 et+2 sin(2(t + 2))u(t + 2)
L
a(t + 2) ←−−−→ e2s A(s) 2e−2 X(s) = e2s (s − 1)2 + 4 ROC: Re(s) > 1 −2t d (f) x(t) = et dt e u(−t) L
a(t) = e−2t u(−t) ←−−−→ A(s) = b(t) =
−1 s+2
L d a(t) ←−−−→ B(s) = sA(s) dt L
x(t) = et b(t) ←−−−→ X(s) = B(s − 1) =
1−s s+1
ROC: Re(s) < 1
6.42. Use the tables of transforms and properties to determine the time signals that correspond to the following bilateral Laplace transforms: 19
1 (a) X(s) = e5s s+2 with ROC Re(s) < −2
(leftsided) L 1 ←−−−→ a(t) = −e−2t u(−t) A(s) = s+2 L
X(s) = e5s A(s) ←−−−→ x(t) = a(t + 5) = −e−2(t+5) u(−(t + 5))
(b) X(s) =
d2 ds2
1 s−3
with ROC Re(s) > 3
(rightsided) L 1 ←−−−→ a(t) = e3t u(t) A(s) = s−3 L d2 X(s) = 2 A(s) ←−−−→ x(t) = t2 e3t u(t) ds (c) X(s) = s
1 s2
−
e−s s2
−
e−2s s
with ROC Re(s) < 0
(leftsided) x(t) x(t)
d (d) X(s) = s−2 ds
e−3s s
d (−tu(−t) + tu(−t − 1) + u(−t − 2)) dt = −u(−t) + u(−t − 1) − δ(t + 2) =
with ROC Re(s) > 0
(rightsided) L 1 ←−−−→ a(t) = u(t) A(s) = s L
B(s) = e−3s A(s) ←−−−→ b(t) = a(t − 3) = u(t − 3) L d C(s) = B(s) ←−−−→ c(t) = −tb(t) = −tu(t − 3) ds t L 1 ←−−−→ d(t) = c(τ )dτ D(s) = s −∞ t L 1 ←−−−→ d(t) = − τ dτ = − (t2 − 9) 2 3 t L 1 X(s) = D(s) ←−−−→ x(t) = d(τ )dτ s −∞ L 1 t 2 ←−−−→ x(t) = − (τ − 9)dτ 2 3 L 1 3 9 ←−−−→ x(t) = − (t − 27) + (t − 3) u(t − 3) 6 2
20
6.43. Use the method of partial fractions to determine the time signals corresponding to the following bilateral Laplace transforms: (a) X(s) = s2−s−4 +3s+2
X(s)
=
−3 2 + s+1 s+2
(i) with ROC Re(s) < −2
(leftsided) x(t)
=
3e−t − 2e−2t u(−t)
(ii) with ROC Re(s) > −1
(rightsided) x(t)
=
−3e−t + 2e−2t u(t)
(iii) with ROC −2 < Re(s) < −1
(two sided) x(t)
(b) X(s) =
=
3e−t u(−t) + 2e−2t u(t)
4s2 +8s+10 (s+2)(s2 +2s+5)
X(s)
=
2(s + 1) 2 −2 + + 2 2 s + 2 (s + 1) + 2 (s + 1)2 + 22
(i) with ROC Re(s) < −2
(leftsided) x(t)
=
−2e−2t − 2e−t cos(2t) + e−t sin(2t) u(−t)
(ii) with ROC Re(s) > −1
(rightsided) x(t)
=
2e−2t + 2e−t cos(2t) − e−t sin(2t) u(t) 21
(iii) with ROC −2 < Re(s) < −1
(two sided) x(t) (c) X(s) =
=
2e−2t u(t) + −2e−t cos(2t) + e−t sin(2t) u(−t)
5s+4 s2 +2s+1
X(s)
=
1 5 − s + 1 (s + 1)2
(i) with ROC Re(s) < −1
(leftsided) x(t)
=
−5e−t + te−t u(−t)
(ii) with ROC Re(s) > −1
(rightsided) x(t) (d) X(s) =
=
5e−t − te−t u(t)
2s2 +2s−2 s2 −1
X(s)
=
2+
1 1 + s+1 s−1
(i) with ROC Re(s) < −1
(leftsided) x(t)
=
2δ(t) − e−t + et u(t)
(ii) with ROC Re(s) > 1
(rightsided) x(t)
=
2δ(t) + e−t + et u(t)
(iii) with ROC −1 < Re(s) < 1
(two sided) x(t)
=
2δ(t) + e−t u(t) − et u(−t) 22
6.44. Consider the RCcircuit depicted in Fig. P6.44. (a) Find the transfer function assuming y1 (t) is the output. Plot the poles and zeros and characterize the system as lowpass, highpass, or bandpass.
x(t)
=
y1 (t)
=
d y1 (t) dt
=
sY1 (s)
=
d x(t) dt
=
sX(s)
=
Y1 (s) X(s)
=
H(s) =
1 t i(t)R + i(τ )dτ C −∞ 1 t i(τ )dτ C −∞ 1 i(t) C 1 = I(s) C d 1 R i(t) + i(t) dt C 1 I(s) Rs + C 1 1 1 RC s + RC
1 Low pass ﬁlter with a pole at s = − RC RC = 10−3 :
Pole−Zero Map 1 0.8 0.6 0.4
Imag Axis
0.2 0 −0.2 −0.4 −0.6 −0.8 −1
−1000
−800
−600 Real Axis
−400
−200
Figure P6.44. (a) PoleZero Plot of H(s)
H(jω)
=
H(j0)
= 23
1 RC
jω + 1
1 RC
0
H(j∞)
=
0
The ﬁlter is low pass. (b) Repeat part (a) assuming y2 (t) is the system output.
y2 (t)
= i(t)R
Y2 (s)
= I(s)R
x(t) x(t) d x(t) dt
1 = y2 (t) + C = i(t)R +
1 C
t
i(τ )dτ −∞ t
i(τ )dτ −∞
d 1 i(t) + i(t) dt C 1 sX(s) = I(s) sR + C s H(s) = 1 s + RC = R
1 High pass ﬁlter with a zero at s = 0, and a pole at s = − RC
RC = 10−3 : Pole−Zero Map 1 0.8 0.6 0.4
Imag Axis
0.2 0 −0.2 −0.4 −0.6 −0.8 −1
−1000
−900
−800
−700
−600
−500 −400 Real Axis
−300
Figure P6.44. (b) PoleZero Plot of H(s) 24
−200
−100
0
H(jω)
jω 1 jω + RC 0
=
H(j0) → H(j∞)
=
1
The ﬁlter is high pass. (c) Find the impulse responses for the systems in parts (a) and (b). For part (a), h(t) = For part (b):
1 1 − RC t u(t). RC e
A(s) =
1 1 s + RC
L
←−−−→ a(t) = e− RC t u(t) 1
L
1 − 1 t d a(t) = − e RC u(t) + δ(t) dt RC
H(s) = sA(s) ←−−−→ h(t) =
6.45. A system has transfer function H(s) as given below. Determine the impulse response assuming (i) that the system is causal, and (ii) that the system is stable. 2 (a) H(s) = 2s s+2s−2 2 −1 H(s)
=
2+
1 1 + s+1 s−1
(i) system is causal h(t)
2δ(t) + e−t + et u(t)
=
(ii) system is stable h(t)
(b) H(s) =
=
2δ(t) + e−t u(t) + −et u(−t)
2s−1 s2 +2s+1
H(s)
−3 2 + s + 1 (s + 1)2
=
(i) system is causal h(t)
=
2e−t − 3te−t u(t)
25
(ii) system is stable h(t)
(c) H(s) =
=
2e−t − 3te−t u(t)
s2 +5s−9 (s+1)(s2 −2s+10)
2(s − 1) −1 3 + + s + 1 (s − 1)2 + 32 (s − 1)2 + 32
H(s) =
(i) system is causal h(t)
=
−e−t + 2et cos(3t) + et sin(3t) u(t)
(ii) system is stable h(t)
= −e−t u(t) − 2et cos(3t) + et sin(3t) u(−t)
2 (d) H(s) = e−5s + s−2 (i) system is causal
h(t)
= δ(t − 5) + 2e−2t u(t)
h(t)
= δ(t − 5) − 2e2t u(−t)
(ii) system is stable
6.46. A stable system has input x(t) and output y(t) as given below. Use Laplace transforms to determine the transfer function and impulse response of the system. (a) x(t) = e−t u(t), y(t) = e−2t cos(t)u(t) X(s)
=
Y (s)
=
H(s)
= =
h(t)
=
1 s+1 s+2 (s + 1)2 + 1 Y (s) s2 + 3s + 2 = 2 X(s) s + 4s + 5 −1 −(s + 2) + 1+ 2 (s + 2) + 1 (s + 2)2 + 1 δ(t) − e−2t cos(t) + e−2t sin(t) u(t)
26
(b) x(t) = e−2t u(t), y(t) = −2e−t u(t) + 2e−3t u(t) 1 s+2 −2 2 Y (s) = + s+1 s+3 −4(s + 2) Y (s) = H(s) = X(s) (s + 1)(s + 3) −2 −2 = + s+1 s+3 h(t) = −2e−t − 2e−3t u(t) X(s)
=
6.47. The relationship between the input x(t) and output y(t) of a causal system is described by the diﬀerential equation given below. Use Laplace transforms to determine the transfer function and impulse response of the system. d (a) dt y(t) + 10y(t) = 10x(t)
sY (s) + 10Y (s) = Y (s) H(s) = = X(s) h(t)
(b)
d2 dt2 y(t)
d + 5 dt y(t) + 6y(t) = x(t) +
=
10X(s) 10 s + 10 10e−10t u(t)
d dt x(t)
Y (s)(s2 + 5s + 6)
= X(s)(1 + s) s+1 H(s) = (s + 3)(s + 2) −1 2 = + s+2 s+3 h(t) = 2e−3t − e−2t u(t)
(c)
d2 dt2 y(t)
−
d dt y(t)
d − 2y(t) = −4x(t) + 5 dt x(t)
Y (s)(s2 − s − 2)
= X(s)(5s − 4) 5s − 4 H(s) = (s − 2)(s + 1) 3 2 = + s+1 s−2 h(t) = 3e−t + 2e2t u(t)
6.48. Determine a diﬀerential equation description for a system with the following transfer function. 1 (a) H(s) = s(s+3) 27
H(s) =
Y (s) X(s)
=
1 s(s + 3)
Y (s)(s2 + 3s) = X(s) d d2 y(t) + 3 y(t) = x(t) 2 dt dt
(b) H(s) =
6s s2 −2s+8
Y (s)(s2 − 2s + 8) = X(s)(6s) d2 d d y(t) − 2 y(t) + 8y(t) = 6 x(t) 2 dt dt dt
(c) H(s) =
2(s−2) (s+1)2 (s+3)
Y (s)(s3 + 5s2 + 7s + 3) = X(s)(3s2 + 6s + 3) d3 d d2 d y(t) + 5 2 y(t) + 7 y(t) + 3y(t) = 2 x(t) − 4x(t) 3 dt dt dt dt
6.49. (a) Use the timediﬀerentiation property to show that the transfer function of a LTI system is expressed in terms of the statevariable description as shown by H(s) = c(sI − A)−1 b + D d q(t) = Aq(t) + bx(t) dt sQ(s) = AQ(s) + bX(s) Q(s) y(t)
=
(sI − A)−1 bX(s)
= cq(t) + Dx(t)
Y (s)
= cQ(s) + DX(s) Y (s) = c(sI − A)−1 b + D H(s) = X(s) (b) Determine the transfer function, impulse response, and diﬀerential equation descriptions for a stable LTI system represented by the following state variable descriptions: −1 1 3 (i) A = , b= , c = 1 2 , D = [0] 0 −2 −1 H(s)
28
= c(sI − A)−1 b + D s+3 = 2 s + 3s + 1 −1 2 + = s+1 s+2
(ii) A =
1 1
2 −6
h(t)
=
d d2 y(t) + 3 y(t) + y(t) dt2 dt
=
, b=
1 2
, c=
2e−t − e−2t u(t)
d x(t) + 3x(t) dt
0
1
, D = [0]
2s − 1 s2 + 5s − 8 2(s + 2.5) − 6 = (s + 2.5)2 − 14.25 2(s + 2.5) −6 = + (s + 2.5)2 − 14.25 (s + 2.5)2 − 14.25 √ √ 6 −2.5t −2.5t e cos(t 14.25) − √ sin(t 14.25) u(t) h(t) = 2e 14.25 d d2 d y(t) + 5 y(t) − 8y(t) = 2 x(t) − x(t) dt2 dt dt H(s)
=
6.50. Determine whether the systems described by the following transfer functions are(i) both stable and causal, and (ii) whether a stable and causal inverse system exists: (s+1)(s+2) (a) H(s) = (s+1)(s 2 +2s+10) H(s)
=
zero at:
s+2 s2 + 2s + 10 −2 −1 ± 3j
poles at:
(i) All poles are in the LHP, and with ROC: Re(s) > 1, the system is both stable and causal. (ii) All zeros are in the LHP, so a stable and causal inverse system exists. (b) H(s) =
s2 +2s−3 (s+3)(s2 +2s+5)
H(s)
=
zero at:
s−1 s2 + 2s + 5 1 −1 ± 2j
poles at:
(i) All poles are in the LHP, and with ROC: Re(s) > 1, the system is both stable and causal. (ii) Not all zeros are in the LHP, so no stable and causal inverse system exists. (c) H(s) =
s2 −3s+2 (s+2)(s2 −2s+8)
H(s)
= 29
s2 − 3s + 2 s3 + 4s + 16
zeros at:
1, 2
√ −2, 1 ± j 7
poles at:
(i) Not all poles are in the LHP, so the system is not stable and causal. (ii) No zeros are in the LHP, so no stable and causal inverse system exists. (d) H(s) =
s2 +2s (s2 +3s−2)(s2 +s+2)
H(s) zeros at:
=
s4
+ 0, −2
s2 + 2s + 3s2 + 4s − 4
4s3
√ −3 ± 17 7 , −0.5 ± j 2 4
poles at:
(i) Not all poles are in the LHP, so the system is not stable and causal. (ii) There is a zero at s = 0, so no stable and causal inverse system exists.
6.51. The relationship between the input x(t) and output y(t) of a system is described by the diﬀerential equation d2 d2 d d y(t) + 5y(t) = y(t) + x(t) − 2 x(t) + x(t) 2 2 dt dt dt dt (a) Does this system have a stable and causal inverse? why? Y (s)(s2 + s + 5) H(s)
= X(s)(s2 − 2s + 1) (s − 1)2 = s2 + s + 5
Since H(s) has zeros in the RHP, this system does not have a causal and stable inverse. (b) Find a diﬀerential equation description for the inverse system. H inv (s)
= =
Inverse system: d2 d y(t) − 2 y(t) + y(t) 2 dt dt
=
1 H(s) s2 + s + 5 s2 − 2s + 1 d2 d x(t) + x(t) + 5x(t) 2 dt dt
6.52. A stable, causal system has a rational transfer function H(s). The system satisﬁes the following conditions: (i) The impulse response h(t) is real valued; (ii) H(s) has exactly two zeros, one of which d2 d is at s = 1 + j; (iii) The signal dt 2 h(t) + 3 dt h(t) + 2h(t) contains an impulse and doublet of unknown 30
strengths and a unit amplitude step. Find H(s).
d2 d h(t) + 3 h(t) + 2h(t) = bδ (1) (t) + aδ(t) + u(t) dt2 dt 1 H(s) s2 + 3s + 2 = bs + a + s bs + a + 1s H(s) = s2 + 3s + 2 bs2 + as + 1 = s(s + 2)(s + 1)
One zero is 1 + j and h(t) is real valued, which implies zeros occur in conjugate pairs, so the other zero is 1 − j.
H(s)
=
−s+1 s(s + 2)(s + 1) 1 2 2s
6.53. Sketch the magnitude response for the systems described by the following transfer functions using the relationship between the pole and zero locations and the jω axis in the splane. s (a) H(s) = s2 +2s+101
H(jω) =
(b) H(s) =
ω j(ω − 10) + 1j(ω + 10) + 1
s2 +16 s+1
H(jω) =
(c) H(s) =
j(ω + 4)j(ω − 4) jω − 1
s−1 s+1
H(jω) = 31
jω − 1 jω + 1
Magnitude response for 6.53 (a) 0.8 0.6 0.4 0.2 0 −15
−10
−5
0 5 Magnitude response for 6.53 (b)
10
15
−10
−5
0 5 Magnitude response for 6.53 (c)
10
15
−10
−5
10
15
20 15 10 5 0 −15 2 1.5 1 0.5 0 −15
0 w
5
Figure P6.53. Magnitude response for the systems
6.54. Sketch the phase response for the systems described by the following transfer functions using the relationship between the pole and zero locations and the jω axis in the splane. (a) H(s) = s−1 s+2
(b) H(s) =
H(jω)
s+1 s+2
(c) H(s) =
ω = π − arctan(ω) − arctan( ) 2
H(jω)
=
ω arctan(ω) − arctan( ) 2
1 s2 +2s+17
H(s)
=
H(jω)
=
H(jω)
1 (s + 1)2 + 42
1 (jω + 1 + j4)(jω + 1 − j4) = − [arctan(ω + 4) + arctan(ω − 4)]
32
(d) H(s) = s2
H(jω)
= −ω 2
H(jω)
= π
6.54 (a)
6.54 (b)
7
0.4
6 0.2 Phase
Phase
5 4 3 2
0
−0.2
1 0 −20
−10
0 ω
10
20
−0.4 −20
−10
6.54 (c)
20
10
20
4.5 4 Phase
2 Phase
10
6.54 (d)
4
0
−2
−4 −20
0 ω
3.5 3 2.5
−10
0 ω
10
20
2 −20
−10
0 ω
Figure P6.54. Phase Plot of H(s)
6.55. Sketch the Bode diagrams for the systems described by the following transfer functions. 50 (a) H(s) = (s+1)(s+10) (b) H(s) = s20(s+1) 2 (s+10) 5 (c) H(s) = (s+1) 3 s+2 (d) H(s) = s2 +s+100 s+2 (e) H(s) = s2 +10s+100 33
Bode Diagram Gm = Inf, Pm = 78.63 deg (at 4.4561 rad/sec)
Magnitude (dB)
20
0
−20
−40
−60 0
Phase (deg)
−45
−90
−135
−180 −1 10
0
1
10
10
2
10
Frequency (rad/sec)
Figure P6.55. (a) Bode Plot of H(s) Bode Diagram Gm = Inf, Pm = 52.947 deg (at 2.1553 rad/sec)
Magnitude (dB)
50
0
−50
−100
Phase (deg)
−120
−150
−180 −1 10
0
1
10
10 Frequency (rad/sec)
Figure P6.55. (b) Bode Plot of H(s) 34
2
10
Bode Diagram Gm = 4.0836 dB (at 1.7322 rad/sec), Pm = 17.37 deg (at 1.387 rad/sec)
Magnitude (dB)
20
0
−20
−40
−60
Phase (deg)
0
−90
−180
−270 −1 10
0
1
10 Frequency (rad/sec)
10
Figure P6.55. (c) Bode Plot of H(s) Bode Diagram Gm = Inf, Pm = 157.6 deg (at 10.099 rad/sec) 10
Magnitude (dB)
0 −10 −20 −30 −40
Phase (deg)
90
0
−90
−180 −1 10
0
1
10
10 Frequency (rad/sec)
Figure P6.55. (d) Bode Plot of H(s) 35
2
10
Bode Diagram Gm = Inf, Pm = Inf −15
Magnitude (dB)
−20 −25 −30 −35 −40
Phase (deg)
45
0
−45
−90 −1 10
0
1
10
2
10
10
Frequency (rad/sec)
Figure P6.55. (e) Bode Plot of H(s)
6.56. The output of a multipath system y(t) may be expressed in terms of the input x(t) as y(t) = x(t) + ax(t − Tdif f ) where a and Tdif f respectively represent the relative strength and time delay of the second path. (a) Find the transfer function of the multipath system.
Y (s)
= X(s) + ae−sTdif f X(s)
H(s)
=
1 + ae−sTdif f
(b) Express the transfer function of the inverse system as an inﬁnite sum using the formula for summing a geometric series.
H −1 (s) =
1 H(s)
= =
1 1 + ae−sTdif f ∞ n −ae−sTdif f n=0
(c) Determine the impulse response of the inverse system. What condition must be satisﬁed for the inverse system to be both stable and causal? 36
hinv (t)
=
∞
n
(−a) δ(t − nTdif f )
n=0
a < 1 for the system to be both stable and causal. (d) Find a stable inverse system assuming the condition determined in part (c) is violated. The following system is stable, but not causal.
H inv (s)
=
1 1 + ae−sTdif f
Use long division in the positive powers of esTdif f , i.e., divide 1 by ae−sTdif f + 1. This yields:
H inv (s) hinv (t)
n ∞ 1 ensTdif f − a n=1 n ∞ 1 − = − δ(t + nTdif f ) a n=1 = −
6.57. In Section 2.12 we derived blockdiagram descriptions for systems described by linear constantcoeﬃcient diﬀerential equations by rewriting the diﬀerential equation as an integral equation. Consider the secondorder system with the integral equation description y(t) = −a1 y (1) (t) − a0 y (2) (t) + b2 x(t) + b1 x(1) (t) + b0 x(2) (t) Recall that v (n) (t) is the nfold integral of v(t) with respect to time. Use the integration property to take the Laplace transform of the integral equation and derive the direct form I and II block diagrams for the transfer function of this system. Y (s)
= −a1
Y (s) Y (s) X(s) X(s) − a0 2 + b2 X(s) + b1 + b0 2 s s s s
x(t)
y(t)
b1
−a1
b0
−a0 37
Figure P6.57. Direct Form I
By the properties of linear systems, H1 (s) can be interchanged with H2 (s), which leads to direct form II after combining the integrators.
X(s) X(s)
H1(s)
H2(s)
H2(s)
H1(s)
Y(s) Y(s)
Figure P6.57. Properties of Linear Systems
x(t)
y(t)
−a1
b1
−a0
b0
Figure P6.57. Direct Form II
Solutions to Advanced Problems 6.58. Prove the initial value theorem by assuming x(t) = 0 for t < 0 and taking theLaplace transform of the Taylor series expansion of x(t) about t = 0+ . Assume that there exists function f (t) such that x(t) = f (t)u(t). The Taylor series expansion of f (t) is
f (t)
=
∞ f (n) (a) (t − a)n n! n=0
= f (0+ ) +
f (0+ ) f (0+ ) (t − 0+ ) + (t − 0+ )2 + ... 1! 2!
Assuming the expansion of t is around a = 0+ . The Laplace transform for x(t) is thus: x(t)
= f (0+ )u(t) +
f (0+ ) f (0+ ) (t − 0+ )u(t) + (t − 0+ )2 u(t) + ... 1! 2! 38
X(s) sX(s)
f (0+ ) f (0+ ) 2f (0+ ) + − + ... s s2 s3 f (0+ ) 2f (0+ ) − + ... = f (0+ ) + s s2
=
Thus: lim sX(s) = f (0+ )
s→∞
6.59. The system with impulse response h(t) is causal and stable and has a rational transfer function. Identify the conditions on the transfer function so that the system with impulse response g(t) is stable and causal, where d (a) g(t) = dt h(t)
g(t) =
L d h(t) ←−−−→ G(s) = sH(s) − h(0− ) = sH(s) dt
All poles of H(s) are in the left half plane, so no conditions are needed. t (b) g(t) = −∞ h(τ )dτ
t
g(t) =
h(τ )dτ −∞
L
1 ←−−−→ G(s) = s
0−
h(τ )dτ + −∞
H(s) H(s) = s s
H(s) must have at least one zero at s = 0 for the transfer function to be stable.
6.60. Use the continuoustime representation xδ (t) for the discretetime signal x[n] introduced in Section 4.4 to determine the Laplace transforms of the following discretetime signals.
xδ (t) =
∞
L
x[n]δ(t − nTs ) ←−−−→ Xδ (s) =
n=−∞
(a) x[n] =
∞ n=−∞
1, −2 ≤ n ≤ 2 0, otherwise Xδ (s) = e2sTs + esTs + 1 + e−sTs + e−2sTs
(b) x[n] = (1/2)n u[n]
Xδ (s)
= =
∞
x[n]e−snTs
n=−∞ ∞ n=0
39
1 2
n
e−snTs
x[n]e−snTs
=
∞ 1
2
n=0
=
−sTs
n
e
1 1−
1 −sTs 2e
(c) x[n] = e−2t u(t)t=nT Xδ (s)
= =
∞
e−2nTs e−snTs
n=0 ∞
e−Ts (2+s)
n
n=0
=
1 1−
e−Ts (2+s)
6.61. The autocorrelation function for a signal x(t) is deﬁned as ∞ r(t) = x(τ )x(t + τ )dτ −∞
(a) Write r(t) = x(t) ∗ h(t). Express h(t) in terms of x(t). The system with impulse response h(t) is called a matched ﬁlter for x(t).
let: h(t)
= x(−t) ∞ h(t) ∗ x(t) = h(τ )x(t − τ )dτ −∞ ∞ x(−τ )x(t − τ )dτ = −∞
let γ = −τ −∞ = − x(γ)x(t + γ)dγ ∞ ∞ x(γ)x(t + γ)dγ = −∞
(b) Use the result from part (a) to ﬁnd the Laplace transform of r(t).
x(−t)
L
←−−−→ X(−s) L
r(t) = x(t) ∗ x(−t) ←−−−→ R(s) = X(s)X(−s)
(c) If x(t) is real and X(s) has two poles, one of which is located at s = σp + jωp , determine the location of all the poles of R(s). Since x(t) is real, then the poles of X(s) are conjugate symmetric, thus s = σp ± jωp . Therefore the poles of X(−s) are s = −σp ± jωp , which implies that the poles of R(s) are at s = σp ± jωp , − σp ± jωp . 40
6.62. Suppose a system has M poles at dk = αk + jβk and M zeros at ck = −αk + jβk . That is, the pole and zero locations are symmetric about the jωaxis. (a) Show that the magnitude response of any system that satisﬁes this condition is unity. Such a system is termed an allpass system since it passes all frequencies with unit gain.
M H(s)
=
k=1 (s
− ck )
k=1 (s
− dk )
M M
H(jω) =
k=1
jω − αk − jβk 
k=1
jω + αk − jβk 
k=1
j(ω − βk ) − αk 
M M
H(jω) =
M
k=1 j(ω − βk ) + αk  M (ω − βk )2 − αk2 H(jω) = k=1 M (ω − βk )2 + αk2 k=1 H(s) = 1
(b) Evaluate the phase response of a single real polezero pair, that is, sketch the phase response of where α > 0.
π 2,
=
H(jω)
=
s−α s+α jω − α jω + α
ω ω − arctan α α ω H(jω) = π − 2 arctan α H(jω)
For α = 1, then H(jα) =
H(s)
= π − arctan
and H(−jα) =
3π 2 .
41
s−α s+α
Plot of ∠ H(jω) 7
6
5
4
3
2
1
0 −10
−8
−6
−4
−2
0 ω
2
4
6
8
10
Figure P6.62. Phase Plot of H(jω)
6.63. Consider the nonminimum phase system described by the transfer function H(s) =
(s + 2)(s − 1) (s + 4)(s + 3)(s + 5)
(a) Does this system have a stable and causal inverse system? The zeros of H(s) : s = −2, 1. Since one of the zeros is in the right half plane, the inverse system can not be stable and causal. (b) Express H(s) as the product of a minimum phase system, Hmin (s), and an allpass system, Hap (s) containing a single pole and zero. (See Problem 6.62 for the deﬁnition of an allpass system.) (s + 2)(s − 1) (s + 4)(s + 3)(s + 5) s−1 Hap (s) = s+1 H(s) = Hmin (s)Hap (s)
Hmin (s)
=
inv inv (c) Let Hmin (s) be the inverse system for Hmin (s). Find Hmin (s). Can it be both stable and causal?
inv Hmin (s)
=
(s + 4)(s + 3)(s + 5) (s + 2)(s − 1)
inv The poles of Hmin (s) are: s = −1, −2. All are in the left half plane, so hinv min (t) can be both causal and stable.
42
inv (d) Sketch the magnitude response and phase response of the system H(s)Hmin (s).
s−1 s+1 = Hap (s) jω − 1 Hap (jω) = jω + 1 Hap (jω) = 1
inv H(s)Hmin (s)
Hap (jω)
=
= π − arctan(ω) − arctan(ω) = π − 2 arctan(ω)
Magnitude and phase plot of Hap(ω) 2
Magnitude
1.5
1
0.5
0 −10
−8
−6
−4
−2
0
2
4
6
8
10
−8
−6
−4
−2
0 ω
2
4
6
8
10
7 6
Phase
5 4 3 2 1 0 −10
Figure P6.63. Magnitude and Phase Plot of Hap (jω) (e) Generalize your results from parts (b) and (c) to an arbitrary nonminimum phase system H(s) and inv (s). determine the magnitude response of the system H(s)Hmin Generalization for H(s) = H (s)(s − c) where H (s) is the minimum phase part, assume Re(c) > 0 Hmin (s) Hap (s) inv (s) Hmin
= H (s)(s + c) s−c = s+c 1 = H (s)(s + c)
inv (s) H(s)Hmin
= Hap (s) jω − c Hap (jω) = jω + c Hap (jω) = 1 43
Hap (jω)
= π − arctan
ω c
6.64. An N th order lowpass Butterworth ﬁlter has squared magnitude response H(jω)2 =
1 1 + (jω/jωc )2N
The Butterworth ﬁlter is said to be maximally ﬂat because the ﬁrst 2N derivatives of H(jω)2 are zero at ω = 0. The cutoﬀ frequency, deﬁned as the value for which H(jω)2 = 1/2, is ω = ωc . Assuming the impulse response is real, then the conjugate symmetry property of the Fourier transform may be used to write H(jω)2 = H(jω)H ∗ (jω) = H(jω)H(−jω). Noting that H(s)s=jω = H(jω), we conclude that the Laplace transform of the Butterworth ﬁlter is characterized by the equation H(s)H(−s) =
1 1 + (s/jωc )2N
(a) Find the poles and zeros of H(s)H(−s) and sketch them in the splane. The roots of the denominator polynomial are located at the following points in the splane:
1
s = jωc (−1) 2N = ωc ejπ
(2k+N −1) 2N
for k = 0, 1, ..., 2N − 1
Assuming N = 3 and ωc = 1: Pole−Zero Map 1 0.8 0.6 0.4
Imag Axis
0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1.5
−1
−0.5
0 Real Axis
0.5
1
1.5
Figure P6.64. (a) PoleZero Plot of H(s)H(−s) (b) Choose the poles and zeros of H(s) so that the impulse response is both stable and causal. Note that if sp is a pole or zero of H(s), then −sp is a pole or zero of H(−s). From the graph in part (a), picking the poles in the left half plane makes the impulse response stable 44
and causal. This implies selecting the poles for k = 0, 1, ..., N − 1. (c) Note that H(s)H(−s)s=0 = 1. Find H(s) for N = 1 and N = 2. For N = 1, the poles are at:
s0
= ωc
s1
= −ωc
H(s)
which implies: 1 = s + ωc
For N = 2, the poles are at: 1
s0
= ωc ejπ 4
s1
= ωc ejπ 4
s2
= ωc ejπ 4
s3
= ωc ejπ 4
3
5
7
H(s) =
which implies: 1 3
5
(s − ωc ejπ 4 )(s − ωc ejπ 4 )
(d) Find the thirdorder diﬀerential equation that describes a Butterworth ﬁlter with cutoﬀ frequency ωc = 1. For N = 3, the 2N = 6 poles of H(s)H(−s) are located on a circle of unit radius with angular spacing of 60 degrees. Hence the left half plane poles of H(s) are:
s1 s2 s3
√ 3 1 = − +j 2 2 = −1 √ 3 1 = − −j 2 2
The transfer function is therefore:
H(s)
= =
1 1 2
(s + 1)(s + + j 1 s3 + 2s2 + 2s + 1
√
3 2 )(s
−
1 2
−j
√ 3 2 )
Which implies d2 d d3 y(t) + 2 y(t) + 2 y(t) + y(t) dt3 dt2 dt
45
= x(t)
6.65. It is often convenient to change the cutoﬀ frequency of a ﬁlter, or change a lowpass ﬁlter to a highpass ﬁlter. Consider a system described by the transfer function H(s) =
1 (s + 1)(s2 + s + 1)
(a) Find the poles and zeros and sketch the magnitude response of this system. Determine whether this system is lowpass or highpass, and ﬁnd the cutoﬀ frequency (the value of ω for which H(jω) = 1/ (2). −1 ± = −1, 2
poles at s :
√
3
This system is lowpass with cutoﬀ frequency at ωc = 1. (a) Magnitude response of H(jω) 0
−10
H(jω) (dB)
−20
−30
−40
−50
−60
−70 −10
−8
−6
−4
−2
0 ω rads/sec
2
4
6
8
10
Figure P6.65. (a) Magnitude response of H(s), 20log10 H(jω). (b) Perform the transformation of variables in which s is replaced by s/10 in H(s). Repeat part (a) for the transformed system.
H(s) poles at s :
=
1 s ( 10
+
s2 1)( 10
+ √ = −10, −5 ± 5
This system is lowpass with cutoﬀ frequency at ωc = 10. 46
s 10
+ 1)
(b) Magnitude response of H(jω) 0
−5
−10
H(jω) (dB)
−15
−20
−25
−30
−35
−40 −40
−30
−20
−10
0 ω rads/sec
10
20
30
40
Figure P6.65. (b) Magnitude response of H(s), 20log10 H(jω). (c) Perform the transformation of variables in which s is replaced by 1/s in H(s). Repeat part (a) for the transformed system.
1 ( 1s + 1)( s12 + √ −1 ± 3 poles at s : = −1, 2 Three zeros at s : = 0 H(s)
=
This system is a high pass ﬁlter with cutoﬀ frequecy ωc = 1. 47
1 s
+ 1)
(c) Magnitude response of H(jω) 0
−10
−20
−30
H(jω) (dB)
−40
−50
−60
−70
−80
−90
−100 −3
−2
−1
0 ω rads/sec
1
2
3
Figure P6.65. (c) Magnitude response of H(s), 20log10 H(jω). (d) Find the transformation that converts H(s) to a highpass system with cutoﬀ frequency ω = 100. Replace s by 100 s :
H(s)
=
1 100 ( 100 + 1)( s s2 + 48
100 s
+ 1)
(d) Magnitude response of H(jω) 0
−10
−20
−30
H(jω) (dB)
−40
−50
−60
−70
−80
−90
−100 −200
−150
−100
−50
0 ω rads/sec
50
100
150
200
Figure P6.65. (d) Magnitude response of H(s), 20log10 H(jω).
Solutions to Computer Experiments 6.66. Use the MATLAB command roots to determine the poles and zeros of the following systems: s2 +2 (a) H(s) = s3 +2s 2 −s+1 (b) H(s) = (c) H(s) =
s3 +1 s4 +2s2 +1 4s2 +8s+10 2s3 +8s2 +18s+20
P6.66 : ======= Part (a) : ———ans = 0 + 1.4142i 0  1.4142i ans = 2.5468 0.2734 + 0.5638i 0.2734  0.5638i Part (b) : 49
———ans = 1.0000 0.5000 + 0.8660i 0.5000  0.8660i ans = 0.0000 + 1.0000i 0.0000  1.0000i 0.0000 + 1.0000i 0.0000  1.0000i Part (c) : ———ans = 1.0000 1.0000 ans = 1.0000 1.0000 2.0000
+ 1.2247i  1.2247i + 2.0000i  2.0000i
6.67. Use the MATLAB command pzmap to plot the poles and zeros for the following systems: s3 +1 (a) H(s) = s4 +2s 2 +1 1 2 1 (b) A = , b= , c = 0 1 , D = [0] 1 −6 2 50
P6.67(a) 1.5
double pole
1
Imag Axis
0.5
0
−0.5
double pole
−1
−1.5 −1
−0.5
0
0.5
Real Axis
Figure P6.67. (a) PoleZero Plot of H(s) P6.67(b) 1 0.8 0.6 0.4
Imag Axis
0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −7
−6
−5
−4
−3
−2 Real Axis
−1
0
1
2
Figure P6.67. (b) PoleZero Plot
6.68. Use the MATLAB command freqresp evaluate and plot the magnitude and phaseresponses for 51
Examples 6.23 and 6.24. P6.68 : Ex 6.23 5
H(w)
4 3 2 1
−10
−8
−6
−4
−2
0 w:rad/s
2
4
6
8
10
2
4
6
8
10
P6.68 : Ex 6.24 3
arg(H(w)):rad
2 1 0 −1 −2 −10
−8
−6
−4
−2
0 w:rad/s
Figure P6.68. Magnitude and phase plot for Ex 6.23 & 6.24
6.69. Use the MATLAB command freqresp evaluate and plot the magnitude and phase responses for Problem 6.53. 52
P6.69(a) 0.5
H(w)
0.4 0.3 0.2 0.1 0 −50
−40
−30
−20
−10
0 w:rad/s
10
20
30
40
50
−40
−30
−20
−10
0 w:rad/s
10
20
30
40
50
1.5
< (H(w)):rad
1 0.5 0 −0.5 −1 −1.5 −50
Figure P6.69. Magnitude and phase plot for Prob 6.53 (a) P6.69(b) 50
H(w)
40 30 20 10 0 −50
−40
−30
−20
−10
0 w:rad/s
10
20
30
40
50
−40
−30
−20
−10
0 w:rad/s
10
20
30
40
50
1.5
< (H(w)):rad
1 0.5 0 −0.5 −1 −1.5 −50
Figure P6.69. Magnitude and phase plot for Prob 6.53 (b)
53
P6.69(c)
H(w)
1.5
1
0.5 −50
−40
−30
−20
−10
0 w:rad/s
10
20
30
40
50
−40
−30
−20
−10
0 w:rad/s
10
20
30
40
50
3
< (H(w)):rad
2 1 0 −1 −2 −50
Figure P6.69. Magnitude and phase plot for Prob 6.53 (c)
6.70. Use your knowledge of the eﬀect of poles and zeros on the magnitude response to design systems having the speciﬁed magnitude response.Place poles and zeros in the splane, and evaluate the corresponding magnitude response using the MATLAB command freqresp. Repeat this process until you ﬁnd pole and zero locations that satisfy the speciﬁcations. (a) Design a highpass ﬁlter with two poles and two zeros that satisﬁes H(j0) = 0, 0.8 ≤ H(jω) ≤ 1.2 for ω > 100π, and has real valued coeﬃcients. Two conjugate poles are needed around the transition, which implies one possible solution is :
H(s)
=
s2 (s + 25 + j10π)(s + 25 − j10π)
(b) Design a lowpass ﬁlter with real valued coeﬃcients that satisﬁes 0.8 ≤ H(jω) ≤ 1.2 for ω < π, and H(jω) < 0.1 for ω > 10π. One possible solution is:
H(s)
=
(s − j50)(s + j50) (s + 2 + jπ)(s + 2 − jπ)
54
P6.70(a) : HPF 1
H(w)
0.8 0.6 0.4 0.2 0
0
50
100
150
200
250 w:rad/s
300
350
400
450
500
30
35
40
45
50
P6.70(b) : LPF
H(w)
0.8 0.6 0.4 0.2 0
0
5
10
15
20
25 w:rad/s
Figure P6.70. Magnitude response
6.71. Use the MATLAB command bode to ﬁnd the bode diagrams for the systems in Problem 6.55. Bode Diagram Gm = Inf, Pm = 78.63 deg (at 4.4561 rad/sec)
Magnitude (dB)
20
0
−20
−40
−60 0
Phase (deg)
−45
−90
−135
−180 −1 10
0
1
10
10 Frequency (rad/sec)
55
2
10
Figure P6.71. (a) Bode Plot of H(s) Bode Diagram Gm = Inf, Pm = 52.947 deg (at 2.1553 rad/sec)
Magnitude (dB)
50
0
−50
−100
Phase (deg)
−120
−150
−180 −1 10
0
1
10
10
2
10
Frequency (rad/sec)
Figure P6.71. (b) Bode Plot of H(s) Bode Diagram Gm = 4.0836 dB (at 1.7322 rad/sec), Pm = 17.37 deg (at 1.387 rad/sec)
Magnitude (dB)
20
0
−20
−40
−60
Phase (deg)
0
−90
−180
−270 −1 10
0
1
10 Frequency (rad/sec)
10
Figure P6.71. (c) Bode Plot of H(s) 56
Bode Diagram Gm = Inf, Pm = 157.6 deg (at 10.099 rad/sec) 10
Magnitude (dB)
0 −10 −20 −30 −40
Phase (deg)
90
0
−90
−180 −1 10
0
1
10
10
2
10
Frequency (rad/sec)
Figure P6.71. (d) Bode Plot of H(s) Bode Diagram Gm = Inf, Pm = Inf −15
Magnitude (dB)
−20 −25 −30 −35 −40
Phase (deg)
45
0
−45
−90 −1 10
0
1
10
10
2
10
Frequency (rad/sec)
Figure P6.71. (e) Bode Plot of H(s)
6.72. Use the MATLAB command ss to ﬁnd statevariable descriptions for the systems in Problem 6.48. 57
P6.72 : ======= Part (a) : ========== a= x1 x2 x1 3 0 x2 1 0 b= u1 x1 1 x2 0 c= x1 x2 y1 0 1 d= u1 y1 0 Continuoustime model. Part (b) : ========== a= x1 x2 x1 2 2 x2 4 0 b= u1 x1 2 x2 0 c= x1 x2 y1 3 0 d= 58
u1 y1 0 Continuoustime model. Part (c) : ========== a= x1 x2 x3 x1 5 0.875 0.09375 x2 8 0 0 x3 0 4 0 b= u1 x1 0.5 x2 0 x3 0 c= x1 x2 x3 y1 0 0.5 0.25 d= u1 y1 0 Continuoustime model.
6.73. Use the MATLAB command tf to ﬁnd transfer function descriptions for the systems in Problem 6.49. P6.73 : ======= Part (a) : ========== Transfer function: s+3 s2 +3s+2
Part (b) : ========== Transfer function: 2s−1 s2 +5s−8
59
60
Solutions to Additional Problems 7.17. Determine the ztransform and ROC for the following time signals: Sketch the ROC, poles, and zeros in the zplane. (a) x[n] = δ[n − k], k > 0
X(z)
=
∞
x[n]z −n
n=−∞
= z −k , z = 0
Im
k multiple
Re
Figure P7.17. (a) ROC (b) x[n] = δ[n + k], k > 0
X(z)
= z k , all z
Im
k multiple
Re
Figure P7.17. (b) ROC
1
(c) x[n] = u[n]
X(z)
=
∞
z −n
n=0
=
1 , z > 1 1 − z −1
Im
Re
1
Figure P7.17. (c) ROC (d) x[n] =
1 n 4
(u[n] − u[n − 5])
X(z)
=
4 1
z −1
n
4 1 −1 5 1 − 4z 1 − 14 z −1 5 z 5 − 14 , all z z 4 (z − 14 ) 4 poles at z = 0, 1 pole at z = 0
n=0
=
=
5 zeros at z = 14 ejk
2π 5
k = 0, 1, 2, 3, 4
Note zero for k = 0 cancels pole at z = 2
1 4
Im
0.25
Re
Figure P7.17. (d) ROC (e) x[n] =
1 n 4
u[−n]
X(z)
n 0 1 −1 = z 4 n=−∞ =
∞
n
(4z)
n=0
=
1 1 , z < 1 − 4z 4
Im
0.25
Re
Figure P7.17. (e) ROC (f) x[n] = 3n u[−n − 1]
X(z)
= =
−1 n=−∞ ∞ n=1
3
3z −1
1 z 3
n
n
= =
1
1 3z − 13 z
−1 , z < 3 1 − 3z −1 Pole at z = 3 Zero at z = 0
Im
3
Re
Figure P7.17. (f) ROC (g) x[n] =
2 n 3
X(z)
=
n n −1 ∞ 3 −1 2 −1 z z + 2 3 n=−∞ n=0
=
−1 1 3 −1 + 1 − 2z 1 − 23 z −1
=
− 56 z 3 2 < z < , 2 (z − 32 )(z − 23 ) 3
Im
2 3
3 2
Re
Figure P7.17. (g) ROC (h) x[n] =
1 n 2
u[n] +
1 n 4
u[−n − 1] 4
X(z)
=
∞ 1
2
n=0
=
z −1
n +
n −1 1 −1 z 4 n=−∞
1 1 1 1 + , z > and z < 2 4 1 − 12 z −1 1 − 23 z −1 No region of convergence exists.
7.18. Given the following ztransforms, determine whether the DTFT of the corresponding time signals exists without determining the time signal, and identify the DTFT in those cases where it exists: (a) X(z) = 1+ 15z−1 , z > 13 3
ROC includes z = 1, DTFT exists. X(ejΩ )
(b) X(z) =
5 , 1+ 13 z −1
z
2 x[n] is rightsided. √ (2) z < 2 x[n] is twosided. (c) Fig. P7.19 (c)
X(z)
=
1 9 (z − )(z + 1)(z 2 + )C, z < ∞ 2 16
x[n] is stable and leftsided. 7.20. Use the tables of ztransforms and ztransform properties given in Appendix E to determine the ztransforms of the following signals: n (a) x[n] = 12 u[n] ∗ 2n u[−n − 1] n z 1 1 1 u[n] ←−−−→ A(z) = , z > 2 2 1 − 12 z −1 z 1 , z < 2 b[n] = 2n u[−n − 1] ←−−−→ B(z) = 1 − 2z −1 a[n] =
x[n] = a[n] ∗ b[n] X(z)
(b) x[n] = n
1 n 2
u[n] ∗
1 n 4
a[n] =
z
←−−−→ X(z) = A(z)B(z) 1 1 1 < z < 2 , = 1 − 2z −1 2 1 − 12 z −1
u[n − 2] n z 1 1 1 u[n] ←−−−→ A(z) = , z > 2 2 1 − 12 z −1 6
b[n] =
n z 1 1 1 u[n] ←−−−→ B(z) = , z > 4 4 1 − 14 z −1
z −2 1 − 14 z −1 z d x[n] = n[a[n] ∗ b[n]] ←−−−→ X(z) = −z A(z)B(z) dz 2z −2 − 34 z −3 1 , z > X(z) = 2 1 − 34 z −1 + 18 z −2 )2 z
c[n] = b[n − 2] ←−−−→ C(z) =
(c) x[n] = u[−n]
X(z)
= =
1 1 −1
1− z 1 , z < 1 1−z
(d) x[n] = n sin( π2 n)u[−n] π = −n sin(− n)u[−n] 2 −1 d z X(z) = −z dz 1 + z −2 z= 1 z
−z −2 z −1 (−2z −3 ) = −z − 1 + z −2 (1 + z −2 )2 z= 1 x[n]
z
=
z 2z 3 + 2 1+z (1 + z 2 )2
(e) x[n] = 3n−2 u[n] ∗ cos( π6 n + π/3)u[n]
1 z 1 n 9 3 u[n] ←−−−→ A(z) = 9 1 − 3z −3 π π π π b[n] = cos( n + )u[n] = cos( n) cos(π3) − sin( n) sin(π3) u[n] 6 3 6 6 z cos( π3 )(1 + z −1 cos( π6 )) sin( π3 )(z −1 sin( π6 )) b[n] ←−−−→ B(z) = − 1 − 2z −1 cos( π6 ) + z −2 1 − 2z −1 cos( π6 ) + z −2 1 cos( π3 )(1 + z −1 cos( π6 )) − sin( π3 ) sin( π6 )z −1 9 X(z) = A(z)B(z) = 1 − 3z −3 1 − 2z −1 cos( π6 ) + z −2
a[n] =
z
2
7.21. Given the ztransform pair x[n] ←−−−→ z2z−16 with ROC z < 4, use the ztransform properties to determine the ztransform of the following signals: (a) y[n] = x[n − 2] 7
z
y[n] = x[n − 2] ←−−−→ Y (z) = z −2 X(z) =
z2
1 − 16
(b) y[n] = (1/2)n x[n] z 1 z2 y[n] = ( )n x[n] ←−−−→ Y (z) = X(2z) = 2 2 z −4
(c) y[n] = x[−n] ∗ x[n] z 1 z2 y[n] = x[−n] ∗ x[n] ←−−−→ Y (z) = X( )X(z) = z 257z 2 − 16z 4 − 16
(d) y[n] = nx[n] z
y[n] = nx[n] ←−−−→ Y (z) = −z
32z 2 d X(z) = 2 dz (z − 16)2
(e) y[n] = x[n + 1] + x[n − 1] z
y[n] = x[n + 1] + x[n − 1] ←−−−→ Y (z) = (z 1 + z −1 )X(z) =
z3 + z z 2 − 16
(f) y[n] = x[n] ∗ x[n − 3] y[n] = x[n] ∗ x[n − 3]
z
←−−−→ Y (z) = X(z)z −3 X(z) =
z (z 2 − 16)2
z
7.22. Given the ztransform pair n2 3n u[n] ←−−−→ X(z), use the ztransform properties to determine the timedomain signals corresponding to the following z transforms: (a) Y (z) = X(2z)
Y (z) = X(2z)
z 1 1 ←−−−→ y[n] = ( )n x[n] = ( )n n2 3n u[n] 2 2
(b) Y (z) = X(z −1 ) z 1 Y (z) = X( ) ←−−−→ y[n] = x[−n] = n2 3−n u[−n] z
8
(c) Y (z) =
d dz X(z)
z d d −1 X(z) = −z −z X(z) ←−−−→ y[n] = −(n − 1)x[n − 1] = −(n − 1)3 3n−1 u[n − 1] Y (z) = dz dz
(d) Y (z) =
z 2 −z −2 X(z) 2
Y (z) =
z 2 − z −2 X(z) 2 y[n]
z
←−−−→ y[n] = =
1 (x[n + 2] − x[n − 2]) 2
1 (n + 2)2 3n+2 u[n + 2] − (n − 2)2 3n−2 u[n − 2] 2
(e) Y (z) = [X(z)]2 z
Y (z) = X(z)X(z) ←−−−→ y[n] = x[n] ∗ x[n] n y[n] = u[n] k 2 3k (n − k)2 3n−k k=0
3n u[n]
=
n 2 2
k n − 2nk 3 + k 4 k=0
7.23. Prove the following ztransform properties: (a) Time reversal z 1 x[n] ←−−−→ X( ) z y[n] = x[−n] ∞ Y (z) = x[−n]z −n n=−∞
let l = −n
∞
=
l=−∞
1 x[l]( )−l z
1 X( ) z
=
(b) Time shift z
x[n − no ] ←−−−→ z −no X(z) y[n]
=
Y (z)
=
x[n − no ] ∞ x[n − no ]z −n n=−∞
let l = n − no 9
∞
=
x[l]z −(l+no )
l=−∞
= =
z
∞
x[l]z
−l
z −no
l=−∞ −no
X(z)
(c) Multiplication by exponential sequence z z αn x[n] ←−−−→ X( ) α y[n] = αn x[n] ∞ Y (z) = αn x[n]z −n n=−∞ ∞
z x[n]( )−n α n=−∞ z X( ) α
= =
(d) Convolution Let c[n] = x[n] ∗ y[n] z
x[n] ∗ y[n] ←−−−→ X(z)Y (z) ∞ C(z) = (x[n] ∗ y[n]) z −n n=−∞ ∞
=
n=−∞ ∞
=
=
x[p]
X(z)Y (z)
(e) Diﬀerentiation in the zdomain.
nx[n] X(z)
z
d X(z) dz ∞ x[n]z −n
←−−−→ −z =
n=−∞
10
y[n − p]z −(n−p)
n=−∞
x[p]z
p=−∞
∞
∞
X(z)
=
x[p]y[n − p] z −n
p=−∞
p=−∞
∞
−p
Y (z)
Y (z)
z −p
Diﬀerentiate with respect to z and multiply by −z. ∞ z d nx[n]z −n −z X(z) ←−−−→ dz n=−∞ Therefore nx[n]
z
←−−−→ −z
d X(z) dz
7.24. Use the method of partial fractions to obtain the timedomain signals corresponding to the following ztransforms: 1+ 7 z −1 (a) X(z) = (1− 1 z−16)(1+ 1 z−1 ) , z > 12 2
3
x[n] is rightsided
(b) X(z) =
1+ 76 z −1 , (1− 12 z −1 )(1+ 13 z −1 )
X(z)
=
1 7 6
=
X(z)
=
x[n]
=
z
0 12
x[n]
(b) X(z) =
10
1 −k , k=5 k z
= δ[n] + 2δ[n − 6] + 4δ[n − 8]
z > 0
x[n] =
10 1 δ[n − k] k
k=5
(c) X(z) = (1 + z −1 )3 , z > 0
X(z) x[n]
=
(δ[n] + δ[n − 1]) ∗ (δ[n] + δ[n − 1]) ∗ (δ[n] + δ[n − 1])
= δ[n] + 3δ[n − 1] + 3δ[n − 2] + δ[n − 3]
(d) X(z) = z 6 + z 2 + 3 + 2z −3 + z −4 , z > 0
= δ[n + 6] + δ[n + 2] + 3δ[n] + 2δ[n − 3] + δ[n − 4]
x[n]
7.26. Use the following clues to determine the signal x[n] and rational ztransform X(z). (a) X(z) has poles at z = 1/2 and z = −1, x[1] = 1, x[−1] = 1, and the ROC includes the point z = 3/4. Since the ROC includes the point z = 3/4, the ROC is X(z)
=
A 1 −1 2z n
+
1 2
< z < 1.
B 1 + z −1
1− 1 x[n] = A u[n] − B(−1)n u[−n − 1] 2 1 x[1] = 1 = A 2 A = 2 x[−1] = 1 B x[n]
= −1B(−1) =
1 n 1 u[n] − (−1)n u[−n − 1] = 2 2
(b) x[n] is rightsided, X(z) has a single pole, and x[0] = 2, x[2] = 1/2.
x[n] x[0] = 2
= c(p)n u[n] where c and p are unknown constants. = c (p)
0
13
c = 2 1 2 x[2] = = 2 (p) 2 1 p = 2 n 1 u[n] x[n] = 2 2 (c) x[n] is twosided, X(z) has one pole at z = 1/4, x[−1] = 1, x[−3] = 1/4, and X(1) = 11/3. X(z)
A
=
1 −1 4z n
1− 1 x[n] = A 4
+
B 1 − cz −1
u[n] − B(c)n u[−n − 1]
x[−1] = 1 = −Bc−1 1 = −Bc−3 x[−3] = 4 c = 2 = −2 −2 A = 1 + 1−2 1− 4 5 A = 4 n 5 1 u[n] + 2(2)n u[−n − 1] x[n] = 4 4
B 11 X(1) = 3
7.27. Determine the impulse response corresponding to the following transfer functions if (i) the system is stable, or (ii) the system is causal: 2− 32 z −1 (a) H(z) = (1−2z−1 )(1+ 1 −1 z ) 2
H(z)
(i) h[n] is stable, ROC
1 2
=
1 1 + 1 − 2z −1 1 + 12 z −1
< z < 2, ROC includes z = 1. 1 h[n] = −(2)n u[−n − 1] + (− )n u[n] 2
(ii) h[n] is causal, ROC z > 2
1 n n h[n] = 2 + (− ) u[n] 2
14
(b) H(z) =
5z 2 z 2 −z−6
H(z)
=
3 2 + 1 − 3z −1 1 + 2z −1
(i) h[n] is stable, ROC z < 2, ROC includes z = 1.
h[n]
=
[−(3)n − (−2)n ] u[−n − 1]
(ii) h[n] is causal, ROC z > 3
h[n]
(c) H(z) =
[3n + (−2)n ] u[n]
=
4z 1 z 2 − 14 z+ 16
H(z)
=
4z −1 (1 − 14 z −1 )2
(i) h[n] is stable, ROC z > 14 , ROC includes z = 1.
(ii) h[n] is causal, ROC z >
h[n] =
1 16n( )n u[n] 4
h[n] =
1 16n( )n u[n] 4
1 4
7.28. Use a power series expansion to determine the timedomain signal corresponding to the following ztransforms: (a) X(z) = 1− 11z−2 , z > 14 4
X(z)
=
∞ 1 ( z −2 )k 4
k=0
15
∞ 1 ( )k δ[n − 2k] 4 k=0 n 1 2 n even and n ≥ 0 4 = 0 n odd
x[n]
(b) X(z) =
1 , 1− 14 z −2
z
0 Note:
cos(α)
=
∞ (−1)k k=0
X(z)
= =
x[n]
=
(2k)!
∞ (−1)k k=0 ∞ k=0 ∞ k=0
(2k)!
(α)2k
(z −3 )2k
(−1)k −6k z (2k)! (−1)k δ[n − 6k] (2k)!
(d) X(z) = ln(1 + z −1 ), z > 0 Note:
ln(1 + α)
∞ (−1)k−1
=
k
k=0
X(z)
=
x[n] =
∞ (−1)k−1 k=1 ∞ k=1
k
(α)k
(z −1 )k
(−1)k−1 δ[n − k] k
16
7.29. A causal system has input x[n] and output y[n]. Use the transfer function to determine the impulse response of this system. (a) x[n] = δ[n] + 14 δ[n − 1] − 18 δ[n − 2], y[n] = δ[n] − 34 δ[n − 1] 1 1 1 + z −1 − z −1 4 8 3 −1 Y (z) = 1 − z 4 Y (z) H(z) = X(z) 5 − 23 3 = 1 −1 + 1 − 4z 1 + 12 z −1
1 n 1 1 n 5(− ) − 2( ) u[n] h[n] = 3 2 4 X(z)
=
n
(b) x[n] = (−3) u[n], y[n] = 4(2)n u[n] −
X(z)
1 n 2
u[n] 1 1 + 3z −1
=
3 (1 − 2z −1 )(1 − 12 z −1 ) Y (z) H(z) = X(z) −7 10 + = 1 − 2z −1 1 − 12 z −1
1 n n h[n] = 10(2) − 7( ) u[n] 2 Y (z)
=
7.30. A system has impulse response h[n] = is given by
1 n
H(z)
2
u[n]. Determine the input to the system if the output
1
=
1 − 12 z −1
(a) y[n] = 2δ[n − 4]
Y (z) X(z)
2z −4 Y (z) = H(z)
=
= x[n] =
2z −4 − z −5 2δ[n − 4] − δ[n − 5]
17
(b) y[n] = 13 u[n] +
2 3
−1 n 2
u[n] 1 3
Y (z) =
1−
z −1
+
1+
2 3 1 −1 2z
Y (z) H(z)
X(z) =
1 4 1 6 3 + = − + 2 1 − z −1 1 + 12 z −1 1 4 1 1 x[n] = − δ[n] + u[n] + (− )n u[n] 2 6 3 2
7.31. Determine (i) transfer function and (ii) impulse response representations for the systems described by the following diﬀerence equations: (a) y[n] − 12 y[n − 1] = 2x[n − 1]
1 Y (z) 1 − z −1 2
=
2z −1 X(z)
Y (z) X(z) 2z −1 = 1 − 12 z −1 1 h[n] = 2( )n−1 u[n − 1] 2
H(z)
=
(b) y[n] = x[n] − x[n − 2] + x[n − 4] − x[n − 6] Y (z)
=
H(z)
=
1 − z −2 + z −4 − z −6 X(z) Y (z) X(z)
= 1 − z −2 + z −4 − z −6 h[n]
(c) y[n] − 45 y[n − 1] −
16 25 y[n
= δ[n] − δ[n − 2] + δ[n − 4] − δ[n − 6]
− 2] = 2x[n] + x[n − 1]
16 4 Y (z) 1 − z −1 − z −2 5 25
=
(2 + z −1 )X(z)
H(z)
=
Y (z) X(z)
=
2
+
13 −1 5 z − 45 z −1 )2
1 − 4 z −1 (1
5 4 n 13 4 n h[n] = 2( ) + n( ) u[n] 5 4 5
18
7.32. Determine (i) transfer function and (ii) diﬀerenceequation representations for the systems with the following impulse responses: n (a) h[n] = 3 14 u[n − 1]
h[n]
=
H(z)
= =
n−1 1 1 3 u[n − 1] 4 4 3 −1 4z − 14 z −1
1 Y (z) X(z)
Taking the inverse ztransform yields: 1 y[n] − y[n − 1] 4
(b) h[n] =
1 n 3
u[n] +
1 n−2 2
=
3 x[n − 1] 4
u[n − 1] n−1 n 1 1 u[n] + 2 u[n − 1] h[n] = 3 2 H(z)
= =
1 + 32 z −1 − 23 z −2 1 − 56 z −1 + 16 z −2 Y (z) X(z)
Taking the inverse ztransform yields: 1 5 y[n] − y[n − 1] + y[n − 2] 6 6
(c) h[n] = 2
2 n 3
u[n − 1] +
1 n
H(z)
4
= =
3 2 = x[n] + x[n − 1] − x[n − 2] 2 3
[cos( π6 n) − 2 sin( π6 n)]u[n]
1 1 1
4 −1 1 − 14 z −1 cos( π6 ) − 12 z −1 sin( π6 ) 3z + 1 −2 − 23 z −1 1 − z −1 21 cos( π6 ) + 16 z √ √ 5 1 1 1 1 −3 −1 −2 + 12 − 8 3 z + 6 − 4 3 z + 12 z √ 2 1√ 1 1 1 −3 −1 −2 − 3 + 4 3 z + 16 + 6 3 z − 24 z
Taking the ztransform yields: √ √ 1 1 + 16 3 y[n − 2] − 24 y[n − 3] y[n] − 23 + 14 3 y[n − 1] + 16 √ √ 5 1 1 1 1 = x[n] + 12 − 8 3 x[n − 1] + 6 − 4 3 x[n − 2] + 12 x[n − 3]
19
(d) h[n] = δ[n] − δ[n − 5] H(z)
1 − z −5
=
Taking the ztransform yields:
y[n]
= x[n] − x[n − 5]
7.33. (a) Take the ztransform of the stateupdate equation Eq. (2.62) using the timeshift property Eq. (7.13) to obtain ˜ (z) = (zI − A)−1 bX(z) q where
˜ (z) = q
Q1 (z) Q2 (z) .. . QN (z)
is the ztransform of q[n]. Use this result to show that the transfer function of a LTI system is expressed in terms of the statevariable description as H(z) = c(zI − A)−1 b + D q[n + 1] ˜ (z) q ˜ (z)(zI − A) q ˜ (z) q
= Aq[n] + bx[n] = A˜ q(z) + bX(z) = bX(z) =
(zI − A)−1 bX(z)
y[n]
= cq[n] + Dx[n]
Y (z)
= c˜ q(z) + DX(z)
Y (z)
= c(zI − A)−1 bX(z) + DX(z)
H(z)
= c(zI − A)−1 b + D
(b) Determine transfer function and diﬀerenceequation representations for the systems described by the following statevariable descriptions. Plot the pole and zero locations in the zplane. 1 0 −2 0 , b= , c = 1 −1 , D = [1] (i) A = 1 0 2 2 H(z)
= c(zI − A)−1 b + D z 2 − 2z − 54 = z 2 − 14 20
1.5
1
Imaginary Part
0.5
0
−0.5
−1
−1.5 −1
−0.5
0
0.5
1
1.5
2
2.5
Real Part
Figure P7.33. (b)(i) PoleZero Plot 1 1 1 − 2 2 , b = , c = (ii) A = 2 0 − 12 − 14 H(z)
=
1
z2
, D = [0]
2z − 14 z −
3 8
1 0.8 0.6
Imaginary Part
0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1
−0.5
0 Real Part
0.5
Figure P7.33. (b)(ii) PoleZero Plot 21
1
(iii) A =
− 14 − 72
1 8 3 4
, b=
2 2
, c=
0
H(z)
=
1
, D = [0]
2z − 13 2 z 2 − 12 z +
1 4
1.5
1
Imaginary Part
0.5
0
−0.5
−1
−1.5
−1
−0.5
0
0.5
1 1.5 Real Part
2
2.5
3
Figure P7.33. (b)(iii) PoleZero Plot
7.34. Determine whether each of the systems described below are (i) causal and stable and (ii) minimum phase. (a) H(z) = z22z+3 +z− 5 16
3 2 5 1 z=− , 4 4
z=−
zero at: poles at:
(i) Not all poles are inside z = 1, the system is not causal and stable. (ii) Not all poles and zeros are inside z = 1, the system is not minimum phase. (b) y[n] − y[n − 1] − 14 y[n − 2] = 3x[n] − 2x[n − 1] 2 3√ 1± 2 z= 2
zeros at:
z = 0,
poles at:
22
(i) Not all poles are inside z = 1, the system is not causal and stable. (ii) Not all poles and zeros are inside z = 1, the system is not minimum phase. (c) y[n] − 2y[n − 2] = x[n] − 12 x[n − 1]
H(z)
=
zeros at: poles at:
z(z − 12 ) z2 − 2 1 z = 0, √2 z=± 2
(i) Not all poles are inside z = 1, the system is not causal and stable. (ii) Not all poles and zeros are inside z = 1, the system is not minimum phase.
7.35. For each system described below, identify the transfer function of the inverse system, and determine whether it can be both causal and stable. −1 +16z −2 (a) H(z) = 1−8z 1− 1 z −1 + 1 z −2 2
4
H(z)
=
H inv (z)
=
poles at:
(z − 4)2 (z − 12 )2 (z − 12 )2 (z − 4)2 z = 4 (double)
For the inverse system, not all poles are inside z = 1, so the system is not causal and stable. (b) H(z) =
81 z 2 − 100 z 2 −1
H inv (z)
=
poles at:
z2 − 1 81 z 2 − 100 9 (double) z= 10
For the inverse system, all poles are inside z = 1, so the system can be causal and stable. (c) h[n] = 10
−1 n 2
u[n] − 9
−1 n 4
u[n]
H(z)
=
z(z − 2) (z + 12 )(z + 14 )
H inv (z)
=
(z + 12 )(z + 14 ) z(z − 2) 23
poles at:
z = 0, 2
For the inverse system, not all poles are inside z = 1, so the system cannot be both causal and stable. (d) h[n] = 24
1 n 2
u[n − 1] − 30
1 n 3
u[n − 1]
H(z)
=
H inv (z)
=
2(z + 12 ) (z − 12 )(z − 13 ) (z − 12 )(z − 13 ) 2(z + 12 ) 1 z=− 2
pole at:
For the inverse system, all poles are inside z = 1, so the system can be both causal and stable. (e) y[n] − 14 y[n − 2] = 6x[n] − 7x[n − 1] + 3x[n − 2] H inv (z)
= =
poles at:
X(z) Y (z) (z − 12 )(z + 12 ) 6z 2 + 7z + 3 √ 7 ± j 23 z= 12
For the inverse system, all poles are inside z = 1, so the system can be both causal and stable. (f) y[n] − 12 y[n − 1] = x[n]
H inv (z)
=
pole at:
z − 12 z z=0
For the inverse system, all poles are inside z = 1, so the system can be both causal and stable.
7.36. A system described by a rational transfer function H(z) has the following properties: 1) the system is causal; 2) h[n] is real; 3) H(z) has a pole at z = j/2 and exactly one zero; 4) the inverse system ∞ has two zeros; 5) n=0 h[n]2−n = 0; 6) h[0] = 1.
By 2) and 3) 24
z=±
poles at
j 2
By 4) H inv
=
H has two poles. H(z)
1 H 1 ±j π e 2 2 A(1 − Cz −1 ) 1 − z −1 cos( π2 ) + 14 z −2
z= = =
A(1 − Cz −1 ) 1 + 14 z −2
By 5) ∞
h[n]2
−n
=
n=0
∞
h[n]z
−n
n=0
Since H(z)
=
= H(z) z=2
A(1 − Cz −1 ) , 1 + 14 z −2
H(z) = 0 implies C h[n] h[0] = 1 H(z) h[n]
=
2 n n 1 1 π π = A cos( n) − 2A sin( n) u[n] 2 2 2 2 = A 1 − 2z −1 = 1 + 14 z −2
n n 1 π π 1 = cos( n) − 2 sin( n) u[n] 2 2 2 2
(a) Is this system stable? The poles are inside z = 1, so the system is stable. (b) Is the inverse system both stable and causal? No, the inverse system has a pole at z = 2, which is not inside z = 1. (c) Find h[n].
h[n] =
n
n 1 1 π π cos( n) − 2 sin( n) u[n] 2 2 2 2
(d) Find the transfer function of the inverse system. H inv (z)
= =
25
1 H(z) 1 + 14 z −2 1 − 2z −1
7.37. Use the graphical method to sketch the magnitude response of the systems having the following transfer functions: −2 (a) H(z) = 1+z49 z−2 64
H(z)
=
H(ejΩ )
=
1 − j 78 ) 1 7 jΩ (e + j 8 )(ejΩ − j 78 ) (z +
j 78 )(z
Im 7 8
1
Re
Figure P7.37. (a) Graphical method. P7.37 (a) Magnitude Response 4.5
4
3.5
H(ejΩ)
3
2.5
2
1.5
1
0.5 −4
−3
−2
−1
0 Ω
1
2
Figure P7.37. (a) Magnitude Response −1 −2 (b) H(z) = 1+z 3+z 26
3
4
H(z) H(ejΩ ) poles at:
z2 + z1 + 1 3z 2 j2Ω + ejΩ + 1 e = 3ej2Ω z = 0, (double)
=
z = e±j
zeros at:
2π 3
Im
3 1
Re
Figure P7.37. (b) Graphical method. P7.37 (b) Magnitude Response 1
0.9
0.8
0.7
H(ejΩ)
0.6
0.5
0.4
0.3
0.2
0.1
0 −4
−3
−2
−1
0 Ω
1
2
Figure P7.37. (b) Magnitude Response −1 (c) H(z) = 1+(18/10) cos(1+z π )z −1 +(81/100)z −2 4
27
3
4
H(z)
=
H(ejΩ )
=
1 + z −1 (1 −
9 j 10 e
3 4π
z −1 )(1 −
9 −j 34 π −1 z ) 10 e jΩ
ej2Ω + e + (18/10) cos( π4 )ejΩ + (81/100) z = −1 9 ±j π jπ z= e 4e 10 ejΩ
zeros: poles:
Im 3 4 1
Re 9 10 Figure P7.37. (c) Graphical method. P7.37 (c) Magnitude Response 6
5
H(ejΩ)
4
3
2
1
0 −4
−3
−2
−1
0 Ω
1
2
3
4
Figure P7.37. (c) Magnitude Response
7.38. Draw blockdiagram implementations of the following systems as a cascade of secondorder sec28
tions with realvalued coeﬃcients: π
(a) H(z) =
π
π
π
(1− 14 ej 4 z −1 )(1− 14 e−j 4 z −1 )(1+ 14 ej 8 z −1 )(1+ 14 e−j 8 z −1 ) π
π
(1− 12 ej 3 z −1 )(1− 12 e−j 3 z −1 )(1− 34 ej
7π 8
z −1 )(1− 34 e−j
7π 8
z −1 )
H(z)
= H1 (z)H2 (z) 1 −2 1 − 12 cos( π4 )z −1 + 16 z H1 (z) = π −1 1 −2 1 − cos( 3 )z + 4 z H2 (z)
1 −2 1 + 12 cos( π8 )z −1 + 16 z 3 7π −1 9 −2 1 − 2 cos( 8 )z + 16 z
=
X(z)
Y(z) z −1 cos(
3
)
z −1 7 3 cos( ) 2 8
−1 cos( ) 4 2
z −1
z −1 1 16
−1 4
1 16
−9 16
Figure P7.38. (a) Block diagram. π
(b) H(z) =
1 cos( ) 8 2
π
(1+2z −1 )2 (1− 12 ej 2 z −1 )(1− 12 e−j 2 z −1 )
π (1− 38 z −1 )(1− 38 ej 3
π z −1 )(1− 38 ej 3
z −1 )(1+ 34 z −1 )
H(z)
= H1 (z)H2 (z) 1 + 4z −1 + 4z −2 H1 (z) = 9 −2 1 + 38 z −1 − 32 z H2 (z)
=
1 + 14 z −2 1 − 34 cos( π3 )z −1 +
9 −2 64 z
X(z)
Y(z) z −1 −3 8
z −1 3 cos( ) 4 3
4
z −1 9 32
z −1 −9 64
4
Figure P7.38. (b) Block diagram.
29
1 4
7.39. Draw block diagram implementations of the following systems as a parallel combination of secondorder sections with realvalued coeﬃcients: n n n n u[n] + −1 u[n] (a) h[n] = 2 12 u[n] + 2j u[n] + −j 2 2
H(z)
=
2 1 − 12 z −1
+
1 1 1 + + 1 + 12 z −1 1 − 2j z −1 1 + 2j z −1
3 + 12 z −1 2 1 −2 + 1 − 4z 1 + 14 z −2 = H1 (z) + H2 (z)
= H(z)
X(z)
3 z −1 0.5
Y(z)
z −1 0.25 2
z −1
z −1 −0.25 Figure P7.39. (a) Block diagram. (b) h[n] = 2
n jπ 4 2e
1
H(z)
u[n] +
= =
H(z)
n jπ 3 4e
1
u[n] +
n −j π 3 4e
1
1
2
1− 1−
+ 1 jπ 4 z −1 2e √ −1
u[n] + 2
1−
1 jπ 3 z −1 4e
4 − 2z + √1 z −1 + 1 z −2 1− 4 2
+
n −j π 4 2e
1
1 1−
1 −j π 3 z −1 4e
2 − 14 z −1 1 −1 1 −2 + 16 z 4z
= H1 (z) + H2 (z)
30
u[n]
+
2 1−
1 −j π 4 z −1 2e
X(z)
4 1 2
z
−1
2
Y(z)
z −1 −0.25 2
z
−1
0.25
−0.25 z −1
−1/16 Figure P7.39. (b) Block diagram.
7.40. Determine the transfer function of the system depicted in Fig. P7.40.
X(z)
Y(z) H1(z)
H2(z)
H3(z) Figure P7.40. System diagram
H1 (z)
=
1 − 2z −1 1 − 14 z −2
z −2 1 + 34 z −1 − 18 z −2 1 H3 (z) = 1 + 23 z −1 H(z) = H1 (z)H2 (z) + H1 (z)H3 (z) H2 (z)
=
31
7.41. Let x[n] = u[n + 4]. (a) Determine the unilateral ztransform of x[n].
zu
x[n] = u[n + 4] ←−−−→ X(z) =
∞
x[n]z −n
n=0
X(z)
∞
=
z −n
n=0
1 1 − z −1
=
(b) Use the unilateral ztransform timeshift property and the result of (a) to determine the unilateral ztransform of w[n] = x[n − 2]. zu
w[n] = x[n − 2] ←−−−→ W (z) = x[−2] + x[−1]z −1 + z −2 X(z) z −2 W (z) = 1 + z −1 + 1 − z −1 7.42. Use the unilateral ztransform to determine the forced response, the natural response, and the complete response of the systems described by the following diﬀerence equations with the given inputs and initial conditions. n (a) y[n] − 13 y[n − 1] = 2x[n], y[−1] = 1, x[n] = ( −1 2 ) u[n]
X(z)
=
1 1 + 12 z −1
1 −1 z Y (z) + 1 = 2X(z) 3 1 1 Y (z) 1 − z −1 + 2X(z) = 3 3 1 1 1 +2 X(z) Y (z) = 1 −1 3 1 − 3z 1 − 13 z −1
Y (z) −
Y (n) (z)
Y (f ) (z)
Natural Response Y (n) (z)
=
y (n) [n] =
1 1 3 1 − 13 z −1 n 1 1 u[n] 3 3
Forced Response Y (f ) (z)
6 5 1 −1 2z
=
+
4 5 − 13 z −1 n
1+ 1
n 6 4 1 y (f ) [n] = + − 5 2 5 32
1 3
u[n]
Complete Response y[n] =
(b) y[n] − 19 y[n − 2] = x[n − 1], X(z) Y (z) −
n n
1 6 17 1 − u[n] + 5 2 15 3
y[−1] = 1, y[−2] = 0, x[n] = 2u[n]
=
2 1 − z −1
1 −2 z Y (z) + z −1 = z −1 X(z) 9 1 −1 1 −2 = Y (z) 1 − z z + z −1 X(z) 9 9 1 z −1 z −1 X(z) Y (z) = + 9 1 − 19 z −2 1 − 19 z −2 Y (n) (z)
Y (f ) (z)
Natural Response Y (n) (z)
= =
y (n) [n]
=
1 z −1 9 1 − 19 z −2 1 1 1 1 − 6 1 − 13 z −1 6 1 + 13 z −1
n n 1 1 1 u[n] − − 6 3 3
Forced Response Y (f ) (z) y (f ) [n]
−
1 − z −1 1
9 3 1 = − − 4 4 3
Complete Response y[n]
9 4
=
=
9 3 − 4 4
−
1 3
3 4 + 13 z −1 n
3 2 1 −1 3z
−
1− n 3 1 u[n] − 2 3
n −
3 2
n
n n 1 1 1 1 u[n] + u[n] − − 3 6 3 3
(c) y[n] − 14 y[n − 1] − 18 y[n − 2] = x[n] + x[n − 1], y[−1] = 1, y[−2] = −1, x[n] = 3n u[n] X(z) Y (z) −
=
1 1 − 3z −1
1 −2 1 −1 z Y (z) + 1 − z Y (z) + z −1 − 1 = X(z) + z −1 X(z) 4 8 1 −1 1 −2 1 1 −1 Y (z) 1 − z − z + z + (1 + z −1 )X(z) = 4 8 8 8 1 (1 + z −1 )X(z) 1 + z −1 Y (z) = + 8 (1 − 12 z −1 )(1 + 14 z −1 ) (1 − 12 z −1 )(1 + 14 z −1 ) Y (n) (z)
Natural Response Y (n) (z)
= 33
1 1 1 1 − 4 1 − 12 z −1 8 1 + 14 z −1
Y (f ) (z)
y (n) [n]
n n 1 1 1 2 u[n] − − 8 2 4
=
Forced Response 1 − 25 − 13 + 1 − 3z −1 1 − 12 z −1 1 + 14 z −1 n n
1 96 2 1 1 n (3) − − u[n] = − 65 5 2 13 4
Y (f ) (z)
96 65
=
y (f ) [n] Complete Response
y[n] =
+
3 96 n (3) − 65 20
n n 21 1 1 u[n] − − 2 104 4
Solutions to Advanced Problems 7.43. Use the ztransform of u[n] and the diﬀerentiation in the zdomain property to derive the formula for evaluating the sum ∞ n2 an n=0
assuming a < 1.
X(z) =
∞
x[n]z −n
n=0
let x[n] = an u[n] ∞ d −nx[n]z −(n−1) X(z) = dz n=−∞ d2 X(z) = dz 2 = d2 dz 2
1 1 − az −1
Evaluate at z = 1 d2 1 2 −1 dz 1 − az z=1
=
∞ n=−∞ ∞
n(n − 1)x[n]z −(n−2) n2 x[n]z −(n−2) −
n=−∞ ∞ 2 n −(n−2)
n a z
∞ n=0
−
∞
n2 an
=
n=0
=
nan z −(n−2)
n=0
n2 an −
∞
nan
n=0
 1 1 d2 d − 2 −1 −1 dz 1 − az dz 1 − az z=1 z=1 d dz X(z) z=1
∞
nx[n]z −(n−2)
n=−∞
n=0
=
∞
2a a + 3 (1 − a) (1 − a)2 34
=
3a − a2 (1 − a)3
7.44. A continuoustime signal y(t) satisﬁes the ﬁrstorder diﬀerential equation d y(t) + 2y(t) = x(t) dt d Use the approximation dt y(t) ≈ [y(nTs ) − y((n − 1)Ts )] /Ts to show that the sampled signal y[n] = y(nTs ) satisﬁes the ﬁrstorder diﬀerence equation
y[n] + αy[n − 1] = v[n] Express α and v[n] in terms of Ts and x[n] = x(nTs ).
y[n] − y[n − 1] + 2y[n] = x[n] Ts 1 1 y[n] + 2 − y[n − 1] = x[n] Ts Ts 1 Ts y[n] − y[n − 1] = x[n] 1 + 2Ts 1 + 2Ts 1 α = − 1 + 2Ts Ts v[n] = x[n] 1 + 2Ts
7.45. The autocorrelation signal for a realvalued causal signal x[n] is deﬁned as rx [n] =
∞
x[l]x[n + l]
l=0
Assume the ztransform of rx [n] converges for some values of z. Find x[n] if 1 Rx (z) = (1 − αz −1 )(1 − αz) where α < 1.
rx [n] = x[n] ∗ x[−n] Let y[n] = x[−n] rx [n] = =
∞ k=−∞ ∞ k=−∞
let p = −k 35
y[k]x[n − k] x[−k]x[n − k]
rx [n] = = =
∞ p=−∞ ∞
x[p]x[n + p] x[l]x[n + l]
l=−∞ ∞
x[l]x[n + l]
l=0
since x[l] = 0, for l < 0 1 Rx (z) = X( )X(z) z 1 1 = 1 − αz 1 − αz −1 Implies 1 X(z) = 1 − αz −1 x[n] = αn u[n]
7.46. The crosscorrelation of two realvalued signals x[n] and y[n] is expressed as rxy (n) =
∞
x[l]y[n + l]
l=−∞
(a) Express rxy [n] as a convolution of two sequences.
rxy [n]
= x[n] ∗ y[−n], see previous problem
(b) Find the ztransform of rxy [n] as a function of the ztransforms of x[n] and y[n].
Rxy (z)
1 = X(z)Y ( ) z
7.47. A signal with rational ztransform has even symmetry, that is, x[n] = x[−n]. (a) What constraints must the poles of such a signal satisfy?
z
←−−−→ X(z) z 1 x[−n] ←−−−→ X( ) z Implies 1 X(z) = X( ) z x[n]
36
This means if there is a pole at zo , there must also be a pole at
1 zo .
Hence poles occur in reciprocal pairs.
∞ (b) Show that the ztransform corresponds to a stable system if and only if n=−∞ x[n] < ∞. The reciprocal poles are zo , z1o assume they take the following form: zo 1 zo
= ro ejθo 1 −jθo = e ro
If zo is inside z = 1, its ztransform is rightsided and stable. For the pole at z1o , its corresponding ztransform is either rightsided unstable, or leftsided stable. For convergence, the ROC must include the unit circle, z = 1, which means the ztransforms are exponentially decaying as they approach ∞, − ∞ respectively. (c) Suppose X(z) =
2 − (17/4)z −1 (1 − (1/4)z −1 )(1 − 4z −1 )
Determine the ROC and ﬁnd x[n]. For the system to be stable, the pole at z = 14 must be rightsided, and the pole at z = 4 must be left sided so their ztransforms are exponentially decaying as they approach ∞, − ∞ respectively. This implies the ROC is 14 < z < 4
7.48. Consider a LTI system with transfer function H(z) =
1 − a∗ z , z−a
a < 1
Here the pole and zero are a conjugate reciprocal pair. (a) Sketch a polezero plot for this system in the zplane.
Let a = a ej
Im
j then 1 = 1 e a * a a1
1 a
Re
Figure P7.48. (a) PoleZero plot.
37
(b) Use the graphical method to show that the magnitude response of this system is unity for all frequency. A system with this characteristic is termed an allpass system.
H(ejΩ )
1 − a∗ ejΩ = jΩ e −a = 1 − a∗ ejΩ
1 − a
ejΩ
As shown below, H(ejΩ ) = 1 for all Ω.
H1(ej )
Im
1+a 1−a
a1
1 a
Re
+
H2(ej )
Im 1 1−a
1 1+a
a1
1 a
Re
Figure P7.48. (b) Magnitude Response.
38
+
H1(ej ) H2(ej ) 1
+
Figure P7.48. (b) Magnitude Response. (c) Use the graphical method to sketch the phase response of this system for a = 12 .
H(ejΩ ) ! arg H(ejΩ )
1 − 12 ejΩ ejΩ − 12 # # " " 1 1 = arg 1 − ejΩ − arg ejΩ − 2 2 " # ! 1 = π + arg ejΩ − 2 − arg ejΩ − 2
=
Im
Im pole
zero
0.5
Re
Figure P7.48. (c) Pole/Zero graphical method.
39
2
Re
zero
2
2
Figure P7.48. (c) Zero phase response.
pole
Figure P7.48. (c) Pole phase response.
H(e j )
Figure P7.48. (c) Phase response. (d) Use the result from (b) to prove that any system with a transfer function of the form H(z) =
P $ 1 − a∗k z z − ak
k=1
40
ak  < 1
corresponds to a stable and causal allpass system. Since ak  < 1, if the system is causal, then the system is stable. P $ 1 − a∗k ejΩ ejΩ − ak k=1 1 − a∗ ejΩ 1 − a∗ ejΩ ∗ jΩ 1 − a e p 2 = jΩ 1 · · · jΩ e − a1 ejΩ − a2 e − ap
H(ejΩ )
=
H(ejΩ )
P $ 1 − a∗k ejΩ ejΩ − ak k=1 1 − a∗ ejΩ 1 − a∗1 ejΩ 1 − a∗2 ejΩ p · · · jΩ = jΩ e − ap e − a1 ejΩ − a2
H(ejΩ )
=
H(ejΩ )
= 1
The system is allpass. (e) Can a stable and causal allpass system also be minimum phase? Explain. For a stable and causal allpass system, ak  < 1 for all k. Using P=1:
H(z) =
1 − a∗1 z z − a1
1 a∗1 Which implies
The zero is zz =
zz  =
1 >1 a∗1 
This system cannot also be minimum phase. 7.49. Let H(z) = F (z)(z − a) and G(z) = F (z)(1 − az) where 0 < a < 1 is real. (a) Show that G(ejΩ ) = H(ejΩ ).
G(z) G(ejΩ )
1 − az z−a allpass term 1 − az = H(ejΩ ) z−a = H(ejΩ ) = H(z)
41
1 − az Since z−a
=
1, see prob 7.48
(b) Show that g[n] = h[n] ∗ v[n] where
z −1 − a 1 − az −1 V (z) is thus the transfer function of an allpass system (see Problem P7.48). V (z) =
V (z)
= =
1 − az z−a z −1 − a 1 − az −1
z
←−−−→ g[n] = h[n] ∗ v[n]
G(z) = H(z)V (z)
(c) One deﬁnition of the average delay introduced by a causal system is the normalized ﬁrst moment ∞ kv 2 [k] d = k=0 ∞ 2 k=0 v [k] Calculate the average delay introduced by the allpass system V (z). V (z) v[k]
a z −1 − 1 − az −1 1 − az −1 = ak−1 u[k − 1] − aak u[k]
=
for k = 0 v[0]
= −a
v 2 [0]
= a2
for k ≥ 1 v[k] v 2 [k]
= ak−1 − ak+1 = a2k−2 + a2k+2 − 2a(k−1)+(k+1)
= a2k (a−2 + a2 − 2) ∞ ∞ kv 2 [k] = (a−2 + a2 − 2) k(a2 )k k=0
k=0
= ∞
v 2 [k]
1 + a4 − 2a2 (1 − a2 )2
= a2 + (a−2 + a2 − 2)
k=0
∞
(a2 )k
k=1
1 + a − 2a 1 − a2 2 1−a = 1 − a2 = 1, which also follows from Parseval’s Theorem. ∞ kv 2 [k] = k=0 ∞ 2 k=0 v [k] = a2 +
d
4
2
42
1 + a4 − 2a2 (1 − a2 )2
=
7.50. The transfer function of a LTI system is expressed as %M b0 k=1 (1 − ck z −1 ) H(z) = %N −1 ) k=1 (1 − dk z where dk  < 1, k = 1, 2, . . . , N , ck  < 1, k = 1, 2, . . . , M − 1, and cM  > 1. (a) Show that H(z) can be factored in the form H(z) = Hmin (z)Hap (z) where Hmin (z) is minimum phase and Hap (z) is allpass (see Problem P7.48).
H(z)
=
=
(1 − cM z
b0
−1
%M −1 b0 k=1 (1 − ck z −1 ) ) %N −1 ) k=1 (1 − dk z
%M −1 k=1
(1 − ck z −1 )(1 − cM1 ∗ z −1 ) 1 − cM z −1 %N 1 − 1 z −1 −1 ) (1 − d z k cM ∗ k=1 Hmin (z)
Hap (z)
= Hmin (z)Hap (z)
(b) Find a minimum phase equalizer with transfer function Heq (z) chosen so that H(ejΩ )Heq (ejΩ ) = 1 and determine the transfer function of the cascade H(z)Heq (z).
Heq (z) so H(z)Heq (z)
1 Hmin (z) = Hap (z) =
H(z)Heq (z) = Hap (z) = 1 1 − cM z −1 Hap (z) = 1 − cM1 ∗ z −1 7.51. A very useful structure for implementing nonrecursive systems is the socalled lattice stucture. The lattice is constructed as a cascade of two input, two output sections of the form depicted in Fig. P7.51 (a). An M th order lattice structure is depicted in Fig. P7.51 (b). (a) Find the transfer function of a secondorder (M = 2) lattice having c1 = 12 and c2 = − 14 .
X(z)
A(z)
z −1 0.5
z −1 −0.25 −0.25
0.5
B(z) Figure P7.51. (a) Lattice Diagram.
43
Y(z)
1 = − A(z)z −1 + B(z) 4 1 A(z) = X(z)(z −1 + ) 2 1 −1 B(z) = X(z)( z + 1) 2 1 −2 3 −1 Y (z) = − z + z + 1 X(z) 4 8 Y (z) H(z) = X(z) 1 3 = 1 + z −1 − z −2 8 4 Y (z)
(b) We may determine the relationship between the transfer function and lattice structure by examining the eﬀect of adding a section on the transfer function, as depicted in Fig. P7.51(c). Here we have deﬁned Hi (z) as the transfer function between the input and the output of the lower ˜ i (z) as the transfer function between the input and the output of the branch in the ith section and H th upper branch in the i section. Write the relationship between the transfer functions to the (i − 1)st and ith stages as ˜ i (z) ˜ i−1 (z) H H = T(z) Hi (z) Hi−1 (z) where T(z) is a twobytwo matrix. Express T(z) in terms of ci and z −1 . ˜ i−1 (z)z −1 + ci Hi−1 (z) = H ˜ i−1 (z)z −1 ci + Hi−1 (z) Hi (z) = H ˜ i−1 (z) ˜ i (z) z −1 ci H H = z −1 ci 1 Hi (z) Hi−1 (z) z −1 ci T(z) = z −1 ci 1 ˜ i (z) H
˜ i (z) = z −i Hi (z −1 ). (c) Use induction to prove that H
i=1
˜ i−1 (z) = Hi−1 (z) = 1 H ˜ 1 (z) = z −1 + c1 H H1 (z) = z −1 c1 + 1 H1 (z −1 ) = 1 + zc1 ˜ 1 (z) = z −1 H1 (z −1 ) H
i=k i=k+1
˜ k (z) = z −k Hk (z −1 ) assume H ˜ k+1 (z) = z −1 H ˜ k (z) + ck+1 Hk (z) H 44
˜ k+1 (z) = z −1 ck+1 H ˜ k (z) + Hk (z) H ˜ k (z) = z −k Hk (z −1 ) substitute H ˜ k+1 (z) = z −(k+1) Hk (z −1 ) + ck+1 Hk (z) H Hk+1 (z) = z −(k+1) ck+1 Hk (z −1 ) + Hk (z) Hk+1 (z −1 ) = z (k+1) ck+1 Hk (z) + Hk (z −1 ) ˜ k+1 (z) = z −(k+1) z (k+1) ck+1 Hk (z) + Hk (z −1 ) H ˜ k+1 (z) = z −(k+1) Hk+1 (z −1 ) H Therefore ˜ i (z) = z −i Hi (z −1 ) H
(d) Show that the coeﬃcient of z −i in Hi (z) is given by ci .
˜i H Hi
=
Hi ˜ i−1 (z) H
z −1 z −1 ci
ci 1
˜ i−1 H Hi−1
˜ i−1 + Hi−1 = z −1 ci H = z −(i−1) + . . . + ci−1
The highest order of (z −1 ) in Hi (z) is (i), and the coeﬃcients of z −i is (ci ), since Hi−1 (z) does not contribute to z −i , therefore the coeﬃcient of z −i is Hi (z) is given by (ci ). (e) By combining the results of (b)  (d) we may derive an algorithm for ﬁnding the ci required by the lattice structure to implement an arbitrary order M nonrecursive transfer function H(z). Start with i = M so that HM (z) = H(z). The result of (d) implies that cM is the coeﬃcient of z −M in H(z). By decreasing i, continue this algorithm to ﬁnd the remaining ci . Hint: Use the result of (b) to ﬁnd a twobytwo matrix A(z) such that ˜ i (z) ˜ i−1 (z) H H = A(z) Hi−1 (z) Hi (z)
˜ i (z) H Hi (z)
˜ i−1 (z) H Hi−1 (z)
= T
= A
A
=
˜ i−1 (z) H Hi−1 (z) ˜ i−1 (z) H Hi−1 (z)
−1 where A = T 1 1 z −1 (1 − c2i ) −z −1 ci
let H(z) =
M
bk z −k
k=0
45
(1)
−ci
= HM (z)
Where H(z) is given. From this we have cM = bM . The following is an algorithm to obtain all ci ’s. (1) cM = bM (2) for i = M to 2, descending ˜ i−1 (z) and Hi−1 (z) from (1) compute H get ci−1 from Hi−1 (z) end Thus we will have c1 , c2 , . . . cM . 7.52. Causal ﬁlters always have a nonzero phase response. One technique for attaining zero phase response from a causal ﬁlter involves ﬁltering the signal twice, once in the forward direction and the second time in the reverse direction. We may describe this operation in terms of the input x[n] and ﬁlter impulse response h[n] as follows. Let y1 [n] = x[n] ∗ h[n] represent ﬁltering the signal in the forward direction. Now ﬁlter y1 [n] backwards to obtain y2 [n] = y1 [−n] ∗ h[n]. The output is then given by reversing y2 [n] to obtain y[n] = y2 [−n]. (a) Show that this set of operations is equivalently represented by a ﬁlter with impulse response ho [n] as y[n] = x[n] ∗ ho [n] and express ho [n] in terms of h[n].
y[n]
= y1 [n] ∗ h[n] =
(x[n] ∗ h[n]) ∗ h[−n]
= x[n] ∗ (h[n] ∗ h[−n]) = x[n] ∗ ho [n] ho [n] = h[n] ∗ h[−n]
(b) Show that ho [n] is an even signal and that the phase response of any system with an even impulse response is zero.
ho [−n]
= h[−n] ∗ h[n] = h[n] ∗ h[−n]
ho [−n]
= ho [n]
Which shows that ho [n] is an even signal. Since ho [n] is even, the phase response can be found from the following:
Ho (ejΩ ) Ho∗ (ejΩ )
∞
= =
n=−∞ ∞ n=−∞
46
ho [n]e−jΩn ho [n]ejΩn
∞
= =
n=−∞ ∞
ho [−n]e−jΩn ho [n]e−jΩn
n=−∞
Ho∗ (ejΩ )
= Ho (ejΩ )
! arg Ho (ejΩ )
=
0
(c) For every pole or zero at z = β in h[n], show that ho [n] has a pair of poles or zeros at z = β and z =
1 β.
= h[n] ∗ h[−n]
ho [n] so
= H(z)H(z −1 )
Ho (z) let H(z)
=
Ho (z)
= =
Therefore, Ho (z) has a pair of poles at z = β,
1 β
let H(z)
=
Ho (z)
=
z−β z−p z − β z −1 − β z − p z −1 − p 1 β (z − β)(z − β )
=
Therefore, Ho (z) has a pair of zeros at z = β,
z−c z−β z − c z −1 − c z − β z −1 − β (z − c)(1 − cz) (z − β)( β1 − z)β
p (z − p)(z − p1 )β 1 β
7.53. The present value of a loan with interest compounded monthly may be described in terms of the ﬁrstorder diﬀerence equation where ρ = 1 +
r/12 100 th
y[n] = ρy[n − 1] − x[n] , r is the annual interest rate expressed as a percent, x[n] is the payment credited
at the end of the n month, and y[n] is the loan balance at the beginning of the n + 1st month. The beginning loan balance is the initial condition y[−1]. If uniform payments of $c are made for L consecutive months, then x[n] = c{u[n] − u[n − L]}. 47
(a) Use the unilateral ztransform to show that Y (z) = Hint: Use long division to show that
Y (z) X(z)
L−1 y[−1]ρ − c n=0 z −n 1 − ρz −1
L−1 1 − z −L = z −n 1 − z −1 n=0
y[−1]ρ − X(z) 1 − ρz −1 1 − z −L = c 1 − z −1 = c 1 + z −1 + z −2 + . . . + z −L+1 =
= c
L−1
z −n
n=0
which implies Y (z)
L−1 y[−1]ρ − c n=0 z −n 1 − ρz −1
=
(b) Show that z = ρ must be a zero of Y (z) if the loan is to have zero balance after L payments.
Y (z)
=
L−1 y[−1]p + c n=0 z −n 1 − ρz −1
The pole at z = p results in an inﬁnite length y[n] in general. If the loan reaches zero after the Lth payment, we have:
y[n] = 0, n ≥ L − 1 So, Y (z) =
L−2
y[n]z −n
n=0
Thus we must have: L−1 y[−1]ρ − c z −n
=
(1 − ρz −1 )
n=0
L−2
y[n]z −n
n=0
to cancel the pole at z = p
From polynomial theory, the ﬁrst term is zero if f (z = ρ) = 0, or ρ = 1 +
r 12
is a zero of Y (z)
(c) Find the monthly payment $c as a function of the intitial loan value y[−1] and the interest rate r assuming the loan has zero balance after L payments. y[−1]ρ − c
L−1
ρ−n
n=0
48
=
0
c
1 − ρ−L 1 − ρ−1
= y[−1]ρ
c = y[−1]
ρ−1 1 − ρ−L
Solutions to Computer Experiments
7.54. Use the MATLAB command zplane to obtain a polezero plot for the following systems: −2 (a) H(z) = 2+z−1 −1+z 1 −2 z + 1 z −3 2
4
P7.54(a) 1 0.8 0.6
Imaginary Part
0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1
−0.5
0 Real Part
0.5
Figure P7.54. (a)PoleZero plot of H(z) 1+z −1 + 32 z −2 + 12 z −3 (b) H(z) = 1+ 3 z−1 + 1 z −2 2
2
49
1
P7.54(b)
1 0.8 0.6
Imaginary Part
0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1
−1.5
−1
−0.5
0 Real Part
0.5
1
1.5
Figure P7.54. (b) PoleZero plot of H(z)
7.55. Use the MATLAB command residuez to obtain the partial fraction expansions required to solve Problem 7.24 (d)  (g). P7.55 : ======= Part (d) : ========== r= 2 1 p= 2.0000 0.5000 k= 0 Part (e) : 50
========== r= 1.4688 1.5312 p= 4 4 k= 0 Part (f) : ========== r= 2 1 p= 1.0000 0.5000 k= 0
1
Part (g) : ========== r= 1 1 p= 1 51
1 k= 2
0
0
7.56. Use the MATLAB command tf2ss to ﬁnd statevariable descriptions for the systems in Problem 7.27. P7.56 : ======= Part (a) : ========== A= 1.5000 1.0000
1.0000 0
B= 1 0 C= 1.5000
2.0000
D= 2 Part (b) : ========== A= 1 1
6 0
B= 1 0
52
C= 5
30
D= 5 Part (c) : ========== A= 0.2500 1.0000
0.0625 0
B= 1 0 C= 4
0
D= 0
7.57. Use the MATLAB command ss2tf to ﬁnd the transfer functions in Problem 7.33. P7.57 : ======= Part (b)(i) : ========== Num = 1.0000
2.0000
1.2500
Den = 1.0000
0
0.2500 53
Part (b)(ii) : ========== Num = 0
2
0
Den = 1.0000
0.2500
0.3750
Part (b)(iii) : ========== Num = 0
2.0000
6.5000
Den = 1.0000
0.5000
0.2500
7.58. Use the MATLAB command zplane to solve Problem 7.35 (a) and (b). 54
P7.58(a) Poles−Zeros of the inverse system 2
1.5
1
Imaginary Part
0.5
2
0
−0.5
−1
−1.5
−2 −1
−0.5
0
0.5
1
1.5 Real Part
2
2.5
3
3.5
Figure P7.58. (a) PoleZero plots of the inverse system. P7.58(b) Poles−Zeros of the inverse system 1 0.8 0.6
Imaginary Part
0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1
−0.5
0 Real Part
0.5
Figure P7.58. (b) PoleZero plots of the inverse system.
55
1
4
7.59. Use the MATLAB command freqz to evaluate and plot the magnitude and phase response of the system given in Example 7.21. P7.59 12
Magnitude
10 8 6 4 2 −3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
2
Phase (rad)
1 0 −1 −2
Figure P7.59. Magnitude Response for Example 7.21
7.60. Use the MATLAB command freqz to evaluate and plot the magnitude and phase response of the systems given in Problem 7.37. 56
P7.60(a) 4
Magnitude
3.5 3 2.5 2 1.5 1 −3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
3
Phase(rad)
2 1 0 −1 −2 −3
Figure P7.60. (a) Magnitude and phase response P7.60(b) 1
Magnitude
0.8 0.6 0.4 0.2
−3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
2
Phase(rad)
1
0
−1
−2
Figure P7.60. (b) Magnitude and phase response
57
P7.60(c) 5
Magnitude
4 3 2 1
−3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
1.5
Phase(rad)
1 0.5 0 −0.5 −1 −1.5
Figure P7.60. (c) Magnitude and phase response
7.61. Use the MATLAB commands ﬁlter and ﬁltic to plot the loan balance at the start of each month n = 0, 1, . . . L + 1 for Problem 7.53. Assume that y[−1] = $10, 000, L = 60, r = 0.1 and the monthly payment is chosen to bring the loan balance to zero after 60 payments. From Problem 7.53:
ρ−1 1 − ρ−L = 10, 000
c = y[−1] y[−1]
L =
60
c =
r = 1.00833 12 0.02125
b
[y[−1]ρ − c,
ρ
=
=
a =
1+
[1,
ρ]
58
− c(ones(1, 59))]
P7.61 10000
9000
8000
7000
Balance
6000
5000
4000
3000
2000
1000
0
0
10
20
30 Month
40
50
60
Figure P7.61. Monthly loan balance.
7.62. Use the MATLAB command zp2sos to determine a cascade connection of secondorder sections for implementing the systems in Problem 7.38. ======= Part (a) : ========== sos = 1.0000 1.0000
0.3536 0.4619
0.0625 0.0625
1.0000 1.0000
0.5000 0.2500 1.3858 0.5625
4.0000 0.2500
1.0000 1.0000
0.3750 0.2813 0.3750 0.1406
Part (b) : ========== sos = 1.0000 1.0000
4.0000 0.0000
59
7.63. A causal discretetime LTI system has the transfer function
H(z) =
0.0976(z − 1)2 (z + 1)2 (z − 0.3575 − j0.5889)(z − 0.3575 + j0.5889)(z − 0.7686 − j0.3338)(z − 0.7686 + j0.3338)
(a) Use the pole and zero locations to sketch the magnitude response.
Im
a 1 = tan−1 a 2 = tan−1
a1
double
a
2
0.3338 0.7686 0.5889 0.3575
double
Re
Figure P7.63. (a) Pole Zero plot
symmetric
a1 a2
2
Figure P7.63. (a) Sketch of the Magnitude Response. (b) Use the MATLAB commands zp2tf and freqz to evaluate and plot the magnitude and phase response.
60
P7.63(b)
H(Omega)
0.8 0.6 0.4 0.2 0
−3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
3
< H(Omega) : rad
2 1 0 −1 −2 −3
Figure P7.63. (b) Plot of the Magnitude Response. (c) Use the MATLAB command zp2sos to obtain a representation for this ﬁlter as a cascade of two secondorder sections with realvalued coeﬃcients.
H(z)
=
0.1413(1 + 2z −1 + z −2 ) 1 − 0.715z −1 + 0.4746z −2
0.6907(1 − 2z −1 + z −2 ) 1 − 1.5372z −1 + 0.7022z −2
(d) Use the MATLAB command freqz to evaluate and plot the magnitude response of each section in (c).
61
P7.63(d) 0.6 H1(Omega)
0.5 0.4 0.3 0.2 0.1 0
−3
−2
−1
0 Omega
1
2
3
−3
−2
−1
0 Omega
1
2
3
H2(Omega)
1.5
1
0.5
0
Figure P7.63. (d) Plot of the Magnitude Response for each section. (e) Use the MATLAB command ﬁlter to determine the impulse response of this system by obtaining the output for an input x[n] = δ[n]. P7.63(e) 0.2 0.15 0.1
Impulse Resp
0.05 0 −0.05 −0.1 −0.15 −0.2 −0.25 −0.3 0
5
10
15
20
25
30
Figure P7.63. (e) Plot of the Impulse Response. 62
35
40
45
(f) Use the MATLAB command ﬁlter to determine the system output for the input π π 3π x[n] = 1 + cos( n) + cos( n) + cos( n) + cos(πn) u[n] 4 2 4 Plot the ﬁrst 250 points of the input and output. P7.63(f) 6 5
Input
4 3 2 1 0
0
50
100
150
200
0
50
100
150
200
1.5 1
Output
0.5 0 −0.5 −1 −1.5
Figure P7.63. (f) System output for given input. The system eliminates “1” and “cos(πn)” terms. It also greatly attenuates the “cos( 3π 4 n)” term. The π π “cos( 2 n)” term is also attenuated, so the output is dominated by “cos( 4 n)”.
63
CHAPTER 8 Additional Problems 8.16
Provided that the channel bandwidth is not smaller than the reciprocal of the transmitted pulse duration T, the received pulse is recognizable at the channel output. With T = 1µs, a small enough value for the channel bandwidth is (1/T) = 106 Hz = 1 MHz.
8.17
For a lowpass filter of the Butterworth type, the squared magnitude response is defined by 2 1 (1) H ( jω ) = 2N 1 + ( ω ⁄ ωc ) At the edge of the passband, ω = ωp, we have (by definition) H ( jω p ) = 1 ∈ We may therefore write 2 1 (1  ∈ ) = 2N 1 + ( ω p ⁄ ωc )
(2)
Define 2
∈ 0 = 1 – (1  ∈ ) = 2∈  ∈
2
Then solving Eq. (2) for ωp: ∈0 ω p =  ω c 1 ∈ 0 Next, by definition, at the edge of the stopband, ω = ωs, we have H ( jω s ) = δ Hence 2 1 δ = 2N 1 + ( ωs ⁄ ωc )
(3)
Define δ0 = δ
2
Hence, solving Eq. (3) for ωs: 1 – δ0 ω s =  ω c δ0
1
8.18
We start with the relation 2
H ( jω ) jω=s = H ( s )H ( – s ) For a Butterworth lowpass filter of order 5, the 10 poles of H(s)H(s) are uniformly distributed around the unit circle in the splane as shown in Fig. 1. jω splane x
x x
x
σ
x
x x
x x
x Figure 1
Let D(s)D(s) denote the denominator polynomial of H(s)H(s). Hence D ( s )D ( – s ) = ( s + 1 )(s + cos 144° + j sin 144°) ( s + cos 144° – j sin 144° ) × (s + cos 108° + j sin 108°) ( s + cos 108° – j sin 108° ) × ( s – 1 )(s – cos 36° + j sin 36°) ( s – cos 36° – j sin 36° ) × (s – cos 72° + j sin 72°) ( s – cos 72° – j sin 72° ) Identifying the zeros of D(s)D(s) in the lefthalf plane with D(s) and those in the righthalf plane with D(s), we may express D(s) as D ( s ) = ( s + 1 )(s + cos 144° + j sin 144°) ( s + cos 144° – j sin 144° ) × (s + cos 108° + j sin 108°) ( s + cos 108° – j sin 108° ) 5
4
3
2
= s + 3.2361s + 5.2361s + 5.2361s + 3.2361s + 1 Hence, 1 H ( s ) = D(s) 1 = 5 4 3 2 s + 3.2361s + 5.2361s + 5.2361s + 3.2361s + 1
8.19
(a) For filter order N that is odd, the transfer function H(s) of the filter must have a real pole in the lefthalf plane. Let this pole be s = a where a > 0. We may then write 1 H ( s ) = ( s + a )D′ ( s )
2
where D′ ( s ) is the remainder of the denominator polynomial. For a Butterworth lowpass filter of cutoff frequency ωc, all the poles of H(s) lie on a circle of radius ωc in the lefthalf plane. Hence, we must have a = ωc. (b) For a Butterworth lowpass filter of even order N, all the poles of the transfer function H(s) are complex. They all lie on a circle of radius ωc in the lefthalf plane. Let s = a jb, with a > 0 and b > 0, denote a complex pole of H(s). All the coefficients of H(s) are real. This condition can only be satisfied if we have a complex conjugate pole at s = a + jb. We may then express the contribution of this pair of poles as 1 1  = 2 2 ( s + a + jb ) ( s + a – jb ) (s + a) + b whose coefficients are all real. We therefore conclude that for even filter order N, all the poles of H(s) occur in complexconjugate pairs.
The transfer function of a Butterworth lowpass filter of order 5 is 1 H ( s ) = (1) 2 2 ( s + 1 ) ( s + 0.618s + 1 ) ( s + 1.618s + 1 ) The lowpass to highpass transformation is defined by 1 s → s where it is assumed that the cutoff frequency of the highpass filter is unity. Hence replacing s with 1/s in Eq. (1), we find that the transfer function of a Butterworth highpass filter of order 5 is 1 H ( s ) = 1 0.618 1 0.618 1 + 1 +  + 1 2 +  + 1 s 2 s s s s 5
s = 2 2 ( s + 1 ) ( s + 0.618s + 1 ) ( s + 1.618s + 1 ) The magnitude response of this highpass filter is plotted in Fig. 1. 1.4
1.2
1 Magnitude Response
8.20
0.8
0.6
Figure 1
0.4
0.2
0
0
0.5
1
1.5
2 2.5 3 Normalized Frequency w/wc
3.5
4
3
4.5
5
8.21
We are given the transfer function 1 H ( s ) = 2 ( s + 1 ) ( s + 0.618s + 1 ) ( s + 1.618s + 1 )
(1)
To modify this lowpass filter so as to assume a cutoff frequency ωc, we use the transformation s s → ωc Hence, replacing s with s/ωc in Eq. (1), we obtain: 1 H ( s ) =  s2 s2 s s s  + 1 2 + 0.618  + 1 2 + 1.618  + 1 ωc ωc ωc ωc ωc 5
ωc = 2 2 2 2 ( s + ω c ) ( s + 0.618ω c s + ω c ) ( s + 1.618ω c s + ω c )
8.22
We are given the transfer function 1 H ( s ) = 2 (s + 1)(s + s + 1)
(1)
To transform this lowpass filter into a bandpass filter with bandwidth B centered on ω0, we use the following transformation: 2
2
s + ω0 s → Bs With ω0 = 1 and B = 0.1, the transformation takes the value 2
s +1 s → 0.1s
(2)
Substituting Eq. (2) into (1): 1 H ( s ) = 2 2 2 2 s + 1 s + 1 s +1   + 1  +  + 1 0.1s 0.1s 0.1s 3
0.001s = 2 4 2 3 2 ( s + 0.1s + 1 )(s + 2s + 1 + 0.1s + 0.1s + 0.01s )
4
3
0.001s = 2 4 3 2 ( s + 0.1s + 1 ) ( s + 0.1s + 2.01s + 0.1s + 1 )
(3)
The magnitude response of this bandpass filter of the Butterworth type, obtained by putting s = jω in Eq. 3, is plotted in Fig. 1. Bandpass Response 1.4
1.2
Squared Magnitude Response
1
0.8
0.6
0.4
Figure 1 0.2
0
8.23
0
0.2
0.4
0.6
0.8 1 1.2 Normalized Frequency w/wc
1.4
1.6
1.8
2
For the counterpart to the lowpass filter of order one in Fig. 8.14(a), we have 1Ω o
v1(t)
o
+ 
o + v2(t)
IF
o
o
Input Filter
5
o Output
For the counterpart to the lowpass filter of order three in Fig. 8.14(b), we have 1Ω o
v1(t)
1H o
+ 
o
+ v2(t)
IF IF o
o
o
o
Input
o Output
Filter
Note that in both of these figures, the 1Ω resistance may be used to account for the source resistance. Note also that the load resistance is infinitely large.
8.24
(a) For Fig. 8.14(a), the capacitor has the value 1 C = F 4 5 ( 10 ) ( 2π × 10 ) 3
10 =  pF 2π = 159 pF (b) For Fig. 8.14(a), the two capacitors are 159 PF each. The inductor has the value 4
( 10 ) H L = 5 ( 2π × 10 ) 100 =  mH 2π = 15.9 mH
8.25
Let the transfer function of the FIR filter be defined by –1
H (z) = (1 – z ) A(z)
(1)
where A(z) is an arbitrary polynomial in z1. The H(z) of Eq. (1) has a zero at z = 1 as prescribed. Let the sequence a[n] denote the inverse ztransform of A(z). Expanding Eq. (1): –1
H (z) = A(z) – z A(z)
(2)
6
which may be represented by the block diagram: ≡
H(z)
.
A(z)
z1
−
Σ +
From Eq. (2) we readily find that the impulse response of the filter must satisfy the condition h[n] = a[n] – a[n – 1] where a[n] is the inverse ztransform of an arbitrary polynomial A(z).
8.26
Let the transfer function of the filter be defined by M⁄2
H′ ( e
jΩ
∑
) =
h d [ n ]e
 jnΩ
n=M ⁄ 2
Replacing n with n  M/2 so as to make the filter causal, we may thus write M
H′ ( e
jΩ
) =
∑ hd n=0
= e
M n –  e 2 M
jMΩ ⁄ 2
∑ hd
M  j n –  Ω 2
M  jnΩ n –  e 2
n=0
(1)
We are given that M M for –  ≤ n ≤ 2 2
h [ n ] = hd [ n ]
This condition is equivalent to M h d n –  = h [ n ] 2
for 0 ≤ n ≤ M
Hence, we may rewrite Eq. (1) in the form H′ ( e
jΩ
) = e
jMΩ ⁄ 2
N
∑ h [ n ]e
 jnΩ
n=0
= e
jMΩ ⁄ 2
H (e
jΩ
)
Equivalently we have jΩ
– j MΩ ⁄ 2
jΩ
H (e ) = e H′ ( e ) which is the desired result.
7
8.27
According to Eqs. (8.64) and (8.65), the magnitude r = z and phase θ = arg{z} are defined by ( 1 + σ ) 2 + ω 2  r =  ( 1 – σ ) 2 + ω 2
1⁄2
(1)
–1 –1 ω ω θ = tan  – tan  1 + σ 1 – σ
(2)
These two relations are based on the transformation jθ 1+s s = σ + jω z = re =  , 1–s For the more general case of a sampling rate 1/Ts for which we have 1 z–1 s =  Ts z + 1 or Ts 1 + s 2 z = Ts 1 – s 2 Ts Ts we may rewrite Eqs. (1) and (2) by replacing ω with ω and σ with σ , obtaining 2 2 2 2 2  + σ + ω  Ts r = 2 2 2  – σ + ω Ts
1⁄2
– 1 ω ω θ = tan  – tan  2 2   + σ  – σ Ts Ts – 1
8.28
(a) From Section 1.10, we recall the inputoutput relation y [ n ] = x [ n ] + ρy [ n – 1 ] Taking ztransforms: –1
Y ( z ) = X ( z ) + ρz Y ( z ) The transfer function of the filter is therefore
8
Y (z) 1 H ( z ) =  = –1 X (z) 1 – ρz
(1)
For ρ = 1, we have 1 H ( z ) = –1 1–z
(2)
For z = jΩ, the frequency response of the filter is defined by jΩ 1 H ( e ) = – jΩ 1–e with H (e
jΩ
1 ) = – jΩ 1–e 1 = 1⁄2 2 2 [ ( 1 – cos Ω ) + sin Ω ] 1 = 1⁄2 ( 2 – 2 cos Ω )
which is plotted in Fig. 1 for 0 ≤ Ω ≤ π . From this figure we see that the filter defined in Eq. (2) does not deviate from the ideal integrator by more than 1% for 0 ≤ Ω ≤ 0.49 . ρ=1 4
3.5
3
Magnitude
2.5 1 Percent error occurs at w = 0.49 2
1.5
1 Filtered 0.5
Figure 1
Ideal 0
0
0.5
1
1.5 Normalized Frequency
2
2.5
9
3
(b) For ρ = 0.99, the use of Eq. (1) yields 1 H ( z ) = –1 1 – 0.99z
(3)
for which the frequency response is defined by jΩ 1 H ( e ) = – jΩ 1 – 0.99e That is, H (e
jΩ
1 ) = – jΩ 1 – 0.99e 1 = 2 2 1⁄2 [ ( 1 – 0.99 cos Ω ) + ( 0.99 sin Ω ) ] 1 ≈ 1⁄2 ( 1.98 – 1.98 cos Ω )
which is plotted in Fig. 2. From this second figure we see that the usable range of the filter of Eq. (3) as an integrator is reduced to 0.35. ρ = 0.99 4
3.5
3 1 Percent error occurs at w = 0.36
Magnitude
2.5
2
1.5
Figure 2
1 Filtered 0.5 Ideal 0
8.29
0
0.5
1
1.5 Normalized Frequency
2
2.5
3
The transfer function of the digital IIR filter is 3
0.0181 ( z + 1 ) H ( z ) = 2 ( z – 0.50953 ) ( z – 1.2505z + 0.39812 ) Expanding the numerator and denominator polynomials of H(z) in ascending powers of
10
z1, we may write –1
–2
–3
0.0181 ( 1 + 3z + 3z + z ) H ( z ) = –1 –2 –3 1 – 1.7564z + 1.0308z – 0.2014z
.
Hence, the filter may be implemented in direct form II using the following configuration: Input x[n]
0.0205
+ Σ
−
+
Σ
+
Output y[n]
z1 Σ
1.7564
.
3
1.0308
.
3
Σ
z1
Σ
Σ
z1 0.2014
8.30
.
(a) The received signal, ignoring channel noise, is given by y ( t ) = x ( t ) + 0.1x ( t – 10 ) + 0.2x ( t – 15 ) where x(t) is the transmitted signal, and time t is measured in microseconds. Suppose y(t) is sampled uniformly with a sampling period of 5 µs, yielding y [ n ] = x [ n ] + 0.1x [ n – 2 ] + 0.2x [ n – 3 ] (1) (b) Taking the ztransforms of Eq. (1): –2
–3
Y ( z ) = X ( z ) + 0.1z X ( z ) + 0.2z X ( z ) which, in turn, yields the transfer function for the channel: –2 –3 Y (z) H ( z ) =  = 1 + 0.1z + 0.2z X (z) The corresponding equalizer is defined by the transfer function 1 1 H eq ( z ) =  = – 2 –3 H (z) 1 + 0.1z + 0.2z which is realized by the IIR filter: y[n]
+
.
Σ 
z2 Σ
.
0.1
z1 0.2
11
Equalizer output
For H eq ( z ) to be stable, all of its three poles or, equivalently, all three roots of the cubic equation 1 + 0.1z
–2
+ 0.2z
–3
= 0 , or equivalently,
3
z + 0.1z + 0.2 = 0 must lie inside the unit circle in the zplane. Using MATLAB, we find the roots are z = 0.2640 + j0.5560 z = 0.2640  j0.5560 z = 0.5280 which confirm stability of the IIR equalizer. [Note: The stability of H eq ( z ) may also be explored uysing an indirect approach, namely, the RouthHurwitz criterion which avoids having to compute the roots. First, we use the bilinear transformation 1+s z = 1–s and then construct the Routh array as described in Section 9.12. The stability of H eq ( z ) is confirmed by examining the coefficients of the first column of the Routh array. The fact that all these coefficients are found to be positive assures the stability of H eq ( z ) .] Realizing the equalizer by means of an FIR structure, we have –3 –1
–2
+ 0.2z )
= 1 – ( 0.1z
–2
+ 0.2z ) + ( 0.1z
= 1 – ( 0.1z
–2
+ 0.2z ) + ( 0.01z
H eq ( z ) = ( 1 + 0.1z
– ( 0.001z
–3
–2
–3
–6
+ 0.006z
–7
–3 2
+ 0.2z ) – ( 0.1z
–4
+ 0.04z
+ 0.012z
–8
–5
–2
–3 3
+ 0.2z ) + …
–6
+ 0.04z ) –9
+ 0.008z ) + …
Ignoring coefficients smaller than 1% as specified, we have the approximate result: –2
–3
–4
–5
H eq ( z ) ≈ 1 – 0.1z – 0.2z + 0.01z + 0.04z + 0.04z which is realized using the following FIR structure:
.
y[n]
z2
.
z1
.
z1
.
z1
.
z1
–6
– 0.012z
.
–8
z2
0.1
0.2
0.01
0.04
0.04
0.012
Σ
Σ
Σ
Σ
Σ
Σ
12
Equalizer output
Advanced Problems The integrator output is y(t ) =
t
∫tT
x ( τ ) dτ
(1)
0
FT
Let x ( t ) ↔ X ( jω ) . We may therefore reformulate the expression for y(t) as jωt 1 ∞ X ( jω )e dω y ( t ) = ∫ ∫ 2π –∞ tT 0 x(τ) t
dτ
8.31
Interchanging the order of integration: jωt ∞ 1 t y ( t ) = ∫  X ( jω ) ∫ e dτ dω tT 0 – ∞ 2π
T 0
T0 ωT 0 jω t – 2 1 =  ∫ X ( jω ) ⋅  sin c  e dω 2π 2π –∞ 2π ∞
(2)
(a) Invoking the formula for the inverse Fourier transform, we immediately deduce from Eq. (2) that the Fourier transform of the integrator output y(t) is given by ωT 0 – j ωT 0 ⁄ 2 T0 Y ( jω ) =  sin c  e (3) 2π 2π Examining this formula, we also readily see that y(t) can be equivalently obtained by passing the input signal x(t) through a filter whose frequency response is defined by ωT 0 – j ωT 0 ⁄ 2 T0 H ( jω ) =  sin c  e 2π 2π The magnitude response of the filter is depicted in Fig. 1: 0.2
0.18 T/2*pi 0.16 Ideal Low Pass Filter
0.14
0.12
0.1
0.08
Figure 1
0.06
0.04
0.02
0 −6*pi/T
−4*pi/T
−2*pi/T
0
2*pi/T
4*pi/T
13
6*pi/T
(b) Figure 1 also includes the magnitude response of an “approximating” ideal lowpass filter. This latter filter has a cutoff frequency ω0 = 2π/T0 and passband gain of T0/2π. Moreover, the filter has a constant delay of T0/2. The response of this ideal filter to a step function applied at time t = 0 is given by T0 y′ ( t ) = π
2π T 0  t – T0 2
∫–∞
sin λ  dλ λ
At t = T0, we therefore have T 0 π sin λ y′ ( T 0 ) =  ∫  dλ π –∞ λ T 0 0 sin λ π sin λ =  ∫  dλ + ∫  dλ π –∞ λ 0 λ T0 =  ( S i ( ∞ ) + S i ( π ) ) π = 1.09T 0 From Eq. (1) we find that the ideal integrator output at time t = T0 in response to the step function x(t) = u(t) is given by y(T 0) =
T0
∫0
u ( τ ) dτ
It follows therefore that the output of the “approximating” ideal lowpass filter exceeds the output of the ideal integrator by 9%. It is noteworthy that this overshoot is indeed a manifestation of the Gibb’s phenomenon. 8.32
To transform a prototype lowpass filter into a bandstop filter of midband rejection frequency ω0 and bandwidth B, we may use the transformation Bs s → 2 2 s + ω0
(1)
An as illustrative example, consider the lowpass filter: 1 H ( s ) = s+1 Using the transformation of Eq. (1) in Eq. (2), we obtain a bandstop filter defined by 1 H ( s ) = Bs +1 2 2 s + ω0
14
(2)
2
2
s + ω0 = 2 2 s + B s + ω0 which is characterized as follows: H (s) s = 0 = H (s) s = ∞ = 1 H (s)
8.33
s = ± jω 0
= 0
An FIR filter of type 1 has an even length M and is symmetric about M/2 in that its coefficients satisfy the condition h[n] = h[ M – n] for n = 0, 1, …, M The frequency response of the filter is M
H (e
jΩ
) =
∑ h [ n ]e
– j nΩ
n=0
which may be reformulated as follows: M  1 2
H (e
jΩ
) =
∑ h [ n ]e
– j nΩ
n=0 M  1 2
=
∑ h [ n ]e
– j nΩ
n=0 M  1 2
=
∑ h [ n ]e
n=0
– j nΩ
M – j MΩ ⁄ 2 + h  e + 2
M – j MΩ ⁄ 2 + h  e + 2 M – j MΩ ⁄ 2 + h  e + 2
M
∑ M n=  +1 2 M  1 2
n=0 M  1 2
∑ h [ n ]e
n=0
M k = 1, 2, …, 2
M a [ 0 ] = h 2 and let M n =  – k 2 We may then rewrite Eq. (1) in the equivalent form:
15
– j nΩ
∑ h [ M – n ]e
Define M a [ k ] = 2h  – k , 2
h [ n ]e
– j Ω ( Mn )
– j Ω ( Mn )
(1)
M
2 M jΩ – j MΩ ⁄ 2 jkΩ – j kΩ M H (e ) = e  – k (e +e ) + h  ∑ h 2 k=1 2 M
2 – j MΩ ⁄ 2 jkΩ – j kΩ +e ) + a(0) = e ∑ 2a [ k ] ( e k=1 M
2 – j MΩ ⁄ 2 = e ∑ a [ k ] cos ( kΩ ) k=0
(2)
From Eq. (2) we may make the following observations for an FIR filter of type I: 1. The frequency response H ( e exponential e
– j MΩ ⁄ 2
jΩ
) has a linear phase component exemplified by the
.
2. At Ω = 0, M⁄2
∑ a[k ]
j0
H (e ) =
k=0
At Ω = π, M⁄2 jπ
H (e ) =
∑ a [ k ] ( –1 )
M  + k 2
k=0
The implications of these two results are that there are no restrictions on H ( e Ω = 0 and Ω = π.
8.34
jΩ
) at
For an FIR filter of type II, the filter length M is even and it is antisymmetric in that its coefficients satisfy the condition M h [ n ] = –h [ M – n ] , 0 ≤ n ≤  – 1 2 The frequency response of the filter is M
H (e
jΩ
) =
∑ h [ n ]e
– jn Ω
n=0
which may be reformulated as follows:
16
M  1 2
H (e
jΩ
) =
∑ h [ n ]e
– j nΩ
n=0
M – j MΩ ⁄ 2 + h  e + 2
M  1 2
∑ h [ n ]e
=
– j nΩ
n=0 M  1 2
∑ h [ n ]e
=
– j nΩ
n=0
M
∑
h [ n ]e
M n=  +1 2 M  1 2
M – j MΩ ⁄ 2 + h  e + 2 M – j MΩ ⁄ 2 + h  e – 2
– j nΩ
∑ h [ M – n ]e
n=0 M  1 2
∑ h [ n ]e
– j ( M – n )Ω
– j ( M – n )Ω
(1)
n=0
Define M k = 1, 2, …, 2
M a [ k ] = 2h  – k , 2 M a [ 0 ] = h  , 2 and let M k =  – n 2
We may then rewrite Eq. (1) in the equivalent form H (e
jΩ
) = e
– j MΩ ⁄ 2
M⁄2
∑
M⁄2
M – j MΩ ⁄ 2 – j MΩ ⁄ 2 jkΩ – j kΩ M M h  – k e + h  e –e h  – k e ∑ 2 2 2 k=1
n=0 M⁄2
= e
– j MΩ ⁄ 2
2 ∑ a[k ](e k=1
= e
– j MΩ ⁄ 2
= e
jkΩ
–e
– j kΩ
) + a[0]
M⁄2
j ∑ a [ k ] sin ( kΩ + a [ 0 ] ) k=1
– j MΩ ⁄ 2
M⁄2
∑ a [ k ] sin ( kΩ )
(2)
k=0
From Eq. (2) we may make the following observations on the frequency response of an FIR filter of Type II 1. The phase response includes a linear component exemplified by the exponential e
– j MΩ ⁄ 2
.
2. At Ω = 0,
17
j0
H (e ) = 0 At Ω = π, sin(kπ) = 0 for integer k and therefore, jπ
H (e ) = 0
8.35
An FIR filter of type III is characterized as follows: • The filter length M is an odd integer. • The filter is symmetric about the noninteger midpoint n = M/2 in that its coefficients satisfy the condition h[n] = h[ M – n] for 0 ≤ n ≤ M The frequency response of the filter is M
H (e
jΩ
) =
∑ h [ n ]e
– j nΩ
n=0
which may be reformulated as follows: M1 2
H (e
jΩ
) =
∑ h [ n ]e
M
– j nΩ
∑
+
M1 2
∑ h [ n ]e
– j nΩ
+
∑
h [ M – n ]e
– j ( M – n )Ω
n=0
n=0 M1 2
=
– j nΩ
M+1 n= 2 M1 n= 2
n=0
=
[ n ]e
∑ h[n](e
– j nΩ
+e
– j ( M – n )Ω
)
n=0 M1 2
M
M
– j  – n Ω 2 j 2 – n Ω – j MΩ ⁄ 2 = e h [ n ] e + e ∑ n=0
Define M+1 b [ k ] = 2h  – k 2 and let
M+1 for k = 1, 2, …, 2
18
(1)
M+1 n =  – k 2 We may then rewrite Eq. (1) in the equivalent form H (e
jΩ
) = e
– j MΩ ⁄ 2
= e
M+1 2
∑
1
1
– j Ω k –  jΩ k – 2 2 1  b ( k ) e +e 2
k=1 M+1 2 – j MΩ ⁄ 2
∑ b ( k ) cos Ω k – 2 1
(2)
k=1
From Eq. (2) we may make the following observations on the frequency response of an FIR filter of type III: 1. The phase response of the filter is linear as exemplified by the exponential factor e
– j MΩ ⁄ 2
.
2. At Ω = 0, M+1 2 j0
H (e ) =
∑ b(k )
k=1 j0
which shows that there is no restriction on H ( e ) . At Ω = π, jπ
H (e ) = e
= e
– j Mπ ⁄ 2
– j Mπ ⁄ 2
M+1 2
∑ b ( k ) cos π k – 2 1
k=1 M+1 2
∑ b ( k ) sin ( πk )
k=1
which is zero since sin(πk) = 0 for all integer values of k.
8.36
An FIR filter of type IV is characterized as follows: • The filter length M is an odd integer. • The filter is antisymmetric about the noninteger midpoint n = M/2 in that its coefficients satisfy the condition h [ n ] = –h [ M – n ] for 0 ≤ n ≤ M The frequency response of the filter is
19
M
H (e
jΩ
) =
∑ h [ n ]e
 jnΩ
n=0
which may be reformulated as follows: M1 2
H (e
jΩ
) =
∑ h [ n ]e
M
 jnΩ
+
M1 2
∑ h [ n ]e
 jnΩ
+
∑ h [ n ]e
 jnΩ
–
= e
∑ h [ n ]e
 j ( M – n )Ω
 j ( M – n )Ω
n=0
n=0 – j MΩ ⁄ 2
 jnΩ
∑ h [ M – n ]e
n=0 M1 2
n=0 M1 2
=
h [ n ]e
M+1 n= 2 M1 2
n=0
=
∑
M1 2
M
M
j n –  Ω – j n – 2 Ω 2 –e ∑ h [ n ] e n=0
(1)
Define M+1 for k = 1, 2, …, 2
M+1 b [ k ] = 2h  – k 2 and let M+1 k =  – n 2
We may then rewrite Eq. (1) in the equivalent form H (e
jΩ
) = e
– j MΩ ⁄ 2
= je
= je
M+1 2
1
1
j k – 2 Ω  j k – 2 Ω M+1  – k e –e ∑ h 2 k=1
– j MΩ ⁄ 2
– j MΩ ⁄ 2
M+1 2
∑ 2h
k=1 M+1 2
M+1 1  – k sin k –  Ω 2 2
∑ b [ k ] sin k – 2 Ω 1
k=1
From Eq. (2) we may make the following observations on the FIR filter of type IV:
20
(2)
1. The phase response of the filter includes a linear component exemplified by the exponential e
– j MΩ ⁄ 2
.
2. At Ω = 0, j0
H (e ) = 0 At Ω = π, jπ
H ( e ) = je
= je
– j MΩ ⁄ 2
– j MΩ ⁄ 2
M+1 2
∑ b ( k ) sin k – 2 π 1
k=1 M+1 2
∑ ( –1 )
k+1
b[k ]
k=1 jπ
which shows that H ( e ) can assume an arbitrary value.
8.37
The FIR digital filter used as a discretetime differentiator in Example 8.6 exhibits the following properties: • The filter length M is an odd integer. • The frequency response of the filter satisfies the conditions: 1. At Ω = 0, j0
H (e ) = 0 2. At Ω = π, jΩ
H (e ) = 0 These properties are basic properties of an FIR filter of type III discussed in Problem 8.34. We therefore immediately deduce that the FIR filter of Example 8.6 is antisymmetric about the noninteger point n = M/2.
8.38
For a digital IIR filter, the transfer function H(z) may be expressed as N (z) H ( z ) = D(z) where N(z) and D(z) are polynomials in z1. The filter is unstable if any pole of H(z) or, equivalently, any zero of the denominator polynomial D(z) lies outside the unit circle in the zplane. According to the bilinear transform, H (z) = H a(s) z1 s = z+1
21
where Ha(s) is the transfer function of an analog filter used as the basis for designing the digital IIR filter. The poles of H(z) outside the unit circle in the zplane correspond to certain poles of H(s) in the right half of the splane. Conversely, the poles of H(s) in the right half of the splane are mapped onto the outside of the unit circle in the zplane. Now if any pole of Ha(s) lies in the righthalf plane, the analog filter is unstable. Hence if any such filter is used in the bilinear transform, the resulting digital filter is likewise unstable.
8.39
We are given an analog filter whose transfer function is defined by N
H a(s) =
∑ k=1
Ak s – dk
Recall the Laplace transform pair d t L 1 e k ↔ s – dk It follows therefore that the impulse response of the analog filter is N
ha ( t ) =
∑ Ak e
dkt
(1)
k=1
According to the method of impulse invariance, the impulse response of a digital filter derived from the analog filter of Eq. (1) is defined by h [ n ] = T s h a ( nT s ) where Ts is the sampling period. Hence, from Eq. (1) we find that N
h[n] =
∑ T s Ak e
nd k T s
(2)
k=1
Now recall the ztransform pair: z nd k T s 1 ↔ e d k T s –1 1–e z Hence, the transfer function of the digital filter is deduced from Eq. (2) to be N
H (z) =
∑ k=1
8.40
T s Ak d T –1 1 – e k sz
Consider a discretetime system whose transfer function is denoted by H(z). By definition,
22
Y (z) H ( z ) = X (z) where Y(z) and X(z) are respectively the ztransforms of the output sequence y[n] and input sequence x[n]. Let Heq(z) denote the ztransform of the equalizer connected in cascade with H(z). Let x′ [ n ] denote the equalizer output in response to y[n] as the input. Ideally, x′ [ n ] = x [ n – n 0 ] where n0 is an integer delay. Hence, –n
–n
z 0 z 0X (z) X′ ( z ) H eq ( z ) =  =  = Y (z) H (z) Y (z) Putting z = e
, we may thus write
– jn Ω
e 0 ) = jΩ H (e ) The phase delay of an FIR filter of even length M and antisymmetric impulse response is linear with frequency Ω as shown by θ ( Ω ) = – MΩ H eq ( e
8.41
jΩ
jΩ
Hence, such a filter used as an equalizer introduces a constant delay ∂θ ( Ω ) τ ( Ω ) = –  = M samples ∂Ω The implication of this result is that as we make the filter length M larger, the constant delay introduced by the equalizer is correspondingly increased. From a practical perspective, such a trend is highly undesirable.
23
Computer Experiments
%Solution to 8.42 b = fir1(22,1/3,hamming(23)); subplot(2,1,1) plot(b); title(’Impulse Response’) ylabel(’Amplitude’) xlabel(’Time (s)’) grid [H,w] = freqz(b,1,512,2*pi); subplot(2,1,2) plot(w,abs(H)) title(’Magnitude Response’) ylabel(’Magnitude’) xlabel(’Frequency (w)’) grid Impulse Response 0.4
Amplitude
0.3 0.2 0.1 0 −0.1
0
5
10
15
20
25
Time (s) Magnitude Response 1.4 1.2 Magnitude
1 0.8 0.6 0.4 0.2 0
0
0.5
1
1.5 2 Frequency (w)
2.5
%Solution to problem 8.43 clear; M = 100; n = 0:M; f = (nM/2);
24
3
3.5
%Integration by parts (see example 8.6) h = cos(pi*f)./f  sin(pi*f)./(pi*f.^2); k = isnan(h); h(k) = 0; h_rect = h; h_hamm = h .* hamming(length(h))’; [H,w] = freqz(h_rect,1,512,2*pi); figure(1) subplot(2,1,1) plot(h_rect); title(’Rectangular Windowed Differentiator’) xlabel(’Step’) ylabel(’Amplitude’) grid subplot(2,1,2) plot(w,abs(H)) title(’Magnitude Response’) ylabel(’Magnitude’) xlabel(’Frequency (w)’) grid [H,w] = freqz(h_hamm,1,512,2*pi); figure(2); subplot(2,1,1) plot(h_hamm); title(’Hamming Windowed Differentiator’) xlabel(’Step’) ylabel(’Amplitude’) grid subplot(2,1,2) plot(w,abs(H)) title(’Magnitude Response’) ylabel(’Magnitude’) xlabel(’Frequency (w)’) grid
25
Rectangular Windowed Differentiator 1
Amplitude
0.5
0
−0.5
−1
0
20
40
60 Step
80
100
120
Magnitude Response 4
Magnitude
3
2
1
0
0
0.5
1
1.5 2 Frequency (w)
2.5
3
3.5
Hamming Windowed Differentiator 1
Amplitude
0.5
0
−0.5
−1
0
20
40
60 Step
80
100
120
Magnitude Response 3.5 3 Magnitude
2.5 2 1.5 1 0.5 0
8.44
0
0.5
1
1.5 2 Frequency (w)
2.5
3
3.5
For a Butterworth lowpass filter of order N, the squared magnitude response is H ( jω )
2
1 = ω 2N 1 +  ω c
(1)
where ωc is the cutoff frequency. (a) We are given the following specifications: (i) ωc = 2π x 800 rad/s (ii) At ω = 2π x 1,500 rad/s, we have
26
10 log 10 H ( jω ) or, equivalently H ( jω )
2
2
= – 15 dB
1 = 31.6228
Substituting these values in Eq. (1): 2π × 1, 200 2N 31.6228 = 1 +  2π × 800 = 1 + ( 1.5 )
2N
Solving for the filter order N: 1 N =  ( log 30.6228 ⁄ log 1.5 ) 2 = 4.2195 So we choose N = 5.
%Solution to Problem 8.44 omegaC = 0.2; N = 5; wc = tan(omegaC/2); coeff = [ 1 3.2361 5.2361 5.2361 3.2361 1]; %(see table 8.1) ns = wc^N; ds = coeff .* (wc.^[0:N]); [nz,dz]=bilinear(ns,ds,0.5); [H,W]=freqz(nz,dz,512); subplot(2,1,1) plot(W,abs(H)) title(’Magnitude Response of IIR lowpass filter’) xlabel(’rad/s’) ylabel(’Magnitude’) grid phi = 180/pi *angle(H); subplot(2,1,2) plot(W,phi) title(’Phase Response’) xlabel(’rad/s’) ylabel(’degrees’) grid
27
%set(gcf,’name’,[’Low Pass: order=’ num2str(N) ’ wc=’num2str(omegaC)]) Magnitude Response of IIR low−pass filter 1.4 1.2 Magnitude
1 0.8 0.6 0.4 0.2 0
0
0.5
1
1.5
2
2.5
3
3.5
2.5
3
3.5
rad/s Phase Response 200
degrees
100
0
−100
−200
0
0.5
1
1.5
2 rad/s
%Solution to Problem 8.45 omegaC = 0.6; N = 5; wc = tan(omegaC/2); coeff = [ 1 3.2361 5.2361 5.2361 3.2361 1]; %(see table 8.1) ns = [1/wc^N zeros(1,N)]; ds = fliplr(coeff ./ (wc.^[0:N])); [nz,dz]=bilinear(ns,ds,0.5); [H,W]=freqz(nz,dz,512); subplot(2,1,1) plot(W,abs(H)) title(’Magnitude Response of IIR highpass filter’) xlabel(’rad/s’) ylabel(’Magnitude’) grid phi = 180/pi *angle(H); subplot(2,1,2) plot(W,phi) title(’Phase Response’) xlabel(’rad/s’) ylabel(’degrees’)
28
grid
Magnitude Response of IIR high−pass filter 1
Magnitude
0.8 0.6 0.4 0.2 0
0
0.5
1
1.5
2
2.5
3
3.5
2.5
3
3.5
rad/s Phase Response 200
degrees
100
0
−100
−200
0
0.5
1
1.5
2 rad/s
%Solution to problem 8.46 w = 0:pi/511:pi; den = sqrt(1+2*((w/pi).^2) + (w/pi).^4); Hchan = 1./den; taps = 95; M = taps 1; n = 0:M; f = nM/2; %Term 1 hh1 = fftshift(ifft(ones(taps,1))).*hamming(taps); %Term 2 h = cos(pi*f)./f  sin(pi*f)./(pi*f.^2); k=isnan(h); h(k)=0; hh2 = 2*(hamming(taps)’.*h)/pi; %Term 3 hh3a = (1./(pi*f)) .* sin(pi*f); hh3b = (2./(pi*f).^2) .* cos(pi*f); hh3c = (2./(pi*f).^3) .* sin(pi*f); hh3 = hamming(taps) .* (hh3a + hh3b + hh3c)’; hh = hh1’ + hh2 + hh3’; hh(48)=2/3;
29
[Heq,w]=freqz(hh,1,512,2*pi); p = 0.7501; Hcheq = (p*abs(Heq)).*Hchan’; plot(w,p*abs(Heq),’b’) hold on plot(w,abs(Hchan),’g.’) plot(w,abs(Hcheq),’r’) legend(’Heq’, ’Hcan’, ’Hcheq’,1) hold off xlabel(’Frequency (\Omega)’) ylabel(’Magnitude Response’)
1.5 Heq Hcan Hcheq
Magnitude Response
1
0.5
0
0
0.5
1
1.5 2 Frequency (Ω)
2.5
30
3
3.5
CHAPTER 9 Additional Problems 9.21
(a) The closedloop gain of the feedback amplifier is given by A (1) T = 1 + βA We are given the following values for the forward amplification A and feedback factor β: A = 2,500 β = 0.01 Substituting these values in Eq. (1): 2500 2500 T =  =  = 92.15 1 + 0.01 × 2500 26 (b) The sensitivity of the feedback amplifier to changes in A is given by A ∆T ⁄ T 1 1 S T =  =  = ∆A ⁄ A 1 + βA 26 With (∆A/A) = 10% = 0.10, we thus have A ∆A ∆T  = S T  A T
1 =  ( 0.10 ) = 0.0038 = 0.38% 26
9.22
(a) The closedloop gain of the feedback system is Ga G p T = 1 + H Ga G p (b) The return difference of the system is F = 1 + H Ga G p Hence the sensitivity of T with respect to changes in Gp is Gp ∆T ⁄ T S T = ∆G p ⁄ G a
1 = F = 1 + H Ga G p G
(c) We are given: H = 1 and Gp = 1.5. Hence for S T p = 1% = 0.01, we require F = 100
1
The corresponding value of Ga is therefore F–1 G a = HGp 99 =  = 66 1.5
9.23
The local feedback around the motor has the closedloop gain Gp/(1 + HGp). The closedloop gain of the whole system is therefore K r Gc G p ⁄ ( 1 + H G p ) T = 1 + K r Gc G p ⁄ ( 1 + H G p ) K r Gc G p = 1 + H G p + K r Gc G p
9.24
The closedloop gain of the operational amplifier is Z 2(s) V 2(s)  = – Z 1(s) V 1(s) We are given Z1(s) = R1 and Z2(s) = R2. The closedloop gain or transfer function of the operational amplifier in Fig. P9.24 is therefore R2 V 2(s)  = – R1 V 1(s)
9.25
(a) From Fig. P9.25, we have 1 Z 1 ( s ) = R 1 + sC 1 Z 2 ( s ) = R2 The transfer function of this operational amplifier is therefore Z 2(s) V 2(s)  = – Z 1(s) V 1(s) R2 = – 1 R 1 + sC 1 sC 1 R 2 = – 1 + sC 1 R 1
(1)
(b) For positive values of frequency ω that satisfy the condition
2
1  > >R 1 ωC 1 we may approximate Eq. (1) as V 2(s)  ≈ – sC 1 R 2 V 1(s) That is, the operational amplifier acts as a differentiator.
9.26
Throughout this problem, H(s) = 1, in which case the openloop transfer equals G(s). In any event, the openloop transfer function of the feedback control system may be expressed as the rational function P(s)/(spQ1(s)), where neither the polynomial P(s) nor Q1(s) has a zero at s = 0. Since 1/s is the transfer function of an integrator, it follows that p is the number of free integrators in the feedback loop. The order p is referred to as the type of the feedback control system. (a) For the problem at hand, we are given 15 G ( s ) = (s + 1)(s + 3) The control system is therefore type 0. The steadystate error for unit step is 1 ∈ ss = 1 + Kp where K p = lim G ( s ) H ( s ) s→0
That is, 15 K p = lim ( s + 1 )(s + 3) s→0 15 =  = 5 3 Hence, 1 1 ∈ ss =  = 1+5 6 For both ramp and parabolic inputs, the steadystate error is infinitely large. (b) For 5 G ( s ) = s(s + 1)(s + 4) the control system is type 1. The steadystate error for a step input is zero. For a ramp of unit slope, the steadystate error is 1 ∈ ss = Kv where
3
5 K v = lim s → 0 (s + 1)(s + 4) 5 = 4 The steadystate error is therefore 4 ∈ ss = 5 For a parabolic input the steadystate error is infinitely large. (c) For 5(s + 1) G ( s ) = 2 s (s + 3) the control system is type 2. Hence, the steadystate error is zero for both step and ramp inputs. For a unit parabolic input the steadystate error is 1 ∈ ss = Ka where 5(s + 1) K a = lim s+3 s→0 5 = 3 The steadystate error is therefore 3/5. For a unit parabolic input the steadystate error is infinitely large. (d) For 5(s + 1)(s + 2) G ( s ) = 2 s (s + 3) the control system is type 2. The steadystate error is therefore zero for both step and ramp inputs. For a unit parabolic input the steadystate error is 1 ∈ ss = Kn where 5(s + 1)(s + 2) K n = lim s+3 s→0 10 = 3 The steadystate error is therefore 3/10.
9.27
For the results on steadystate errors calculated in Problem 9.26 to hold, all the feedback control systems in parts (a) to (d) of the problem have to be stable.
4
15 (a) G ( s )H ( s ) = (s + 1)(s + 3) The characteristic equation is A ( s ) = ( s + 1 ) ( s + 3 ) + 15 2
= s + 4s + 18 both roots of which are in the lefthalf plane. The system is therefore stable.
5 (b) G ( s )H ( s ) = s(s + 1)(s + 4) 5 = 3 2 s + 5s + 4s 3
2
A ( s ) = s + 5s + 4s + 5 Applying the RouthHurwitz criterion: s3
1
4
s2
5
5
s1
15/5
0
s0
5
0
There are no sign changes in the first column coefficients of the array; the system is therefore stable. 5(s + 1) (c) G ( s )H ( s ) = 2 s (s + 3) 3
2
A ( s ) = s + 3s + 5s + 5 The RouthHurwitz array is: s3
1
5
s2
3
5
s1
10/3
0
s0
5
0
Here again there are no sign changes in the first column of coefficients, and the control system is therefore stable.
5
5(s + 1)(s + 2) (d) G ( s )H ( s ) = 2 s (s + 3) 3
2
3
2
2
A ( s ) = s + 3s + 5s + 15s + 10 = s + 8s + 15s + 10 The RouthHurwitz array is s3
1
15
s2
8
10
s1
110/8 0
s0
10
0
Here again there are no sign changes in the first column of array coefficients, and the control system is therefore stable. 9.28
4
2
(a) s + 2s + 1 = 0 Equivalently, we may write 2
2
(s + 1) = 0 which has double roots at s = +j. The system is therefore on the verge of instability. 4
3
(b) s + s + s + 0.5 = 0 By inspection, we can say that this feedback control system is unstable because the term s2 is missing from the characteristic equation. We can verify this observation by constructing the Routh array: s4
1
0
0.5
s3
1
1
0
s2
1
0.5
0
s1
+1.5
0
s0
9.5
0
There are two sign changes in the first column of array coefficients, indicating that the characteristic equation has a pair of complexconjugate roots in the righthalf of the splane. The system is therefore unstable as previously observed. 4
3
2
(c) s + 2s + 2s + 2s + 4 = 0
6
The Routh array is s4
1
3
4
s3
2
2
0
s2
2
4
0
s1
2
0
s0
4
0
There are two sign changes in the first column of array coefficients, indicating the presence of two roots of the characteristic equation in the righthalf of the splane. The control system is therefore unstable.
9.29
The characteristic equation of the control system is 3
2
s +s +s+K = 0 Applying the RouthHurwitz criterion: s3
1
1
s2
1
K
s1
1K
0
s0
K
0
The control system is therefore stable provided that the parameter K satisfies the condition: 0 0, a 0 > 0 and a2 a1 – a3 a0  > 0 a2
(1)
(b) For the characteristic equation 3
2
s +s +s+K = 0 we have a3 = 1, a2 = 1, and a1 = 1 and a0 = K. Hence, applying the condition (1): 1×1–1×K  > 0 1 or K 0 since we must have a0 > 0. Hence, K must satisfy the condition 0 > K < 1 for stability.
9.31
(a) We are given the loop transfer function K L ( s ) = , K>0 2 s(s + s + 2) 7 1 L(s) has three poles, one at s = 0 and the other two at s = –  ± j  . 2 2 Hence, the root locus of L(s) has 3 branches. It starts at these 3 pole locations and terminate at the zeros of L(s) at s = ∞. The straightline asymptotics of the root locus are defined by the angles (see Eq. (9.69)) ( 2k + 1 )π k = 0, 1, 2 θ k = , 3 or π/3, π, and 5π/3. Their location point on the real axis of the splane is defined by (see Eq. (9.70)) 7 7 1 1 0 + –  + j  –  – j  2 2 2 2 σ 0 = 3 1 = – 3 Since L(s) has a pair of complexconjugate poles, we need (in addition to the rules described in the text) a rule concerning the angle at which the root locus leaves a complex pole (i.e., angle of departure). Specifically, we wish to determine the angle θdep indicated in Fig. 1 for the problem at hand:
8
jω
splane
x
θdep
θ1
σ
0 θ2 x
Figure 1 The angle θ1 and θ2 are defined by –1 7⁄2 –1 θ 1 = tan  = tan ( 2.6458 ) = 111.7° –1 ⁄ 2 θ 2 = 90° Applying the angle criterion (Eq. (9..........)): θ dep + 90° + 111.7° = 180 Hence, θ dep = – 21.70° We may thus sketch the root locus of the given system as in Fig. 2:
jω
x
splane
x
σ
0
x
Figure 2 (b) The characteristic equation of the system is 3
2
s + s + 2s + K = 0
9
Applying the RouthHurwitz criterion: s3
1
2
s2
1
K
s1
2K
0
s0
K
0
The system is therefore on the verge of instability when K = 2. For this value of K the root locus intersects the jωaxis at s2 + K = 0 or s = ± j K = ± j 2 , as indicated in Fig. 2.
9.32
The closedloop transfer function of a control system with unity feedback is L(s) T ( s ) = 1 + L(s) We are given K K L ( s ) =  = 2 s(s + 1) s +3 Hence K T ( s ) = 2 s +3+K (a) For K = 0.1 the system is overdamped 0.1 T ( s ) = 2 s + s + 0.1 where ω n =
0.1 and ζ = 0.5 ⁄ 0.1 .
(b) For K = 0.25 the system is critically damped: 0.25 T ( s ) = 2 s + s + 0.25 where ω n = 0.5 and ζ = 1 . (c) For K = 2.5 the system is underdamped: 0.25 T ( s ) = 2 s + s + 2.5 where ω n =
2.5 and ζ = 0.5 ⁄ 2.5 .
10
Figure 1 plots the respective step responses of these three special cases of the feedback system. %Solution to problem 9.32 t = 0:0.1:40;clc; %Underdamped System K = 0.1; w = sqrt(K); z = 0.5/w; fac = w*sqrt(1z^2)*t + atan(sqrt(1z^2)/z); y1 = 1  1/sqrt(1z^2).*exp(z*w*t).*sin(fac); figure(1);clf; plot(t,y1,’’) %ylim([0 1.5]) xlabel(’Time (s)’) ylabel(’Response’) %title(’Underdamped K = 0.1’) hold on %Critically Damped System K = 0.25; w = sqrt(K); z = 0.5/w tau = 1/w y2 = 1  exp(t/tau)  t.*exp(t/tau); plot(t,y2,’.’) %Over Damped System K = 2.50000; w = sqrt(K); z = 0.5/w; t1 = 1 / (z*w  w*sqrt(z^21)); t2 = 1 / (z*w + w*sqrt(z^21)); k1 = 0.5 * (1 + z/sqrt(z^2+1)); k2 = 0.5 * (1  z/sqrt(z^2+1)); y3 = 1  k1*exp(t/t1)  k2*exp(t/t2); plot(t,y3,’’) hold off legend (’K = 0.1’,’K = 0.25’,’K = 2.5’)
11
1.4 K = 0.1 K = 0.25 K = 2.5
1.2
1
Response
0.8
0.6
0.4
0.2
0
Figure 1 −0.2
−0.4
0
5
10
15
20 Time (s)
25
12
30
35
40
9.33
We are given the loop transfer function K ( s + 0.5 ) L ( s ) = 4 (s + 1) Figure 1 shows the root locus diagram of the closedloop feedback system for varying positive values of K.
%Solution to problem 9.33 num = [1 0.5]; den = [1 4 6 4 1]; rlocus(num,den) axis([1.5 .1 2.2 2.2])
Root Locus
2
1.5
1
Imag Axis
0.5
0
−0.5
−1
−1.5
−2 −1.5
−1
−0.5 Real Axis
13
0
9.34
We are given the loop transfer function K L ( s ) = 2 (s + 1) (s + 5) (a) Figure 1 shows the root locus diagram of the closedloop feedback system for varying K. Next, putting s = jω, K L ( jω ) = 2 ( jω + 1 ) ( jω + 5 ) Parts (a), (b) and (c) of Fig. 2 show plots of the Nyquist locus of the system for K = 50, 72, and 100. On the basis of these figures we can make the following statements: 1. For K = 50 the locus does not encircle the critical point (1, 0) and the system is stable. 2. For K = 100 the locus encloses the critical point (1, 9) and the system is unstable. 3. For K = 72 the Nyquist locus passes through the critical point (1,0) and the system is on the verge of instability. (b) The critical value K = 72 is determined in accordance with the condition K critical L ( jω p ) = 1⁄2 2 2 ( ω p + 1 ) ( ω p + 25 ) where ωp is the phase crossover frequency defined by –1 –1 ω p 180° = 2 tan ( ω p ) + tan  5
For a gain margin of 5 dB: 20 log 10K = 20 log 10K critical 5 Hence 72 72 K =  =  = 40.49 antilog(0.25) 1.7783 (c) For K = 40.49 the gaincrossover frequency is 2 2
2
2
( 40.49 ) = ( 1 + ω g ) ( 25 + ω g ) 2
Let ω g = x , so rewrite this equation in the form of a cubic equation in x: 3
2
x + 27x + 51x – 1614 = 0 The only positive root of x is 6.2429. Hence ωg =
6.2429 = 2.4986
14
The phase of L(jω) at ω = 2.4986 is – 1 2.4986 –1 2 tan ( 2.4986 ) + tan  = 2 × 68.2° + 26.5° 2 = 162.9° The phase margin is therefore ϕ m = 180° – 162.9° = 17.1°
%Solution to problem 9.34 clear; clc; num = [1]; den = [1 7 11 5]; figure(1);clf; rlocus(num,den) figure(2);clf; nyquist(num,den) xlim([1.1 0.22])
Root Locus
10
5
Imag Axis
K = 73.56
0
−5
−10
Figure 1 −15
−10
−5
0
Real Axis
15
5
Nyquist Diagram
0.1
Imaginary Axis
0.05
0
−0.05
−0.1
−1
−0.8
−0.6
−0.4
−0.2
Real Axis
16
0
0.2
Figure 2
9.35
We are given K L ( s ) = s(s + 1) For s = jω, K L ( jω ) = jω ( jω + 1 ) from which we deduce: K L ( jω ) = 1⁄2 2 ω(ω + 1)
(1) –1
arg { L ( jω ) } = – 90° – tan ( ω )
(2)
From Eq. (2) we observe that the phase response arg{L(jω)} is confined to the range [90o, 180o]. The value of 180o is attained only at ω = ∞. At this frequency we see from Eq. (1) that the magnitude response L(jω) is zero. Hence, the Nyquist locus will never encircle the critical point (1,0) for all positive values of K. The feedback system is therefore stable for all K > 0. 9.36
We are given K L ( s ) = 2 s (s + 1) For s = jω, K L ( jω ) = 2 ω ( jω + 1 ) from which we deduce: K L ( jω ) = 1⁄2 2 2 ω (ω + 1) –1
arg { L ( jω ) } = 180° – tan ( ω ) Hence the Nyquist locus will encircle the critical point (1,0) for all K > 0; that is, the system is unstable for all K > 0. We may also verify this result by applying the RouthHurwitz criterion: s3
1
0
s2
1
K
s1
K
0
17
s0
K
0
There are 2 sign changes in the first column of coefficients in the array, assuming K > 0. Hence the system is unstable for all K > 0.
9.37
We are given the loop transfer function K L ( s ) = s(s + 1)(s + 2) (a) For s = jω, K L ( jω ) = jω ( jω + 1 ) ( jω + 2∂ ) Hence K L ( jω ) = 1⁄2 1⁄2 2 2 ω(ω + 1) (ω + 4)
(1)
–1 ω –1 arg { L ( jω ) } = – 90° – tan ( ω ) – tan  2
(2)
Setting K = 6 in Eq. (1) under the condition L(jωp) = 1: 2
2
2
36 = ω p ( ω p + 1 ) ( ω p + 4 ) 2
where ωp is the phasecrossover frequency. Putting ω p = x and rearranging terms: 3
2
x + 5x + 4x – 36 = 0 Solving this cubic equation for x we find that it has only one positive root at x = 2. Hence, ωp =
2
Substituting this value in Eq. (2): –1
2 arg { L ( jω p ) } = – 90° – tan ( 2 ) – tan  2 = – 90° – 54.8° – 35.2° = – 180° –1
This result confirms ωp as the phasecrossover frequency, and K = 6 as the critical value of the scaling factor for which the Nyquist locus of L(jω) passes through the critical point (1,0).
18
We may also verify this result by applying the RouthHurwitz criterion to the characteristic equation: 3
2
s + 3s + 2s + K = 0 Specifically, we write s3
1
2
s2
3
K
s1
6K 3 K
0
s0
The system is therefore on the verge of instability for K = 6. (b) For K = 2 the gain margin is 20 log 106 – 20 log 102 = 20 log 103 = 9.542 dB To calculate the phase margin for K = 2 we need to know the gaincrossover frequency ωg. At ω = ωg, L(jω) = 1. Hence for the problem at hand 2 1 = 1 ⁄2 1⁄2 2 2 ωg ( ωg + 1 ) ( ωg + 4 ) 2
Let ω g = x , for which we may then rewrite this equation as 3
2
x + 5x + 4x – 4 = 0 The only positive root of this equation is x = 0.5616. That is, ω g = 0.5616 = 0.7494 The phase margin is therefore –1
–1
180° – ( 90° + tan ( 0.7494 ) + tan ( 0.3749 ) ) = 90° – 36.9° – 20.5° = 32.6° (c) Let ωg denote the gaincrossover frequency for the required phase margin ϕ m = 20° . We may then write L ( jω g ) = 1 –1 ω –1 g arg { L ( jω g ) } = – 90° – tan ( ω g ) – tan  2 With arg { L ( jω g ) } = 180° – 20° we thus have
19
ωg 70° = tan ( ω g ) + tan  2 Through a process of trial and error, we obtain the solution ω g = 0.972 –1
–1
For L ( jω g ) = 1 we thus require 2
K = ωg ( ωg + 1 )
1⁄2
2
( ωg + 4 )
1⁄2
1⁄2
= 0.972 ( 1.9448 ) ( 4.9448 ) = 3.0143 The gain margin is therefore
1⁄2
6 20 log 10  ≈ 20 log 102 = 6dB 3.0143
9.38
For the purpose of illustration, suppose the loop frequency response L(jω) has a finite magnitude at ω = 0. Suppose also the feedback system represented by L(jω) is stable. We may then sketch the Nyquist locus of the system for 0 < ω < ∞, including gain margin and phase margin provisions, as follows Im{L(jω)} (ω = ωp)
. (1,0) critical point
L(jωp)
.
(ω = ∞)
.
L(0) 0 ϕm
Re{L(jω)} (ω = 0)
(ω = ωg)
The phase margin of the system is ϕm (measured at the gaincrossover frequency ωg). The 1 gain margin is 20 log 10 (measured at the phase crossover frequency ωp). L ( jω p )
9.39
We are given the loop frequency response 6K L ( jω ) = ( jω + 1 ) ( jω + 2 ) ( jω + 3 ) 6K = 3 2 ( jω ) + 6 ( jω ) + ( jω ) + 6
20
(a) Figure 1 on the next page plots the Bode diagram of L(jω) for K = 1. Changing the value of K merely shifts the loop gain response by a constant amount equal to 20log10K. This constant gain is tabulated as follows: K
20log10K, dB
7 8 9 10 11
16.90 18.06 10.08 20.00 20.83
Applying the gain adjustments (as shown in this table) to the loop response of Fig. 1, we may make the following observations: 1. The feedback system is stable for K = 7, 8, 9. 2. It is on the verge of instability for K = 10. 3. It is unstable for K = 11. Hence the system is on the verge of instability for K = 10. (b) For K = 7 the gain margin is 10 20 log  = 3.098 dB 10 7 To calculate the phase margin we need to know the gaincrossover frequency ωg. At ω = ωg the following condition is satisfied 42 1 = 1 ⁄ 2 1⁄2 2 2 2 ( ωg + 1 ) ( ωg + 4 ) ( ωg + 9 ) 2
Let ω g = x and so rewrite this equation as 3
2
x + 14x + 49x – 1728 = 0 The only positive root of this cubic equation is at x = 7.8432. Hence, ω g = 7.8432 = 2.8006 The phase margin for K = 7 is therefore –1
–1
–1
180° – tan ( 2.8006 ) – tan ( 1.4003 ) – tan ( .9335 ) = 180° – 70.4° – 54.5° – 43° = 12.2° Proceeding in a similar fashion we get the following results: • For K = 8, gain margin = 1.938 dB phase margin = 7.39o
21
•
For K = 9,
gain margin = 0.915 dB phase margin  3.41o
%Solution to problem 9.39 clear; clc; for K = 7:11, figure(K6) num = [6*K]; den = [1 6 11 6]; subplot(1,2,1) margin(num,den) subplot(1,2,2) nyquist(num,den) text(0.3,0.3,[’K = ’ num2str(K)]) end Bode Diagram Nyquist Diagram
Gm = 3.098 dB (at 3.3166 rad/sec), Pm = 12.154 deg (at 2.8005 rad/sec) 20
5
0
3 −40 −60
2
−80
1 Imaginary Axis
Magnitude (dB)
4 −20
−100
Phase (deg)
0
K=7 0
−45
−1
−90
−2
−135 −3 −180 −4 −225 −270 −1 10
−5 0
10
1
10
2
0
10
2 Real Axis
Frequency (rad/sec)
Figure 1
22
4
6
Bode Diagram Nyquist Diagram
Magnitude (dB)
Gm = 1.9382 dB (at 3.3166 rad/sec), Pm = 7.3942 deg (at 2.9889 rad/sec) 20 0
5
−20
4
−40
3
−60
2
Imaginary Axis
−80 −100 0 −45
K=8 0
−1
−2
−90 Phase (deg)
1
−135
−3
−180
−4
−225
−5
−270 −1 10
0
10
1
10
2
0
10
2
4 Real Axis
Frequency (rad/sec)
Figure 2
23
6
8
Bode Diagram Nyquist Diagram
Gm = 0.91515 dB (at 3.3166 rad/sec), Pm = 3.408 deg (at 3.1598 rad/sec) 20 6
Magnitude (dB)
0 −20 4 −40 −60 2 Imaginary Axis
−80 −100 0
K=9
0
−45 −2 Phase (deg)
−90 −135 −4 −180 −225 −6 −270 −1 10
0
10
1
10
2
−2
10
0
2
4 Real Axis
Frequency (rad/sec)
Figure 3
24
6
8
Bode Diagram Nyquist Diagram
Gm = 4.7762e−006 dB (at 3.3166 rad/sec), Pm = 3.3141e−006 deg (at 3.3166 rad/sec) 20
Magnitude (dB)
0
6
−20 4
−40 −60
2 Imaginary Axis
−80 −100 0
K = 10
0
−45 −2 Phase (deg)
−90 −135
−4
−180 −6
−225 −270 −1 10
0
10
1
10
2
10
−2
0
2
4 Real Axis
Frequency (rad/sec)
Figure 4
25
6
8
10
Bode Diagram Nyquist Diagram
Gm = −0.82785 dB (at 3.3166 rad/sec), Pm = −2.9625 deg (at 3.462 rad/sec) 50
8
Magnitude (dB)
6 0
4 −50
Imaginary Axis
2
−100 0 −45
K = 11
0
−2
Phase (deg)
−90 −4
−135 −180
−6 −225 −270 −1 10
−8 0
10
1
10
2
−2
10
0
2
4 Real Axis
Frequency (rad/sec)
Figure 5
26
6
8
10
9.40
Figures 1, 2, 3, and 4 on the next four pages plot the Bode diagrams for the following loop transfer functions: 50 (a) L ( s ) = (s + 1)(s + 2) 5 (b) L ( s ) = (s + 1)(s + 2)(s + 5) 5 (c) L ( s ) = 3(s + 1) 10 ( s + 0.5 ) (d) L ( s ) = (s + 1)(s + 2)(s + 5)
%Solution to problem 9.40 clear; clc; figure(1) num = [50]; den = [1 3 2]; margin(num,den) figure(2) num = [5]; den = [1 8 17 10]; margin(num,den) figure(3) num = [5]; den = [1 3 3 1]; margin(num,den) figure(4) num = [10 5]; den = [1 8 17 10]; margin(num,den)
figure(4) num = [10 5]; den = [1 8 17 10]; margin(num,den)
27
Bode Diagram Gm = Inf, Pm = 24.432 deg (at 6.8937 rad/sec) 30 25
Magnitude (dB)
20 15 10 5 0 −5 −10 0
Phase (deg)
−45
−90
−135
−180 −1 10
0
1
10
10
Frequency (rad/sec)
Figure 1
28
Bode Diagram Gm = 28.036 dB (at 4.1248 rad/sec), Pm = Inf 0
Magnitude (dB)
−20 −40 −60 −80 −100 −120 0 −45
Phase (deg)
−90 −135 −180 −225 −270 −1 10
0
1
10
10 Frequency (rad/sec)
Figure 2
29
2
10
Bode Diagram Gm = 4.0836 dB (at 1.7322 rad/sec), Pm = 17.37 deg (at 1.387 rad/sec) 20 10
Magnitude (dB)
0 −10 −20 −30 −40 −50 −60 0 −45
Phase (deg)
−90 −135 −180 −225 −270 −1 10
0
10
Frequency (rad/sec)
Figure 3
30
1
10
Bode Diagram Gm = Inf, Pm = Inf 0 −10
Magnitude (dB)
−20 −30 −40 −50 −60 −70 −80 45
Phase (deg)
0
−45
−90
−135
−180 −1 10
0
1
10
10 Frequency (rad/sec)
Figure 4
31
2
10
9.41
From Example 9.9 we have the loop transfer function 0.5K ( s + 2 ) L ( s ) = s ( s + 12 ) ( s – 4 ) For s = jω, 0.5K ( jω + 2 ) L ( jω ) = jω ( jω + 12 ) ( jω – 4 ) Figures 1, 2, and 3 on the next three pages plot the Bode diagrams of L(jω) for K = 100, 128, and 160, respectively. From these figures we make the following observations: (a) The phasecrossover frequency ωp =
for all K.
(b) For K = 128 the Bode diagram exactly satisfies the conditions L(jω) = 1 and arg{L(jω)} = 180o. Hence the system is on the verge of instability for K = 128. (c) For K = 100 the gain margin is 2.14 dB, which is negative. The system is therefore unstable for K = 100. (d) For K = 160 the gain margin is 1.94 dB, which is positive. The system is therefore stable for K = 160. These observations reconfirm the conclusions reached on the stability performance of L(s) in Example 9.9. %Solution to problem 9.41 clear; clc; figure(1) K=128; num = 0.5*K*[1 2]; den = [1 8 48 0]; [Gm128,Pm128,Wcg128,Wcp128] = margin(num,den) figure(2) K=100; num = 0.5*K*[1 2]; den = [1 8 48 0]; [Gm100,Pm100,Wcg100,Wcp100] = margin(num,den) figure(3) K=160;
32
num = 0.5*K*[1 2]; den = [1 8 48 0]; [Gm160,Pm160,Wcg160,Wcp160] = margin(num,den)
Bode Diagram Gm = −1.9287e−015 dB (at 4 rad/sec), Pm = 0 (unstable closed loop) 40
Verge of Instability
Magnitude (dB)
20
0
−20
−40
−60 225
Phase (deg)
180
135
90 −1 10
0
1
10
10 Frequency (rad/sec)
Figure 1
33
2
10
Bode Diagram Gm = 2.1442 dB (at 4 rad/sec), Pm = −12.476 deg (at 2.8831 rad/sec) 40
Stable
Magnitude (dB)
20
0
−20
−40
−60 225
Phase (deg)
180
135
90 −1 10
0
1
10
10 Frequency (rad/sec)
Figure 2
34
2
10
Bode Diagram Gm = −1.9382 dB (at 4 rad/sec), Pm = 7.9361 deg (at 5.1942 rad/sec) 40
Unstable
Magnitude (dB)
20
0
−20
−40
−60 225
Phase (deg)
180
135
90 −1 10
0
1
10
10 Frequency (rad/sec)
Figure 3
35
2
10
9.42
Let r(t) denote the feedback (return) signal produced by the sensor H(s). Then, following the material presented in Section 9.16, we may, insofar as the computation of the Laplace transform R(s) = L{r(t)} is concerned, replace the sampleddata system of Fig. P9.42 with that shown in Fig. 1 below: X(s)
Impulse sampler
Xδ(s) +
Eδ(s)
Σ
Aδ(s)
B(s)
_
Rδ(s) Impulse sampler
R(s)
Figure 1 The blocks labelled Aδ(s) and B(s) are defined as follows, respectively, 1. Aδ(s) = transfer function of the discretetime components of the system – sT
s ) = Dδ(s) ( 1 – e where τ is the sampling period and Dδ ( s ) = D ( z ) sT s
z=e
2. B(s) = transfer function of the continuoustime components of the system 1 =  G c ( s )G p ( s )H ( s ) s where Gc(s), Gp(s), and H(s) are as indicated in Fig. P9.42. Adapting Eq. (9.94) of the text to the problem at hand: Rδ ( s ) Lδ ( s )  = X δ(s) 1 + Lδ ( s )
(1)
where L δ ( s ) = A δ ( s )B δ ( s )
(2)
The Bδ(s) is itself defined in terms of B(s) by the formula 1 B δ ( s ) = Ts
∞
∑ B ( s – jkωs ) k=∞
where ωs = 2π/Ts.
36
From Fig. P9.42 we note that E ( s ) = H ( s )Y ( s ) Hence, E δ ( s ) = H δ ( s )Y δ ( s )
(3)
where 1 H δ ( s ) = Ts
∞
∑ H ( s – jkωs ) k=∞
Eliminating E δ ( s ) between Eqs. (1) and (3): Lδ ( s ) ⁄ H δ ( s ) Y δ ( s ) =  X δ ( s ) 1 + Lδ ( s ) Changing this result to the ztransform: L(z) ⁄ H (z) Y ( z ) =  X ( z ) 1 + L(z) where Y (z) = Y δ(s)
e
sT s
=z
and so on for L(z), H(z), and X(z).
9.43
From the material presented in Section 9.16 we note that (see Eq. (9.94)) Lδ ( s ) Y δ ( s ) = X δ ( s ) 1 + Lδ ( s )
(1)
where L δ ( s ) = A δ ( s )B δ ( s )
(2)
For the problem at hand, A δ ( s ) and B δ ( s ) are defined as follows, respectively:
37
1. A δ ( s ) = ( 1 – e
– sT s
)D δ ( s )
(3)
where Dδ ( s ) = D ( z )
1 2. B δ ( s ) = Ts
z=e
sT s
∞
∑ B ( s – jkωs ) ,
2π ω s = Ts
k=∞
where 1 B ( s ) =  G ( s ) s K = 3 s We are given z1 D ( z ) = K 1.5 +  z –1
= K ( 2.5 – z ) Hence, D δ ( s ) = K ( 2.5 – e
– sT s
)
The use of Eq. (3) thus yields Aδ ( s ) = K ( 1 – e
– sT s
– sT s
) ( 2.5 – e
= K ( 2.5 – 3.5e
– sT s
+e
)
– 2sT s
)
Correspondingly, we may write A ( z ) = K ( 2.5 – 3.5z
–1
–2
+z )
Next, with B(s) = K/s3 we may write
38
–1
b(t ) = L { B(s)} 2
Kt = u ( t ) 2 Hence, ∞
Bδ ( s ) =
∑ b [ n ]e
– nsT s
k=∞
K = 2
∞
∑n
2 – nsT s
e
k=0
which may be summed to yield B ( z ) = Bδ ( s )
e
Ts
=z
2 KTs
2
–1
–1
z(z + 1) K T s z (1 + z ) =  3 =  –1 3 4 (z – 1) 4 (1 – z ) The use of Eq. (2) yields L ( z ) = Lδ ( s )
e
sT s
=z
= A ( z )B ( z ) 2
= K ( 2.5 – 3.5z
–1
–1 –1 –2 T s z ( 1 + z ) + z )  –1 3 4 (1 – z )
2 2
K T s z –1 ( 2.5 – z –1 ) ( 1 + z –1 ) =  –1 2 4 (1 – z ) Finally, the closedloop transfer function of the system is L(z) T ( z ) = 1 + L(z) where L(z) is defined by Eq. (4).
9.44
Here again following the material presented in Section 9.16 we may state that Y (z) L(z)  = X (z) 1 + L(z)
39
where L ( z ) = Lδ ( s )
e
sT s
=z
L δ ( s ) = A δ ( s )B δ ( s ) Aδ ( s ) = ( 1 – e 1 B δ ( s ) = Ts
– sT s
)
∞
∑ B ( s – jkωs ) ,
2π ω s = Ts
k=∞
1 B ( s ) =  G ( s ) s 5 = 2 s (s + 2) 5⁄4 5⁄2 5⁄4 = –  +  + 2 s+2 s s To calculate B δ ( s ) we first note that 5 5 5 –2t b ( t ) = –  u ( t ) +  t u ( t ) +  e u ( t ) 4 2 4 Hence, ∞
Bδ ( s ) =
∑ b [ n ]e n=∞ ∞
– nsT s
∞
∞
n=0
n=0
– nsT s 5 – nsT s 5 – 2nsT s – nsT s 5 = –  ∑ e +  ∑ ne +  ∑ e e 4 2 4 n=0
Correspondingly, we may write –1
T sz 5 5 1 5 1  +   +  B ( z ) = –  – 1 2 – –1 2 4 1–z 4 1 – e 2T s z –1 (1 – z ) From the expression for A δ ( s ) : –1
A(z) = (1 – z ) The loop transfer function of the system is therefore
40
L ( z ) = A ( z )B ( z ) –1
T sz 1 5 5 1 5  +  = ( 1 – z ) –  + – 1 2 – –1 4 1–z 2 4 1 – e 2T s z –1 (1 – z ) –1
Putting this expression on a common denominator: – 2T
–1
– 2T
–1
s ) + ( 1 – e s ( 1 + 2τ )z ) ) 5 z ( ( – 1 + 2T s + e L ( z ) =  – 2T s – 1 –1 4 (1 – z )(1 – e z )
The closedloop transfer function is L(z) T ( z ) = 1 + L(z) (a) For sampling period Ts = 0.1s, the loop transfer function takes the value –1
– 0.2
–1
–1
– 0.2
–1
5 z ( ( – 1 + 0.2 + e ) + ( 1 – e ( 1.2 )z ) ) L ( z ) =  –1 – 0.2 – 1 4 (1 – z )(1 – e z ) 5 z ( 0.019 + 0.017z ) =  4 ( 1 – z –1 ) ( 1 – 0.819z –1 ) The closedloop transfer function is therefore –1
–1
5 z ( 0.019 + 0.017z )  4 ( 1 – z –1 ) ( 1 – 0.819z –1 ) T ( z ) = –1 –1 5 z ( 0.019 + 0.017z ) 1 +  4 ( 1 – z –1 ) ( 1 – 0.819z –1 ) 5 –1 –1  z ( 0.019 + 0.017z ) 4 = 5 –1 –1 –1 –1 ( 1 – z ) ( 1 – 0.819z ) +  z ( 0.019 + 0.017z ) 4 5 –1 –1  z ( 0.019 + 0.017z ) 4 = –1 –2 1 – 1.795z + 0.840z T(z) has a zero at z = 0.895 and a pair of complex poles at z = 0.898 + j0.186. The poles are inside the unit circle and the system is therefore stable.
41
(b) For sampling period Ts = 0.05s, L(z) takes the value 5 –1 – 0.1 – 0.1 –1  z ( ( – 1 + 0.1 + e ) + ( 1 – e ( 1.1 ) )z ) 4 L ( z ) = –1 – 0.1 – 1 (1 – z )(1 – e z ) –1
–1
5 z ( 0.005 + 0.005z ) =  4 ( 1 – z –1 ) ( 1 – 0.996z –1 ) The closedloop transfer function is therefore –1
–1
5 z ( 0.005 + 0.005z )  4 ( 1 – z –1 ) ( 1 – 0.996z –1 ) T ( z ) = –1 –1 5 z ( 0.005 + 0.005z ) 1 +  4 ( 1 – z –1 ) ( 1 – 0.996z –1 ) 5 –1 –1  z ( 0.005 + 0.005z ) 4 = 5 –1 –1 –1 –1 ( 1 – z ) ( 1 – 0.996z ) +  z ( 0.005 + 0.005z ) 4 –1
–1
0.006z ( 1 + z ) = –1 –2 1 – 1.989z + 1.001z T(z) has a zero at z ≈ 1 and a pair of complex poles at z = 0.995 + j0.109. The poles lie inside the unit circle in the zplane and the sampleddata system remains stable. Note, however, the reduction in the sampling period has pushed the closedloop poles much closer to the unit circle.
42
Advanced Problems 9.45
(a) We are given the loop transfer function L ( s ) = G ( s )H ( s ) βA = s ω 1 + Q  + 0 ω0 s βAω 0 s = 2 2 Qs + ω 0 s + Qω 0 where ω0 is the center frequency and Q is the quality factor. The poles of L(s) are given by the roots of the quadratic equation: 2
2
Qs + ω 0 s + Qω 0 = 0 That is, 1 2 2 2 s =  ( – ω 0 − + ω 0 – 4Q ω 0 ) 2Q ω0 2 =  ( – 1 ± j 4Q – 1 ) 2Q With Q assumed to be large, we may approximate the poles as ω0 ω s ≈  ( – 1 ± j2Q ) = – 0 ± jω 0 2Q 2Q which lie inside the left half of the splane and very close to the jωaxis. The loop transfer function L(s) has a single zero at s = 0. Hence, the root locus of the closedloop feedback amplifier starts at the two poles, tracing out the path illustrated in Fig. 1 for Q = 100 and ω0 = 1. %Solution to problem 9.45 w = 1; Q = 100; num = [w 0]; den = [Q w Q*w^2]; figure(1) rlocus(num,den)
43
Root Locus 1
0.8
0.6
0.4
Imag Axis
0.2
0
−0.2
−0.4
−0.6
−0.8
Figure 1
−1 −1.6
−1.4
−1.2
−1
−0.8
−0.6
−0.4
−0.2
0
Real Axis
(b) The closedloop transfer function is G(s) T ( s ) = 1 + G ( s )H ( s ) which may be expressed in the form A  1 + βA T ( s ) = ω Q s Q   + 1 +  0 1 + βA ω 0 1 + βA s From this expression we immediately deduce that the application of feedback produces two effects: 1. It reduces the gain of the amplifier by the factor 1 +βA. 2. It reduces the quality factor Q by the same amount 1 + βA. By definition, we have ω Q = 0B where B is the bandwidth. Therefore, for fixed ω0, if the quality factor Q is reduced by the amount 1 + βA as a result of applying the feedback, the closedloop bandwidth is increased by the same amount.
9.46
(a) The “sensor” transfer function is 1 H ( s ) = s
44
Hence the closedloop gain (i.e., transfer function) of the phaselocked loop model shown in Fig. P9.46 is V (s) G(s)  = Φ1 ( s ) 1 + G ( s )H ( s ) K 0H (s)(1 ⁄ K v) = 1 1 +  K 0 H ( s ) s sK 0 H ( s ) ( 1 ⁄ K v ) = s + K 0H (s)
(1)
(b) For positive frequencies that satisfy the condition K0H(jω) >> ω, we may approximate Eq. (1) as V ( s ) sK 0 H ( s ) ( 1 ⁄ K v )  ≈ K 0H (s) Φ1 ( s ) s = Kv Correspondingly, we may write 1 d v ( t ) ≈   φ 1 ( t ) K v dt
9.47
The error signal e(t) is defined as the difference between the actuating or target signal yd(t) and the controlled signal (i.e., actual response) y(t). Expressing these signals in terms of their respective Laplace transforms, we may write E(s) = Y d(s) – Y (s) = [ 1 – T ( s ) ]Y d ( s )
(1)
G ( s )H ( s ) = 1 –  Y d ( s ) 1 + G ( s )H ( s ) 1 =  Y d ( s ) 1 + G ( s )H ( s )
(2)
The steadystate error of a feedback control system is defined as the value of the error signal e(t) when time t approaches infinity. Denoting this quantity by ∈ ss, we may thus write (3) ∈ ss = lim e ( t ) s→0
Using the final value theorem of Laplace transform theory described in Chapter 6, we may redefine the steadystate error ∈ ss in the equivalent form
45
∈ ss = lim sE ( s )
(4)
s→0
Hence, substituting Eq. (2) into (4), we get sY d ( s ) ∈ ss = lim s → 0 1 + G ( s )H ( s )
(5)
Equation (5) shows that the steadstate error ∈ ss of a feedback control system depends on two quantities: • The openloop transfer function G(s)H(s) of the system • The Laplace transform Yd(s) of the target signal yd(t) However, for Eq. (5) to be valid, the closedloop control system of Fig. 9.14 must be stable. In general, G(s)H(s) may be written in the form of a rational function as follows: P(s) G ( s )H ( s ) = (6) p s Q1 ( s ) where neither the polynomial P(s) nor Q1(s) has a zero at s = 0. Since 1/s is the transfer function of an integrator, it follows that p is the number of free integrators in the loop transfer function G(s)H(s). The order p is referred to as the type of the feedback control system. We thus speak of a feedback control system being of type 0, type 1, type 2, and so on for p = 0,1,2,..., respectively. In light of this classification, we next consider the steadystate error for three different input functions. Step Input For the step input yd(t) = u(t), we have Yd(s) = 1/s. Hence, Eq. (5) yields the steadystate error 1 ∈ ss = lim 1 + G ( s )H ( s ) s→0 1 = (7) 1 + Kp where Kp is called the position error constant, defined by K p = lim G ( s ) H ( s ) s→0
P(s) = lim p s → 0 s Q (s) 1
(8)
For p > 1, Kp is unbounded and therefore ∈ ss = 0. For p = 0, Kp is finite and therefore ∈ ss ≠ 0 . Accordingly, we may state that the steadystate error for a step input is zero for a feedback control system of type 1 or higher. On the other hand, for a system of type 0, the steadystate error is not zero, and its value is given by Eq. (7).
46
Ramp Input For the ramp input yd(t)  tu(t), we have Yd = 1/s2. In this case. Eq. (5) yields 1 ∈ ss = lim s → 0 s + sG ( s )H ( s ) 1 = Kv where Kv is the velocity error constant, defined by
(9)
K v = lim sG ( s )H ( s ) s→0
P(s) = lim s → 0 s p1 Q ( s ) 1
(10)
For p > 2, Kv is unbounded and therefore ∈ ss = 0. For p = 1, Kv is finite and ∈ ss ≠ 0 . For p = 0, Kv is zero and ∈ ss = ∞ . Accordingly, we may state that the steadystate error for a ramp input is zero for a feedback control system of type 2 or higher. For a system of type 1, the steadystate error is not zero, and its value is given by Eq. (9). For a system of type 0, the steadystate error is unbounded. Parabolic Input For the parabolic input yd(t) = (t2/2)u(t), we have Yd = 1/s3. The use of Eq. (5) yields 1 ∈ ss = lim 2 2 s → 0 s + s G ( s )H ( s ) 1 = Ka where Ka is called the acceleration error constant, defined by P(s) K a = lim s → 0 s p2 Q ( s ) 1
(11)
(12)
For p > 3, Ka is unbounded and therefore ∈ ss = 0. For p = 2, Ka is finite and therefore ∈ ss = 0. For p = 0,1, Ka is zero and therefore ∈ ss = ∞ . Accordingly, we may state that for a parabolic input, the steadystate error is zero for a type 3 system or higher. For a system of type 2, the steadystate error is not zero, and its value is given by Eq. (11). For a system of type 0 or type 1, the steadystate error is unbounded. In Table 1, we present a summary of the steadystate errors according to system type as determined above.
47
Additional Notes A type 0 system is referred to as a regulator. The object of such a system is to maintain a physical variable of interest at some prescribed constant value despite the presence of external disturbances. Examples of regulator systems include: • Control of the moisture content of paper, a problem that arises in the papermaking process • Control of the chemical composition of the material outlet produced by a reactor • Biological control system, which maintains the temperature of the human body at approximately 37oC despite variations in the temperature of the surrounding environment Table 1: SteadyState Errors According to System Type
Type 0 Type 1 Type 2
Step
Ramp
Parabolic
1/(1 + Kp) 0 0
∞ 1/Kv 0
∞ ∞ 1/Ka
Type 1 and higher systems are referred to as servomechanisms. The objective here is to make a physical variable follow, or track, some desired timevarying function. Examples of servomechanisms include: • • •
Control of a robot, in which the robot manipulator is made to follow a preset trajectory in space Control of a missile, guiding it to follow a specified flight path Tracking of a maneuvering target (e.g., enemy aircraft) by a radar system
Example: Figure P9.47 shows the block diagram of a feedback control system involving a dc motor. The dc motor is an electromechanical device that converts dc electrical energy into rotational mechanical energy. Its transfer function is approximately defined by K G ( s ) = s ( τL s + 1 ) where K is the gain and τL is the load time constant. Here it is assumed that the field time constant of the dc motor is small compared to the load time constant. The transfer function of the controller is ατs + 1 H ( s ) = τs + 1 The requirement is to find the steadystate errors of this control system. The loop transfer function of the control system in Fig. P9.47 is K ( ατs + 1 ) (13) G ( s )H ( s ) = s ( τ L s + 1 ) ( τs + 1 ) which represents a type 1 system. Comparing Eq. (13) with Eq. (6), we have for the example at hand
48
αK 1 P ( s ) =  s +  τL ατ 1 1 Q 1 ( s ) = s +  s +  τ τ L p = 1 Hence the use of these values in Eqs. (8), (10), and (12) yields the following steadystate errors for the control system of Fig. P9.47: (a) Step input: ∈ ss = 0. (b) Ramp input: ∈ ss = 1/K = constant. (c) Parabolic input: ∈ ss = ∞ . The effect of making the gain K large is to reduce the steadystate error of the system due to a ramp input. This, in turn, improves the behavior of the system as a velocity control system. +
Input
Controller ατs + 1 τs + 1
Σ

DC motor K s(τLs + 1)
.
Output
Figure P9.47
9.48
From Fig. P9.20 the closedloop transfer function of the feedback system is 2
K ⁄ ( s + 2s ) T ( s ) = 2 1 + K ⁄ ( s + 2s ) K = 2 s + 2s + K
(1)
In general, we may express T(s) in the form 2
ωn T ( 0 ) T ( s ) = 2 2 s + 2ζω n s + ω n
(2)
Hence, comparing Eqs. (1) and (2): T (0) = 1
(3)
ωn =
(4)
K
ζ = 1 ⁄ ωn = 1 ⁄ K
(5)
We are given K = 20, for which the use of Eqs. (4) and (5) yields ωn =
20 = 4.472 rad/second
ζ = 0.224
49
The time constant of the system is 1 τ =  = 1 second ζω n 9.49
The loop transfer function of the feedback control system is L ( s ) = G ( s )H ( s ) 0.2K P = (s + 1)(s + 3) The closedloop transfer function of the system is L(s) T ( s ) = 1 + L(s) 0.2K P = 2 s + 4s + ( 3 + 0.2K P )
(1)
For a secondorder system defined by 2
T ( 0 )ω n T ( s ) = 2 2 s + 2ζω n s + ω n the damping factor is ζ and the natural frequency is ωn. Comparing Eqs. (1) and (2):
(2)
2ζω n = 4 2
ω n = 3 + 0.2K P 2
T ( 0 )ω n = 0.2K P Hence, ωn =
(3)
3 + 0.2K P
ζ = 2 ⁄ 3 + 0.2K P
(4)
T ( 0 ) = 0.2K p ⁄ ( 3 + 0.2K P ) We are required to have ω n = 2 rad/s
(5)
Hence the use of Eq. (3) yields 3 + 0.2K P = 2 That is, 4–3 K p =  = 5 0.2
50
Next, the use of Eq. (4) yields 2 ζ = 3 + 0.2 × 5 2 =  = 1 4 which means that the system is critically damped. The time constant of the system is defined by 1 τ = ζω n 1 =  = 0.5 seconds 1×2
(Note: The use of Eq. (5) yields 0.2 × 5 1 T ( 0 ) =  = 3 + 0.2 × 5 4
9.50
The loop transfer function of the system is K 0.25 L ( s ) = K P + I  s s + 1 K I ⁄ K P 1 = 0.25K P 1 +   s s + 1 With KI/KP = 0.1 we have 0.25K P ( s + 0.1 ) L ( s ) = s(s + 1)
(1)
The root locus of the system is therefore as shown in Fig. 1: jω
splane
x
1
0 0.1
x
σ 0
Figure 1
51
The closedloop transfer function of the system is L(s) T ( s ) = 1 + L(s) Substituting Eq. (1) into (2): 0.25K P ( 5 + 0.1 ) T ( s ) = s ( s + 1 ) + 0.25K P ( s + 0.1 )
(2)
(3)
For a closedloop at s = 5 we require that the denominator of T(s) satisfy the following equation (4) (s + 5)(s + a) = 0 where s = a is the other closedloop pole of the system. Comparing the denominator of Eq. (3) with Eq. (4): 5 + a = 1 + 0.25K P 5a = 0.1 × 0.25K P Solving this pair of equations for KP and a, we get 20 K P =  = 16.33 1.225 a = 0.005 × K P = 0.08165
9.51
The loop transfer function of the PD controller system is 1 L ( s ) = ( K P + K D s )  s ( s + 1 ) K P 1 = K D s +  K D s ( s + 1 ) We are given KP/KD = 4. Hence, K D(s + 4) L ( s ) = s(s + 1) We may therefore sketch the root locus of the system as follows:
52
%Solution to problem 9.51 clear; clc; figure(1) K = 1; num = K*[1 4]; den = [1 (1+K) K*4]; rlocus(num,den) figure(2) K = 3; num = K*[1 4]; den = [1 (1+K) K*4]; rlocus(num,den) Root Locus
3
2 + j2sqrt(2) 2
Imag Axis
1
0
−1
−2
2 − j2sqrt(2) −3
−14
−12
−10
−8
−6
−4
Real Axis
53
−2
0
The closedloop transfer function of the system is L(s) T ( s ) = 1 + L(s) K D(s + 4) = 2 s + s + K D(s + 4)
(1)
We are required to choose KD so as to locate the closedloop poles at s = – 2 ± j2 2 . That is, the characteristic equation of the system is to be ( s + 2 + j2 2 ) ( s + 2 – j2 2 ) = 0 or 2
(2)
s + 4s + 12 = 0 Comparing the denominator of Eq. (1) with Eq. (2): 1 + KD = 4 4K D = 12 Both of these conditions are satisfied by choosing KD = 3
9.52
(a) For a feedback system using PI controller, the loop transfer function is K L ( s ) = K P + I L′ ( s ) s where L′ ( s ) is the uncompensated loop transfer function. For s = jω we may write K 1 K P + I =  ( K I + jK P ω ) jω jω The contribution of the PI controller to the loop phase response of the feedback system is –1 K ω P φ = – 90° + tan  KI For all positive values of KP/KI, the angle tan1(KPω/KI) is limited to the range [0,90o]. It follows therefore that the use of a PI controller introduces a phase lag into the loop phase response of the feedback system, as shown in Fig. 1. (b) For a feedback system using PD controller, the loop transfer function of the system may be expressed as L ( s ) = ( K P + K D s )L′ ( s ) where L′ ( s ) is the uncompensated loop transfer function. For s = jω, the contribution of the PD controller to the loop phase response of the system is tan1(KDω/KP). For all
54
positive values of KD/KP, this contribution is limited to the range [0,90o]. We therefore conclude that the use of PD controller has the effect of introducing a phase lead into the loop phase response of the system, as illustrated in Fig. 2. %Solution to problem 9.52 clear; clc; w = 0:0.01:5*pi; Kp = 1.2; Ki = 1; y = pi/2 + atan(w*Kp/Ki); plot(w,y) xlabel(’\omega’) ylabel(’Phase hi’) title(’i/2 + tan^{1}(\omega Kp/Ki)’) figure(2) y = pi/2 + atan(w*Kp/Ki); plot(w,y) xlabel(’\omega’) ylabel(’Phase hi’) title(’tan^{1}(\omega Kp/Ki)’) −1
−π/2 + tan (ω Kp/Ki) 0
−0.2
−0.4
Kp/Ki = 1.2
Phase φ
−0.6
−0.8
−1
−1.2
−1.4
Figure 1 −1.6
0
2
4
6
8 ω
10
12
55
14
16
tan−1(ω Kp/Ki) 0
−0.2
−0.4 Kp/Ki = 1.2
Phase φ
−0.6
−0.8
−1
−1.2
−1.4
−1.6
9.53
Figure 2 0
2
4
6
8 ω
10
12
14
16
K
I The transfer function of the PID controller is defined by K P + + K D s . The requirement s is to use this controller to introduce zeros at s = 1 +j2 into the loop transfer function of the feedback system. Hence, 2
2
s + ( K P ⁄ K D )s + ( K I ⁄ K D ) = ( s + 1 ) + ( 2 )
2
2
= s + 2s + 5 Comparing terms: KP = 2 KD K I = 5 D The loop transfer function of the compensated feedback system is therefore K K P + I + K D s s L ( s ) = (s + 1)(s + 2) 2
K D ( s + 2s + 5 ) = s(s + 1)(s + 2)
56
Figure 1 displays the root locus of L(s) for varying KD. From this figure we see that the root locus is confined to the left half of the splane for all positive values of KD. The feedback system is therefore stable for KD > 0. %Solution to problem 9.53 clear; clc; num = [1 2 5]; den = [1 3 2 0]; rlocus(num,den) Root Locus 2
1.5
1
Imag Axis
0.5
0
−0.5
−1
−1.5
−2 −3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
Figure 1
Real Axis
9.54
Let KP denote the action of the proportional controller. We may then express the loop transfer function of the controlled inverted pendulum as K P ( s + 3.1 ) ( s – 3.1 ) L ( s ) = 2 s ( s + 4.4 ) ( s – 4.4 ) which has zeros at s = +3.1 and poles at s = 0 (order 2) and s = +4.4. The resulting root locus is sketched in Fig. 1. This diagram shows that for all positive values of KP, the
57
closedloop transfer function of the system will have a pole in the righthalf plane and a pair of poles on the jωaxis. splane double pole
x
0 3.1
4.4
x
0 3.1
0
x
σ
4.4
Figure 1 Accordingly, the inverted pendulum cannot be stabilized using a proportional controller for KP > 0. The reader is invited to reach a similar conclusion for KP < 0. The stabilization of an inverted pendulum is made difficult by the presence of two factors in the loop transfer function L(s): 1. A double pole at s = 0. 2. A zero at s = 3.1 in the righthalf plane. We may compensate for (1) but nothing can be done about (2) if the compensator is itself to be stable. Moreover, we have to make sure that the transfer function of the compensator is proper for it to be realizable. We may thus propose the use of a compensator (controller) whose transfer function is 2
K s ( s – 4.4 ) C ( s ) = ( s + 1 ) ( s + 2 ) ( s + 3.1 ) The compensated loop transfer function is therefore (after performing polezero cancellation) L′ ( s ) = C ( s )L ( s ) K ( s – 3.1 ) = ( s + 1 ) ( s + 2 ) ( s + 4.4 ) Figure 2 shows a sketch of the root locus of L′ ( s ) . From this figure we see that the compensated system is stable provided that the gain factor K satisfies the condition 4×2 0 < K <  = 2.839 3.1
58
%Solution to problem 9.54 clear; clc; num = [1 0 961/100]; den = [1 0 484/25 0 0]; rlocus(num,den) figure(2); clc; num = [1 3.1]; den = [1 37/5 76/5 44/5]; rlocus(num,den) Root Locus
4
3
2
Imag Axis
1
0
−1
−2
−3
−4
−4
−3
−2
−1
0
1
2
3
4
Real Axis
Root Locus 30
20
Imag Axis
10
0
−10
−20
−30 −5
−4
−3
−2
−1
0
Real Axis
59
1
2
3
9.55
For linear control systems, the transient response is usually measured in terms of the step response. Typically, the step response, denoted by y(t), is oscillatory as illustrated in Fig. P9.55. In describing such a response, we have two conflicting criteria: swiftness of the response, and closeness of the response to the desired response. Swiftness of the response is measured in terms of the rise time and peak time. Closeness of the response to the desired response is measured in terms of the percentage overshoot and settling time. these four quantities are defined as follows: • • •
•
Rise time, Tr, is defined as the time taken by the step response to rise from 10% to 90% of its final value y(∞). Peak time, Tp, is defined as the time taken by the step response to reach the overshoot (overall) maximum value ymax. Percentage overshoot, P.O., is defined in terms of the maximum value ymax and final value y(∞) by y max – y ( ∞ ) P.O. =  × 100 y(∞) Settling time, Ts, is defined as the time required by the step response to settle within +δ% of the final value y(∞), where δ is user specified.
Figure P9.55 illustrates the definitions of these four quantities, assuming that y(∞) = 1.0. They provide an adequate description of the step response y(t). Most importantly, they lend themselves to measurement. Note that in the case of an overdamped system, the peak time and percentage overshoot are not defined. In such a case, the step response of the system is specified simply in terms of the rise time and settling time. For reasons that will become apparent later, the underdamped response of a secondorder system to a step input often provides an adequate approximation to the step response of a linear feedback control system. Accordingly, it is of particular interest to relate the abovementioned quantities to the parameters of a secondorder system. Example: Consider an underdamped secondorder system of damping ratio ζ and natural frequency ωn. Determine the rise time, peak time, percentage overshoot, and settling time of the system. For settling time, use δ = 1. Solution: Unfortunately, it is difficult to obtain an explicit expression for the rise time Tr in terms of the damping ratio ζ and natural frequency ωn. Nevertheless, it can be determined by simulation. Table 9.2 presents the results of simulation for the range 0.1 < ζ < 0.9. In this table we have also included the results obtained by using the approximate formula: 1 T r ≈  ( 0.60 + 2.16ζ ) ωn This formula yields fairly accurate results for 0.3 < ζ < 0.8, as can be seen from Table 1.
60
To determine the peak time Tp, we may differentiate Eq. (...) with respect to time t and then set the result equal to zero. We thus obtain the solutions t = ∞ and nπ t = , n = 0, 1, 2, … 2 ωn 1 – ζ Table 1: Normalized Rise Time as a Function of the Damping Ratio ζ for a SecondOrder System ωnTr ζ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Simulation
Approximate Value
1.1 1.2 1.3 1.45 1.65 1.83 2.13 2.5 2.83
0.82 1.03 1.25 1.46 1.68 1.90 2.11 2.33 2.54
The solution t = ∞ defines the maximum of the step response y(t) only when ζ > 1 (i.e., the critically damped or overdamped case). We are interested in the underdamped case. The first maximum of y(t) occurs for n = 1; hence the peak time is π T p = 2 ωn 1 – ζ The first maximum of y(t) also defines the percentage overshoot. Thus, putting t = Tp in Eq. (...) and simplifying the result, we get – πζ ⁄ ( 1 – ζ ) 2
y max = 1 + e The percentage overshoot is therefore – πζ ⁄ ( 1 – ζ ) 2
P.O. = 100e Finally, to determine the settling time Ts we seek the time t for which the step response y(t) defined in Eq. (...) decreases and stays within +δ% of the final value y(∞) = 1.0. For δ = 1, this time is closely approximated by the decaying exponential factor e shown by – ζω n T
e or
≈ 0.01
4.6 T s ≈ ζω n
61
– ζω n t
, as
9.56
We often find that the poles and zeros of the closedloop transfer function T(s) of a feedback system are grouped in the complex splane roughly in the manner illustrated in Fig. P9.56. In particular, depending on how close the poles and zeros are to the jωaxis we may identify two groupings: 1. Dominant poles and zeros, which are those poles and zeros of T(s) that lie close to the jωaxis. They are said to be dominant because they exert a profound influence on the frequency response of the system. Another way of viewing this situation is to recognize that poles close to the jωaxis correspond to large time constants of the system. The contributions made by these poles to the transient response of the system are slow and therefore dominant. 2. Insignificant poles and zeros, which are those poles and zeros of T(s) that are far removed from the jωaxis. They are said to be insignificant because they have relatively little influence on the frequency response of the system. In terms of timedomain behavior, poles that are far away from the jωaxis correspond to small time constants. The contributions of such poles to the transient response of the system are much faster and therefore insignificant. Let s = a and s = b define the boundaries of the dominant poles and the insignificant poles, as indicated in Fig. P9.56. As a rule of thumb, the grouping of poles into dominant poles and insignificant poles is justified if the ratio b/a is greater than 4. Given that we have a highorder feedback system whose closedloop transfer function fits the picture portrayed in Fig. P9.56, we may then approximate the system by a reducedorder model simply by retaining the dominant poles and zeros of T(s). The use of a reducedorder model in place of the original system is motivated by the following considerations: • •
Loworder models are simple; they are therefore intuitively appealing in system analysis and design. Loworder models are less demanding in computational terms than highorder ones.
A case of particular interest is when the use of a firstorder or secondorder model as the reducedorder model is justified, for then we can exploit the wealth of information available on such loworder models. When discarding poles and zeros in the derivation of a reducedorder model, it is important to rescale the fain of the system. Specifically, the reducedorder model and the original system should have the same gain at zero frequency. Example: Consider again the linear feedback amplifier. Assuming that K = 8, approximate this system using a secondorder model. Use the step response to assess the quality of the approximation. Solution: For K = 8, the characteristic equation of the feedback amplifier is given by
62
3
2
s + 6s + 11s + 54 = 0 Using the computer, the roots of this equation are found to be s = – 5.7259, s = – 0.1370 ± j3.0679 The locations of these three roots are plotted in Fig. 2. We immediately observe from the polezero map of Fig. 2 that the poles at s = 0.1370 + j3.0679 are the dominant poles, and the poles at s = 5.7259 is an insignificant pole. The numerator of the closedloop transfer function T(s) consists simply of 6K. Accordingly, the closedloop gain of the feedback amplifier may be approximated as 6K′ T ′ ( s ) ≈ ( s + 0.1370 + j3.0679 ) ( s + 0.1370 – j3.0679 ) 6K′ = 2 s + 0.2740s + 9.4308 To make sure that T′ ( s ) is scaled properly relative to the original T(s), the gain K′ is chosen as follows: 9.4308 K′ =  × 8 = 1.3972 54 for which we thus have T′ ( 0 ) = T(0). In light of Eq. (9.41), we set 2ζω n = 0.2740 2
ω n = 9.4308 from which we readily find that ζ = 0.0446 ω n = 3.0710 The step response of the feedback amplifier is therefore underdamped. The time constant of the exponentially damped response is from Eq. (9.44) 1 1 τ =  =  = 8.2993s ζω n 0.137 The frequency of the exponentially damped response is 2
ω n 1 – ζ = 3.0710 1 – ( 0.0446 )
2
= 3.0679 rad/s Figure 3 shows two plots, one displaying the step response of the original thirdorder feedback amplifier with K = 8, and the other displaying the step response of the
63
approximating secondorder model with K′ = 1.3972. The two plots are very similar, which indicates that the reducedorder model is adequate for the example at hand. Imaginary Dominant x pole
splane
Insignificant pole
x
Real Dominant x pole
Figure 1: Polezero map for feedback amplifier with gain K = 8.
Root Locus 6
4
Imag Axis
2
0
−2
−4
−6 −9
−8
−7
−6
−5
−4
−3
−2
−1
0
Real Axis
Figure 2: Root locus of original 3rd order system
64
1
1.8 3rd Order Reduced 2nd Order 1.6
1.4
Response
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
35
40
45
Time
Figure 3: Step response for thirdorder system shown as dashed curve, and step response for (reduced) secondorder system shown as solid curve.
%Solution to problem 9.56 %Original 3rd order system num = 6*8; den = [1 6 11 54]; figure(2) rlocus(num,den) figure(3) [order3,T] = step(tf(num,den)); %Reduced 2nd order system num = 6*1.3972; den = [1 0.2740 9.4308]; [order2,T] = step(tf(num,den)); plot(T,order3,’b’,T,order2,’r’) legend(’3rd Order’,’Reduced 2nd Order’) xlabel(’Time’) ylabel(’Response’)
65
9.57
Consider first the original system described by the transfer function 48 T ( s ) = 2 ( s + 5.7259 ) ( s + 0.2740s + 9.4308 ) The corresponding Bode diagram, obtained by putting s = jω, is plotted in Fig. 1 on the next page. Consider next the reducedorder model described by the transfer function 8.3832 T′ ( s ) = 2 s + 0.2740s + 9.4308 which is obtained from T(s) by ignoring the distant pole at s = 5.7259 and readjusting the constant gain factor. Figure 2, on the page after the next one, plots the Bode diagram of the approximating system. Comparing these two figures, 1 and 2, we see that the frequency responses of the original feedback system and its reducedorder approximate model are close to each other over the frequency range 0 < ω < 4.0.
%Solution to problem 9.57 clear; clc; num = [8.3832]; den = [1 0.2740 9.4308]; margin(num,den) figure(2); clc; num = [48]; den = [1 6 11 54]; margin(num,den) Bode Diagram Gm = Inf, Pm = 7.9113 deg (at 4.2112 rad/sec) 20
Second Order
Magnitude (dB)
10
0
−10
−20
−30 0
Figure 1: Frequency response of original system
Phase (deg)
−45
−90
−135
−180 0 10
1
10 Frequency (rad/sec)
66
Bode Diagram Gm = −12.04 dB (at 3.3167 rad/sec), Pm = −25.849 deg (at 4.0249 rad/sec) 20
Third Order
Magnitude (dB)
0 −20 −40 −60 −80 −100
Figure 2: Frequency response of reducedorder model
0
Phase (deg)
−45 −90 −135 −180 −225 −270 0 10
1
2
10
10
Frequency (rad/sec)
9.58
The guidelines used in the classical (frequencydomain) approach to the design of a linear feedback control system are usually derived from the analysis of secondorder servomechanism dynamics. Such an approach is justified on the following grounds. First, when the loop gain is large the closedloop transfer function of the system develops a pair of dominant complexconjugate poles. Second, a secondorder model provides an adequate approximation to a higherorder model whose transfer function is dominated by a pair of complexconjugate poles. Consider then a secondorder system whose loop transfer function is given by K T ( s ) = s ( τs + 1 ) This system was studied in Section 9.11. When the loop gain K is large enough to produce an underdamped step response (i.e., the closedloop poles form a complexconjugate pair), the damping ratio and natural frequency of the system are defined by, respectively (see Drill Problem 9.8), 1 ζ = 2 τK
and
ωn =
K τ
Accordingly, we may redefine the loop transfer function in terms of ζ and ωn as follows: 2
ωn L ( s ) = s ( s + 2ζω n )
(1)
By definition, the gain crossover frequency ωg is determined from the relation L ( jω g ) = 1 Therefore, putting s = jωg in Eq. (1) and solving for ωg, we get
67
ωg = ωn
4
4ζ + 1 – 2ζ
2
(2)
The phase margin, measured in degrees, is therefore –1 ω g φ m = 180° – 90° – tan  2ζω n – 1 2ζω n = tan  ωg – 1 2ζ = tan  4 2 4ζ + 1 – 2ζ
(3)
Equation (3) provides an exact relationship between the phase margin φ m , a quantity pertaining to the openloop frequency response, and the damping ratio ζ , a quantity pertaining to the closedloop step response. This exact relationship is plotted in Fig. 1. For values of the damping ratio in the range 0 < ζ < 0.6, Eq. (3) is closely approximated by φ m ≈ 100ζ , 0 < ζ < 0.6 Equivalently, we may write φm , ζ = 0 < ζ < 0.6 (4) 100 where, as mentioned previously, the phase margin φ m is measured in degrees. This approximation is included as the dashed line in Fig. 1. Once we have obtained a value for the damping ratio ζ , we can use Eq. (2) to determine the corresponding value of the natural frequency ωn in terms of the gain crossover frequency ωg. With the values of ζ and ωn at hand, we can then go on to determine the rise time, peak time, percentage overshoot, and settling time as descriptors of the step response using the formulas derived in ..........
68
80
70
Phase Margin φm (degrees)
60
Exact
50
Approximate
40
30
Figure 1
20
10
0
0
0.1
0.2
0.3
0.4
0.5 0.6 Damping ratio ξ
0.7
0.8
0.9
1
Computer Experiments 9.59
We are given K L ( s ) = 3(s + 1) The MATLAB command for finding the value of K corresponding to a desired pair of closedloop poles, and therefore damping factor ζ , is given below. The root locus of the unity feedback system is presented in Fig. 1. A damping factor ζ = 0.5 corresponds to a pair of dominant closedloop poles lying on the angles + 120o., Hence using the command rlocfind, we obtain the value K = 1. The corresponding values of the complex closedloop poles are s = 0.5 + j0.866. Hence the natural frequency ω n =
2
2
0.5 + 0.866 = 1 .
%Solution to Problem 9.59 num = [1]; K = 0; den = [1 3 3 1]; rlocus(num,den) K = rlocfind(num,den) den = [1 3 3 1]; [Wn z]=damp(den)
69
Root Locus
0.8
K=1 0.6
0.4
Imag Axis
0.2
0
−0.2
−0.4
−0.6
Figure 1 −0.8 −2
−1.8
−1.6
−1.4
−1.2
−1
−0.8
−0.6
−0.4
−0.2
0
Real Axis
9.60
The MATLAB command for this experiment is presented below. Figure 1 displays the root locus for the loop transfer function K L ( s ) = 2 s(s + s + 2) The value of K, corresponding to a pair of complex poles with damping factor ζ = 0.0707, is computed to be K = 1.46 using rlocfind. Figure 2 presents the Bode diagram of L(s) for K = 1.5. From this figure we find the following: (1) Gain margin  6.1 dB Phase crossover frequency = 1.4 rad/sec (2) Phase margin = 71.3o Gain crossover frequency = 0.6 rad/sec
%Solution to Problem 9.60 figure(1); clf; num = [1]; den = [1 1 2 0]; rlocus(num,den) axis([1 1 2 2]) K = rlocfind(num,den)
70
figure(2); clf; K = 1.5; num = K*[1]; margin(num,den) Root Locus 2
1.5
1
Imag Axis
0.5
0
−0.5
−1
Figure 1
−1.5
−2 −1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Real Axis
Bode Diagram Gm = 2.4988 dB (at 1.4142 rad/sec), Pm = 34.487 deg (at 1.1118 rad/sec) 20 10
Magnitude (dB)
0 −10 −20 −30 −40 −50 −60 −90
Phase (deg)
−135
−180
−225
Figure 2 −270 −1 10
0
1
10
10
Frequency (rad/sec)
9.61
The MATLAB code for this experiment is presented below. Figure 1 displays the root locus of the feedback system defined by the loop transfer function K (s – 1) L ( s ) = 2 (s + 1)(s + s + 1)
71
Using the command rlocfind, we find that the value of K for which the only real closedloop pole lies on the jωaxis (i.e., the system is on the verge of instability) is K = 1. We may verify this result by constructing the Routh array for the characteristic equation 2
(s + 1)(s + s + 1) + K (s – 1) = 0 as shown by 3
s 1 s s s
2 1 0
2+K
2 –K 4 + 3K  0 2 –K
0
For K = 1 there is only one sign change in the first column of array coefficients. Figure 2 displays the Nyquist locus of L(jω) for K = 0.8. We see that the critical point is not enclosed, and therefore the feedback system is stable. Figure 3 displays the Bode diagram of L(jω) for K = 0.8. From this figure we obtain the following values: Gain margin = 2.0 dB Phasecrossover frequency = 0 rad/sec. Note that the gain margin is exactly equal to 1 – 20 log 10K = 20 log 10 = 2dB 0.8 %Solution to Problem 9.61 figure(1); clf; K = 1; num = K*[1 1]; den = [1 2 2 1]; rlocus(num,den) ylim([5 5]) K = rlocfind(num,den) figure(2); clf; K = 0.8; num = K*[1 1]; nyquist(num,den) return
72
figure(3); clf; K = 0.8; num = K*[1 1]; margin(num,den) Root Locus 5
4
3
2
Imag Axis
1
0
−1
−2
Figure 1
−3
−4
−5
−1
−0.5
0
0.5
1
Real Axis
Nyquist Diagram
8
6
4
Imaginary Axis
2
0
−2
−4
−6
Figure 2
−8
−8
−6
−4
−2
0
2
4
6
8
Real Axis
9.62
The MATLAB code for this experiment is below. We are given a feedback system with the loop transfer function K (s + 1) L ( s ) = 4 3 2 s + 5s + 6s + 2s – 8 This L(s) is representative of a conditionally stable system in that for it to be stable, K must lie inside a certain range of values. The characteristic equation of the system is 4
3
2
s + 5s + 6s + ( K + 2 )s + ( K – 8 ) = 0
73
Constructing the Routh array: s s s s s
4 3 2 1 0
1
6
K–8
5
K+2
0
5.6 – 0.2K
K–8
0
2
51.2 + 0.2K – 0.2K 5.6 – 0.2K
0
K–8
0
From this array we can make the following deductions: 1. For stability, K > 8, which follows from the last entry of the first column of array coefficients. For K = 8, the system is on the verge of instability. From the fourth row, it follows that this occurs for the last row: s0. 2. From the fourth row of the array, it also follows that K must satisfy the condition 2
51.2 + 0.2K – 0.2K > 0 for the system to be stable. That is, ( K – 16.508 ) ( K + 15.508 ) < 0 from which it follows that for stability K < 16.508 If this condition is satisfied, then 5.6  0.2K > 0. When K = 16.508 the system is again on the verge of instability. From the third row of the Routh array, this occurs when 2
( 5.6 – 0.2K )s + ( K – 8 ) = 0 or 8.508 s = ± j  = ± j1.923 2.302 Accordingly, we may state that the feedback system is conditionally stable provided that K satisfies the condition 8 < K 16.5, the complex closedloop poles of the system move to the righthalf plane. Hence for stability, we must have 8 < K < 16.5. (b) Bode diagram. Figure 2(a) shows the Bode diagram of L(s) for K max + K min K = 2 16.5 + 8 = 2 = 12.25 From this figure we obtain the following stability margins: Gain margin = 3.701 dB; phasecrossover frequency = 0 rad/s Phase margin = 14.58o, gaincrossover frequency = 1.54 rad/s Figure 2(b) shows the Bode diagram for K = 8. From this second Bode diagram we see that the gain margin is zero, and the system is therefore on the verge of instability. Figure 2(c) shows the Bode diagram for K = 16.508. From this third Bode diagram we see that the phase margin is zero, and the feedback system is again on the verge of instability. (c) Nyquist diagram. Figure 3(a) shows the Nyquist diagram of L(jω) for K = 12.25, which lies between Kmin = 8 and Kmax = 16.5. The critical point (1,0) is not encircled for this value of K and the system is therefore stable. Figure 3(b) shows the Nyquist locus of L(jω) for K = 8. The locus of Fig. 3(b) passes through the critical point (1,0) exactly and the system is therefore on the verge of instability. Figure 3(c) shows the Nyquist locus for K < 8, namely K = 5. The locus of Fig. 3(c) encircles the critical point (1,0) and the system is again unstable. Figure 3(d) shows the Nyquist locus for K > 16.508, namely, K = 20. The locus of Fig. 3(d) encircles the critical point (1,0) and the system is therefore unstable. Finally, Fig. 3(c) shows the Nyquist locus for K = 16.508. Here again we see that the locus passes through the critical point (1,0) exactly, and so the system is on the verge of instability. Care has to be exercised in the interpretation of these Nyquist loci, hence the reason for the use of shading. %Solution to Problem 9.62 figure(1); clf; K = 1; num = K*[1 1]; den = [1 5 6 2 8]; rlocus(num,den); Kmin = rlocfind(num,den) Kmax = rlocfind(num,den) figure(2); clf; Kmid = (Kmin+Kmax)/2; num = Kmid*[1 1]; margin(num,den)
75
figure(3); clf; nyquist(num,den) Root Locus
4
3
2
Imag Axis
1
0
−1
−2
−3
Figure 1
−4 −7
−6
−5
−4
−3
−2
−1
0
1
Real Axis
Bode Diagram Gm = 2.6725 dB (at 1.924 rad/sec), Pm = 15.132 deg (at 1.5257 rad/sec) 20 0
Magnitude (dB)
−20 −40 −60 −80 −100 −135
Phase (deg)
−180
−225
−270 −1 10
0
1
10
2
10
10
Figure 2
Frequency (rad/sec)
Nyquist Diagram 0.8
0.6
0.4
Imaginary Axis
0.2
0
−0.2
−0.4
−0.6
Figure 3 −0.8 −2
−1.8
−1.6
−1.4
−1.2
−1
−0.8
−0.6
−0.4
Real Axis
76
−0.2
0
%Solution to Problem 9.63 figure(1); clf; K = 1; num1 = 0.5*K*[1 2]; den1 = [1 16 48 0]; subplot(3,1,1) rlocus(num1,den1) %Make figure 9.22 K = 1; num2 = K; den2 = [1 1 0]; subplot(3,1,2) rlocus(num2,den2) %Make figure 9.28 K = 1; num3 = 0.5*K*[1 2]; den3 = [1 8 48 0]; subplot(3,1,3) rlocus(num3,den3) figure(2); clf; subplot(2,2,1) nyquist(num1,den1) subplot(2,2,3) nyquist(num2,den2) subplot(2,2,4) nyquist(num3,den3)
Imag Axis
Root Locus
10 0 −10 −12
−10
−8
−6
−4
−2
0
Real Axis Root Locus
Imag Axis
0.5
0
−0.5 −1
−0.9
−0.8
−0.7
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
Real Axis
Imag Axis
Root Locus
Figure 1
20 0 −20 −12
−10
−8
−6
−4
−2
0
2
Real Axis
77
4
Nyquist Diagram
Imaginary Axis
2
1
0
−1
−2 −1
−0.8
−0.6
−0.4
−0.2
0
Real Axis
Nyquist Diagram
Nyquist Diagram
20
2
Imaginary Axis
Imaginary Axis
10
0
−10
−20 −1
1
0
−2 −0.8
−0.6
−0.4
−0.2
0
−1
Real Axis
9.64
Figure 2
−1
−0.8
−0.6
−0.4
−0.2
0
Real Axis
We are given a unity feedback system with loop transfer function K L ( s ) = s(s + 1) K = 2 s +s The closedloop transfer function of the system is L(s) T ( s ) = 1 + L(s) K = 2 s +s+K (a) With K = 1, T(s) takes the value 1 T ( s ) = 2 s +s+1 In general, the transfer function of a secondorder system is described as
(1)
(2)
(3)
2
ωn T ( s ) = 2 2 s + 2ζ n ω n s + ω n Comparing Eqs. (3) and (4): ωn = 1
(4)
ζ n = 0.5 The closedloop poles of the uncompensated system with K = 1 are located at 1 3 s = –  ± j  . 2 2
78
Figure 1 shows the root locus of the uncompensated system. jω closedloop pole for K=1 splane
. .
x
−1
j√3/2 σ
x0 − j√3/2
Figure 1 (b) The feedback system is to be compensated so as to produce a dominant pair of closedloop poles with ωn = 2 and ζ n = 0.5 . That is, the damping factor is left unchanged but the natural frequency is doubled. This set of specifications is equivalent to pole locations at s = – 1 ± j 3 , as indicated in Fig. 2, shown below. It is clear from the root locus of Fig. 1 that this requirement cannot be satisfied by a change in the gain K of the uncompensated loop transfer function L(s). Rather, we may have to use a phaselead compensator as indicated in the problem statement. Closedloop pole
.
jω j√3 splane
x  1 τ
θ2
o
120ο
θ1
1
 1 ατ
x
.
σ
x 0
− j√3
Figure 2 We may restate the design requirement: • Design a phaselead compensated system so that the resulting root locus has a dominant pair of closedloop poles at s = – 1 ± j 3 as indicated in Fig. 2. For this to happen, the angle criterion of the root locus must be satisfied at s = – 1 ± j 3 . With the openloop poles of the uncompensated system at s = 0 and s = 1, we readily see from Fig. 2 that the sum of contributions of these two poles to the angle criterion is – 120° – 90° = – 210° We therefore require a phase advance of 30o to satisfy the angle criterion.
79
Figure 2 also includes the polezero pattern of the phaselead compensator, where the angles θ1 and θ2 are respectively defined by
3 θ 1 = tan  1  – 1 ατ – 1
(5)
3 θ 2 = tan 1  – 1 τ where the parameters α and τ pertain to the compensator. To realize a phase advance of 30, we require θ 1 – θ 2 = 30° or θ 1 = 30° + θ 2 – 1
(6)
Taking the tangents of both sides of this equation: tan 30° + tan θ 2 tan θ 1 = 1 – tan 30° tan θ 2 1  + tan θ 2 3 = 1 1 –  tan θ 2 3
(7)
Using Eqs. (5) and (6) in (7) and rearranging terms, we find that with α > 1 the time constant τ is constrained by the equation 2
τ – 0.95τ + 0.025 = 0 Solving this equation for τ: τ = 0.923 or τ = 0.027 The solution τ = 0.923 corresponds to a compensator whose transfer function has a zero to the right of the desired dominant poles and a pole on their left. On the other hand, for the solution τ = 0.027 both the pole and the zero of the compensator lie to the left of the dominant closedloop poles, which conforms to the picture portrayed in Fig. 2. Se we choose τ = 0.027, as suggested in the problem statement. (c) The loop transfer function of the compensated feedback system for α = 10 is L c ( s ) = G c ( s )L ( s ) ατs + 1 K =  τs + 1 s ( s + 1 )
80
K ( 0.27s + 1 ) = s ( s + 1 ) ( 0.027s + 1 ) Figure 3(a) shows a complete root locus of L(s). An expanded version of the root locus around the origin is shown in Fig. 3(b). Using the RLOCFIND command of MATLAB, the point – 1 + j 3 is located on the root locus. The value of K corresponding to this closedloop pole is 3.846. %Solution to Problem 9.64 figure(1); clf; K = 1; t = 0.027; num = K*[10*t 1]; den = [t 1+t 1 0]; rlocus(num,den) figure(2); clf; rlocus(num,den) axis([5 1 4 4]) K = rlocfind(num,den) Root Locus
40
30
20
Imag Axis
10
0
−10
Figure 1
−20
−30
−40
−35
−30
−25
−20
−15
−10
−5
0
Real Axis
Root Locus 4
3
2
K=3.897
Imag Axis
1
0
Figure 2
−1
−2
−3
−4 −5
−4
−3
−2
−1
0
1
Real Axis
81
9.65
The loop transfer function of the compensated feedback system is L ( s ) = G ( s )H ( s ) 10K ( ατs + 1 ) =  , α