Chapter 2 MATHEMATICAL FOUNDATION 2-1 (a)
Poles: s = 0, 0, Zeros: s =
−1, −10;
(b)
Poles: s =
−2, ∞, ∞, ∞.
−2, −2...
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Chapter 2 MATHEMATICAL FOUNDATION 2-1 (a)
Poles: s = 0, 0, Zeros: s =
−1, −10;
(b)
Poles: s =
−2, ∞, ∞, ∞.
−2, −2;
Zeros: s = 0. The pole and zero at s =
(c)
Poles: s = 0,
−1 + j, −1 − j;
Zeros: s =
(d)
Poles: s = 0,
−1, −2, ∞.
−2.
2-2 (a)
(b) G (s) =
5
( s + 5)
(c) G ( s) =
2
(d)
(s
4s 2
(e) G ( s)
=
−1 cancel each other.
1 s
2
∞
G (s)
+4
+4
=
)
∑e
+
1
G (s)
s+ 2
kT ( s + 5 )
1
= 1
k =0
−e
2-3 (a) g ( t ) = u s ( t ) − 2u s (t − 1) + 2 u s( t − 2) − 2 u s ( t − 3) + L G (s ) =
1 s
(1 − 2e − s + 2e−2 s − 2e −3s + L ) =
gT (t ) = u s (t ) − 2us (t − 1) + us (t − 2) GT (s ) =
1 s
(1 − 2e − s + e −2s ) = ( 1 − e − s ) 1
1−e
(
−s
s 1+ e
−s
0≤ t ≤ 2 2
s
1
)
−T ( s+5 )
=
4 s
2
+ 4s +8
∞
g(t )
∞
∑
=
g
T
− 2k )us (t − 2k )
(t
G (s)
∑s
=
k =0
1
−e
(1
−s
2
) e
−2 ks
=
k =0
−s
1− e
s (1 + e
−s
)
(b) g ( t) = 2tu s ( t ) − 4(t − 0.5) u s (t − 0.5) + 4(t − 1) us (t − 1) − 4(t − 1.5)us (t − 1.5) + L G ( s) = g
T
(1 − 2e
2 2
s
−0.5 s
+ 2e
−s
− 2e
−1.5 s
− 0.5 s ( ) + L) = 2 −0.5 s s (1 + e )
2 1−e
= 2 tu s ( t ) − 4 ( t − 0 . 5) u s ( t − 0 . 5) + 2( t − 1 ) u s ( t − 1 )
(t )
(1 − 2e−0.5 s + e− s ) = s 2 (1 − e−0.5 s )
2
GT ( s ) =
s
2
2
∞
g (t ) =
∑
k=0
G(s ) =
≤ t ≤1
2
∞
g T ( t − k )us ( t − k )
0
∑ s2 ( 2
1 −e
−0.5 s
k=0
)
2
e
− ks
=
( −0.5 s ) 2 −0.5s s (1 + e ) 2 1−e
2-4 g(t )
= ( t + 1 ) u s ( t ) − ( t − 1 ) u s ( t − 1 ) − 2 u s ( t − 1 ) − ( t − 2 ) u s ( t − 2 ) + ( t − 3) u s ( t − 3) + u s ( t − 3)
G ( s) =
s
2-5 (a)
(1 − e − s − e −2 s + e −3 s ) + s (1 −2e − s + e −3 s )
1
1
2
Taking the Laplace transform of the differential equation, we get 1 1 2 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)( s + 2 )( s f (t )
=
1
e
−4 t
+
6
(b)
sX ( s ) 1
1
e
−t
−
3
− x1( 0 ) =
1
e
−2 t
t
+4)
=
1 6( s
+ 4)
=
X (s)
=
2
2
s
X (s)
=1
x (0)
2
s( s
+ 3 s +1
+ 1 )( s + 2 ) −1
(s
+ 1 )( s + 2 )
1
= 0 .5 + e
−t
− 0 .5 e
3( s
+ 1)
−
1 2( s
+ 2)
≥0
1
sX
2
(s)
− x 2 ( 0 ) = −2 X 1 ( s ) − 3 X 2 ( s ) +
1
= =
2s
+
−1 s
+1
1 s
+
+1
−
1 2( s
+ 2)
1 s
+2
Taking the inverse Laplace transform on both sides of the last equation, we get x (t )
1
2
Solving for X1 (s) and X2 (s), we have X 1 ( s)
+
−2 t
t
≥0
x (t ) 2
= −e
2
−t
+e
−2 t
t
≥0
1 s
x (0) 2
=0
2-6 (a) G (s)
(b) G (s)
=
(c) G (s ) =
=
1 3s
−2 . 5
(
50
5
−
s
(s
20
=
1 s
−
s s
2
g(t )
= 0 .5 t
2
e
−0.5 t
2 .5
+
s
2
=
1 s
g(t )
+3
s +4
−1
g (t )
+ 3)
3( s
30s + 20
+s+2
g (t ) = 1 + 1.069e
(e)
−
s +1
(d) G (s)
+ 1)
2
1
+
+ 2)
2( s
+
+1
s
1
−
+
)
e
s
1
−
1
e
3
2
= −2 . 5 e
−t
[
−s
−2 t
+s +2
−
+
1
e
−3 t
t
≥0
3
+ 5 te
g (t ) = 50 − 20e
1 2
=
−t
− (t −1)
+ 2 .5 e
−3 t
t
≥0
] us (t − 1)
− 30cos2(t − 1) − 5sin2(t − 1)
s s
2
Taking the inverse Laplace transform,
+s+2
[ sin1.323t + sin (1.323t − 69.3o ) ] = 1 + e−0.5 t (1.447sin1.323t − cos1.323t )
−t
t
≥0
2-7
−1 2 0 A = 0 −2 3 −1 −3 −1
2-8
0 0 B = 1 0 0 1
u (t ) =
(a)
u1( t) u ( t) 2
(b) Y (s )
=
R (s )
3s + 1 3
Y (s)
2
s + 2 s +5s + 6
=
R (s )
(c)
5 4
2
s + 10 s + s + 5
(d)
Y (s ) R (s )
=
s ( s + 2) 4
3
Y (s )
2
s + 10 s + 2 s + s + 2
R (s )
3
=
1+ 2e 2
−s
2s + s + 5
t≥0
4
5
6
7
8
9
10
11
12
13
Chapter 4 MATHEMATICAL MODELING OF PHYSICAL SYSTEMS
4-1 (a) Force equations: 2
f ( t) = M 1
d y1 dt
dy1 − dy 2 + K ( y − y ) 1 2 dt dt
dy1
+ B1
2
+ B3
dt
2 dy1 dy2 d y2 dy2 B3 − + K ( y 1 − y 2 ) + M 2 2 + B2 dt dt dt dt
Rearrange the equations as follows: 2
d y1 dt
=−
2
2
d y2 dt (i) State diagram:
2
(B
1
+ B 3 ) dy1
M1
B3 dy1
=
dt −
(B
M 2 dt
Since y
1
− y2
2
B3 dy2
+
M 1 dt
+ B3 ) dy2
M
1
dt
dx2 dt
=
K
x1 −
M2
(ii) State variables:
x
1
(B
=
2
+ B3 )
x2 +
M2 y , 2
x
2
=
dy
B3 M2 2
dx3
x3 x
,
1
dt
1
dt dx dt
3
= x2 =
x
dx dt dx
4
2
dt
4
=− =
K M
K
=
M
x
− y2 )
y ,
x
1
1
−
B
1
+
1
State diagram:
14
B M
2
+ B3
M 3
1
x
2
dy
=
4
1
x
2
K M
2
x
,
dt
B3 M1
x2 −
=
3
(B
1
dy
1
.
dt
+ B3 )
M1
x3 +
.
dt
+
2
−
dy
=
2
x1 +
M1
2
x
K
=−
dt
3
x
2
State equations: dx
M1
appears as one unit, the minimum number of integrators is three.
1
= − x2 + x3
f
− y2 ) +
1
(y
M2
State equations: Define the state variables as x = y − y ,
dx1
(y
M1 K
+
dt
2
K
−
x 1
3
K M
−
x 2
B
1
3
+
B
+ B3 M
1
3
M x
x
4
2
4
+
1 M
f 1
1 M
f
Transfer functions:
Y1 ( s ) F (s ) Y2 ( s ) F ( s)
M 2 s + ( B2 + B3 ) s + K 2
= =
{
3
{
3
s M 1 M 2 s + [( B1 + B 3 ) M 2 + ( B2 + B3 ) M 1 ] s + [K ( M 1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B 1 + B2 ) K 2
B3 s + K
s M1 M 2 s + [( B1 + B3 ) M 2 + ( B2 + B3 ) M1 ] s + [ K ( M1 + M 2 ) + B1 B2 + B2 B3 + B1 B3 ] s + ( B1 + B2 ) K 2
(b) Force equations: 2
d y1 dt
2
=−
(B
+ B2 ) dy1
1
M
+
dt
B2 dy2
+
M dt
1
dy2
f
M
dy1
=
dt
dt
K
−
B2
(i) State diagram:
Define the outputs of the integrators as state variables, x
1
=
y ,
x
2
=
2
dy
1
.
dt
State equations: dx
1
dt
=−
K B
x
1
dx
+ x2
dt
2
(ii) State equations: State variables: dx dt
1
=−
K B
x
1
+ x3
2
2
dx dt
2
x
1
=
=−
K M
y ,
x
2
dx
= x3
dt
Transfer functions:
15
x
3
2
1
−
=
=−
B
1
x
M
y ,
x
1
K M
2
x
1
3
−
1
+ = B
f
M
1
M
dy
1
.
dt x
3
+
1 M
f
y2
} }
Y1 ( s ) B2 s + K = 2 F (s ) s MB2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K
Y2 ( s) F ( s)
(c) Force equations: dy1
dy2
=
dt
2
1
+
dt
d y2
f
B1
dt
=−
2
(B
1
B2
=
M B2 s + ( B1 B2 + KM ) s + ( B1 + B2 ) K 2
+ B2 ) dy2 M
B1 dy2
+
dt
B1 dy1
+
M dt
M dt
(i) State diagram:
State equations: Define the outputs of integrators as state variables. dx
= x2
1
dt
dx
=−
2
dt
(ii) State equations: state variables: dx dt
1
= x3 +
1 B
dx
f
2
dt
1
x
1
K
x
M
1
=
y , 1
−
B
M x
dx
= x3
x
2
=
2
3
dt
2
+
1
f
M
y ,
x
2
=−
K
x
M
2
3
= −
dy
2
.
dt
B
2
M
x
3
State diagram:
Transfer functions:
Y1 ( s ) F (s )
Ms + ( B1 + B 2 ) s + K 2
=
(
B1 s Ms + B2 s + K 2
Y2 ( s )
)
F (s )
4-2 (a) Force equations:
16
=
1 Ms + B 2 s + K 2
+
1 M
f
−
K M
y2
y
=
1
1 K
+ Mg ) + y 2
( f
2
y
dt
2
d
2
=−
2
B dy M
−
2
dt
K
1
+ K2
y
M
+
2
K
2
M
y
1
State diagram:
State equations: Define the state variables as: x = y , 1
dx
dx
= x2
1
dt
=−
2
dt
K
1
x
M
x
2
1
−
B
x
M
2
2
=
+
dy
2
.
dt 1
( f
+ Mg
Ms
2
M
)
Transfer functions: Y (s) 1
F ( s)
s
=
2
+ Bs + K 1 + K 2
K ( Ms 2
2
Y (s) 2
+ Bs + K 1 )
F (s)
1
=
+ Bs + K 1
(b) Force equations: dy1 dt
=
1 B1
[ f ( t) + Mg] +
dy2 dt
−
K1 B1
(y
1
− y2 )
2
d y2 dt
2
=
B 1 dy 1
dy K B B dy − + ( y − y ) − ( y − y )− M dt dt M M M dt 2
1
2
1
2
2
1
2
State diagram: (With minimum number of integrators)
To obtain the transfer functions Y ( s ) / F ( s ) and Y ( s ) / F ( s ), we need to redefine the state variables as: x
1
=
y , 2
x
2
= dy 2
1
/ dt , and x
3
=
2
y . 1
State diagram:
17
2
Transfer functions:
Y1 ( s )
Ms + ( B1 + B 2 ) s + K 1 2
=
F (s )
s
2
[MBs + ( B B 1
4-3 (a) Torque equation: 2 d θ B dθ =− + 2 dt
1
2
Y2 ( s )
+ MK1 )]
=
F (s )
Bs + K 1
s [M B1 s + ( B1 B2 + MK1 ) ] 2
State diagram: 1
J dt
T (t )
J
State equations: dx
dx
= x2
1
dt
2
dt
=−
B
x
J
+
2
1
T
J
Transfer function:
Θ( s )
1
=
T (s)
+ B)
s ( Js
(b) Torque equations: d θ1 2
dt
2
=−
K J
(θ
1
−θ 2 ) +
1
dθ 2
K (θ 1 − θ 2 ) = B
T
J
dt
State diagram: (minimum number of integrators)
State equations: dx
1
dt
=−
K
x
B
1
dx
+ x2
dt
State equations: Let x = θ , 1
dx dt
1
=−
K B
x
1
+
K B
=−
2
x
2
x dx
2
dt
2
2
K
x
J
= θ1, =
x
1
+
1
T
J
and
x
3
dx 3
dt
State diagram:
18
= 3
dθ
1
.
dt
=
K J
x
1
−
K J
x
2
+
1 J
T
Transfer functions:
Θ1 ( s ) T ( s)
=
Bs + K
(
s BJs + JKs + BK 2
Θ2 (s )
)
T ( s)
=
K
(
s BJs + JKs + BK 2
)
(c) Torque equations: d θ1 2
T ( t ) = J1
dt
2
+ K (θ 1 − θ 2 )
d θ2 2
K (θ 1 − θ 2 ) = J 2
2
dt
State diagram:
State equations: state variables: dx dt
1
dx
= x2
dt
2
=−
K J
x
x
1
2
1
+
=θ2, K J
x
x
2
= dx
3
2
dt
dθ
2
x
,
dt 3
=
x
3
= θ1, dx
4
dt
4
x
=
=
4
K J
x
dθ
1
.
dt
1
1
−
K J
1
x
3
+
1 J
T
1
Transfer functions:
Θ1 ( s ) T ( s)
J 2s + K
Θ 2 (s )
2
=
s
2
J1 J2 s + K ( J1 + J 2 ) 2
T ( s)
=
K 2
s
J1 J 2 s + K ( J1 + J 2 ) 2
(d) Torque equations: d θm 2
T ( t) = J m
dt
2
+ K 1 (θ m − θ 1 ) + K 2 (θ m − θ 2 )
K 1 (θ m − θ 1 ) = J 1
19
d θ1 2
dt
2
K2 (θm − θ2 ) = J2
d θ2 2
dt
2
State diagram:
State equations: x = θ 1
dx
1
dt
dx
= −x2 + x3
2
dt
=
K J
1
x
m
− θ1,
dx 1
dt
1
3
x
=−
K J
2
1
dθ
=
x
1
x
,
dt
−
1
m
K J
2
x
3
+
4
m
=
dθ
x
,
dt
1 J
m
T
m
dx dt
4
4
= θm −θ2 ,
=
x
3
− x5
x
dx dt
5
5
=
=
dθ
2
.
dt
K J
x
2
4
2
Transfer functions:
Θ1 ( s ) T ( s) Θ 2 ( s) T (s )
K 1 (J 2 s + K 2 ) 2
=
s
2
s 4 + ( K1 J2 J m + K2 J 1 J m + K1 J 1 J 2 + K 2 J 1 J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 ) K2 ( J1 s + K1 ) 2
=
s
2
s 4 + ( K1 J2 J m + K 2 J 1 J m + K1 J 1J 2 + K 2J 1J 2 ) s 2 + K1 K 2 ( J m + J 1 + J 2 )
(e) Torque equations: d 2θ m dt
2
=−
K1 Jm
( θ m − θ1 ) −
K2 Jm
( θ m − θ2 ) +
1 Jm
d θ1 2
T
dt
2
K1
=
J1
State diagram:
20
(θ
m
− θ1 ) −
B1 d θ1 J1 dt
d 2θ 2 dt
2
=
K2 J2
(θ
m
−θ 1 ) −
B 2 dθ 2 J 2 dt
State variables:
x
1
= θ m − θ1,
x
2
=
dθ
1
x
,
dt
dθ
=
3
m
x
,
dt
= θm −θ2 ,
4
x
dθ
=
5
.
2
dt
State equations: dx 1 dt
= −x 2 + x 3
dx
2
dt
K1
=
J
x
1
B1
−
J
1
dx 3
x
2
dt
1
Transfer functions:
Θ1 ( s )
(
K1 J 2 s + B2 s + K 2
=
K1
=−
2
J
x
1
−
K
m
J
2
)
J
Θ2 ( s )
dx
T
dt
m
=
(
dx
= x3 − x5
4
=
5
dt
K 2 J1 s + B1 s + K1 2
K2 J
x
4
−
2
B2 J
d θm 2
dt
2
Output equation:
+ Bm e
=
o
dθ m dt Eθ 20
∆( s )
3
m
2
1
1
2
m
1
2
2
1
2
2
1
2
1
1
m
1
1
2
m
2
1
1
2
1
2
m
2
+ K (θ m − θ L )
K ( θm − θ L ) = J L
d θL 2
dt
2
+ Bp
dθ L dt
L
π
State diagram:
Transfer function:
Θ L (s )
=
Tm ( s) Eo ( s )
=
Tm ( s)
(
)
K
(
)
s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3
(
)
2
KE / 2 0π
(
)
s J m J L s + Bm J L + Bp J m s + Jm K + J L K + Bm B p s + Bm K 3
2
4-5 (a) d θ1 2
Tm ( t) = Jm
θ3 =
N1 N3 N2N4
dt θ1
2
+ T1
T2 =
T1 = N3 N4
N1 N2
T4 =
T3 =
T2
N3 N4
2
)
4-4 System equations: Tm ( t) = Jm
x
(B + B ) s + [ ( K J + K J ) J + ( K + K ) J J + B B J ] s + B K ) J + B K J + B K J ] s + K K ( J + J + J )} 4
+ [( B1K 2
1
+
T (s )
∆ ( s ) = s { J 1 J2 Jm s + J 2
4
m
∆( s )
T ( s)
x
d θ3 2
JL
dt
2
N3 N4
d θ3 2
T4 = J L
T4
dt
2
T2 = T3
θ2 =
N1 N2
θ1
2 N1 N3 d 2θ1 Tm = J m 2 + T4 = J m + JL 2 dt N 2N 4 N 2 N 4 dt
d θ1 2
21
N1 N 3
2
5
(b) d θ1 2
Tm = Jm θ2 =
dt
N1
2
+ T1
2
θ1
N2
d θ2
T2 = J2 N 1N 3
θ3 =
N2N4
dt
T4 = ( J3 + J L )
+ T3
2
d θ2 2
θ1
T2 = J 2
dt
2
2
N1
dt
T4 = J 2
N4
2 d θ3 d 2θ 2 N3 Tm ( t) = Jm + J 2 dt 2 + N ( J3 + J 4 ) dt 2 = Jm 2 dt N2 4
d θ1
2
T1 =
2
d θ2 2
N3
+
d θ3
dt
2
+
N3 N4
(J
N1 N2
T3 =
T2
+ JL ) 3
N3 N4
d θ3 2
dt
2
2 2 d 2θ1 N1 N1 N3 + J2 + N N ( J3 + J L ) dt 2 N2 2 4
4-6 (a) Force equations:
dy1 − dy 2 dt dt
2 dy1 − dy 2 = M d y 2 + B dy 2 t 2 dt dt dt dt
f ( t) = K h ( y1 − y2 ) + Bh (b) State variables:
x
1
=
y
1
− y2,
x
dy
=
2
K h ( y1 − y2 ) + Bh 2
dt
State equations: dx
=−
1
dt
K B
h
x
1
1
+
B
h
dx
f (t )
2
dt
h
=−
B
t
x
M
2
+
1
f (t )
M
4-7 (a) T
m
= Jm
2
d
θm
dt
2
d θm
+T1
2
Tm = Jm
Set
∂α L ∂n
dt
2
= 0.
T
=JL
d θL
2
2
θL
dt
2
d
(T
m
dt
2
T
N
1
T
2
= nT 2
(
2
)
(
θm N 1 = θ L N 2
2
nTm − n TL 2
Thus, α L =
)
− 2 nTL ) J m + n J L − 2nJL nTm − n J L = 0
n
∗
=−
J T m
2
L
2J T
L
2
Or, n
Jm + n JL 2
2
+
J T m
L
J T
+
2
J mTL
+ 4 J m J LT m
Torque equation about the motor shaft:
∗
=
L m
L
Relation between linear and rotational displacements:
22
−
J J
m L
where the + sign has been chosen.
2J T
Jm / J
n
2
= 0 , the optimal gear ratio is n
4-8 (a)
N
+ nTL =
L m
When T
=
L m
Optimal gear ratio:
(b)
1
J m + nJ α + nT L L L n
2
+ nJL
+TL
=0
T4
T
(b)
m
2
d
= Jm
θm
dt
2
+ Mr
2
d
2
θm
dt
dθ
+ Bm
2
m
= r θm
y
dt
Taking the Laplace transform of the equations in part (a), with zero initial conditions, we have
(
Tm ( s) = Jm + Mr
)sΘ
2
2
m
( s) + Bm sΘ m (s )
Y ( s) = rΘ m ( s)
Transfer function:
Y ( s) Tm ( s)
=
r
(
s J m + Mr
r
)s + B
m
4-9 (a) d θm 2
Tm = Jm
dt
2
(
+ r ( T1 − T2 )
2
T1 − T2 = M
dt
d θm 2
d y
Thus, Tm = J m
2
(c) State equations: dx1 = rx3 − x2 dt
dx2
=
Tm ( s)
=
dt
2
K1 + K 2
dt
(d) Transfer function: Y ( s)
)
T1 = K2 rθ m − rθ p = K 2 ( rθ m − y )
M
+ r ( K1 + K2 )( rθ m − y )
dx3
x1
=
dt
T2 = K1 ( y − rθ m ) 2
M
− r ( K1 + K 2 ) Jm
d y dt
x1 +
= ( K1 + K2 )( rθ m − y )
2
1 Jm
Tm
r ( K1 + K 2 )
s
2
Jm Ms + ( K1 + K2 ) ( Jm + rM ) 2
(e) Characteristic equation:
s J m Ms + ( K1 + K2 ) ( Jm + rM ) = 0 2
2
4-10 (a) Torque equations: d θm 2
Tm ( t) = Jm
dt
2
+ Bm
dθ m dt
+ K (θ m − θ L )
K ( θm − θ L ) = J L
State diagram:
23
d θL 2
dt
2
+ BL
dθ L dt
(b) Transfer functions: Θ L ( s) Tm ( s)
=
K
Θ m ( s)
∆ ( s)
Tm ( s)
=
J L s2 + BL s + K
∆ ( s) = s J mJ L s3 + ( Bm J L + BL J m ) s 2 + ( KJ m + KJ L + Bm BL ) s + Bm K
∆ (s )
(c) Characteristic equation:
∆( s ) = 0
(d) Steady -state performance:
T (t ) m
= T m = consta
T (s)
nt.
m
=
T
m
.
s
J L s + BL s + K 2
lim ω m ( t) = lim sΩ m ( s) = lim t →∞
(e)
s →0
s →0
Thus, in the steady state,
ω m = ω L.
The steady-state values of
ωm
4-11 (a) State equations: dθ L =ωL
dω
dt
dθ m dt
= ωm
L
dt
dω m dt
=
and
K J
=−
2
J m J L s + ( Bm J L + BL J m ) s + ( KJ m + KJ L + Bm BL ) s + Bm K
ωL
θm −
L
Bm Jm
3
2
do not depend on J
K J
ωm −
2
dθ
θL
L
(K
1
dt
+ K2 ) Jm
(b) State diagram:
(c) Transfer functions:
24
m
t
and J . L
= ωt
θm +
K1 Jm
θt +
dω dt
K2 Jm
t
=
θL +
K J
1
θm −
t
1 Jm
Tm
K J
1
t
θt
=
1 Bm
ΘL ( s )
=
(
K 2 J t s + K1 2
)
Θ t(s )
∆ ( s)
Tm ( s)
=
(
K1 J L s + K 2 2
)
Θ m (s )
∆ ( s)
T m( s)
J t J L s + ( K1 J L + K 2 J t ) s + K1 K 2 4
=
2
∆ ( s)
Tm ( s )
∆ ( s ) = s [ J mJ Ls + B mJ LJ t s + ( K1 J L J t + K 2J L J t + K 1Jm J L + K 2J m J t ) s 5
4
3
+ Bm J L ( K1 + K 2 ) s + K1 K2 ( J L + J t + J m ) s + BmK 1K 2 ] = 0 2
(d) Characteristic equation: ∆ ( s ) = 0 .
4-12 (a) K H (s ) −K 1 H i (s ) 1 + K1 H e ( s ) + 1 i H e (s) + B + Js Ra + L a s B + Js Ra + La s = ≅ =0 −1
Ω m (s ) TL ( s)
∆( s)
ωr = 0
∆( s)
Thus,
H e (s ) = −
H i (s )
H i (s)
Ra + La s
H e (s)
Ω m (s )
(b)
=
Ω r ( s) ∆ ( s ) = 1 + K1 H e ( s ) + = 1+
(R
a
Ω m (s ) Ω r ( s)
= TL =0
(R
(R
(R
+ La s ) ( B + Js )
K1 K b
K1 K b
+ La s )( B + Js )
+ La s )( B + Js )
+
R a + La s
d
2
θ 2
+
(R
K1 K i K b H e( s) a
+ La s ) ( ( B + Js )
K 1 Ki
(R
+ La s ) ( B + Js )
a
+ La s ) ( B + Js ) + Ki Kb + K1 Ki Kb H e ( s)
a
dt
F d a
1
= T s d 2 δ + K F d 1θ
= J α α1 = J
d
2
dt
θ 2
K
F
1
≅
Kb H e ( s)
d
1
θ
sin
δ≅δ
− K F d 1θ = T s d 2δ
Θ( s ) − K F d 1 Θ( s ) = T s d 2 ∆ ( s )
(b)
Js
(c)
With C and P interchanged, the torque equation about C is:
2
K1 H i ( s)
+
K1 K i
dt
J
a
∆( s)
4-13 (a) Torque equation: (About the center of gravity C) 2 d θ J = T s d 2 sin δ + F d d 1 2 Thus,
K1 K i
TL =0
a
= − ( Ra + L a s )
Ts ( d1 + d 2 ) δ + Fα d 2 = J
d θ 2
dt
2
Ts ( d 1 + d 2 ) δ + K F d 2θ = J
Js Θ(s ) − K F d 2 Θ( s ) = Ts ( d1 + d 2 ) ∆( s ) 2
25
Θ( s ) ∆ (s )
=
d θ 2
dt
Ts ( d1 + d 2 ) Js − K F d 2 2
2
θe = θr −θo
4-14 (a) Cause-and-effe ct equations: dia
Ra
=−
dt
d θm
La
dt
d θo
=−
2
dt
2
(θ
JL
a
1
+
J m dt
KL
=
(e
La
Bm dθ m
2
2
1
ia +
J
− eb )
e
= Ke
a
Tm = K ii a nKL
Tm −
= K sθ e
e
( nθ
Jm
m
− θo )
Tm
T2 =
θ 2 = nθ m
n
− θo )
2
State variables:
x
1
= θo,
x
= ωo ,
2
x
=θm,
3
x
4
= ωm,
x
5
= ia
State equations: dx
1
dt dx
4
dt
dx
= x2
=−
=−
2
dt nK J
K J
L
x
x
1
−
n KL J
m
x
3
nK J
L
2
L
+ 1
Bm
−
J
m
m
x
4
L
x
dx 3
dt
L
+
Ki J
x
3
= dx
5
x
5
dt
m
4
=−
KK L
s
x
1
−
a
Kb L
x
4
a
−
Ra L
x
5
+
KK
a
L
s
a
(b) State diagram:
(c) Forward-path transfer function: Θ o ( s) KK s K i nK L = 4 3 2 2 Θe ( s ) s J mJ L La s + J L ( Ra J m + Bm J m + Bm La ) s + n K LL a J L + K L J m La + Bm R a J L s +
(
(n R K J 2
a
L
L
)
)
+ R a KL J m + B mK L L a s + K i K bK L + R a B mK L
Closed-loop transfer function:
Θo ( s) Θr ( s )
=
L
)
(R J
(n R K J
L
+ Ra KL J m + Bm KL L a s + ( K i K b K L + Ra BmK L ) s + nKK s K i K L
a
K
(
L
2
(d)
KK s K i nK L
J mJ LL a s + J 5
L
a
= ∞, θ o = θ 2 = n θ m .
+ Bm J m + B m La ) s + n K LL a J L + K L J m L a + B mR a J L s + 4
m
)
J
L
2
2
is re flecte d to m otor s ide so
State equations:
26
J
T
=
J
m
+n
2
J . L
3
θr
dω dt
m
=−
B J
m
ωm +
T
K J
i
i
dθ a
dt
T
di
= ωm
m
a
dt
=−
R L
a
i
a
a
+
KK L
s
θr −
a
KK L
s
nθ
m
a
−
K L
b
ωm
a
State diagram:
Forward-path transfer function:
Θ o ( s) Θe ( s )
=
KKs K i n s J TL a s + ( Ra J T + Bm L a ) s + Ra Bm + K i K b 2
Closed-loop transfer function:
Θo ( s) Θr ( s )
=
KK sK in
JT L a s + ( R a J T + Bm La ) s + ( Ra Bm + Ki Kb ) s + KK s K i n 3
2
From part (c), when K
L
= ∞,
all t he ter ms wit hout K
L
in
Θo (s ) / Θe ( s )
and
Θ o ( s ) / Θr ( s )
can b e negl ected.
The same results as above are obtained.
4-15 (a) System equations:
f = K ii a = M T (b)
dv dt
+ B Tv
ea = Ra ia + ( La + Las )
dia dt
− Las
di s dt
0 = Rsis + ( L s + L as )
+ eb
di s dt
− L as
di a dt
Take the Laplace transform on both sides of the last three equations, with zero initial conditions, we have
Ea ( s ) = [ Ra + ( La + Las ) s ] I a ( s) − Las sI s ( s ) + K b V( s)
Ki I a ( s ) = ( MT s + BT ) V ( s )
0 = − Las sI a ( s) + [ Rs + s ( Ls + Las ) ] I s ( s ) Rearranging these equations, we get
V (s ) = Ia (s ) =
Ki M T s + BT
Y (s ) =
I a (s )
1 Ra + ( La + L as ) s
[E
a
V (s ) s
=
Ki
s ( M T s + BT )
( s ) + Las sI s ( s ) − KbV ( s ) ]
Block diagram:
27
Ia (s)
Is (s ) =
L as s
Ra + ( L a + L as ) s
I a ( s)
(c) Transfer function: Y ( s)
K i [R s + ( L s + L as ) s ]
=
s [R a + ( L a + L as ) s ][ R s + ( L s + L as ) s ]( M T s + BT ) + K i K b [R s + ( L a + L as ) s] − Lass 2
E a (s )
(M
2
T
s + BT )
4-16 (a) Cause-and-effect equations:
θe = θ r − θ L
e = K sθ e dω m
Tm = K i ia
dt
=
1
Bm
Tm −
Jm K
K s = 1 V/rad
b
Jm
ω−
= 15 . 5 V
KL Jm
(θ
m
ea = Ke dω L
−θ L )
=
dt 15 . 5
=
/ KRPM
ia =
× 2 π / 60
1000
KL
(θ
JL
m
ea − eb Ra
−θ L )
= 0 .148
V / rad / sec
Bm
KL
eb = Kbω m
State equations:
d θL dt
= ωL
d ωL
=
dt
KL JL
θm −
KL JL
θL
d θm dt
d ωm
= ωm
=−
dt
Jm
ωm −
Jm
θL+
1 Ki Jm Ra
( KK θ s
e
− K bω m )
(b) State diagram:
(c) Forward-path transfer function: G ( s) =
J m Ra J L
K i K Ks KL s J mJ L Ra s + ( Bm Ra + K i K b ) J L s + R a K L ( J L + J 3
= 0 .03 × 1 .15 × 0 . 05 = 0 . 001725
2
Bm Ra J L
= 10 × 1 .15 × 0 .05 = 0 . 575
28
m
)s
+ K L ( Bm Ra + K i K b )
Ki K bJ L
=
21 × 0 .148
× 0 . 05 = 0 .1554
R K J a
L
L
= 1.15 × 50000 × 0 . 05 =
R K J
2875
a
L
m
= 1 .15 × 50000 × 0 . 03 = 1725
K KK K i
K L ( Bm Ra + K i K b ) = 50000(10 × 1.15 + 21 × 0.148) = 730400
s
=
L
608.7 × 10 K
21
× 1 × 50000 K = 105000
0K
6
G ( s) =
(
s s + 423.42 s + 2.6667 × 10 s + 4.2342 × 10 3
2
6
8
)
(d) Closed-loop transfer function:
M (s ) =
Θ L ( s) Θ r ( s)
=
G( s) 1 + G ( s) M (s)
K i K Ks K L
=
=
J mJ LR a s + ( B m R a + K i K b ) J L s + R a K L ( J L + J m ) s2 + K L ( Bm Ra + K iK b ) s + K i K Ks K L 4
3
6 . 087 s
+ 423
4
.42 s
3
+ 2 .6667 × 10
Characteristic equation roots: K
4-17
=1
K
s
= − 1.45
s
= − 159
s
= − 131 . 05 ±
. 88
6
s
× 10 2
8
K
+ 4.2342 × 10
8
= 2738
s
+ 6 . 087 × 10
K
s
= ± j 1000
s
= −211 . 7 ±
j 1273 . 5
s
=
s
= −617
405
±
j1223 .4
.22
(a) Block diagram:
Tr ( s )
=
KM KR
( 1 + τ s ) (1 + τ s ) + K c
s
4-19 (a) Block diagram:
29
= m
KR
3.51 20 s + 12 s + 4.51 2
K
= 5476
j 1614. 6
(b) Transfer function: TAO ( s )
8
±
j 1275
(b) Transfer function: Ω( s ) α ( s)
=
K1 K 4 e
−τ D s
Js + ( JK L + B ) s + K 2 B + K3 K 4 e 2
− τ Ds
(c) Characteristic equation:
Js + ( JK L + B ) s + K 2 B + K 3 K 4e 2
(d) Transfer function: Ω( s ) α ( s)
− τD s
=0
K1 K 4 ( 2 − τ D s )
≅
∆ ( s)
Characteristic equation:
∆ ( s ) ≅ J τ D s + ( 2 J + JK 2τ D + Bτ D ) s + ( 2 JK 2 + 2B − τ D K 2 B − τ D K 3 K4 ) s + 2 ( K 2 B + K 3 K 4 ) = 0 3
2
4-19 (a) Transfer function: Ec ( s)
G ( s) =
=
E (s )
1 + R2C s
1 + ( R1 + R 2 ) Cs
(b) Block diagram:
(c) Forward-path transfer function: Ωm ( s)
=
E (s )
[1 + ( R
1
K (1 + R2C s )
+ R 2 ) Cs ] ( K b Ki + Ra JL s )
(d) Closed-loop transfer function: Ωm ( s) Fr ( s ) (e)
=
Gc ( s) =
[1 + ( R
1
E c ( s)
=
Kφ K (1 + R2C s )
+ R2 ) Cs] ( K b K i + Ra J L s ) + Kφ KK e N (1 + R2C s )
(1 + R C s )
E (s )
Forward-path transfer function:
Ωm ( s) E (s )
2
RCs 1
=
K (1 + R2C s )
RCs ( K b K i + Ra J L s ) 1
Closed-loop transfer function:
30
Ωm ( s) Fr ( s ) (f)
Ke
= 36
pulse s / rev
f
= 120
pulse s / sec
=
e
f
r
ω
NK
ω m = 120
Thus, N = 1. For
K φ K (1 + R2C s )
=
R1C s (K b K i + Ra J L s ) + Kφ KKe N (1 + R2C s )
= 36
/ 2 π pulse s / rad
ωm = pulse s / sec
ω m = 1800
=
= 5 . 73 pul
200 RPM
= 200(
ses / rad. 2 π / 60 ) rad / sec
N ( 36 / 2 π ) 200( 2 π / 60 )
RPM,
120
=
= 120
N pulse s / sec
N ( 36 / 2 π ) 1800( 2 π / 60 )
= 1080
N.
Thus,
N
= 9.
4-20 (a) Differential equations:
dθ m − dθ L dt dt dt dt 2 dθ dθ L = J d θ L + B dθ L + T K (θ m − θ L ) + B m − L 2 L L dt dt dt dt d θm 2
Ki ia = Jm
(b)
2
+ Bm
dθ m
+ K ( θm − θ L ) + B
Take the Laplace transform of the differential equations with zero initial conditions, we get
(
)
Ki I a ( s ) = J m s + Bm s + Bs + K Θm ( s ) + ( Bs + K ) Θ L ( s )
( Bs + K ) Θ Solving for
Θm ( s )
and
2
m
ΘL ( s )
Θ m (s ) = Θ L (s ) =
(
)
( s ) − ( Bs + K ) Θ L ( s ) = J L s + BL sΘ L (s ) + TL ( s) 2
from the last two equations, we have
Ki
J m s + ( Bm + B ) s + K 2
Bs + K J L s + ( BL + B ) s + K 2
I a (s ) + Θ m (s ) −
Bs + K J m s + ( Bm + B ) s + K 2
Θ L (s )
TL ( s )
J L s + ( BL + B ) s + K 2
Signal flow graph:
(c) Transfer matrix: 2 1 Ki J L s + ( BL + B ) s + K Θ m ( s ) = Θ (s ) ∆ (s ) Ki ( Bs + K ) L o
31
Ia ( s ) 2 Jm s + (B m + B ) s + K −TL ( s ) Bs + K
∆ o ( s ) = J L Jm s + [ J L 3
(B
+ B) + J m ( BL + B ) ] s + [ BL Bm + ( BL + BM
) B + (J
3
m
+J
m
) K]s
L
+ K ( BL + B ) s
2
4-21 (a) Nonlinear differential equations: dx ( t ) dt With R
a
= 0,
φ (t ) =
=
dv ( t )
v(t )
= −k ( v ) − g ( x ) +
dt e(t )
=K
K v(t )
f
=
i (t ) f
=
K i (t ) f
f
K
= − Bv
f (t )
i (t )
(t )
f (t )
i (t )
Then,
f a
+ a
=
b
=
K
φ ( t ) ia ( t ) = i
(b) State equations:
K e (t ) i
K
2
b
K
.
2
f
dv ( t ) Thus,
v (t )
dt
= − Bv
+
(t )
K 2
K K b
i
2
e (t )
2
f
v (t )
i ( t ) as input. a
dx ( t ) dt
dv ( t )
= v (t )
= − Bv
dt
(t ) + K K
2
i (t )
i
f a
(c) State equati ons: φ ( t ) as input. f (t )
= K iK
dx ( t ) dt
2
i (t )
ia ( t )
f a
dv ( t )
= v (t )
= if
= − Bv
dt
(t )
(t ) +
φ( t )
=
K
K K
f
φ
i
2
(t )
f
4-22 (a) Force and torque equations: Broom: vertical direction:
horizontal direction:
f
f
x
=
M
Car: horizontal direction:
Eliminating f
x
=
Define the state variables as x
1
v 2
d
x (t )
f
x
d b
2
θ
= θ,
x
dx1 dt dx3 dt
= x2
dx2
= x4
dx4
=
dt
dt
(M
( L cos
b
y
θ−
x (t )
=
≅
,
f L cos x
J
2
1
x
=
dt x and
cos x
1
b
=
1
u ( t) + M b Lx2 x1 − 3 M b gx1 / 4
(M
32
b
+ M c ) − 3M b / 4
2
b
x
4
+ M b ) gx1 − M b Lx2 x1 − u ( t )
L [ 4 (M b + M c ) / 3 − M b ]
(c) Linearization:
M L
≅ 1.
2
c
θ
3
x , and
3
2
=
θ)
dt
f L sin
dt dθ
2
and f y from the equations above, and sin x
2
d
+ Mc
2
d
2
=
2
dt
M
+ L sin θ
dt J
u(t )
− M bg =
b
rotation about CG:
(b) State equations:
K K b
2
f (t )
e(t 0
=
dx . dt
f
v (t )
∂ f1
∂ f1
=0
∂ x1 ∂ f2
∂ x2
=
c
b
∂ f2
∂f 3 ∂x 3
∂f 3
=0
M b Lx2 − 3 M b g / 4
∂ f4
∂ x1 ∂ f4
(M
+ M c ) − 3M b / 4
b
∂ f4
=0
∂ x3
∂ x2
(M
∂ f4
=0
∂ x4
=
∂ x3
=0
∂u
=0
2 M b Lx1 x2
+ M c ) − 3M b / 4
b
1
=
∂u
∂f 3
=0
∂x 4
2
=
L ( M b + M c − 3M b / 4 )
L [ 4 (M b + M c ) / 3 − M b ]
=0
∂x 2
∂ f2
=0
−1
=
∂u ∂f 3
=0
∂u
−2 M b x1 x2
=
∂ x2
∂ f1
=0
∂ x4
∂ f2
2
b
∂ f2
=0
∂ f4
b
c
=0
∂ x4 ∂x 1
∂ x3
(M + M ) g − M x = 0 L ( M + M − 3M / 4) b
∂ f1
=0
2
∂ x1
∂f 3
∂ f1
=1
(M
b
+ M c ) − 3M b / 4
Linearized state equations:
0 ∆x& 1 3 ( M b + M c ) g ∆x& L ( M + 4 M ) b c 2 = ∆x& 3 0 & − 3M b g ∆x 4 M + 4M b c
=
(b)
i
eq
E
=
Ki
2
y (t )
1 eq
dt dy
eq
=0
dt x
1
eq
= i, x
R
x
2 eq
2
=
y
=
y , and x
E
eq
R
=
3
E
=
eq
+
dt di ( t )
eq
E
dy
At equili brium,
R
=
dL ( y ) dy ( t )
Ri ( t ) + i ( t )
(t )
2
Define the state variables as
Then, x
=
dt
= Mg −
My ( t )
Thus,
d L ( y ) i( t )
+
Ri ( t )
0
0
L( y) = L y
4-23 (a) Differential equations: e(t )
0 ∆x −3 1 0 0 0 ∆x2 L ( M b + 4 M c ) + ∆u 0 0 1 ∆x 3 0 ∆x 4 4 0 0 0 M + 4M b c 1
L di ( t ) y
dt
dy ( t )
= 0,
=
dt
−
Ri ( t ) d
= 0,
2
L y
2
y (t )
dt
2
i( t )
dy ( t ) dt
=0
K
eq
R
Mg
dy . dt
K
x
Mg
3 eq
=0
The differential equations are written in state equation form: dx dt
1
=−
R L
x x 1
2
+
x x 1
x
2
3
+
x
2
L
e
=
f
dx 1
2
dt
(c) Linearization:
33
=
x
3
=
f
dx 2
dt
2
3
=
g
−
K x1 M x
2 2
=
f
3
+
L di ( t ) y
dt
∂f 1 ∂x 1
=−
∂f 1 ∂e
R
x
L
=
∂f 3 ∂x 1
x
2 eq
+
2 eq
x
=
L
=−
x
3 eq
E
1
K
L
Mg
x
∂f 2
eq
∂x 1
x
L
E
∂x 2
2 K x 1 eq M
x
x x 1
x
3
2
2 eq
K
∂e
2 Rg
Mg
Eeq
K
x
1 eq
Mg
=
K
2 eq
=0
∗
K 0 0
0
x
∆x& = A ∆x + B ∆e ∗
Eeq K RL Mg ∗ B = 0 0
Mg
0
=
=0
∂e ∂f 3
The linearized state equations about the equilibrium point are written as:
Eeq K − L Mg ∗ A = 0 2 Rg − E eq
∂x 3
∂f 2
Mg
eq
∂f 1
=0
=1
∂x 3
E
eq
L
2
2 Rg
=
E
+
3
∂f 2
=0
∂x 2
=
−
1 eq
2
∂f 3
eq
R
∂f 2
=0
2 Rg
2 eq
=−
∂x 2
Mg
R
=−
2
∂f 1
K
eq
L
2 eq
2 K x 1 eq M
E
=−
4-24 (a) Differential equations: d
M1
M
2
y1 (t ) 2
dt d
2
2
=
y 2 (t ) 2
dt
Define the state variables as x
1
=
=
M 2g
y ,
x
1
dy 1 ( t )
−B
M 1g
2
dt dy
−B
=
2
−
dy
(t )
2
dt 1
x
,
dt
3
2
Ki
(t )
2
y1 ( t )
2
1
(t ) y2 (t )
Ki
− =
+ Ki 2
2
x
4
2
(t )
y2 (t ) − y1 (t ) y ,
− y1 ( t )
=
dy
2
2
.
dt
The state equations are:
dx1 dt
= x2
M1
dx2 dt
Ki
= M 1 g − Bx2 − dx
At equilibrium,
1
dt
= 0,
M 1g − Solving for I, with X
1
dx
2
2
x1
dt
(x
3
= 0,
2
Ki
+
dx
2
KI
2
X1
+
3
dt
KI
(X
3
2
dx3
− x1 )
2
dt
dx
= 0,
4
dt
= 0.
= x4
Thus , x
M2
2 eq
=0
2
− X 1)
=0
2
M 2g −
dx4 dt
3
4 eq
=0
2
= 1 , we have
M + M2 Y2 = X 3 = 1 + 1 M2
1/2
(b) Nonlinear state equations:
34
= 0.
2
− X1)
Ki
(x
3
and x
KI
(X
= M 2 g − Bx4 −
( M1 + M 2) g I= K
1/2
2
− x1 )
2
dx1 dt
dx2
= x2
B
= g−
dt
M1
(c) Linearization: ∂ f1 =0 ∂ x1 ∂ f2
=
∂ x1
∂ f2 ∂i
∂ f4 ∂ x1
=
=
2 KI
x2 −
3
M 1 x1
2
∂ x2
2 KI
M2 ( X3 − X1 )
3
∂ f1 ∂ x3
∂ x2
dx3
∂ f3
∂ f4
=0
∂ x3
Linearized state equations:
M
1
B
∂ f2
M1
∂ x3
∂ x2
=
= 2,
∂ f3
=0
2 KI
∂ x3
= 1,
2
g
∂ f1 ∂i
=
−2 KI
= 32 .2,
32.2(1 + 2) X = 96.6 X = 9.8285X 1 1 1 1
B
= 0 .1,
1/2
I =
(
X1 =
)
X 3 = 1 + 1 + 2 X 1 = 2.732 X 1 = Y2 = 2.732 0 2 1 2 KI 1 + 3 3 M1 X1 ( X 3 − X 1 ) ∗ A = 0 2 −2 KI 3 M 2 ( X 3 − X1 )
1
0 −2 KI
M1
M 1 ( X 3 − X1 )
0
0 2 KI
2
M 2 ( x3 − x1 )
∂ f2 ∂ x4
3
∂ f3
= 1
∂i
∂ f4
M2
∂i
=
=0
= 0
−2 KI M2 ( X3 − X1)
= 1.
1 9.8285
2 3
2
M 2 ( X 3 − X 1)
3
=1
X 3 − X 1 = 1.732
1 0 0 115.2 −0.05 = 1 0 0 0 − B − 37.18 M 2
0 1 0 2 KI −1 + M 1 X 12 ( X 3 − X 1 ) 2 − 6.552 ∗ B = = 0 0 − 6.552 −2 KI 2 M 2 ( X 3 − X 1)
4-25 (a) System equations:
35
2
=0
B
K
Ki
0
−B
0
∂ x4
=−
x4 −
2
∂ f3
= 0
∂ x4
3
M2
M1 ( X3 − X1 )
∂ f4
2
M 2 ( X 3 − X1 ) M
dt
B
= g−
=0
∂ x4
∂ f3
=0
∂ x1
dt
dx4
= x4
∂ f1
=−
∂ x2
3
2
=0
∂ f2
M1 ( X 3 − X 1 )
∂ f4
2
M 1 ( x 3 − x1 )
2
−1 1 2+ 2 M 1 X1 ( X 3 − X 1 ) 2
Ki
=0
2 KI
−2 KI
i + 2
M 1x1
∂ f1
2
+
K
−18.59 0 0 1 37.18 − 0.1 0
0
2
Tm = K i ia = ( J m + JL ) T
D
=
dω m dt
d (sec) V
+ B mω m e
ea = Ra ia + La
= r −b
b
dia
= Ksy
dt
+ K b ωm
E ( s) a
=
y = nθ m
KG ( s ) E ( s ) c
Block diagram:
(b) Forward-path transfer function: Y (s )
=
E (s )
K Kin G c ( s ) e
− TD s
s {( Ra + La s ) [( Jm + J L ) s + Bm ] + K b Ki }
Closed-loop transfer function:
Y (s ) R (s )
=
− TD s
K Kin G c ( s ) e
s ( Ra + La s ) [( Jm + J L ) s + Bm ] + K bK i s + KG c ( s )K i ne
36
− TD s
y = y ( t − TD )
Chapter 5 STATE VARIABLE ANALYSIS OF LINEAR DYNAMIC SYSTEMS 5-1 (a) State variables:
x
1
=
y,
x
2
=
dy dt
State equations:
Output equation:
dx1
dt 0 1 x1 0 = + r dx2 −1 −4 x2 5 dt (b) State variables:
x
1
=
y,
x
2
=
y = [1
dy
x
, dt
3
2
d
=
dt
x1 = x1 x2
0]
y 2
State equations:
Output equation:
dx1
dt 1 0 x1 0 dx 2 = 0 0 1 x2 + dt dx −1 −2.5 −1.5 x3 3 dt x1 =
(c) State variables:
∫
t
0
y (τ ) d τ ,
0 0 r 0.5
x2 =
dx1
x& 2 = x&3 x& 4
0
dt
0
0
x1
0
x
1
=
y,
x
2
=
dy ,
x
dt
3
State equations:
x& 2 = x&3 x& 4 5-2
dt
d y dt
2
x1 x 2 y = [1 0 0 0] = x1 x 3 x 4
0 0 1 0 x 0 2+ r 0 0 0 1 x3 0 − 1 − 1 − 3 −5 x 1 4
0
2
x4 =
,
Output equaton:
1
(d) State variables:
x&1
dy
x3 =
,
State equations:
x&1
x1 y = [1 0 0 ] x2 = x1 x3
=
2
d
dt
3
y 2
x
,
4
=
d y dt
3
Output equation:
1
0
0
x1
0
x1 x 2 y = [1 0 0 0] = x1 x 3 x 4
0 0 1 0 x2 0 + r 0 0 1 x3 0 0 −1 −2.5 0 −1.5 x 1 4
We shall first show that
Φ ( s ) = ( sI − A ) = We multiply both sides of the equation by
I
A
1 A
2
+ 2 + +L 2 s s 2! s ( sI − A ) , and we get I = I. Taking the inverse Laplace transform −1
37
on both sides of the equation gives the desired relationship for
5-3 (a) Characteristic equation: Eigenvalues:
s
∆( s ) =
= −0 . 5 +
− A = s2 + s + 2 = 0
sI
− 0 .5 −
j 1. 323 ,
φ( t ) .
j 1. 323
State transition matrix:
cos1.323t + 0.378sin1.323t
φ ( t) =
(b) Characteristic equation:
−1.512sin1.323 t
∆( s ) =
sI − A
−0.5t e −1.069sin (1.323t − 69.3 ) 0.756sin1.323t
o
= s 2 + 5s + 4 = 0
Eigenvalues:
= − 4,
s
−1
State transition matrix:
1.333 e− t − 0.333e−4 t
φ ( t) =
−t
−1.333 e − 1.333 e
(c) Characteristic equation:
−0.333e + 1.333e −t
0.333 e − 0.333e
−4 t
−t
∆ ( s ) = ( s + 3) = 0 2
−4 t
−4 t
Eigenvalues:
= −3, − 3
s
State transition matrix:
e −3 t φ ( t) = 0 (d) Characteristic equation:
∆( s ) =
−3 t
− 9 = 0 Eigenvalues:
2
s
e 0
s
= −3 ,
3
s
=
−
State transition matrix:
e3 t φ ( t) = 0 (e) Characteristic equation:
e 0
−3 t
∆ ( s ) = s + 4 = 0 Eigenvalues: 2
j2,
j2
State transition matrix:
cos2 t − sin2 t
φ ( t) = (f) Characteristic equation:
∆( s ) =
s
3
s i n 2t
cos2t
+ 5 s + 8 s + 4 = 0 Eigenvalues: 2
s
= − 1, − 2 ,
−2
State transition matrix:
e− t φ ( t) = 0 0 (g) Characteristic equation:
∆( s ) =
s
3
0 e
0
+ 15
e φ ( t) = 0 0
−5 t
5-4 State transition equation: x (t ) = φ (t )x( t ) +
−2 t
+ 75 s + 125 = 0
2
s
te −2t e 0
−2 t
te e
−5 t
−5 t
0
Eigenvalues:
s
= − 5,
− 5,
−5
te −5 t e 0
−5 t
∫ φ (t − τ )Br (τ )d τ t
0
(a)
38
φ (t )
for each part is given in Problem 5-3.
∫
t
0
1 s + 1 1 0 1 1 1 ( sI − A) −1 BR( s ) = L−1 ∆ ( s ) −2 s 1 0 1 s s+2 2 s(s + s + 2) 1 + 0.378sin1.323 t − cos1.323t −1 = =L t≥0 s−2 −1 + 1.134sin1.323 t + cos1.323t s s2 + s + 2 ) ( −1
φ ( t − τ ) Br (τ ) d τ = L
(b)
∫ φ ( t − τ ) Br (τ ) dτ = L ( sI − A) t
−1
0
−1
BR ( s ) = L
−1
1 s + 5 1 1 1 1 ∆( s) −4 s 1 1 s
s+6 1.5 − 1.67 + 0.167 1.5 − 1.67 e− t + 0.167e −4 t s ( s + 1)( s + 2) −1 = L−1 s s + 1 s + 4 = =L −t s−4 −1 + 1.67 − 0.667 −1 + 1.67 e − 0.667 e − 4t s( s + 1)( s + 4) s s + 1 s + 4 (c)
1 t −1 −1 −1 s + 3 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR( s) = L 0 0 0 −1 = =L 1 t≥ −3t 0.333 (1 − e ) s ( s + 3 )
0 1 1 1 s s + 3 0
0
(d)
1 t −1 −1 −1 s − 3 φ ( t − τ ) B r ( τ ) d τ = L ( s I − A ) B R ( s ) = L ∫0 0 0 = =L 1 s ( s + 3 ) −1
0 0.333 (1 − e −3t )
(e)
39
0 1 1 1 s s + 3
t≥ 0
0
t≥0
1 2 t −1 −1 −1 s + 4 φ ( t − τ ) B r ( τ ) d τ = L s I − A B R ( s ) = L ( ) ∫0 −2 s 2 + 4 2 s 2 −1 =L = t ≥0 1 0.5sin2t ( s 2 + 4 )
0 1 s 1 s 2 s + 4 2
(f)
1 s + 1 t −1 −1 −1 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR( s) = L 0 0 0 0 1 −1 −2 t = 0.5 (1 − e ) =L t≥0 s( s + 2) 0 0
0 1 s+ 2 0
0 1 1 1 2 (s + 2 ) s 0 1 s +2 0
(g)
1 s + 5 t −1 −1 −1 ∫0 φ ( t − τ ) Br (τ ) dτ = L ( sI − A) BR (s ) = L 0 0
1
( s + 5) 1 s +5 0
2
0 1 1 0 2 ( s + 5) s 1 1 s+5 0
0 0 0 1 0.04 0.04 0.2 −1 −1 −5 t −5 t =L =L 2 s − s + 5 − ( s + 5 ) 2 = 0.04 − 0.04 e − 0.2 te u s ( t) −5t s ( s + 5) 0.2 − 0.2 e 1 0.2 0.2 − s ( s + 5 ) s s +5 5-5 (a) (b)
Not a state transition matrix, since
φ( 0 ) ≠ I
(identity matrix).
Not a state transition matrix, since
φ( 0 ) ≠ I
(identity matrix).
(c) φ ( t )
is a state transition matrix, since
φ ( 0 ) = I and
40
1 [φ ( t ) ] = − t 1 − e (d) φ ( t )
is a state transition matrix, since
0 e
t
= φ ( − t)
φ ( 0 ) = I , and
e2 t [φ ( t) ]−1 = 0 0 5-6 (a) (1) Eigenvalues of A:
−1
1 = −t 1 − et e 0
−1
− te
2 2t
2t
− te
0
e
e
2 .325 ,
t e / 2
2t
− 0 . 3376 +
= φ ( − t)
2t
2t
j 0 . 5623 ,
− 0 .3376 −
j 0 . 5623
(2) Transfer function relation: −1
−1
s 1 −1 X( s) = ( sI − A ) B U ( s) = 0 ∆ (s ) 1
s 2 + 3s + 2 0 U ( s) = 1 −1 ∆ (s ) −s 1
0 0
−1 s + 3
s 2
s +3
1 0
1 1 s ( s + 3 ) s 0 U ( s) = s U (s ) ∆ (s ) 2 2 −2 s − 1 s 1 s
∆( s ) = s + 3 s + 2 s + 1 3
2
(3) Output transfer function:
1 1 s = = C ( s ) ( sI − A) B = [1 0 0 ] 3 2 U (s ) ∆( s ) 2 s + 3 s + 2 s + 1 s Y ( s)
(b) (1) Eigenvalues of A:
1
−1
− 1,
− 1.
(2) Transfer function relation:
1 ( s + 1) 2 1 0 1 s + 1 −1 U (s ) X ( s ) = ( sI − A) BU ( s ) = U (s ) = ∆ ( s ) 0 s + 1 1 1 ( s + 1)
∆ ( s ) = s + 2s + 1 2
(3) Output transfer function:
1 ( s + 1) 2 Y ( s) 1 1 s+2 −1 = = C ( s ) ( sI − A) B = [1 1] + = 2 2 U (s ) 1 ( s + 1) s + 1 ( s + 1) s + 1 (c) (1) Eigenvalues of A:
− 1,
0,
− 1.
(2) Transfer function relation:
s + 2s = 1 2
1 −1 X( s) = ( sI − A) BU ( s ) = ∆ (s )
0 0
s+2
1 0
1 1 s ( s + 2) ) s 0 U ( s) = s U ( s) ∆ (s ) −s s 1 s 2
41
2
(
)
∆( s ) = s s + 2s + 1 2
(3) Output transfer function:
1 s +1 1 = C ( s ) ( sI − A) B = [1 1 0 ] s = = 2 U (s ) s ( s + 1) s ( s + 1) s 2 Y ( s)
5-7
dy We write dt
=
dx
1
dt
+
dx
−1
2
dt
dx1 dt d x dy = = dt dt 2 d y dt2
2
d
= x 2 + x3
dt
y
=
dx
+
2
dt
dx dt
3
= −x1 − 2 x2 − 2 x3 + u
0 1 0 x1 0 0 1 1 x + 0 u 2 −1 −2 −2 x3 1
x1 1 0 0 x = y = 1 1 0 x y& 0 1 1
(1)
1 0 0 x = − 1 1 0 x 1 − 1 1
(2)
Substitute Eq. (2) into Eq. (1), we have
−1 1 0 = A 1 x + B1u = 0 0 1 x = dt −1 0 −2
dx
5-8 (a) s sI
− A = −1 1
a1 M = a2 1
s
−2
0
−2
0
0
a2 1 0
1 0 0
s
0 0 u 1
= s − 3 s + 2 = s + a 2 s + a1 s + a0 3
2
3
2
a
0
= 2,
a
1
= 0,
−1
0 −3 1 = − 3 1 0 1 0 0
S = B
−2 2 0 P = SM = 0 −1 1 −4 −2 1
42
AB
0 2 4 A B = 1 2 6 1 1 − 1 2
a
2
= −3
(b) s sI
− A = −1
−2
0
−1
0
s
−1
1
a1 M = a2 1
(c) s sI
−A =
+2
0
sI
−A =
0 0
1 0
+1
3
2
3
1 0
a
2,
s
2
s
a
3
+ a 2 s + a1 s + a0 2
a
0
3 3 1 = 3 1 0 1 0 0
s
−1 2
= 0,
−1 s
= 12 ,
= 1,
S = B
0 1 −1 AB A B = 1 0 − 1 1 − 1 1 0 1 1 2
(e) −A =
0
+1
2 1 P = SM = 2 3 1 2
sI
1
a
= −3
2
a
1
= 16 ,
a
2
1 −1 0 S = B AB A B = 1 −2 4 1 −6 23 9 6 1 P = SM = 6 5 1 −3 1 1
3
0
0 0
=
2
−1 = s + 3 s + 3 s + 1 =
1
0
2
16 7 1 = 7 1 0 1 0 0
=1
a2
a
= s + 7 s + 16 s + 12 = s + a 2 s + a 1 s + a 0
0
0
2
2
−1 s
0
a1 M = a2 1
1
a2
3
+3
s
2
a1 M = a2 1
2
1 2 6 S = B AB A B = 1 3 8 0 0 1 0 −1 1 P = SM = −1 0 1 1 0 0
0
s
− 3 s + 2 = s + a 2 s + a1 s + a 0
3
0 −3 1 = − 3 1 0 1 0 0
−1
1
s
0 0
0
0
(d)
1
1
+2
s
−1
s
a2
=
+3
=
s
2
+ 2 s −1 = s + a1 s + a 0 2
43
a
0
= − 1,
a
1
=2
a
1
= 3,
a
2
=3
=7
M=
a1 1
1
2 1 0 1 0 P = SM =
5-9 (a)
S = [B
=
AB ] =
0 1 1 −3
1 0 −1 1
From Problem 5-8(a),
0 −3 1 M = − 3 1 0 1 0 0 Then,
C V = CA = 2 CA (b)
0.5 1 3 Q = (MV ) = 0.5 1.5 4 −0.5 −1 −2 −1
From Problem 5-8(b),
C V = CA = 2 CA (c)
1 0 1 − 1 2 1 1 2 1
16 7 1 M = 7 1 0 1 0 0 1 0 1 −1 − 1 3 1 Q = (MV ) = 2 5 1
0.2308 0.3077 1.0769 0.1538 0.5385 1.3846 −0.2308 −0.3077 −0.0769
From Problem 5-8(c),
C V = CA = 2 CA
1 0 0 − 2 1 0 4 −4 0
Since V is singular, the OCF transformation cannot be conducted.
(d)
From Problem 5-8(d),
3 3 1 M = 3 1 0 1 0 0 Then,
C V = CA = 2 CA (e)
1 0 1 − 1 1 −1 1 −2 2
−1 1 0 Q = ( MV ) = 0 1 − 2 1 − 1 1 −1
From Problem 5-8(e),
44
M=
5-10 (a)
2 1 1 0
Eigenvalues of A:
T = [p1
Eigenvalues of A:
2
T = [ p1
1
2
Q = ( MV ) =
p are the eigenvectors. 3
−0.7321
1, 2.7321,
where p , p , and
1 0 1 1
0 0.5591 0.8255 p3 ] = 0 0.7637 −0.3022 1 −0.3228 0.4766
p2
where p , p , and
(b)
C CA 2 =
−0.7321
1, 2.7321,
1
V=
Then,
p2
0 0.5861 0.7546 p3 ] = 0 0.8007 −0.2762 1 0.1239 0.5952
p are the eigenvalues. 3
(c)
Eigenvalues of A:
-3, -2, -2. A nonsingular DF transformation matrix T cannot be found.
(d)
Eigenvalues of A:
−1, −1, −1
The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I.
(e)
Eigenvalues of A:
0.4142,
−2.4142 T = [p2
p2 ] =
0.8629 −0.2811 − 0.5054 0.9597
5-11 (a)
1 −2 0 0
S = [B
AB] =
S = B
1 −1 1 2 AB A B = 2 −2 2 3 −3 3
S = [B
AB] =
S is singular.
(b) S is singular.
(c)
2 2
2 +2 2 2+
2
S is singular.
(d)
45
−1
0 1 1 −3
1 −2 4 AB A B = 0 0 0 1 −4 14
S = B 5-12 (a)
2
S is singular.
Rewrite the differential equations as:
d θm 2
dt
2
B d θm 2
=−
J dt
2
−
K J
θm +
State variables: x = θ ,
x
m
1
Ki
2
J =
dia
ia dθ
=−
Kb dθ m
dt m
x
,
dt
3
La dt
−
Ra La
ia +
K a Ks
(θ
La
r
−θm )
= ia
State equations:
Output equation:
dx1 dt 0 dx2 = − K dt J dx K K 3 − a s dt La
1 −
B J
−
Kb La
0 x1 Ki x2 + 0 θ r J x K K Ra 3 a s − La La 0
y
=
1
0
0 x
(b) Forward-path transfer function:
s Θm ( s ) K G ( s) = = [1 0 0 ] J E (s ) 0
−1 s+
B J
Kb La
0 Ki − J Ra s+ L a
−1
0 KiK a 0 = K ∆ o ( s) a La
∆ o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 3
2
Closed-loop transfer function:
s Θm ( s ) K M ( s) = = [1 0 0 ] J Θr ( s ) KaKs La =
−1 s+
B J
Kb La K i Ka K s
0 Ki − J R s+ a La
−1
0 K s G( s ) 0 = K K 1 + K s (s ) a s La
JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 3
2
5-13 (a)
46
= x1
A=
0 1 −1 0
A = 2
− 1 0 0 −1
A = 3
0 − 1 1 0
A = 4
1 0 0 1
(1) Infinite series expansion: 3 5 t t t2 t4 1 − + − L t − + − L 1 2 2 2! 4! 3! 5! φ ( t) = I + At + A t + L = 3 5 2 4 2! −t + t − t + L 1 − t + t L 3! 5 ! 2! 4 !
cos t = − sin t
sin t cos t
(2) Inverse Laplace transform: −1
1 s 1 s −1 Φ ( s ) = ( sI − A ) = 1 s = s 2 + 1 −1 s −1
φ (t ) =
cos t − sin t
sin t cos t
(b) A=
−1 0 0 −2
A = 2
1 0 0 4
A = 3
−1 0 0 − 8
A = 4
−1 0 0 16
(1) Infinite series expansion: 2 3 4 t t t 1 − t + − + +L 0 1 2 2 2 ! 3! 4 ! φ ( t) = I + At + A t + L = 2 3 2! 4t 8t 0 1 − 2t + − +L 2! 3!
e −t = 0
e 0
−2 t
(2) Inverse Laplace transform: Φ ( s ) = ( sI − A ) = −1
s + 1 0
1 s +1 = s + 2 0
1 s + 2 0
−1
0
e− t
−2 t e 0
φ (t ) =
0
(c) A=
0 1 1 0
A = 2
1 0 0 1
A = 3
0 1 1 0
A = 4
1 0 0 1
(1) Infinite series expansion: 3 5 t t t2 t4 1+ + +L t+ + L 1 2 2 e− t + et 2! 4! 3! 5 ! φ ( t) = I + At + A t + L = = −t t 3 5 2 4 2! t + t + t L 1 + t + t + L −e + e 3! 5 ! 2 ! 4!
e +e −t
−e + e −t
(2) Inverse Laplace transform: −1
1 s 1 s −1 −1 Φ ( s ) = ( sI − A ) = −1 s = s 2 − 1 1 s =
47
0.5 − s +1 −0.5 + s + 1
s − 1 s + 1 s − 1 0.5 0.5 0.5 + s − 1 s + 1 s − 1 0.5
− 0.5
+
0.5
t
t
e− t + et φ ( t) = 0.5 − t t −e + e 5-14 (a)
e = K s (θ r − θ y ia = Solve for i
a
)
ea = e − es
eu − eb
eb = K b
Ra + Rs
in terms of
θy
dθ and
y
e +e −t
−e + e −t
t
es = Rs ia
dθ y
t
eu = Kea
Tm = K i ia = ( J m + J L )
dt
d θy 2
dt
2
, we have
dt
ia =
KKs (θr − θ y ) − Kb
dθ y dt
Rs + Rs + KRs
Differential equation:
d θy 2
dt
2
K ii a
=
Jm + J L
State variables: x = θ ,
x
y
1
dθ y − KK sθ y + KK sθ y −K b ( J m + J L ) (R a + R s + KRs ) dt Ki
=
2
=
dθ
y
dt
State equations:
dx1 0 1 0 dt x1 θ = + − KK s K i − K b Ki − KK s Ki r dx2 ( J + J ) (R + R + KR ) ( J + J ) (R + R + KR ) x2 ( J + J ) (R + R + KR ) m L a s s m L a s s s dt m L a s 1 x1 0 0 = + θr −322.58 −80.65 x2 322.58 We can let v ( t )
= 322
.58
θr,
then the state equations are in the form of CCF.
(b) −1
−1 1 s s + 80.65 1 = 2 322.58 s + 80.65 s + 80.65 s + 322.58 −322.58 s −0.014 0.014 −0.06 − 1.059 s + 76.42 s + 4.22 s + 76.42 + s + 4.22 = 1.0622 0.0587 4.468 − 4.468 s + 76.42 s + 4.22 s + 76.42 − s + 4.22
( sI − A ) = −1
For a unit-step function input, u ( t ) s
=1 /
s.
322.2 1 s ( s + 76.42)( s + 4.22) = ( sI − A)−1 B = s 322.2 s ( s + 76.42)( s + 4.22)
48
1 + 0.0584 − 1.058 s s + 76.42 s + 4.22 − 4.479 + 4.479 s + 76.42 s + 4.22
−76.42t −4.22t −0.014e + 0.01e −0.06e −76.42 t − 1.059e −4.22 t x (t ) = x (0) −76.42 t −4.22t −76.42t −4.22t − 4.468e 1.0622 e − 0.0587 e 4.468 e 1 + 0.0584 e −76.42 t − 1.058e −4.22 t = t≥0 −76.42 t −4.22t + 4.479e −4.479 e
(c) Characteristic equation: (d)
∆( s ) =
s
2
+ 80 . 65 s + 322
From the state equations we see that whenever there is to increase the effective value of
Ra by (1 + K ) Rs .
=0
. 58
Ra there is ( 1 + K ) Rs .
Thus, the purpose of R is s
This improves the time constant of the system.
5-15 (a) State equations: dx1
0 1 0 dt x1 θ + − KKs Ki − Kb Ki KKs Ki = r dx2 J ( R + R + KR ) J ( R + R + KR ) x2 J ( R + R + KR ) s s s s s s dt 1 x1 0 0 = + θ r −818.18 −90.91 x2 818.18
Let v
= 818
.18
θr.
The equations are in the form of CCF with v as the input. −1
s −1 1 s + 90.91 ( sI − A ) = = 818.18 s + 90.91 ( s + 10.128 ) ( s + 80.782) −818.18 −10.128t −80.78t 0.01415 e − 0.0141e 1.143 e−10.128t − 0.142 e−80.78t x1 (0) x (t ) = −10.128 t −80.78 t −10.128 t −80.78t + 0.1433e − 0.1433 e + 1.143e − 11.58 e x2 (0) 11.58 e−10.128 t − 11.58 e−80.78t + t≥0 −10.128 t −80.78t + 0.1433 e 1 − 1.1434 e −1
(b)
(c) Characteristic equati on:
2
. 18
=0
− 10.128, −80.782
Eigenvalues:
(d)
∆ ( s ) = s + 90 . 91 s + 818
Same remark as in part (d) of Problem 5-14.
5-16 (a) Forward-path transfer function: Y ( s) 5 ( K1 + K 2s ) G ( s) = = E ( s ) s [s ( s + 4)( s + 5) + 10 ] M ( s) =
Y (s ) R (s )
=
G (s) 1 + G(s )
=
Closed-loop transfer function:
5 ( K1 + K 2s )
s + 9 s + 20 s + (10 + 5 K 2 ) s + 5 K1 4
3
2
(b) State diagram by direct decomposition:
49
1 s
State equations:
x&1
x& 2 = x&3 x& 4
0 0 0 −5 K
(c) Final value:
Output equation:
x1 0 1 0 x2 0 + r 0 1 x3 0 −20 −9 x4 1
1
0
0 0 − (10 + 5 K 2 )
r(t )
= u s ( t ),
R( s)
0
1
=
5-17
s →0
=
5 K1
5K2
. s
lim y( t) = lim sY ( s ) = lim t →∞
y
s →0
5 ( K1 + K 2s )
s + 9 s + 20 s + ( 10 + 5 K 2 ) s + 5 K1 4
3
2
=1
In CCF form,
0 0 A= M 0 − a0
1
0
0
L
0
1
0
L
M
M
M
O
0
0
0
L
− a1
− a2
− a3
L
s 0 sI − A = M 0 a 0 sI
−A =
s
n
0 M 1 − an −1
−1
0
0
s
−1
0
M
M
M
0
0
0
a1
a2
a3
+ a n −1 s
n −1
0 0 B = M 0 1
0
+ a n−2 s
L
L 0 O M 0 −1 L s + a n n−2
0
+ L + a1 s + a 0
Since B has only one nonzero element which is in the last row, only the last column of going to contribute to the last row of
adj (s I − A ) B
( sI − A ) .
. The last column of
Thus, the last column of
adj (s I − A )
adj (s I − A ) B
50
is 1
adj (s I − A )
is
is obtained from the cofactors of s
s
2
L
s
n −1
'
.
0 x
5-18 (a) State variables:
x
1
=
y,
x
2
=
dy , dt
x
3
=
d
2
y 2
dt
x& ( t ) = Ax ( t ) + B r ( t )
State equ ations:
0 1 0 A= 0 0 1 −1 −3 −3
0 B = 0 1
(b) State transition matrix:
s2 + 3 s + 3 s + 3 1 s −1 0 1 −1 2 Φ ( s ) = ( sI − A ) = 0 s −1 = −1 s + 3s s ∆ (s ) 2 − s 1 3 s + 3 −3 s − 1 s 1 1 1 1 2 1 + + + 2 3 ( s + 1) 2 ( s + 1)3 ( s + 1 )3 s + 1 ( s + 1) ( s + 1) −1 1 1 2 s = + − 2 s + 1 ( s + 1) ( s + 1 )3 ( s + 1) 3 ( s + 1) 3 2 −s −3 2 s + ( s + 1 )3 ( s + 1) 2 ( s + 1)3 ( s + 1)3 −1
∆ (s ) = s3 + 3 s2 + 3s + 1 = ( s + 1)
(1 + t + t 2 / 2 ) e− t 2 −t φ ( t) = −t e / 2 ( −t + t 2 / 2 ) e −t
(t + t ) e (1 + t − t ) e −t
2
2
2
t e
3
( t − t / 2 ) e (1 − 2t + t 2 / 2 ) e −t 2 −t
t e /2
−t
−t
2
−t
(d) Characteristic equation: ∆ ( s ) = s + 3 s + 3 s + 1 = 0 3
− 1, −1, −1
Eigenvalues:
5-19 (a) State variables:
2
x
1
=
y,
x
2
=
dy dt
State equations:
dx1 (t ) dt 0 1 x1 ( t) 0 = + r ( t) dx2 ( t) − 1 − 2 x2 ( t) 1 dt State transition matrix:
s+ 2 ( s + 1) 2 s −1 −1 Φ ( s ) = ( sI − A ) = 1 s + 2 = − 1 ( s + 1) 2 −1
( s + 1) s ( s + 1)2 1
2
51
(1 + t ) e−t
φ (t ) =
− te
−t
(1 − t ) e te
−t
−t
Characteristic equation:
=
y,
−y=
x
(b) State variables:
x
1
∆ (s ) = ( s +1) = 0 2
x
2
=
y
dy
+
dt
State equations: dx dt
1
=
dy dt
=
x
2
dx
− x1
2
2
dt
d
=
2
y
dt
dy
+
2
dt
= −y −
dy
+r = −x2 +r
dt
dx1 dt −1 2 x1 0 = + r dx2 0 −1 x2 1 dt State transition matrix:
1 s + 1 −2 s + 1 Φ (s ) = = 0 s + 1 0
( s + 1) 1 s +1 −2
−1
2
(c) Characteristic equation: ∆ (s ) = ( s +1) = 0
φ (t ) =
e −t 0
− te e
−t
−t
2
5-20 (a) State transition matrix: ω s − σ sI − A = −ω s − σ
( sI − A )
−1
which is the same as in part (a).
s − σ ∆( s) ω 1
=
s − σ −ω
2
cos ω t − sin ω t σ t ( sI − A )−1 = e sin ω t cos ω t
−1
φ ( t) = L
σ + jω , σ − jω
(b) Eigenvalues of A:
5-21 (a) Y (s) 1
U ( s)
=
1
Y ( s) 2
U (s) 2
s 1+ s
−1
+ 2s s
= 1
+s
−1
(b) State equations [Fig. 5-21(a)]:
0 1 0 A1 = 0 0 1 −3 −2 −1 State equations [Fig. 5-21(b)]:
0 0 −3 A 2 = 1 0 −2 0 1 −1
−3 −2
+3s
−3
=
1 s
3
+s +2s+3 2
−3
+2s
−2
+ 3s
−3
=
1 s
3
x& = A x + B u 1
Y ( s) 1
U (s) 1
Output equation: y = C x
1 1
1
0 B1 = 0 1
C1 = [1
x& = A x + B u 2
+ s + 2s +3 2
=
2
1 B2 = 0 0
2
0
1
0]
Output equation: y
C2 = [0
52
0
(
∆ ( s ) = s − 2σ + σ + ω
1]
2
= C 2x
2
2
)
A
Thus,
= A1 '
2
5-22 (a) State diagram:
(b)
State diagram:
5-23 (a) G (s) X (s)
=
Y (s) U ( s)
=
10 s 1 + 8 .5 s
= U ( s ) − 8 .5 s
−1
−1
−3
+ 20 . 5 s
X ( s)
X (s) −2
− 20 . 5 s
+ 15
−2
s
−3
X ( s ) − 15 s
State diagram:
State equation:
x& ( t ) = Ax ( t ) + B u ( t )
53
Y ( s)
X (s) −3
X (s)
= 10
X (s)
1 0 0 A= 0 0 1 −15 −20.5 −8.5 (b) G (s) Y ( s)
=
Y (s) U ( s)
= 10
s
−3
=
10 s
−3
+ 20
1 + 4. 5 s
−1
+ 20
−4
X ( s)
s
0 B = 0 1 s
−4
+ 3 .5 s
A and B are in CCF
X (s) −2
X (s)
X (s)
X ( s)
= −4. 5 s
−1
X ( s ) − 3 .5 s
−2
X ( s) +U ( s)
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 A= 0 0
0 1 0 0 0 1 0 −3.5 −4.5 1
0
0 0 B= 0 1
0
(c) G (s)
Y (s)
=
Y (s) U ( s)
= 5s
−2
State equations:
=
5( s s(s
X (s)
+ 1)
+ 2 )( s + 10 )
+ 5s
−3
X (s)
=
5s
−2
1 + 12 s
+ 5s
−1
X (s)
A and B are in CCF
−3
+ 20
s
X (s) −2
X (s)
= U ( s ) − 12
x& ( t ) = Ax ( t ) + B u ( t )
54
s
−1
X (s)
− 20
s
−2
X ( s)
0 0 1 A= 0 0 1 0 −20 −12
0 B = 0 1
A and B are in CCF
(d) G ( s) = Y ( s)
=
Y (s ) U (s ) s
−4
=
(
1
)
s ( s + 5) s + 2s + 2
X ( s)
2
X (s)
=
= U ( s) − 7s
s
−4
X (s )
−1
−2
1 + 7 s + 12 s + 10 s −1
−2
X ( s ) − 12 s
X (s)
−3
− 10
X (s ) s
−3
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 0 1 0 0 1 0 A= 0 0 0 1 0 −10 −12 −7
0 0 B= 0 1
A and B are in CCF
5-24 (a) G (s)
=
Y (s) U ( s)
=
10 s
3
+ 8 . 5 s + 20 . 5 s + 15 2
=
5 . 71 s
55
+ 15
−
6 . 67 s
+2
+
0 . 952 s
+5
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
−1.5 0 0 A= 0 −2 0 0 −5 0
5.71 B = −6.67 0.952
The matrix B is not unique. It depends on how the input and the output branches are allocated.
(b) G (s)
Y (s)
=
=
U ( s)
10( s 2
s (s
+ 2)
−4. 5
=
= 1 )( s + 3 . 5)
0 .49
+
s
+ 3 .5
s
4
+ s
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equation:
0 0 1 0 0 0 0 0 A= 0 0 −1 0 0 0 0 −3.5 (b) G (s)
=
Y ( s) U (s)
State equations:
=
5( s s( s
+ 1)
+ 2 )( s + 10 )
=
0 1 B= 1 1
2 .5 s
+
0 . 313 s
+2
x& ( t ) = Ax ( t ) + B u ( t )
0 0 0 A = 0 −2 0 0 0 −10
1 B = 1 1
(d)
56
−
0 . 563 s
+ 10
+1
+
5 . 71 s
2
Y (s )
G ( s) =
U (s )
=
(
1
s ( s + 5) s + 2s + 2 2
)
=
0.1 s
−
0.0118 s+ 5
−
0.0882s + 0.235 s + 2s + 2 2
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 0 −5 A= 0 0 0 0
0 1 −2
0
1 1 B= 0 1
0
0 0 −2
5-25 (a) G (s)
=
Y (s) U ( s)
=
10
+ 1. 5)( s + 2 )( s + 5)
(s
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
−5 1 A = 0 −2 0 0
1 −1.5 0
0 B= 0 10
(b) G ( s) =
Y (s ) U (s )
=
10 s + 2 1 2 s ( s + 1)( s + 3.5) s s + 1 s + 3.5 10(s + 2)
2
=
State diagram:
57
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 0 1 0 0 0 1 1 A= 0 0 −1 1 0 0 0 −3.5
0 0 B= 0 10
(c) G ( s) =
Y (s )
=
U (s )
5 s + 1 1 s( s + 2)( s + 10) s s + 2 s + 10 59 s + 1)
=
State diagram:
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
1 0 0 A = 0 − 10 −1 0 0 −2
0 B = 0 5
(d) G ( s) =
Y (s) U (s )
=
(
1
s ( s + 5) s + 2s + 2 2
)
State diagram:
58
x& ( t ) = Ax ( t ) + B u ( t )
State equations:
0 1 0 0 A= 0 −2 0 0
0 1 −5
0
0 0 B= 0 1
0
1 −2 0
5-26 (a) G (s)
=
Y (s) E (s)
=
10 s(s
+ 4 )( s + 5)
=
10 s 1+ 9 s
−1
−3
+ 20
X (s) s
−2
X (s)
(b) Dynamic equations:
x&1 x& = 2 x&3
1 0 0 0 −10 −20
x1 0 1 x2 + 0 r −9 x3 1 0
y = [10
0
0] x
(c) State transition equation:
s −1 (1 + 9s −1 + 20 s −2 ) s −2 (1 + 9 s −1 ) s −3 x1 (0) s −3 X 1 (s ) −3 −1 −1 −2 X (s ) = 1 1 s −2 1 − 10 s s (1 + 9 s ) s x2 (0) + 2 ∆ (s ) ∆ ( s ) −1 s −2 −2 −1 s − 10 s − 20 s s x3 (0) X 3 ( s ) 1 2 s s + 9 s + 20 s+9 1 x1 (0) 1 1 = − 10 s( s + 9) s x2 (0) + 1 ∆ c (s ) ∆ c (s ) 2 s − 10 s − 10 ( 2 s + 1 ) s x3 (0) ∆( s ) = 1 + 9 s
−1
+ 20
s
−2
+ 10
s
−3
∆c( s) =
59
s
3
+ 9 s + 20 s + 10 2
1.612 0.946 0.114 −0.706 −1.117 −0.169 −0.708t x(t ) = −1.14 −0.669 −0.081 e + 1.692 2.678 4.056 e −2.397t 0.807 0.474 0.057 −4.056 −6.420 −0.972
0.0935 0.171 0.055 + −0.551 −1.009 − 0.325 e −5.895t x(0) 3.249 5.947 1.915
0.1 − 0.161e−0.708t + 0.0706 e−2.397t − 0.00935e−5.895 t −0.708 t −2.397 t −5.895 t + − 0.169e + 0.055e 0.114 e −0.087 e−0.708t + 0.406 e− 2.397t − 0.325e− 5.895t
t ≥0
(d) Output:
(
y ( t ) = 10 x1 (t ) = 10 1.612e
−0.708t
(
+ 10 1.141 e
− 0.706e
−0.708 t
−2.397t
− 0.169 e
+ 0.0935e
−2.397 t
−5.897 t
+ 0.0550 e
) x (0) + 10 ( 0.946e − 1.117e + 0.1711e ) x (0) ) x (0) + 1 −1.61e −0.708t + 0.706e −2.397 t − 0.0935e −5.895t −0.708 t
−2.397t
−5.895t
1
2
−5.895 t
3
5-27 (a) Closed-loop transfer function: Y (s)
10
=
R( s)
s
3
+ 9 s + 20 s + 10 2
(b) State diagram:
(c) State equations:
x&1 x& = 2 x&3
(d) State transition equations: [Same answers as Problem 5-26(d)]
1 0 0 0 −10 −20
x1 0 1 x2 + 0 r −9 x3 1 0
(e) Output:
5-28 (a) State diagram:
(b) State equations:
60
[Same answer as Problem 5-26(e)]
t≥0
x&1 x& 2 = x&3 x& 4
−2 20 −1 0 x1 0 −1 0 −10 1 0 x2 0 0 u + −0.1 0 −20 1 x3 0 0 T D 0 0 0 − 5 x4 30 0
(c) Transfer function relations: From the system block diagram, −0.2s 0.3 30 e U ( s) 90U( s) −1 T ( s ) + T ( s ) + + D D ∆( s) s + 2 ( s + 2)( s + 20) ( s + 2)( s + 5)( s + 20) (s + 5)( s + 20)
1
Y ( s) =
∆( s ) = 1 + Y ( s)
−0 . 2 s
+ 2 )( s + 20
(s
− ( s + 19 . 7 )
=
+ 2 )( s + 20 ) + 0 .1e
(s
−( s + 20
Ω( s ) =
−0 . 2 s
)
+ 2 )( s + 20 ) + 0 .1e
(s
5-29 (a)
0 . 1e
−0 . 2 s
=
+ 2 )( s + 20 ) + 0 .1 e
(s
(s
) T
D
30 e
(s)+
+ 5)
(s
T (s) D
+ 2 )( s + 20 −0 . 2 s
(s
(s
+ 5)
(s
)
+ 90(
s
+ 2 )U ( s )
+ 2 )( s + 20 ) + 0 . 1e 30 e
+
− 0. 2 s
−0 . 2 s
−0 . 2 s
U ( s)
+ 2 )( s + 20 ) + 0 .1e
There should not be any incoming branches to a state variable node other than the s
−1
−0 . 2 s
branch. Thus, we
should create a new node as shown in the following state diagram.
Notice that there is a loop with gain −1 after all the s
(b) State equations: dx
1
dt
=
17
x
2
1
+
1
x
2
dx 2
2
dt
=
15 2
x
−
1
1 2
x
2
+
1
r
R (s )
= 2.
Output equation:
Ks + 5 s + 1
(
( s + 1) s + 11s + 2 2
( s + 1) ( s
)
2
y
= 6 .5 x 1 + 0 .5 x 2
K = 1: Y (s) R( s)
=
s s
3
2
+ 12
+ 5s +1 s
2
+ 13 s + 2
State diagram:
61
)
+ 11s + 2 = 0
Roots of characteristic equation: −1, −0.185, −-10.82. These are not functions of K.
(c) When
∆
(b) Characteristic equation: 2
=
branches are deleted, so
2
5-30 (a) Transfer function: Y (s )
−1
(d) When
K = 4:
Y (s ) R (s )
4s + 5s + 1
(s + 1)(4 s + 1)
2
=
( s + 1) ( s
2
+ 11s + 2
=
) ( s + 1) (s
2
+ 11s + 2
)
=
4s + 1 s + 11s + 2 2
State diagram:
(e) Y (s) R( s)
=
Ks (s
2
+ 5s =1
+ 1 )( s + 0 .185)(
s
+ 10 .82
When K = 4, 2.1914, 0.4536, pole-zero cancellation occurs. )
5-31 (a) Gp ( s) =
Y ( s) U (s )
=
1
( 1 + 0.5s ) (1 + 0.2s + 0.02s
2
)
=
100 s + 12 s + 70 s + 100 3
2
State diagram by direct decomposition:
State equations:
x&1 x& = 2 x&3
0 0 −100
x1 0 0 1 x2 + 0 u −70 −12 x3 1 1
0
(b) Characteristic equation of closed-loop system:
Roots of characteristic equation:
62
s
3
+ 12
s
+ 70 s + 200 = 0
2
− 5. 88 ,
− 3 . 06 +
j 4. 965 ,
− 3 . 06 −
5-32 (a) Gp ( s) =
Y ( s) U (s )
≅
1 − 0.066 s
(1 + 0.5s ) (1 + 0.133s + 0.0067s
2
)
=
−20( s − 15) s + 22 s + 190 s + 300 3
2
State diagram by direct decomposition:
State equations:
x&1 x& = 2 x&3
0 0 −300
x1 0 0 1 x2 + 0 −190 −22 x3 1 1
0
Characteristic equation of closed-loop system: s
3
+ 22
s
2
+ 170
5-33 (a) State variables:
x
s
+ 600 = 0
1
= ωm
State equations: dω dt
m
=−
K K b
i
and x
−12, −5 + j5, −5 −j5
2
+ K bRa
JR
a
Roots of characteristic equation:
= ωD
+
K
D
J
ωD +
KK JR
i
e
a
(b) State diagram:
63
dω dt
D
=
K J
D R
ωm −
K J
D R
ωD
j 4. 965
(c) Open-loop transfer function: Ωm ( s)
=
E (s )
KK i ( J R s + K D )
JJ RR a s + ( K b J R Ki + K DR a J R + K D JRa ) s + K DK b Ki 2
Closed-loop transfer function:
Ωm ( s) Ω r ( s)
=
Ks KKi ( J R s + KD )
JJ R Ra s + ( Kb J R Ki + KD R a J R + KD JRa + Ks K Ki J R ) s + KD K b Ki + Ks K Ki KD 2
(d) Characteristic equation of closed-loop system: ∆ ( s ) = JJ R R a s + ( K D J R K i + K D R a J R + K D JRa + K s KK i J R ) s + K D K b K i + K s K Ki KD = 0 2
∆( s ) = Characteristic equation roots:
−19.8,
s
2
+ 1037
+ 20131
.2
=0
−1017.2
5-34 (a) State equations: x& ( t ) = Ax ( t ) + B r ( t ) A=
s
−b d −2 1 c − a = 2 − 1
B=
0 1
S = [B
AB ] =
0 1 1 −1
Since S is nonsingular, the system is controllable.
(b) S = [B
AB ] =
0 d 1 − a
AB
1 −1 1 A B = 1 −1 1 1 −1 1
The system is controllable for d
≠ 0.
5-35 (a) S = B
2
S is singular. The system is uncontrollable.
(b)
64
1 − 1 1 AB A B = 1 − 2 4 1 −3 9
S = B
2
S is nonsingular. The system is controllable.
5-36 (a) State equations: x& ( t ) = Ax ( t ) + B u ( t )
A=
−2 3 1 0
B=
Output equation:
V = C
'
'
AC
'
y
=
1 1
S = [B
= Cx
0 x
1
1 −2 = 0 3
1 1 1 1
C = 1
S is singular. The system is uncontrollable.
0
V is nonsingular. The system is observable.
(b) Transfer function: Y (s)
AB ] =
s
=
U ( s)
s
2
+3
+2s−3
1
= s
−1
Since there is pole-zero cancellation in the input-output transfer function, the system is either uncontrollable or unobservable or both. In this case, the state variables are already defined, and the system is uncontrollable as found out in part (a).
5-37 (a) α = 1, (b)
2 , or 4
.
These values of
α will cause pole-zero cancellation in the transfer function.
The transfer function is expanded by partial fraction expansion, Y (s) α −1 α R( s)
=
3( s
+ 1)
−
2( s
By parallel decomposition, the state equations are: x& ( t )
−1 0 0 A = 0 −2 0 0 −4 0 The system is uncontrollable for
(c)
α = 1,
or
−2 + 2)
+
α −4 6( s
+4)
= Ax ( t ) + B r ( t ) ,
α −1 B = α − 2 α − 4
output equation: y ( t )
D=
= C x ( t ).
1 − 1 3 2
1 6
α = 2, or α = 4.
Define the state variables so that
−1 0 0 A = 0 −2 0 0 0 − 4
The system is unobservable for
1 3 −1 B= 2 −1 6
α = 1, or α = 2, or α = 4.
65
D = [α − 1
α −2
α − 4]
5-38 S = [B
AB] =
b 1 b ab − 1
S = ab − 1 − b ≠ 0 2
The boundary of the region of controllability is described by ab
− 1 − b = 0. 2
Regions of controllability:
5-39 S = [B
AB] =
b1 b1 + b 2 b b 2 2
The system is completely controllable when b
V = C
'
'
AC
'
S = 0 when b1 b2 − b1 b2 − b2 = 0,or b2 = 0 2
≠ 0.
2
d2 d1 = d2 d 1 + d 2
The system is completely observable when d
≠ 0.
2
5-40 (a) State equations: dh 1 K nN K = ( qi − qo ) = I θm − o h dt A A A State variable:
x
1
= h,
x
=θm,
x
x& = Ax + B e
State equations:
−K o A A= 0 0
2
K I nN A 0 0
dθ m
=ω
dt 3
=
dθ dt
m
d ωm m
=−
dt
Ki Kb JRa
ωm +
Ki K a JRa
ei
J
= Jm +n
= ωm
i
1 = KK − i b JRa 0
V = 0 when d 1 ≠ 0 .
0 −1 0.016 0 0 1 0 0 −11.767
State diagram:
66
0 B = 0 = K K i a JRa
0 0 8333.33
2
J
L
(b) Characteristic equation of A: s+ sI − A =
Ko
− K I nN
A
A
0
s
0
0
Eigenvalues of A:
0,
0
s+
= s s +
−1 Ki K b
s + Ki K b = s ( s +1)( s + 11.767) A JRa
Ko
JRa
−1, −11.767.
(c) Controllability: S = B
0 133.33 0 A B = 0 8333.33 − 98058 8333.33 −98058 1153848 2
AB
S ≠ 0.
The system is controllable.
(d) Observability: (1) C =
1
0
V = C
(2) C =
0
(3) C =
0
'
1
V = C
'
'
'
'
'
'
−1 1 1 ( A ) C = 0 0.016 −0.016 0 0.016 0 ' 2
'
V is nonsingular. The system is observable.
:
AC
1
'
'
AC
0
0
V = C
:
0
0 0 0 ( A ) C = 1 0 0 0 1 −11.767 ' 2
'
V is singular. The system is unobservable.
:
AC
5-41 (a) Characteristic equation:
0 0 0 ( A ) C = 0 0 0 1 −11.767 138.46 ' 2
'
∆( s ) = s I − A
∗
Roots of characteristic equation: −5.0912,
=
s
4
− 25 . 92
s
5.0912, 0, 0
(b) Controllability:
67
2
=0
V is singular. The system is unobservable.
S = B
∗
∗
AB
∗
∗2
A B
− 0.0732 0 − 1.8973 0 −0.0732 0 −1.8973 0 ∗3 ∗ A B = 0 0.0976 0 0.1728 0.0976 0 0.1728 0
∗
∗
S is nonsingular. Thus, A ,
B
∗
is controllable.
(c) Observability: (1)
C
∗
=
1
0
V = C
∗'
0
0
∗'
A C
∗'
∗' 2
(A ) C
∗'
1 0 ∗' 3 ∗' ( A ) C = 0 0
0
25.92
1
0
0
0
0
0
25.92 0 0 0
S is singular. The system is unobservable. ∗
(2) C = 0
1
V = C
0
∗'
0
∗'
A C
∗'
∗'
2
(A ) C
∗'
0 671.85 0 25.92 1 0 25.92 0 ∗' 3 ∗' (A ) C = 0 0 0 0 0 0 0 0
S is singular. The system is unobservable.
∗
(3) C = 0
0
1
V = C
∗'
0
∗'
A C
∗'
∗'
2
(A ) C
∗'
0 0 ∗' 3 ∗' (A ) C = 1 0
S is nonsingular. The system is observable.
∗
(4) C = 0
0
0
1
68
0
− 2.36
0
0
0
0
1
0
− 2.36 0 0 0
V = C
∗'
∗'
A C
∗'
∗' 2
(A ) C
0 −61.17 0 −2.36 0 0 −2.36 0 ∗' 3 ∗' ( A ) C = 0 0 0 0 1 0 0 0
∗'
S is singular. The system is unobservable.
5-42
The controllability matrix is
0 − 384 0 −1 0 −16 −1 0 −16 0 −384 0 0 0 0 16 0 512 S= 0 512 0 0 0 16 0 1 0 0 0 0 0 0 0 1 0 0
Rank of S is 6. The system is controllable.
5-43 (a) Transfer function: Θv ( s ) R (s )
=
J vs
2
(J
KI H s + K Ps + K I + K N 2
G
)
State diagram by direct decomposition:
x& ( t ) = Ax ( t ) + B r ( t )
State equations:
0
1
0 0 A = 0 0 0 0
0 1 − KP J G
0
0
1 0 − ( KI + KN
(b) Characteristic equation:
JG
Jvs
2
(J
)
0 0 B= 0 1
)
s + K P s + K I + KN = 0 2
G
69
5-44 (a) State equations: x& ( t ) = Ax ( t ) + B u 1 ( t )
A=
−3 1 0 − 2
B=
S is nonsingular. A ,
0 1
S = [B
(b)
With feedback, u 2
= − kc 2 ,
'
0 1 1 −2
B is controllable.
Output equation: y 2 = C x
V = C
AB ] =
'
AC
'
C = −1
−1 3 = 1 −3
the state equation is: x& ( t )
−3 − 2 k A = 1+ g 0
=
V is singular. The system is unobservable.
Ax ( t )
1+ k − 2 1
1
+ B u1 ( t ) . 1 0 S= 1+k 1 −2
0 B= 1
S is nonsingular for all finite values of k. The system is controllable. State diagram:
y2 = Cx
Output equation:
V = D '
−1 1 + k −1 1 + K ' ' A D = 1 1 + k
1
C=
1 + k
(1 + k ) 3 + 2k − 2 (1 + k ) 3 + 2k
2
V is singular for any k. The system with feedback is unobservable.
5-45 (a) S = [B V = C (b) u = − k 1
1 2 2 − 7 1 − 1 ' ' A C = 1 −2
AB ] = '
S is nonsingular. System is controllable.
V is nonsingular. System is observable.
k2 x
0 1 k1 − −1 −3 2 k1
A c = A − BK =
70
1 − k2 −k1 = 2 k2 − 1 − 2 k1 − 3 − 2 k 2 k2
S = [B
A cB ] =
For controllabillity, k 2
V = C For observability, V
'
≠−
1 − k1 − 2 k2 + 2 2 − 7 − 2k − 4k 1 2
11 2 '
A cC
'
− 1 − 1 − 3 k1 = 1 −2 − 3 k 2
= −1 + 3k 1 − 3 k 2 ≠ 0
71
S = − 11 − 2 k2 ≠ 0
Chapter 6 6-1 (a)
STABILITY OF LINEAR CONTROL SYSTEMS = 0 , − 1. 5 +
Poles are at s
j 1. 6583 ,
− 1. 5 −
One poles at s = 0. Marginally stable .
j 1. 6583
(b) Poles are at s = − 5, − j 2 , j 2 (c) Poles are at s = − 0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344 − j 2 . 3593 (d) Poles are at s = − 5, − 1 + j , − 1 − j (e) Poles are at s = − 1.3387 , 1. 6634 + j 2 . 164, 1. 6634 − j 2 .164 (f) Poles are at s = − 22 . 8487 ± j 22 . 6376 , 21 . 3487 ± j 22 . 6023 6-2 (a)
s
+ 25
3
s
+ 10 s + 450 = 0
2
Two poles on j ω axis. Marginally stabl e . Two poles in RHP. Unstable . All poles in the LHP. Stable . Two poles in RHP. Unstable . Two poles in RHP. Unstable .
Roots: − 25 . 31 , 0 .1537 + j 4.214, 0 .1537 − 4.214
Routh Tabulation: s
3
s
2
s
1
1
10
25
450
− 450
250
Two sign changes in the first column. Two roots in RHP.
= −8
0
25
(b)
s
s
0
3
+ 25
450
+ 10 s + 50 = 0
2
s
Roots: − 24. 6769 , − 0 .1616 + j 1.4142 , − 0 .1616 − j 1.4142
Routh Tabulation: s
3
s
2
s
1
s
0
1
10
25
50
− 50
250
No sign changes in the first column. No roots in RHP.
=8
0
25
(c)
s
50
+ 25
3
s
2
+ 250
s
+ 10 = 0
Roots: − 0 . 0402 , − 12 .48 + j 9 . 6566 , − j 9 . 6566
Routh Tabulation: s
3
s
2
s
1
s
0
1
250
25
10
− 10
6250
= 249
No sign changes in the first column. No roots in RHP. .6
0
25
(d)
2s
4
10
+ 10
s
3
+ 5 . 5 s + 5 . 5 s + 10 = 0 2
Roots: − 4.466 , − 1.116 , 0 .2888 + j 0 . 9611 , 0 .2888 − j 0 . 9611
Routh Tabulation:
71
s
4
s
3
s
2
55
1
10 24.2
s
2
5 .5
10
5 .5
− 11
= 4.4
− 100
10
10
= − 75 . 8
4.4 s
0
10
Two sign changes in the first column. Two roots in RHP.
(e)
s
6
+ 2 s + 8 s + 15 5
4
s
3
+ 20
s
2
+ 16 s + 16 = 0
Roots: − 1.222 ± j 0 .8169 , 0 . 0447 ± j 1.153 , 0 .1776 ± j 2 .352
Routh Tabulation: s
6
s
5
s
4
s
3
s
2
s
1
16
1
8
20
2
15
16
− 15
40
= 0 .5
− 16
2
2
− 33
− 48
−396 + 24
= 11 .27
−33 − 541 .1 + 528
= −1.16
16
= 12
16
0
11 .27 s
0
0
Four sign changes in the first column. Four roots in RHP.
(f)
s
4
+ 2 s + 10 3
s
2
+ 20 s + 5 = 0
Roots: −0 .29 , − 1. 788 , 0 . 039 + j 3 .105 , 0 . 039 − j 3 .105
Routh Tabulation: s
4
s
3
s
2
20
1
10
2
20
− 20
=0
5
5
2
6-3 (a)
s
4
s
2
s
1
s
0
ε 20
5
ε − 10 ε
≅−
Replac e 0 in last row by
10
ε
Two sign changes in first column. Two roots in RHP.
5
+ 25 s + 15 3
s
2
ε
+ 20 s + K = 0
Routh Tabulation:
72
s
4
s
3
s
2
1
K
15 25
20
− 20
375
= 14.2
K
25 s
− 25
284
1
K
= 20 − 1. 76
K
20
− 1. 76
K
>0
>0
K
or K
< 11 . 36
14.2 s
0
K
Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec.
(b)
s
2
+ 11 . 36 = 0 .
The solution of A(s) = 0 is s
2
= − 0 .8 .
The
+ Ks + 2 s + ( K + 1) s + 10 = 0
4
3
2
Routh Tabulation: s
4
s
3
s
2
1
2
K
K
− K −1
2K
K
=
K
−1
10
+1
>0
K K
10
K
−9 K − 1
>1
2
s
1
s
0
−9 K −1 > 0 2
−1
K 10
The conditions for stability are: K > 0, K > 1, and − 9 K − 1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K. Thus, the system is unstable for all values of K. 2
(c)
s
2
+ ( K + 2 ) s + 2 Ks + 10 = 0
3
2
Routh Tabulation: s
3
s
2
s
1
K 2K
1
2K
+2
10
2
+ 4 K − 10 K
s
0
+2
K
> −2
K
2
+ 2 K −5 > 0
10
The conditions for stability are: K > −2 and K + 2 K − 5 > 0 or (K +3.4495)(K − 1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is 2
marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec.
(d)
s
3
+ 20
s
2
+ 5 s + 10
K
= 3.4495
s
=0
Routh Tabulation:
73
2
+ 10 = 0 .
The solution is s
2
= −2 .899
.
s
3
s
2
s
1
1
5 10 K
20
− 10
100
K
= 5 − 0 .5 K
5
− 0 .5 K > 0
K
>0
or K
< 10
20 s
0
10 K
The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is
= 20
marginally stable. The auxiliary equation is A ( s ) equation is s
(e)
s
= −5 .
2
3
2
+ 100 = 0 .
The solution of the auxiliary
The frequency of oscillation is 2.236 rad/sec.
+ Ks + 5 s + 10 s + 10
4
s
2
K
=0
Routh Tabulation: s
4
s
3
s
2
5
K
10
K
10 K
5K
− 10
5K
10 K
1
>0 − 10 > 0
K
− 100
50 K s
K 5K
1
− 10
K
>2
2
=
− 10
or K
50 K
− 100 − 10 5K
K
3
5K
− 10
− 10 − K > 0 3
K s
0
10 K
K
The conditions for stability are: K > 0, K > 2, and 5 K
β K
+ 2 . 9055
γε
K
2
− 2 . 9055
K
+ 3 .4419
ϕ
2 and K < is unstable for all values of K.
(f)
s
4
− 10 − K > 0 . 3
>0 The last condition is written as
The second-order term is positive for all values of K.
−2.9055.
Since these are contradictory, the system
+ 12 . 5 s + s + 5 s + K = 0 3
2
Routh Tabulation: s
4
s
3
s
2
1
1
12 . 5
5
−5
12 . 5
= 0 .6
K
K
12 . 5 s
1
3
− 12 . 5 K
= 5 − 20 .83
K
5
− 20 . 83 K > 0
or K
< 0 .24
0 .6 s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary 0
equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec.
2
6-4
The characteristic equation is Ts
3
+ 0 .24 = 0 .
The solution of the auxiliary equation is s
+ ( 2T +1) s + ( 2 + K ) s + 5 K = 0 2
Routh Tabulation:
74
2
= − 0 .4.
The frequency of
s
3
T
s
2
2T
s
1
( 2T
+1
0
T
>0
5K
T
> −1 /
+ 1 )( K + 2 ) − 5 KT 2T
s
+2
K
K (1 − 3 T )
+1
5K
T > 0, K > 0, and K
4T
+2
3T
−1
+ 24
Ks
0
>0
K
The conditions for stability are:
2
. The regions of stability in the
+ 80
K
=0
Routh Tabulation: s
5
s
4
s
3
3
s
2
214080
1
50000
24 K
600
K
80 K
× 10
7
−K
14320 K
600
3 s
1
0
< 3 × 10
7
600 00 K
−K
2
80 K
× 10 − K 7
− 7 .2 × 10
+ 3 .113256 × 10
16
600( 214080 s
K
00
11
− 14400
K
−K
K
< 214080
K
00
2
K
2
− 2 .162 × 10
)
80 K
K
Conditions for stability:
75
>0
7
K
+ 5 × 10
12
0
row:
Thus, the final condition for stability is: When K
5
(K
or
+ ( K + 2 ) s + 30 2
Ks
+ 200
=0
K
Routh tabulation: s
3
s
2
s
1
s
0
30 K
1
+2
K
2
− 140
K
+2
30 K
200 K K
200 K
Stability Condition:
Characteristic equation: s
3
> −2
K
> 4. 6667
K
>0
K > 4.6667
When K = 4.6667, the auxiliary equation is A ( s ) The frequency of oscillation is 11.832 rad/sec.
(c)
K
+ 30
s
+ 200
2
= 6 . 6667
s
2
+ 933
. 333
=0.
The solution is s
2
+K =0
s
Routh tabulation: s
3
s
2
s
1
1
200
30
K
−K
6000
K
< 6000
K
>0
30 s
0
K
Stabililty Condition:
0
−5
K
> −1
30 s
0
K +1
Stability condition:
K>
−1.
When K =
−1 the zero element occurs in the first element of the
76
= − 140
.
0
= s + ( k 2 − 1 ) s + 20 − 2 k1 − k 2 = 0 2
or k
2
>1
− 2 k1 − k 2 > 0
or
k
< 20 − 2 k 1
2
Parameter plane:
6-7 Characteristic equation of closed-loop system: s −1 0 sI − A + BK = 0 k1
= s + ( k 3 + 3 ) s + ( k 2 + 4 ) s + k1 = 0
s
−1
k2 + 4
s + k3 + 3
3
2
Routh Tabulation:
s
3
s
2
s
1
s
k2 + 4
1 k3 + 3
(k
3
k3 +3>0 or k3 > − 3
k1
+ 3) ( k 2 + 4 ) − k 1
(k
k3 + 3 0
+ 3 )( k + 4) − k > 0 2
1
k >0
k
1
1
Stability Requirements:
k 3 > − 3, 6-8 (a)
3
(k
k 1 > 0,
3
+ 3) ( k 2 + 4 ) − k 1 > 0
Since A is a diagonal matrix with distinct eigenvalues, the states are decoupled from each other. The second row of B is zero; thus, the second state variable, x is uncontrollable. Since the uncontrollable 2
77
state has the eigenvalue at −3 which is stable, and the unstable state x with the eigenvalue at −2 is 3
controllable, the system is stabilizable.
(b) 6-9
Since the uncontrollable state x has an unstable eigenvalue at 1, the system is no stabilizable. 1
The closed-loop transfer function of the sysetm is
Y (s)
=
R (s )
1000 s + 15.6 s + ( 56 + 100 K t ) s + 1000 3
s
The characteristic equation is:
2
3
+ 15 . 6 s + ( 56 + 100 2
K )s t
+ 1000 = 0
Routh Tabulation: s
3
s
2
s
1
1 15 . 6 873 . 6
+ 100
56
+ 1560
K
t
1000 Kt
− 1000
1560 K t
15 .6 s
0
.4
>0
1000
Stability Requirements:
6-10
− 126
K
t
> 0 . 081
The closed-loop transfer function is Y (s) R( s)
=
K(s s
3
s
The characteristic equation:
+ Ks 3
+ Ks
+ 2 )( s + α )
+ ( 2 K + αK − 1 ) s + 2 αK
2
2
+ ( 2 K + α K − 1) s + 2 αK = 0
Routh Tabulation: s
3
s
2
s
1
2 αK
K (2
+ αK − 1
2K
1
K
+ α ) K − K − 2 αK
>0
2
(2 + α )K
K s
0
2 αK
Stability Requirements: α > 0 , K -versus- α Parameter Plane:
α >0 K
> 0,
K
>
1
+ 2α
2
+α
78
.
− 1 − 2α > 0
6-11 (a)
Only the attitude sensor loop is in operation: K
Θ( s) Θr ( s ) If KK
s
If KK
s
t
= 0. The system transfer function is:
G (s) p
=
1+ K G ( s) s
=
p
K s
2
− α + KK
s
>α,
the characteristic equation roots are on the imaginary axis, and the missible will oscillate.
≤α,
the characteristic equaton roots are at the origin or in the right-half plane, and the system
is unstable. The missile will tumble end over end.
(b)
Both loops are in operation: The system transfer function is
Θ( s) Θr ( s ) KK
For stability: When K
=0
t
G (s) p
1 + K sG t
> 0,
t
and KK
=
KK
>α,
s
(s)
p
s
+ K sG p ( s )
=
K s
2
+ KK
t
s
+ KK s − α
−α > 0 .
the characteristic equation roots are on the imaginary axis, and the missile
will oscillate back and forth. For any KK − α if KK < 0, the characteristic equation roots are in the right-half plane, and the system s
If KK
t
t
is unstable. The missile will tumble end over end. > 0 , and KK < α , the characteristic equation roots are in the right-half plane, and the system is t
unstable. The missile will tumble end over end.
6-12
Let s
1
= s + α,
then when s
= −α,
s
1
= 0.
This transforms the
s = −α axis in the s-plane onto the imaginary
axis of the s -plane. 1
(a)
F (s)
=
Or
s
+ 5s +3 = 0
s
2
2
+ 3 s1 − 1 = 0
1
Le t s
= s1 − 1
2
s1
Routh Tabulation:
s s
We get
1
−1) + 5 ( s1 −1) + 3 = 0 2
−1
1
1
(s
3
1
−1
0 1
Since there is one sign change in the first column of the Routh tabulation, there is one root in the region to the right of s = −1 in the s-plane. The roots are at −3.3028 and 0.3028.
(b)
F (s)
=
s
3
+ 3s + 3s +1 = 0 2
Let s
=
s
1
−1
We get
79
( s1 −1)3
+ 3( s1 −1) + 3 ( s1 −1) + 1 = 0 2
(c)
Or
s
F (s)
=
= 0.
3 1
s
Or
3
s
3 1
The three roots in the s -plane are all at s 1
+ 4 s + 3 s + 10 = 0 2
=
Let s
s
1
1
= 0.
Thus, F(s) has three roots at s =
We get
1
−2
1
10
(s
−1
1
−1.
−1) + 4 ( s1 −1) + 3 ( s1 −1) +10 = 0 3
2
+ s 1 − 2 s1 + 10 = 0 2
s s
Routh Tabulation:
3 1 2 1
− 12
1
s1 s
0
10
1
Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.8897, −0.0552 + j1.605, and −0.0552
(d)
F (s) Or
=
s
+4s +4s +4 = 0
3
s
− j1.6025. 2
3 1
Let s
=
s
1
−1
(s
We get
1
−1) + 4 ( s1 −1) + 4 ( s1 −1) + 4 = 0 3
2
+ s1 − s1 + 3 = 0 2
s s
Routh Tabulation:
3 1 2 1 1
s1 s
1
−1
1
3
−4
0
3
1
Since there are two sign changes in the first column of the Routh tabulation, F(s) has two roots in the region to the right of s = −1 in the s-plane. The roots are at −3.1304, −0.4348 + j1.0434, and −0.4348
−j1.04348.
6-13 (a) Block diagram:
(b) Open-loop transfer function: G ( s) =
H ( s)
=
E (s )
K a K i nK I N
s (R a Js + K i K b ) ( As + K o )
=
16.667 N s( s + 1)( s + 11.767)
Closed-loop transfer function: H (s) R( s)
G (s)
= 1
+G ( s)
16 .667 N
= s
3
+ 12 . 767
(c) Characteristic equation:
80
s
2
+ 11 . 767
s
+ 16 . 667
N
s
+ 12 . 767
3
2
s
+ 11 . 767
s
+ 16 . 667
=0
N
Routh Tabulation: s
3
s
2
s
1
1
11 . 767 16 . 667 N
12 .767 150 .22
− 16 . 667
N
− 16.667
1 50.22
N
>0
or
N
NA
>0
12 . 767 s
0
N
16 .667N
Stability condition:
0
0
0
0 . 706 A
+3 > 0
24.92 A
+ 105
.9
− 15
N > 0
250N
From the s row,
(b)
+3
.9
A
35 .3
N < 1.66 + 7.06/A
When A
→∞
N
max
→ 1. 66
For A = 50, the characteristic equation is
3 s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 250 N = 0 3
2
Routh tabulation
81
Thus, N
max
= 1.
− 588
o
. 33
2
N < 0.0000 5653 K o
K
250 N
N
+ 0 . 03323
>0
N = 10, A = 50. The characteristic equation is
s + ( 35.3 + 0.06 K o ) s + 0.706 Ko s + 50 KI = 0 3
2
Routh Tabulution: s
3
s
2
s
1
0 . 706 K
1 35 . 3
+ 0 . 06 2
0 . 04236 K o
0
50 K
6-15 (a) Block diagram:
50 K
o
+ 24. 92
Ko
+ 0 . 06
Ko
35 .3 s
K
− 50
o
K
I
KI
I
o
> −588
. 33
KI
< 0 . 000847
K
>0
I
2
2Ko
+ 0 .498
(b) Characteristic equation: 2 Ms + K s + K + K = 0 D s P 500 s
82
2
+ K D s + 500 + K P = 0
Ko
K
o
(c)
For stability, K
D
> 0,
0 .5
+ K P > 0.
Thu s,
K
P
> − 0 .5
Stability Region:
6-16 State diagram:
∆ = 1 + s + Ks
2
Characteristic equation: s
Stability requirement:
2
+s+K =0
K>0
83
Chapter 7
TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS
7-1 (a) ζ ≥ 0 . 707
(c) ζ ≤ 0 . 5
7-2 (a)
Type 0
7-3 (a)
K
(b)
K
(c)
K
(d)
K
(e)
K
(f)
K
ωn ≥ 2
1
rad / sec
≤ ω n ≤ 5 rad
(b)
/ sec
Type 0
(c)
(d)
Type 1
Type 2
≤ ζ ≤ 0 . 707
(b)
0
(d)
0 .5
(e)
Type 3
ωn ≤ 2
≤ ζ ≤ 0 . 707
(f)
ω n ≤ 0 . 5 rad
Type 3
= lim
G ( s)
= 1000
Kv
= lim
sG ( s )
=0
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=1
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=K
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=∞
Ka
= lim
s G ( s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=1
Ka
= lim
s G (s)
p
= lim
G ( s)
=∞
Kv
= lim
sG ( s )
=∞
Ka
= lim
s G (s)
p
s→0
s→0
s→0
s→0
s→0
s→0
s →0
s →0
s→0
s →0
s →0 s →0
84
s→0
s→0
s→0
s→0
s→0 s→0
rad / sec
2
=0
2
=0
2
=0
2
=1
2
=0
2
=
K
/ sec
7-4 (a)
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
(b)
p
= 1000
v
=0
a
=0
1 1001
∞ ∞
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
p
=∞
0
v
=1
1
a
=0
∞
(c) Input
Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
p
=∞
0
v
=K
1/ K
a
=0
∞
The above results are valid if the value of K corresponds to a stable closed-loop system.
(d)
The closed-loop system is unstable. It is meaningless to conduct a steady-state error analysis.
(e)
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
(f)
p
=∞
0
v
=1
1
a
=0
∞
Input Error Constants Steady -state Error ________________________________________________________________________________ u (t )
K
tu ( t )
K
2
K
s
s
t u (t ) / 2 s
p
=∞
0
v
=∞
0
a
=K
1/ K
The closed-loop system is stable for all positive values of K. Thus the above results are valid.
7-5 (a)
K
H
=
H (0)
=1
M (s) a
0
=
= 3,
G (s) 1+ G ( s)H ( s) a
1
= 3,
a
2
s
=
=
3
s
+1
+ 2 s + 3s +3 = 1, b 1 = 1. 0 2
b
2,
Unit-step Input:
ess = Unit-ramp input: a
(b)
K
H
=
H (0)
2 =3
0
− b0 K H = 3 − 1 = 2 ≠ 0.
0
− b0 K H = 2 ≠ 0
Unit-parabolic Input: a
b0 KH 1 − a KH 0 1
and
a
Thus
1
e
ss
= ∞.
− b1 K H = 1 ≠ 0.
Thus
e
ss
= ∞.
=5 M (s)
=
G (s) 1+ G ( s)H ( s)
=
1 s
2
+5s + 5
85
a
0
= 5,
a
1
= 5,
b
0
= 1,
b
1
= 0.
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
ss
=
a a
− b1 K H
1
e
(c)
K
H
=
H (0)
ss
1
5 1 − = 0 5 5
− b0 K H = 0
0
a K
5
=
i
= 1:
s
+5
a
1
− b1 K H = 5 ≠ 0
1
=
25
H
0
Unit-parabolic Input:
=
1
5
=∞
=1/5 M (s)
=
G (s) 1+ G ( s)H ( s) a
0
= 1,
a
=
+ 15 s + 50 s + s + 1 a = 50 , a = 15 , b = 5 , 2 3 0
s
= 1,
1
4
3
2
The system is stable. b
1
=1
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
ss
=
a a
− b0 K H = 0
0
− b1 K H
1
a K
e
(d)
K
H
=
H (0)
ss
=
= 1:
i
1−1 / 5
a
1
− b1 K H =
4 /5
≠0
=4
1/5
H
0
Unit-parabolic Input:
5/5 = 5 1 − 1 = 0
1
=∞
= 10 M (s)
G (s)
=
1+ G ( s)H ( s)
a
0
= 10 ,
a
=
= 5,
1
1 s
+ 12 s + 5 s + 10 a = 12 , b = 1, 2 0 3
The system is stable.
2
b
= 0,
1
b
=0
2
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
ss
=
a a
1
− b0 K H = 0
0
− b1 K H a K 0
Unit-parabolic Input: e
7-6 (a)
M (s)
=
s s
4
+ 16
s
3
+4
ss
=
5
i
= 1:
a
1
− b1 K H = 5 ≠ 0
= 0 . 05
100
H
=∞
+ 48 s + 4 s + 4 a = 4, a = 4, 0 1 2
1 10 = 10 1 − 10 = 0
1
K a
=1
H
2
= 48 ,
The system is stable. a
3
= 16 ,
b
0
= 4,
b
1
= 1,
b
2
= 0,
Unit-step Input:
ess = Unit-ramp input: i
= 0:
b0K H 1 − a KH 0 1
a
0
4 = 1 − 4 = 0
−b0 KH = 0
i
= 1:
86
a
1
− b1 K H = 4 − 1 = 3 ≠ 0
b
3
=0
e
=
ss
a
− b1 K H
1
a K 0
Unit-parabolic Input: e
(b)
M (s)
s
3
3
=
4
H
4
=∞
+ 3)
K(s
=
ss
−1
4
=
K
+ 3s + ( K + 2)s + 3K a = 3K , a = K + 2, 0 1 2
a
=1
H
The system is stable for K
= 3,
2
b
0
=3K,
> 0.
=K
b
1
Unit-step Input:
ess = Unit-ramp Input:
b0K H 1 − a KH 0
= 0:
i e
=
ss
3K = 1 − 3 K = 0
1
a a
0
− b0 K H = 0
− b1 K H
1
a K 0
Unit-parabolic Input: e
ss
+2 −K
K
=
= 1:
i
=
3K
H
a
1
− b1 K H = K + 2 − K = 2 ≠ 0
2 3K
=∞
The above results are valid for K > 0.
(c)
M (s)
=
s s
4
+ 15
s
3
+5 + 50 s + 10 s a = 0 , a = 10 , 0 1 2
a
2
10 s
=
H ( s)
s+5 50 , a
=
K
3
H
= lim
H (s)
s →0
= 15 ,
b
0
= 5,
s b
1
=2 =1
Unit-step Input:
ess =
a 2 − b1K H a KH 1
Unit-ramp Input: e
ss
=∞
ss
=∞
Unit-parabolic Input: e
(d)
M (s)
=
s
4
+ 17
s
3
K(s
+ 5)
+ 60
s
a
2
K
+ 5 Ks + 5 K
= 5K ,
0
1 50 − 1 × 2 = 2 10 = 2.4
1
a
1
= 5K ,
a
2
H
=1
The system is stable for 0 < K < 204.
= 60 ,
a
3
= 17 ,
b
0
= 5K,
b
1
=
K
Unit-step Input:
ess = Unit-ramp Input: i
= 0:
e
=
ss
b0K H 1 − a KH 0 1
a a
1
0
−b0 KH = 0
− b1 K H a K 0
5K = 1 − 5 K = 0
=
H
Unit-parabolic Input: e
ss
5K
i
−K
5K
= 1:
=
4 5
=∞
The results are valid for 0 < K < 204.
87
a
1
− b1 K H = 5 K − K = 4 K ≠ 0
7-7 G (s)
=
Y (s)
=
E ( s)
KG 1
(b) (c)
r(t )
r(t )
r(t )
K
= u s ( t ):
e
= tu s ( t ): =t
2
e
u ( t ) / 2:
e
s
ss
=
=
+ K tG p( s)
Error constants:
(a)
( s ) 20 s
p
1
p
= ∞,
=
ss
=
ss
K
1 1+ K 1
20 s ( 1 + 0 .2 s
=
5K 1 + 100 K
+ 100 K
,
Type-1 system. K ) t
a
=0
t
=0 p
1 + 100 K
=
K
v
100 K
t
5K
v
=∞
K
a
7-8 G p (s)
G (s)
100
= (1
+ 0 .1 s )( 1 + 0 . 5 s )
(b) (c)
r(t )
r(t )
K
= u s ( t ):
e
= tu s ( t ): =t
2
KG
=
E (s)
+ 0 .1 s )( 1 + 0 . 5 s ) + 100
Error constants:
r(t )
Y (s)
=
20 s 1
p
(s)
+ K tG p ( s )
100 K
= 20 s ( 1
(a)
G (s)
e
u ( t ) / 2:
e
s
ss
=
1 K
p
ss
ss
= ∞,
= =
K
1 1+ K 1 K
=
5K 1 + 100 K
,
K
a
=0
t
=0 p
= v
v
Kt
1 + 100 K
t
5K
=∞ a
Since the system is of the third order, the values of K and K
must be constrained so that the system is
t
stable. The characteristic equation is
s + 12 s + ( 20 + 2000 K t ) s + 100 K = 0 3
2
Routh Tabulation: s
3
s
2
s
1
1
20
+ 24000
K
t
100 K
12 240
+ 2000
Kt
− 100
K
12 s
0
100 K
Stability Conditions:
K>0
12 ( 1+ 100 K t ) − 5 K > 0 or
1 + 100 K t
Thus, the minimum steady-state error that can be obtained with a unit-ramp input is 1/12.
88
5K
>
1 12
7-9 (a)
From Figure 3P-19,
Θ o ( s)
=
Θr ( s )
Θo ( s) Θr ( s )
=
K1 K 2
1+ 1+
K1 K 2 Ra + La s
Ra + La s
+
K i K b + KK1 K i K t
(R
a
+ La s ) ( Bt + J t s )
K i K b + KK1 K i K t
+
(R
a
+ La s ) ( Bt + J t s )
+
KK s K 1K i N
s ( Ra + La s )( Bt + J t s )
s [( Ra + La s ) ( Bt + Jt s ) + K1 K2 ( Bt + Jt s ) + Ki Kb + KK1 K i K t ]
L a J t s + ( L a Bt + Ra J t + K 1 K 2 J t ) s + ( Ra Bt + K i K b + K Ki K1K t + K 1 K 2 Bt ) s + KK s K 1K i N 3
2
θ r ( t ) = u s ( t ),
Θr ( s ) =
1
lim s Θ ( s ) e
s →0
s
=0
Provided that all the poles of s Θ ( s ) are all in the left -half s-plane. e
(b)
For a unit-ramp input,
Θr ( s ) = 1 / e
ss
=
lim
t →∞
2
s .
θ e (t ) =
lim s Θ ( s ) e
s →0
R B
=
a
t
+ K 1 K 2 B t + K i K b + KK KK
s
1
K K i
t
K K N i
1
if the limit is valid.
7-10 (a) Forward-path transfer function:
[n(t) = 0]:
K (1 + 0.02s ) G ( s) =
Y ( s)
=
E (s )
Error Constants:
K
For a unit-ramp input, r ( t )
2 K (1 + 0.02s ) s ( s + 25) = 2 K Kt s s s + 25 s + KKt 1+ 2 s ( s + 25)
(
= ∞,
p
= tu s ( t ),
K
R( s)
=
1
=
v
1 s
2
K ,
K
,
a
)
Type-1 system.
=0
t
e
ss
=
lim e ( t )
t→ ∞
= lim
s →0
sE ( s )
=
1 K
=
K
t
v
Routh Tabulation:
s s
3
2
1
s
1
KK t + 0.02 K
25
K
25 K ( Kt + 0.02) − K 25
s
0
K K >0
Stability Conditions:
(b)
With r(t) = 0, n ( t )
= u s ( t ),
N (s)
=1/
25 ( Kt + 0.02 ) − K > 0 or K t > 0.02
s.
System Transfer Function with N( s) as Input:
K Y (s ) N (s )
K s ( s + 25) = 3 2 K (1 + 0.02 s) K Kt s s + 25 s + K ( K t + 0.02 ) s + K 1+ 2 + 2 s ( s + 25) s ( s + 25) 2
=
89
Steady -State O utput due to n ( t):
=
y ss
7-11 (a)
n(t )
= 0,
r (t )
lim y ( t )
t →∞
= lim
=1
sY ( s )
s→0
if the limit is valid.
= tu s ( t ).
Forward-path Transfer function:
G ( s) =
Y ( s) E (s )
K ( s + α )( s + 3)
=
(
2
Ramp-error constant:
Kv
= lim
Steady -state error:
e
=
Characteristic equation: s + Ks Routh Tabulation: 3
s
3
s
2
s
1
s
0
ss
3K
s →0
1 K
sG ( s )
=− v
= −3 K α
1 3K
v
+ αK − 1 3 αK
K K (3K
Type-1 system.
+ [ K ( 3 + α ) − 1] s + 3α K = 0
2
1
)
s s −1
n =0
+ αK − 1 ) − 3 αK K 3α K
+ αK − 1 − 3α > 0
3K
Stability Conditions:
K
or
1+ 3K
>
3
+α
αK > 0 (b)
When r(t) = 0, n ( t )
= u s ( t ),
N (s)
=1/
s.
K ( s + 3) Y (s )
Transfer Function between n ( t) and y( t):
N (s ) Steady -State Output due to n ( t):
=
y ss
lim y ( t )
t →∞
= 1+
r =0
= lim
s→0
2 Ks ( s + 3) s −1 = K ( s + α )( s + 3) s 3 + Ks 2 + [ K ( s + α ) − 1]s + 3α K
(
s s −1 2
)
=0
sY ( s )
7-12
if the limit is valid.
− πζ
Percen t maxi mum ov
=e
ershoo t = 0 .25
1 −ζ
2
Thus
πζ Solving for
ζ
Peak T ime
1−ζ
2
= − ln0.25 = 1.386
from the last equation, we have t
max
=
π ωn
1−ζ
ζ
= 0 .01
(
π ζ = 1.922 1 − ζ 2
2
2
)
= 0.404. sec.
π
ωn =
Thus,
2
1 − ( 0 .404 )
0 . 01
Transfer Function of the Second-order Prototype System: Y (s) R( s)
ωn 2
=
s
2
+ 2ζω n s + ω n 2
7-13 Closed-Loop Transfer Function:
=
117916 s
2
+ 277
.3 s
+ 117916
Characteristic equation:
90
= 343 2
.4
rad / sec
Y (s )
=
R (s )
25 K
s + ( 5 + 500 Kt ) s + 25 K = 0 2
s + ( 5 + 500 Kt ) s + 25 K 2
For a second-order prototype system, when the maximum overshoot is 4.3%,
ωn =
2 ζω
25 K ,
n
= 5 + 500
K
ζ = 0 . 707
= 1.414
t
.
25 K
Rise Time: [Eq. (7-104)] t
r
=
1 − 0 .4167
ζ + 2 . 917 ζ
K K
With
= =
=
ωn ωn 2
Thus,
2
( 10 . 82 )
=
25 4. 68
K
= 0 .2
ωn
sec
ω n = 10 . 82
Thu s
rad / sec
2
= 4. 68
25 and
2 .164
t
5
= 0 . 0206
,
+ 500
K
t
= 1.414 ω n = 15 . 3
the sy stem t ransfe
Y (s)
t
=
10 . 3
= 0 . 0206
500
r func tion i s
117
=
R( s)
K
Thus
s
2
+ 15 . 3 s + 117
Unit-step Response: y = 0.1 at t = 0.047 sec. y = 0.9 at t = 0.244 sec. t = 0 .244 − 0 . 047 = 0 .197 r
y
7-14 Closed-loop Transfer Function: Y (s )
=
R (s )
25 K
2
πζ 1−ζ
, we get
ζ
( 4. 32% max.
s + ( 5 + 500 Kt ) s + 25 K = 0
s + ( 5 + 500 Kt ) s + 25 K
ζ
= 0 . 0432
Characteristic Equation:
2
When Maximum overshoot = 10%,
Solving for
max
= − ln0.1 = 2.3
(
π ζ = 5.3 1 − ζ 2
2
2
2
)
= 0.59.
The Natural undamped frequency is
ωn =
25 K
Thus,
5 + 500 K
t
= 2 ζω n = 1.18 ω n
Rise Time: [Eq. (7-114)] r
=
K
=
t
1 − 0 .4167
ζ + 2 . 917 ζ ωn
ωn
2
= 0 .1 =
1. 7696 sec.
ωn
Th us
ω n = 17 . 7
rad / sec
2
= 12 . 58
25 With K = 12.58 and K
t
=
sec.
K
t
=
15 . 88
= 0 . 0318 500 0 . 0318 , the system transfer function is Thus
Y (s) R( s)
313
= s
2
+ 20 .88 s + 314.
5
Unit-step Response: y = 0.1 when t = 0.028 sec. y = 0.9 when t = 0.131 sec.
91
overs hoot)
t
y
7-15
= 0 . 131 − 0 . 028 = 0 .103
r
= 1.1
max
( 10%
sec.
max.
overs hoot )
Closed-Loop Transfer Function:
Characteristic Equation:
Y (s)
s + ( 5 + 500 Kt ) s + 25 K = 0
R( s)
=
25 K s
2
+ ( 5 + 500
2
+ 25 K
K )s t
πζ
When Maximum overshoot = 20%,
1−ζ Solving for
ζ
, we get
ζ = 0 .456
The Natural undamped frequency
= − ln0.2 = 1.61
(
π ζ = 2.59 1 − ζ 2
2
2
2
)
.
ωn =
25 K
5
+ 500
K
t
= 2 ζω n = 0 . 912 ω n
Rise Time: [Eq. (7-114)] t
r
=
1 − 0 .4167
ζ + 2 . 917 ζ ωn
2
= 0 . 05 =
1.4165 sec.
ωn
Thus,
ωn
ωn =
1.4165
=
28 .33
0 . 05
2
= 32 .1 5 + 500 K = 0 . 912 ω = 25 . 84 Thus, t n 25 With K = 32.1 and K = 0 . 0417 , the system transfer function is K
=
K
t
= 0 . 0417
t
Y (s) R( s)
802 . 59
=
s
2
+ 25 . 84 s + 802
. 59
Unit-step Response: y = 0.1 when t = 0.0178 sec. y = 0.9 when t = 0.072 sec. t = 0 . 072 − 0 . 0178 = 0 . 0542
sec.
r
y
7-16 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t
d
≅
When Maximum overshoot = 4.3%,
1.1
max
= 1.2
With K = 20.12 and K
max.
overs hoot )
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
+ 0 .125 ζ + 0 .469 ζ
2
= 0 .1
ωn
ζ = 0 . 707
t
.
=
d
1.423
ωn
sec.
= 0 .1
sec.
Thus
ω n = 14.23 = 8.1 5 + 500 K = 2ζω = 1.414 ω = 20.12 K = t n n 5 5 2
( 20%
ω n = 14.23
rad/sec.
2
t
= 0 . 0302
, the system transfer function is Y (s) R( s)
202 . 5
= s
2
+ 20 .1 s + 202
Unit-Step Response:
92
.5
Thus
K
t
=
15 .12 500
= 0 . 0302
When y = 0.5, t = 0.1005 sec. Thus, t = 0 .1005 sec. d
y
7-17 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K From Eq. (7-102), Delay time t
≅
d
1.1
( 4. 3%
max.
overs hoot )
s + ( 5 + 500 Kt ) s + 25 K = 0 2
+ 0 .125 ζ + 0 .469 ζ
2
= 0 . 05 =
ωn
1. 337 Thus,
ωn
ωn =
1. 337
= 26 . 74
0 . 05
2
K =
With K = 28.6 and K
= 1. 043
Characteristic Equation:
ω n = 26.74 = 28.6 5 5 2
max
t
= 0 . 0531
,
5
+ 500
K
t
= 2 ζω n = 2 × 0 . 59 × 26 . 74 = 31 . 55
the sy stem t ransfe
Y (s)
K
t
= 0 . 0531
r func tion i s
715
=
R( s)
Thus
s
2
+ 31 . 55 s + 715
Unit-Step Response: y = 0.5 when t = 0.0505 sec. Thus, t = 0 . 0505 sec. d
y
7-18 Closed-Loop Transfer Fu nction: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For Maximum overshoot = 0.2, ζ = 0.456 From Eq. (7-102), Delay time t
d
=
1.1
max
= 1.1007
( 10 . 07%
max.
overs hoot )
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
.
+ 0 .125 ζ + 0 .469 ζ
2
=
ωn
1.2545
= 0 . 01
ωn
sec.
ω n = 15737.7 = 629.5 25 5 2
Natural Undamped Frequency 5
+ 500
K
t
ωn =
1.2545
t
= 0.2188
.45
rad/sec. Thus,
0 . 01
= 2 ζω n = 2 × 0 .456 × 125
With K = 629.5 and K
= 125 .45
= 114.41
K =
Thus, K
t
= 0.2188
, the system transfer function is Y (s) R( s)
=
15737 . 7 s
2
+ 114.41
Unit-step Response:
s
+ 15737
.7
y = 0.5 when t = 0.0101 sec.
93
Thus, t y
7-19 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K ζ = 0 .6 2 ζω = 5 + 500 n From Eq. (7-109), settling time t
K
t
=
3 .2
≅
s
1.2 ω
n
−5
= 1.2
sec.
( 20% ma
x. overs hoot )
Characteristic Equation:
s + ( 5 + 5000 Kt ) s + 25 K = 0 2
K
=
ζω n
max
= 0 . 0101
d
t
= 1.2 ω n 3.2
0 .6 ω
= 0 .1
sec. Thus,
ωn =
n
ωn
3 .2
= 53 . 33
rad / sec
0 . 06
2
= 0 .118
=
K
500
= 113
. 76
25
System Transfer Function: Y (s) R( s)
=
2844 s
2
+ 64 s + 2844
Unit-step Response:
y(t) reaches 1.00 and never exceeds this value at t = 0.098 sec. Thus, t = 0 . 098 sec. s
7-20 (a) Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K For maximum overshoot = 0.1, ζ = 0 . 59 . Settling time:
t
s
=
K
3 .2
ζω n t
=
3 .2
=
1.18
0 .59
ωn
ωn −5
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
5
+ 500
= 0 . 05
K
t
= 2 ζω n = 2 × 0 . 59 ω n = 1.18 ω n ωn =
sec.
ωn
3 .2 0 . 05
× 0 . 59
= 108
.47
2
= 0 .246
K
=
500
= 470
. 63
25
System Transfer Function: Y (s) R( s)
=
11765 . 74 s
2
+ 128
s
+ 11765
. 74
Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.048 sec. Thus, t = 0 . 048 sec. s
94
(b)
ζ = 0 .456
For maximum overshoot = 0.2, Settling time t
s
=
3 .2
ζω n
.
3 .2
=
0 .456 K
t
=
5
+ 500
= 0 . 01
ωn
0 . 912
K
t
= 2 ζω n = 0 . 912 ω n ωn =
sec.
ωn −5
3 .2
× 0 . 01
0 .456
= 701 . 75
rad / sec
= 1.27
500
System Transfer Function: Y (s) R( s)
=
492453 s
2
+ 640
s
+ 492453
Unit-Step Response: y(t) reaches 1.05 and never exceeds this value at t = 0.0074 sec. Thus, t = 0 . 0074 sec. This is less s
than the calculated value of 0.01 sec.
7-21 Closed-Loop Transfer Function: Y (s ) 25 K = 2 R ( s ) s + ( 5 + 500 Kt ) s + 25 K Damping ratio
ζ = 0 . 707
Characteristic Equation:
s + ( 5 + 500 Kt ) s + 25 K = 0 2
Settling time t
.
s
=
4. 5ζ
ωn
=
3 .1815
ωn
= 0 .1
sec.
Thus,
ω n = 31 .815 ωn 2
5
+ 500
K
t
= 2 ζω n = 44. 986
Thus , K
t
= 0 . 08
K
=
2ζ
= 40 .488
System Transfer Function: Y (s) R( s)
Unit-Step Response:
=
1012 .2 s
2
+ 44.
986 s
+ 1012
.2
The unit-step response reaches 0.95 at t = 0.092 sec. which is the measured t . s
95
rad/sec.
7-22 (a)
When
ζ = 0 . 5 , the rise time is t
r
≅
1 − 0 .4167
ζ + 2 . 917 ζ ωn
2
=
1. 521
=1
ωn
sec.
Thus
ω n = 1. 521
rad/sec.
The second-order term of the characteristic equation is written s
2
+ 2 ζω n s + ω n = 2
The characteristic equation of the system is
For zero remainders,
28 .48 a
=
2
+ 1. 521 s + 2 . 313 = 0
s
3
+ ( a + 30
)s
+ 1. 521 s + 2 . 313
2
s
Dividing the characteristic equation by
s
45 . 63
Thus,
a
+ 30
2
as
+K =0
, we have
= 1. 6
K
= 65 .874 + 2 . 313
a
= 69 . 58
Forward-Path Transfer Function: G (s)
69 . 58
= s( s
+ 1.6 )( s + 30
)
Unit-Step Response: y = 0.1 when t = 0.355 sec. y = 0.9 when t = 1.43 sec. Rise Time: t = 1.43 − 0 . 355 r
(b)
seconds
The system is type 1.
(i)
For a unit-step input, e
(ii)
For a unit-ramp input,
= 1. 075
ss
= 0. K
= lim
v
s→0
sG ( s )
=
K 30 a
=
60 . 58 30
× 1. 6
7-23 (a) Characteristic Equation: s
3
+ 3s + (2 + K ) s − K = 0 2
Apply the Routh-Hurwitz criterion to find the range of K for stability.
Routh Tabulation:
96
= 1.45
e
ss
=
1 K
= 0 .69 v
sec.
s
3
s
2
s
1
s
0
6
+K
1
2
3
−K
+4K 3
−K
Stability Condition:
-1.5 < K < 0 This simplifies the search for K for two equal roots. When K = −0.27806, the characteristic equation roots are: −0.347,
−0.347, and −2.3054. (b) Unit-Step Response: ( K = − 0.27806)
(c) Unit-Step Response ( K = − 1)
The step responses in (a) and (b) all have a negative undershoot for small values of t. This is due to the zero of G(s) that lies in the right-half s-plane.
7-24 (a)
The state equations of the closed-loop system are: dx
= − x1 + 5 x 2
dx
= −6 x 1 dt dt The characteristic equation of the closed-loop system is 1
∆=
2
s +1
−5
6 + k1
s + k2
− k1 x 1 − k 2 x 2 + r
= s + ( 1 + k2 ) s + ( 30 + 5 k1 + k 2 ) = 0 2
97
For
(b)
For
ω n = 10
ζ = 0 . 707
rad / sec,
, 2ζω
2
For
ω n = 10
1
2
k
2
1
+ k 2 = 70
.
1.414
ζ = 0 . 707 an d
k 2 = 59 + 10 k1 2
Thus
, 1
+ k 2 = 2ζω n = 14. 14
k
Th us
2
= 13 .14
The closed-loop transfer function is
Y (s)
=
R (s )
5 s + ( k 2 + 1) s + ( 30 + 5 k1 + k 2 ) 2
lim y ( t )
For a unit-step input,
t →∞
(e)
k
= 30 + 5k1 + k 2
= 11 . 37 .
1
Thu s 5 k
.
2
2
and
1
ωn = 1 +
Th us
+ k 2 = 100
Solving for k , we hav e
(d)
2
(1 + k )
rad / sec 5k
+ 5 k 1 + k 2 = ω n = 100
= 1+ k2.
n
ωn = (c)
30
= lim
sY ( s )
s→0
5
=
=
5 s + 14.14 s + 100 2
= 0 . 05
100
For zero steady-state error due to a unit-step input, 30
+ 5 k1 + k 2 = 5
Thus
5k
1
+ k 2 = −25
Parameter Plane k versus k : 1
2
7-25 (a) Closed-Loop Transfer Function Y (s) 100 ( K P + K D s ) = 2 R ( s ) s + 100 K D s + 100 K P The system is stable for K > 0 and P (b)
For
ζ = 1,
2ζω
ω n = 10
n
= 100 K
K
P
(c)
See parameter plane in part (g) .
(d)
See parameter plane in part (g) .
D
(b) Characteristic Equation: s K
D
> 0.
n
= 100
2
+ 100
K
.
T hus
2ω
KD
= 20
98
KP
K
D
= 0 .2
K
P
D
s
+ 100
K
P
=0
(e)
K
Parabolic error constant
= 1000
a
−2
sec
K a = lim s G ( s ) = lim100 ( K P + K D s ) = 100 K P = 1000
Thus K P = 10
2
s →0
(f)
s →0
Natural undamped frequency
ω n = 50
ω n = 10 (g)
When K
P
K
rad/sec.
= 50
P
K
Th us
=
P
25
= 0, G (s)
=
100 K s
D
s
2
100 K
=
D
(pole-zero cancellation)
s
7-26 (a) Forward-path Transfer Function: Y ( s)
G ( s) = When r ( t )
(b)
= tu s ( t ),
=
E (s ) K
= lim
v
KKi
s [ Js(1 + Ts ) + K i K t ]
s→0
sG ( s )
=
K K
e
=
ss
10 K
(
s 0.001 s + 0.01 s + 10 K t =
t
2
1 K
=
K
)
t
K
v
When r(t) = 0
Y (s ) Td ( s) For T ( s ) d
(c)
=
1 s
=
1 + Ts
=
s [ Js(1 + Ts) + Ki Kt ] + KKi lim y ( t ) t→ ∞
=
lim sY ( s ) s →0
=
1 + 0.1s
(
1 if the system is stable. 10 K
The characteristic equation of the closed-loop system is 0 . 001 s
3
+ 0 . 01 s + 0 .1 s + 10 2
K
=0
The system is unstable for K > 0.1. So we can set K to just less than 0.1. Then, the minimum value of the steady-state value of y(t) is 1 + =1 − 10 K K =0 .1 However, with this value of K, the system response will be very oscillatory. The maximum overshoot will be nearly 100%.
(d)
For K = 0.1, the characteristic equation is 0 . 001 s
3
+ 0 . 01 s + 10 2
K s t
For the two complex roots to have real parts of
+1 = 0
s
or
3
+ 10
s
2
+ 10
4
K s t
+ 1000 = 0
−2/5. we let the characteristic equation be written as
99
)
s 0.001 s + 0.01 s + 10 Kt + 10 K 2
( s + a) ( s
+ 5 = 10 = − a = −5
a
Then,
The three roots are: s
7-27 (a)
K
= 10000
t
)
+ 5s + b = 0
2
s + ( s + 5)s + (5a+ b) s+ ab = 0 3
or
2
=5 ab = 1000 b = 200 s = − a = −5 s = −2 . 5 ± j13 . 92
a
+ b = 10
5a
4
K
t
K
t
= 0 . 0225
oz - in / rad
The Forward-Path Transfer Function:
9 × 10 K 12
G ( s) =
(
s s + 5000 s + 1.067 ×10 s + 50.5 × 10 s + 5.724 × 10 4
3
7
2
9
12
)
9 × 10 K 12
=
s ( s + 116)( s + 4883)( s + 41.68 + j3178.3)( s + 41.68 − j 3178.3)
Routh Tabulation: s
5
s
4
s
3
s
1
1
s
0
× 10
5.7 2. 895
× 10
16 . 6
5
5 . 72
× 10 + 1. 579 × 10 8
13
+ 8 .473 × 10 29
12
7
K
+ 1. 579 × 10
9
7
50.5
× 10
9
12
− 2 . 8422 × 10
K
− 2981
− 58303
.14 K
9
5.7 24 9
− 1. 8 × 10
12
× 10
9
12
9
K
× 10
× 10
12
12
K
0
K
2
K
K
K
1
2
× 10
K
From the s row, the condition of stability is 165710 or
× 10
500 0
2
s
1.067
0 − 3000 . 57 ) < 0 2
K
0 < K < 3000.56
The critical value of K for stability is 3000.56. With this value of K, the roots of the characteristic equation are: −4916.9, −41.57 + j3113.3, −41.57 + j3113.3, −j752.68, and j752.68
(b)
K
= 1000
L
oz-in/rad. The forward-path transfer function is
9 × 10 K 11
G ( s) =
(
s s + 5000 s + 1.582 × 10 s + 5.05 ×10 s + 5.724 × 10 4
3
6
2
9
11
)
9 × 10 K 11
=
s (1 + 116.06)( s + 4882.8)( s + 56.248 + j1005)( s + 56.248 − j1005)
(c) Characteristic Equation of the Closed-Loop System: s
5
+ 5000
s
+ 1. 582 × 10
4
6
3
s
+ 5 . 05 × 10
9
s
2
+ 5 . 724 × 10
11
+ 9 × 10
s
11
K
=0
Routh Tabulation: s
5
s
4
s
3
s
2
1 500 0 5.72 4.6503
× 10
5
× 10 + 1. 5734 × 10 7
5 .724 6
K
1 .582
× 10
6
5. 05
× 10
9
× 10
11
9
100
5.7 24 9
− 1.8 × 10
× 10
11
K
8
K
× 10
× 10 0
11
11
K
s
26 . 618
1
× 10
18
+ 377
15
− 2 . 832 × 10
K
× 10 + 1. 5734 × 10 7
4. 6503 s
× 10
.43
0
× 10
9
11
1
2
K
− 1332
. 73 K
K
2
K
K
From the s row, the condition of stability is Or,
6
14
× 10
26 . 618
− 93990 < 0
(K
or
− 1400
4
)( K
+ 3774. 3 K − 2 .832 + 67 .14 ) < 0
K
2
>0
Stability Condition: 0 < K < 1400 The critical value of K for stability is 1400. With this value of K, the characteristic equation root are: −4885.1, −57.465 + j676, −57.465 − j676, j748.44, and −j748.44
(c)
K
L
= ∞.
Forward-Path Transfer Function:
G ( s) = =
nK s K i K s L a J T s + ( Ra J T + R mL a ) s + R a B m + K i K b 891100 K
(
s s + 5000 s + 566700 2
=
)
891100 K s ( s + 116)( s + 4884)
Characteristic Equ ation of the Closed-Loop System: s
3
+ 5000
s
2
+ 566700
s
+ 891100
K
=0
Routh Tabulation: s
3
s
2
s
1
s
0
1
5667 00
5000
89 1100 K
− 178
5 66700
.22 K
89 91100 K
From the s row, the condition of K for stability is 566700 − 178.22K > 0. 1
Stability Condition :
0 < K < 3179.78
The critical value of K for stability is 3179.78. With K = 3179.78, the characteristic equation roots are
−5000, When the motor shaft is flexible, K becomes stiffer, K K
L
=∞,
L
L
−j752.79.
j752.79, and
is finite, two of the open-loop poles are complex. As the shaft
increases, and the imaginary parts of the open-loop poles also increase. When
the shaft is rigid, the poles of the forward-path transfer function are all real. Similar effects
are observed for the roots of the characteristic equation with respect to the value of K . L
7-28 (a) G (s) c
=1
G (s)
=
+ 2)
100( s s
2
K
−1
p
= lim
G (s)
s→0
= − 200
When d(t) = 0, the steady-state error due to a unit-step input is e
ss
=
1 1+ K
1
= p
1
− 200
=−
1 199
101
JT = J m + n J L 2
2
= − 0 . 005025
(b) Gc (s ) = (c)
s +α
G (s ) =
s
100( s + 2)( s + α )
(
s s −1 2
Kp = ∞
)
α=5
m aximum
overs hoot
= 5.6%
α
= 50
ma ximum
oversh oot
= 22%
α
= 500
max imum o versho ot
e ss = 0
= 54.6%
As the value of α increases, the maximum overshoot increases because the damping effect of the zero at s = −α becomes less effective.
Unit-Step Responses:
(d)
r(t )
=0
and
G (s) c
= 1.
d (t )
= u s(t )
D (s)
1
=
s
System Transfer Function: ( r = 0) Y ( s) D (s)
= r =0
+ 2)
100( s s
3
+ 100
s
2
+ ( 199 + 100 α ) s + 200 α
Output Due to Unit-Step Input: Y ( s)
100( s
= s s
y
(e)
r(t )
= 0,
d (t )
ss
=
3
+ 100
lim y ( t ) t →∞
s
+ ( 199 + 100 α ) s + 200 α
2
= lim
sY ( s )
s→0
= us(t ) G (s) c
=
s
+ 2)
=
200 200
α
=
1
α
+α s
System Transfer Function [ r( t) = 0] Y (s) D ( s)
= r =0
100 s ( s s
3
+ 100
s
2
+ 20
+ ( 199 + 100 α ) s + 200 α
102
D (s)
=
1 s
y ss
=
= lim
sY ( s )
100 s ( s
+ 2)
lim y ( t )
t →∞
s→0
=0
(f) Y (s)
α=5
=
D ( s ) r =0 Y (s)
α = 50
D ( s)
D (s)
s
3
s
3
= r =0
Y ( s)
α = 5000
s
3
= r =0
+ 100
s
2
+ 699
100 s ( s
+ 100
s
2
+ 2)
+ 5199
100 s ( s
+ 100
s
2
+ 1000
s
s
+ 10000
+ 2)
+ 50199
s
+ 100000
Unit-Step Responses:
(g)
As the value of α increases, the output response y(t) due to r(t) becomes more oscillatory, and the overshoot is larger. As the value of α increases, the amplitude of the output response y(t) due to d(t) becomes smaller and more oscillatory.
7-29 (a) Forward-Path Transfer function: G (s)
=
H (s) E (s)
=
Characteristic Equation:
10 N s( s
+ 1 )( s + 10
N
≅
s(s
)
N=1: Characteristic Equation:
s
2
s
+ 1)
+ s +1 = 0
+s+N =0
2
ζ = 0 .5
ωn = 1
rad/sec.
−πζ
Maximu
m over shoot
= e
1 −ζ
2
= 0 .163
N=10: Characteristic Equation:
s
2
(16.3%)
Peak time t
+ s = 10 = 0
max
=
π ωn
1
ζ = 0 . 158
−ζ
= 3 . 628
ω n = 10
rad/sec.
= 1.006
sec.
−πζ
Maximum overshoot
=e
(b) Unit-Step Response:
1 −ζ
2
= 0 . 605
(60.5%)
Peak time t
N=1
103
max
=
π ωn
1
−ζ
2
sec.
2
Second-order System Maximum overshoot Peak time
Third-order System
0.163 3.628 sec.
0.206 3.628 sec.
Unit-Step Response: N = 10
Second-order System Maximum overshoot Peak time
Third-order System
0.605 1.006 sec.
0.926 1.13
sec.
7-30 Unit-Step Responses:
When T
z
is small, the effect is lower overshoot due to improved damping. When T
overshoot becomes very large due to the derivative effect. T
z
z
is very large, the
improves the rise time, since 1
derivative control or a high-pass filter.
104
+Tz s
is a
7-31 Unit-Step Responses
> 0 . 707
less stable. When T p
1
=−
The effect of adding the pole at s
T
to G(s) is to increase the rise time and the overshoot. The system is
p
, the closed-loop system is stable.
7-32 (a) N=1 Closed-loop Transfer Function: M
H
Y ( s)
=
(s)
10
=
R( s)
s
1
=
+ 11 s + 10 s + 10
3
2
1
+ s + 1. 1 s + 0 .1 s 2
3
Second-Order Approximating system: M
L
(s)
=
1 1+ d s
+ d2s
1
M M d e f f
1
2
= m1 =
f
2
d
=
2m
2
(s)
H L
(s)
= m2
l
1
=
1
1+ s
=1
+ d2s
+ 1.1 s + 0 .1 s 2
l
2
= 1.1
l
=
3
6
= 2 l6 − 2 l 1 l 5 + 2 l 2 l 4 − l 3 = − l3 = − 0 . 01
f
2
f M
L
(s)
= 1.2
6
e
2
= −0 . 01
d
1
1
= 1
+ 0 . 9 s + 1. 005
s
2
2
+ l2 s + l3s 2
3
4
4
=
f
=
2l
= 2 m 4 − 2 m 1m 3 + m 2 = m 2 = d 2 2
4
2
− 2 l1 l 3 + l 2 = −2 × 1 × 0 .1 + ( 1.1 ) = 1. 01 2
4
2
2d
− d1
2
2
f
= 2 d 2 − 1.2 = 0 .81 =
s
+ 0 . 8955
s
= 1. 01
Thus d
0 . 995 2
4
+ 0 . 995
1
e
= 1. 01 = d 2
2
4
Thus d
= 0 .9 G L ( s)
=
0 . 995 s( s
+ 0 .895)
Roots of Characteristic Equation: Third-order System
−10.11 −0.4457 + j0.8892
2
2
= 1.2 = 2
+m2s
= 0 .1
3
2
2
1+ l s
e
= 2 l 2 − l1 = 2 × 1.1 − 1 = 1.2 f
1
1
2
2
Thus,
1+ m s
2
− m1 = 2 d 2 − d1 2
2
1+ d s
2
−0.4457 − j0.8892
Second-order System
−0.4478 + j0.8914 −0.4478 − j0.8914
105
2
= 1. 005
The real root at −10.11 is dropped by the second-order approximating system, and the two complex roots are slightly preturbed.
Unit-Step Responses:
(b)
N = 2:
Closed-loop Transfer Function: M
l
= 0 .5
1
e d
l
= d2 = 2
4
=
2 1
M
L
2d
(s)
2
f
2
H
(s)
20
=
s
= 0 . 55
l
3
3
=
+ 11 s + 10 s + 20 2
= 0 . 05
e
2
=
4
2d
+ 0 . 55
− d1 = 2
2
= −2 l1 l 3 + l 2 = −0 . 05 + 0 . 3025 = 0 .2525
1 1
1 + 0 .5 s
2
− e 2 = 2 × 0 . 5025 − 0 .85 = 0 .155
=
1
+ 0 . 3937
s
+ 0 . 5025
s
2
d
1
s
2
+ 0 . 7834
s
2
+ 0 . 05
2
=
2l
Thus
s
3
− l1 = 2 × 0 . 55 − ( 0 . 5) 2
2
d
2 2
=
f
4
+ 1. 99
G L(s)
1. 99
= s(s
Roots of Characteristic Equations: Third-order System
− 10.213
−0.3937 + j1.343 −0.3937 − j1.343
= 0 .2525
= 0 . 3937
1. 99
=
f
s
Second-order System
−0.3917 + j1.3552 −0.3917 − j1.3552
The real root at −10.213 is dropped by the second-order approximating system, and the two complex roots are slightly preturbed.
Unit-Step Responses:
106
+ 0 . 7835)
2
= 0 .85 d
2
= 0 . 5025
(c)
N=3 Closed-Loop Transfer Function: M
l e
= 0 . 3333
1
= 2 d 2 − d1 = 2
2
Thus, d
H
=
2 2
f
4
(s)
l f
=
L
(s)
s
+ 11 s + 10 s + 30 l
2
+ 0 . 333
s
+ 0 . 3667
d 1
2 1
= 2 d2 −
+ 0 .2186
s
f
= 0 .2186
+ 0 . 335
2
s
2
+ 0 . 0333
s
3
s
2
= d2 = 2
4
f
= −2 l1 l 3 + l 2 = 0 .1122 2
4
= 0 . 335
2
2 . 985
=
e
= 2 × 0 . 335 − 0 . 6222 = 0 . 0477 d
2
s
= 0 . 0333
= 2 l 2 − l1 = 0 . 7333 − 0 .1111 = 0 . 6222
1 1
1
2
= 0 . 1122
=
3
1
=
2
= 0 . 3667
2
d
M
30 3
+ 0 . 6524
s
+ 2 .985
G L(s)
2 . 985
= s( s
+ 0 . 6525)
Roots of Characteristic Equation: Third-order System
Second-order System
−10.312 −0.3438 + j1.6707 −0.3438 − j1.6707
−0.3262 + j1.6966 −0.3262 − j1.6966
The real root at −10.312 is dropped by the second-order approximating system, and the complex roots are slightly preturbed.
Unit-Step Responses:
(d)
N= 4
Closed-Loop Transfer Function:
107
M
40
=
(s)
H
s
3
+ 11 s + 10 s + 40 l
e
2
=
Thus,
d
− d1 = 2
2d
2 2
2
= M
f
L
= 0 .25
1
d
2
+ 0 .1225
s
l
=
3
=
2 1
+ 0 . 025
2
s
s
3
= 0 . 025
4
d
2
+ 0 .275
=d2 =
4
= 0 .2513
+ 0 .2513
s
1 + 0 .25 s
e
1 1
1
= 0 .275
2
= 2 l 2 − l 1 = 0 .4875
2
=
(s)
l
2
f
= 0 . 06313
4
=
2
2d
4
= −2 l1 l 3 + l 2 = 0 . 06313
−
f
2
f 2
2
= 0 . 5025 − 0 .4875 = 0 . 015
3 . 98 s
+ 0 .4874
2
s
G
+ 3 .98
L
(s)
=
Roots of Characteristic Equation: Third-order System
Second-order System
− 10.408 −0.2958 + j1.9379 −/2958 − j1.9379
−0.2437 + j1.98 −0.2437 − j1.98
The real root at −10.408 is dropped by the second-order approximating system, but the complex roots are preturbed. As the value of N increases, the gain of the system is increased, and the roots are more preturbed.
Unit-Step Responses:
(e)
N= 5
Closed-loop Transfer Function: M
l e e
H
1
(s)
= 0 .2
2
= 2 d1 − d 2 =
4
= d2 =
2
2
f
=
f
50 s
3
+ 11 s + 10 s + 50 2
l
2
= 0 .22
1
= 1 l
3
= 0 . 02
= 2 l 2 − l1 = 0 .44 − 0 .04 = 0 .4 2
2
= −2 l1 l 3 + l 2 = −0 . 008 + 0 . 0484 = 0 . 0404 2
4
+ 0 .2 s + 0 .22
108
s
2
+ 0 . 02
s
3
d
1
= 0 .1225
3 . 98 s( s
+ 0 .4874
)
Thus, d d M
L
(s)
2 2 2 1
=
f
= 0 .0404
=
2d
4
+ 0 . 0447
1
s
= 0 .201
2
− e 2 = 0 .402 − 0 .4 = 0 . 002
2
1
=
d
+ 0 .201
s
2
=
d
1
1 s
2
+ 0 .2225
s
= 0 . 04472 G L(s)
+ 4. 975
4. 975
=
s(s
+ 0 .2225)
Roots of Characteristic Equation: Third-order System
Second-order System
− 10.501 −0.2494 + j2.678 −0.2494 −j2.678
−0.1113 + j2.2277 −0.1113 − j2.2277
The real root at −10.501 is dropped by the second-order approximating system, and the complex roots are changed, especially the real parts.
Unit-Step Responses:
7-33 (a)
K=1
Forward-path Transfer Function:
G ( s) =
891100
(
s s + 5000 s + 566700 2
)
=
891100 s ( s + 116)( s + 4884)
Closed-Loop Transfer Function : M
H
(s)
891100
= l
s
1
e e
3
+ 5000
s
+ 566700
2
= 0 . 636
l
2
= 2 d 2 − d1 =
4
= d2 =
2
2
f
2
f
s
+ 891100
= 5. 611 × 10
−3
+ 0 . 636 l
s
+ 5. 611 × 10
= 1.122 × 10
3
−3
Thus,
109
−3
s
+ 1.1222 × 10
2
−6
s
3
−6
− 0 .4045 = −0 . 3933
= − 2 l1 l3 + l 2 = −2 × 0 . 636 × 1.1222 × 10 2
4
1
= 2 l 2 − l1 = 1.1222 × 10 2
2
1
=
−6
+ ( 5 . 611 × 10
−3
)
2
= 0 . 000030
06
2
d
2
d
2
M
L
(s)
1
= 0 . 000030 = 2 d2 −
f
d
06 2
1
+ 0 . 6358
s
= 0 .005482
= 0 . 01096 + 0 . 3933 = 0 .4042
1
=
2
+ 0 . 005482
s
=
2
d
1
= 0 . 6358
182 .4 s
+ 115
2
+ 182
. 97 s
G L ( s)
.4
182 .4
=
s( s
+ 115
. 97 )
Roots of Characteristic Equations: Third-order System
Second-order System
−1.595 −114.4 −4884
−1.5948 −114.38
The real root at −4884 is dropped by the second-order approximating system. the other two roots are hardly preturbed.
Unit-Step Responses
(b)
K = 100
M
(s)
H
Closed-loop Transfer Function:
891100
=
s
3
+ 5000 l
e e
s
00
+ 566700 + 891100
2
= 0 .00636
1
2
= 2 d 2 − d1 =
4
= d2 =
2
2
f
f
l
2 2 2
d
M
L
(s)
1
=
f
4
1
+ 0 . 007403
−5
= 5. 611 × 10
−5
l
= 1. 1222 × 10
−8
−5
3
− 4. 045 × 10
4
f
−5 −8
= 2 l 4 − 2 l1 l 3 + l 2 = − 2 × 0 . 00636 × 1.1222 × 10 2
2
−9
= 0 . 000109
d 6
s
+ 0 . 000054
8s
2
− 0 . 000071
1
=
+ 5. 611 × 10
1 + 0 . 00636 s
= 2 l 2 − l1 = 11 .222 × 10
= 3 . 0056 × 10
= 2 d2 −
00
1
2
2
Thus, d
2
=
2
= 0 . 000054 8
s
2
+ 135
= 0 . 000054
.1 s
2
+ 1.1222 × 10
= 0 . 000071
+ 18248
8
d
1
G
+ ( 5 . 611 × 10
−5
)
2
(s)
=
18248 s( s
+ 135
. 1)
Roots of Characteristic Equations: Third-order System
Second-order System
110
s
3
= 3 . 0056 × 10
= 0 . 007403
L
−8
8
8
18248
=
s
−9
−4887.8 −56.106 + j122.81 −56.106 − j122.81
−67.55 + j114.98 −67.55 − j114.98
Unit-Step Responses
(c)
K = 1000 Closed-loop Transfer Function: M
H
(s)
891100
=
s
+ 5000
3
l e e
s
2
+ 566700
= 2 d 2 − d1 =
4
= d2 =
2
f
f
d
M
L
(s)
2 2 2 1
=
f
=
2d
4
−
l
2
1
−6
= 5 . 611 × 10
−6
l
3
f
2
= 0 . 000010
d 965
− 0 . 000010
+ 0 . 000382
9s
+ 0 . 000005
482 s
2
=
818
−6
= 1.1222 × 10
−9
− 0 .000000
2
−11
+ 5 . 611 × 10
s
= − 2 l1 l3 + l 2 = −2 × 0 . 000636 × 1.1222 × 10
4
1
=
1 + 0 . 000636
000
= 2 l 2 − l1 = 11 .222 × 10
= 3 . 0055 × 10 2
+ 891100
1
=
2
2
Thus, d
s
= 0 . 000636
1
2
2
000
2
404 −9
= 0 . 000010
+ ( 5 . 611 × 10
= 0 . 000005
482
= 0 . 000000
147
d
182415 .177 s
2
+ 69 . 8555
s
+ 182415
.177
1
s
−6
)
2
= 3 . 0056 × 10
= 0 . 000382
−34.928 + j425.67
111
3
G L (s)
−34.928 − j425.67
−11
9
=
182415 .177 s( s
Second-order System
− 4921.6 −39.178 + j423.7 −39.178 − j423.7
s
82
Roots of Characteristic Equations: Third-order System
−9
+ 1.1122 × 10
2
+ 69 . 8555)
Unit-step Re sponses
7-34
Forward-path Transfer Function G (s)
=
K(s s(s
Closed-loop Transfer Function
− 1)
M (s)
+ 1 )( s + 2 )
Second-order System: M
L
(s)
=
M L (s )
−s +1 s
+3s + s +1
3
2
+ c1 s
1
1+ d s
+ d2s
1
M H (s )
=
2
( − s + 1) (1 + d s + d s ) 1 + ( d − 1) s + ( d − d ) s − d s = ( s + 3s + s + 1) ( 1+ c s ) 1 + ( c + 1) s + ( c + 3) s + ( 3c + 1) s + c s 2
=
1
3
2
2
1
2
= 1 + c1
1
m
1
l
= d1 −1
3
1
2
2
1
l
2
m
1
2
= 3 + c1
2
= d 2 − d1
l
3
= 1 + 3 c1
3
= −d 2
m
3
1
1
l
1
= c1
4
e2 = f 2 = 2m 2 − m1 = 2 ( d 2 − d 1 ) − ( d 1 − 1) = 2 d 2− d 1 − 1 2
2
2
= 2l2 − l1 = 2 ( 3 + c1 )− ( 1+ c1 ) = 5 − c1 2
2
2
e4 = f 4 = 2m 4 − 2m1 m3 + m 2 = −2 ( d 1 − 1) ( −d 2 ) + ( d 2 − d1 ) = d 2 − 2d 2 + d 1 2
2
2
2
= 2l4 − 2l1l3 + l2 = − 2 ( 1 + c1 ) (1 + 3c1 ) + ( 3 + c1 ) = 7 − 2c1 − 5c1 2
2
2
e6 = f 6 = 2m 6 − 2 m1m 5 + 2 m 2 m4 − m 3 = −m 3 = − ( −d 2 ) = −d 2 2
2
2
2
= 2l6 − 2l1l5 + 2l 2l4 − l3 = 2l2 l4 − l3 = 2 ( 3 + c1 ) c1 − (1 + 3c1 ) = − 1 − 7c1 2
2
2
Simultaneous equations to be solved:
Solutions:
112
2
4
c1
= −1. 0408
= 1 + 7 c1
d
= 0 . 971
− 2 d 2 + d 1 = 5 − c1
d
2d
− d1 = 6 −c1
2
d d
M
L
2 2
(s)
2
2
2
2
2 2
− 1. 0408
1
=
2
s
+ 0 . 971 s + 2 . 93
1
s
−0 .3552(
=
2
s
s
+ 0 . 3314
2
− 0 . 9608 s
1
2
)
= 2 . 93 G L(s)
+ 0 .3413
−0 . 3552(
=
s( s
s
− 0 . 9608
+ 0 . 69655)
Roots of Characteristic Equations: Third-orde r System
Second-order System
−2.7693 −0.1154 + j0.5897 −0.1154 − j0.5897
−0.1657 + j0.5602 −0.1657 −j0.5602
Unit-step Responses
7-35 (a)
K = 10 Closed-loop Transfer Function: M
(s)
H
=
10 s
+ 23
4
s
3
+ 62
s
2
+ 40 s + 10
=
1 1+ 4 s
+ 6 .2 s + 2 . 3 s + 0 .1 s 2
3
Second-order System Approximation: M
L
(s)
=
1 1+ d s 1
l e e e
2
4
6
=
f
=
f
=
f
2
=4
1
=
2d
l
2
= 6 .2
2
l
3
= 2 .3
l
4
− d 1 = 2 l 2 − l1 = 12 .4 − 16 = −3 . 6 2
2
+ d2s
2
4
= d 2 = 2 l 4 − 2 l1 l 3 + l 2 = 0 .2 − 2 × 4 × 2 . 3 + ( 6 .2 ) = 20 .24
6
= 2 d 6 − 2 d 1 d 5 + 2 d 2 d 4 − d 3 = − d 3 = 2 l 2 l 4 − l3 = − 4. 05
2
2
2
2
2
2
Thus, d d
2 2 2 1
=
20 .24
=
2d
2
−
d f
2
= 9 + 3. 6 = 12 . 6
2
d
113
1
= 4. 5 = 3 . 55
= 0 .1
4
)
M
L
1
=
(s)
+ 3 .55 s + 4. 5 s
1
0 .2222
=
2
s
2
+ 0 . 7888
s
G L(s)
+ 0 .2222
0 .2222
=
s( s
+ 0 . 7888
Roots of Characteristic Equations: Fourth -order system
Second-order system
−2.21 −20 −0.3957 + j0.264 −0.3957 −j0.264 (b)
−0.3944 + j0.258 −0.3944 −j0.258
K = 10 Third-order System Approximation:
M
L
(s)
=
1+ d s 1
e e
2
4
=
f
=
f
2
=
2d
M
1
+ d 2s + d3s 2
3
M
− d 1 = −3 . 6 2
2
d
L
(s) (s)
1
= 1
= −f6 =
2 3
+ d1 s + d 2 s + d 3 s 2
2
3
+ 4 s + 6 .2 s + 2 . 3 s + 0 .1 s 2
4. 05
= 2 d 4 − 2 d 1 d 3 + d 2 = − 2 d 1 d 3 + d 2 = − 4. 025 2
4
H
Thus
d
+d2 = 2
1
f
4
3
d
3
4
= 2 . 0125
= 20 .24
Thus, d
= 0 . 5 d 1 − 1.8 2
2
Solving for d ,
d
1
1
= 3 .9528
2
d ,
2
− 3 . 0422
= 0 .25 d 1 − 1. 8 d 1 + 3 .24 = 20 .24 + 4. 025 4
− 0 .4553 +
,
d
Selecting the positive and real solution, we have
M
L
(s)
2
1
= 3 . 9528
1
= 1
+ 3. 9528
s
+ 6 . 0123
s
2
GL ( s) =
+ 2 . 0125
(
s
3
=
− 0 .4553 −
j 2334, d
.
1
j 2334.
= 0 . 5 d 1 − 1.8 = 6 . 0123 2
2
0 .4969 s
3
+ 2 . 9875
0.4969
s s + 2.9875 s + 1.964 2
d
s
2
+ 1. 964
s
+ 0 .4969
)
Roots of Characteristic Equations: Fourth -order System
−2.21 −20 −0.39565 + j0.264 −0.39565 − j0.264
Third-order System
−2.1963 −0.3956 + j0.264 −2.1963 − j0.264
Unit-step Response
114
)
(c)
K = 40 Closed-loop Transfer Function: M
(s)
H
40
=
s
l
1
4
+ 23
s
+ 62
3
=1
l
s
2
+ 40 s + 40
= 1. 55
2
=
l
1 1+ s
+ 1. 55
= 0 . 575
3
s l
2
4
+ 0 . 575
s
+ 0 . 025
3
s
4
= 0 . 025
Second-order System Approximation: M
L
(s)
1
=
1+ d s 1
e e
2
4
=
f
=
f
2
2
= 2 d2 − d1 =
4
= d 2 = 2 l 4 − 2 l1 l3 + l 2 = 0 . 05 − 1.15 + 2 .4025 = 1. 3025
2
2l
− l 1 = 3 .1 − 1 =
+ d2s
2
2
2
2 .1
2
Thus, d d
M
(s)
L
= 1. 3025
2 2
d
= 2 d2 −
f
+ 1.1413
2
2 1
2
1
= 1
+ 0 .4273
s
s
= 2 × 1.1413 − 2 . 1 = 0 .1825
= 1.1413
2
d
1
= 0 .4273
0 .8762
=
s
+ 0 . 3744
2
s
G L ( s)
+ 0 .8762
=
0 .8762 s( s
+ 0 . 3744
Roots of Characteristic Equations: Fourth -order System
Second-order System
−2.5692 −19.994 −0.2183 + j0.855 −0.2183 − j0.855
−0.1872 + j0.9172
−0.1872 − j0.9172
Third-order system Approximation: M
L
(s)
1
= 1
e2
=
f2
= 2 d2 − d1 =
e4
=
f4
= −2 d 1 d 3 + d 2 = −1. 0062
e
=
f
= − d 3 = 2 l 2 l 4 − l3 = − 0 .2531
6
2
2l2
2
− l 1 = 3 .1 − 1 = 2
2
2
6
+ d 1s + d 2 s + d 3 s
d1
3
2 .1
+ d 2 = 2 l 4 − 2 l1 l 3 + l 2 = 1. 3025 2
2
2
Equations to be Solved Simultaneously: 2d d
2 2
− d 1 = 2 .1 2
2
= 0 .25
d
4 1
d
2 2
= −1. 00623
+ 1. 05 d 1 + 1.1025 2
d
1
= 1. 3025
Th us
115
d
4 1
+ 4.2
Thus d
2 1
d
− 4. 0249
= 0 .5 d 1 + 1. 05 2
2
d
1
− 0 .8 = 0
)
d
The roots of the last equation are:
1
= − 0 .1688
Selecting the positive real solution, we have d d
= 1. 00623
2 2
1
1
0 . 9525 ,
= 0 .9525
+ 1.3025 = 2 .261
= 0 .2531
2 3
M
d
d
,
L
(s)
1
= 1
+ 0 . 9525
GL ( s) =
s
(
+ 1. 5037
s
2
+ 0 . 5031
s
1.9876
s s + 2.9886 s + 1.8932 2
d
T hus
d
)
Roots of Characteristic Equati ons: Fourth -order System
−2.5692 −19.994 −0.2183 + j0.855 −0.2183 − j0.855 Unit-step Responses
116
=
j 2 .196 ,
− 0 . 392 −
.
Th us
=
3
− 0 .392 +
2
3
= 1. 5037 = 0 . 5031 1. 9876
s
3
+ 2 . 9886
s
2
+ 1. 8932
s
+ 1.9876
1.9876 s ( s + 2.0772)( s + 0.9114)
Third-order System
−2.552 −0.2183 ± j 0 .8551
j 2 .196
Chapter 8 8-1 (a)
P (s)
ROOT LOCUS TECHNIQUE
= s + 4 s + 4 s +8 s 4
3
2
Q (s)
= s +1
0,
Finite zeros of Q( s):
−1
−3.5098, −0.24512 ± j1.4897
Finite zeros of P( s):
Asymptotes: K > 0: Intersect of Asymptotes:
P( s)
= s + 5s + s 3
2
Q (s)
Finite zeros of P( s) : Finite zeros of Q( s):
,
180
=s
2
Q (s)
Finite zeros of P( s): Finite zeros of Q( s): Asymptotes:
(d)
P (s)
−1
o
90
3
= s + 2 s + 3s 5
4
,
270
o
,
240
o
= −1
K < 0:
3
±
−1
o
0 ,
180
o
= −2
j 1. 5874
o
(
−1± 1, −1, −3 0,
2
K < 0:
0
o
) ( s + 3)
j 1.414
There are no asymptotes, since the number of zeros of P( s) and Q( s) are equal.
3
Q (s) 0,
0,
− 1.5 ±
= s +3s + 5 2
−1 ±
0,
= s + 2 s + 10 2
Finite zeros of P( s): Finite zeros of Q( s):
Q (s)
60
o
,
180
o
,
300
o
K < 0:
−1 − 1 − ( −1. 5) − ( −1. 5) 5
−2
=
o
0 ,
120
o
,
240
,
240
o
1 3
= s +5
− 1.0398 ± −5
Asymptotes: K > 0: Intersect of Asymptotes:
j 1.414
j 1. 6583
σ1 = 4
o
−4. 7913 − 0 .2087 − ( − 1)
0 . 083156 180
Asymptotes: K > 0: Intersect of Asymptotes:
P (s)
120
Q(s ) = s − 1
Finite zeros of P( s): Finite zeros of Q( s):
(f)
−1
0 ,
2
0, 0 − 3.156 ,
2
Finite zeros of Q( s): Asymptotes:
P (s)
o
K < 0:
= s + 3s + 2s +8
K > 0:
Finite zeros of P( s):
(e)
o
−4.7912, −0.20871
0,
= s + 2s + 3s 3
300
4
σ1 = P( s)
,
= s +1
Asymptotes: K > 0: Intersect of Asymptotes:
(c)
o
−3 . 5 − 0 .24512 − 0 .24512 − ( −1 )
σ1 = (b)
o
60
j 1.4426 ,
60
o
,
180
1. 0398 o
,
300
120
± o
j1.4426 K < 0:
o
0 ,
120
o
o
−1. 0398 − 1. 0398 + 1. 0398 + 1. 0398 − ( − 5)
σ1 =
4
8-2 (a) Angles of departure and arrival. K > 0:
−θ 1 − θ 2 − θ 3 + θ 4 = −180 o
o
o
= −180
o
o
=0
o
−θ 1 − θ 2 − θ 3 + θ 4 = −180
o
−θ 1 − 90 − 45 + 90 o
−θ 1 − 90 − 45 + 90 θ 1 = −45
o
o
θ 1 = 135 K < 0:
o
o
(b) Angles of departure and arrival. K > 0: K < 0:
o
−θ 1 − 135
o
− 90 + 90
o
o
=0
o
θ 1 = −135
(c) Angle of departure: K > 0:
−θ 1 − θ 2 − θ 3 + θ 4 = −180 o
−θ 1 − 135 θ 1 = −90
− 90
o
− 45
o
o
= −180
o
o
(d) Angle of departure K > 0:
−θ 1 − θ 2 − θ 3 − θ 4 = −180 −θ 1 − 135
o
θ 1 = −180
− 135
o
− 90
o
o
= −180
o
o
121
−1
=
−5 3
(e) Angle of arrival K < 0:
θ 1 + θ 6 − θ 2 − θ 3 − θ 4 − θ 5 = −360 o
8-3 (a)
θ 1 + 90 − 135
o
θ 1 = −108
o
.435
− 135
o
o
o
− 45 − 26 . 565
o
= −360
(b)
(c)
(d)
122
o
8-4 (a) Breakaway-point Equation:
2s
5
+ 20
s
4
+ 74
− 0 . 7275
Breakaway Points:
(b) Breakaway-point Equation: 3 s + 22 6
s
+ 65
5
+ 110
− 2 . 3887
,
+ 100
4
s
(c) Breakaway-point Equation: 3 s + 54 6
s
+ 347
5
− 2 .5 ,
Breakaway Points:
(d) Breakaway-point Equation: − s − 8 s − 19 6
5
+ 86
3
s
8-5 (a) =
G ( s)H ( s)
90
o
K(s s(s
+ 925
s
+ 8 s + 94
s
.5 s
s
2
+ 44 s + 12 = 0
4
3
+ 867
.2 s
+ 120
s
2
− 781 .25 s − 1953 = 0
1. 09 4
s
3
− 0 . 6428
Breakaway Points:
,
2
+ 48 = 0
2 .1208
+8)
+ 5)( s + 6 )
and
o
270
Intersect of Asymptotes:
σ1 =
0
K < 0:
− 5 − 6 − ( −8 ) 3
Breakaway-point Equation: 2s
Breakaway Points: Root Locus Diagram:
+ 48 s = 0
2
s
− 1, − 2 . 5
Breakaway Points:
Asymptotes: K > 0:
3
s
3
+ 35
− 2 .2178
s ,
2
−1
+ 176
s
123
o
= − 1.5
+ 240 = 0
− 5 . 5724,
8-5 (b)
0
− 9 . 7098
and
180
o
G ( s)H ( s)
Asymptotes: K > 0: Intersect of Asymptotes:
K
=
45
o
s(s ,
+ 1 )( s + 3)( s + 4 )
135
o
,
225
σ1 =
G ( s)H ( s)
=
+4)
2
+ 2) o
Asymptotes: K > 0: 60 , Intersect of Asymptotes:
o
K < 0:
0 ,
K < 0:
0 ,
90
o
,
180
o
,
= −2
4 2
+ 38 s + 12 = 0 − 2 , − 3 . 5811
2
180
o
,
300
σ1 =
0
o
+ 0 − 2 − 2 − ( −4 ) 4
Breakaway-point Equation: Breakaway Points:
o
315
−1 −3 − 4
3
K(s s (s
,
4 s + 24 s −0 .4189 ,
Breakaway-point Equation: Breakaway Points: Root Locus Diagram:
8-5 (c)
0
o
3s 0,
4
−1
=0
+ 24 s + 52 s + 32 s = 0 − 1. 085 , − 2 , − 4. 915 3
2
Root Locus Diagram:
124
o
120
o
,
240
o
270
o
8-5 (d) G ( s)H ( s)
=
K(s s( s
2
+ 2)
+ 2s + 2) o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
270
o
K < 0:
σ1 = 3
0
− 1 − j − 1 − j − ( −2 ) 3
−1
o
0 ,
180
o
=0
+8 s +8 s + 4 = 0
Breakaway-point Equation:
2s
Breakaway Points:
− 2 .8393
2
The other two solutions are not breakaway points.
Root Locus Diagram
125
8-5 (e) G ( s) H( s) =
(
K ( s + 5)
s s + 2s + 2 2
o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
)
o
270
σ1 =
0
K < 0:
− 1 − j − 1 − j − ( −5) 3
Breakaway-point Equation: 2 s + 17 s − 7.2091 3
Breakaway Points:
2
−1
(
K
s ( s + 4 ) s + 2s + 2 2
180
o
= 1. 5
+ 20 s + 10 = 0 The other two solutions are not breakaway points.
8-5 (f) G ( s) H( s) =
o
0 ,
) 126
o
Asymptotes: K > 0: 45 , Intersect of Asymptotes:
135
o
,
σ1 =
o
225 0
315
o
o
K < 0:
− 1 − j −1 + j − 4
0 ,
3
G ( s)H ( s)
Asymptotes:
=
2
s (s
K > 0:
+ 4)
2
+ 8)
2
90
o
,
270
o
2
Intesect of Asymptotes:
Breakaway-point Equation: Breakaway Points:
0,
s
5
,
180
o
+ 20 s + 8 = 0 The other solutons are not breakaway points.
o
K < 0:
σ1 =
o
= −1. 5
8-5 (g) K(s
90
4
4 s + 18 s −3 . 0922
Breakaway-point Equation: Breakaway Point:
,
0
+ 20
0 ,
180
o
+ 0 − 8 − 8 − ( −4 ) − ( −4 ) 4 s
4
+ 160
s
−2 3
−4, −8, −4 − j4, −4 + j4
127
+ 640
s
2
+ 1040
s
=0
,
270
o
8-5 (h) G ( s)H ( s)
=
K 2
s (s o
Asymptotes: K > 0: 45 , Intersect of Asymptotes:
+ 8)
2
135
o
,
225
o
σ1 = Breakaway-point Equation: Breakaway Point:
s 0,
3
,
315
−8 − 8
o
K < 0:
= −4
4
+ 12 s + 32 −4, −8 2
=0
s
8-5 (i) 128
o
0 ,
90
o
,
180
o
,
270
o
G ( s) H( s) =
(
K s + 8 s + 20 2
s ( s + 8) 2
o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
270
)
2
o
o
K < 0:
σ1 = 5
0 ,
−8 − 8 − ( − 4 ) − ( −4 ) 4
+ 20
Breakaway-point Equation:
s
Breakaway Points:
− 4,
o
180
s
4
+ 128
−2 s
−8, − 4 +
3
+ 736
j 4. 9 ,
s
2
= −4 + 1280
−4 −
s
=0
j 4. 9
(j) G ( s) H( s) =
Ks
(s
2
2
−4
)
Since the number of finite poles and zeros of G ( s ) H ( s ) are the same, there are no asymptotes.
Breakaway-point Equation: 8 s = 0 Breakaway Points:
s=0
129
8-5 (k) G ( s) H( s) = Asymptotes:
(s
K > 0:
(
K s −4 2
2
)( s
+1 90
Intersect of Asymptotes: σ
1
Breakaway-point Equation: Breakaway Points:
o
,
=
2
)
+4 270
)
o
−2 + 2 4
−2 6
s 0,
K < 0:
o
0 ,
180
o
=0 − 8 s − 24 4
3 .2132 ,
=0 − 3 .2132
s
2
8-5 (l) 130
,
j 1. 5246 ,
−
j1. 5246
K ( s − 1) 2
G ( s) H( s) =
(s
2
Asymptotes: K > 0:
)( s
+1 90
o
,
2
+4 270
)
o
Intersect of Asymptotes:
σ1 = 5
−1 + 1 4
s
Breakaway Points:
− 2 . 07 ,
G ( s)H ( s)
=
K(s
3
j 1.47 ,
j 1.47
=0
3
2 . 07 ,
−
+ 1 )( s + 2 )( s + 3) s (s
0 ,
− 2s − 9 s = 0
Breakaway-point Equation:
(m)
−2
o
K < 0:
− 1) 131
180
o
Asymptotes: K > 0:
180
o
K < 0:
Breakaway-point Equation: s + 12 s + 27 s 6
5
G ( s)H ( s)
Asymptotes:
=
K (s 3
s (s
K > 0:
+ 5)( s + 40
+ 250 60
+ 2 s − 18 3
s
o
2
− 1.21, −2.4, −9.07,
Breakaway Points:
(n)
4
0
o
,
)( s
o
,
0.683,
0,
0
)
+ 1000
180
=0
)
300
o
K < 0:
132
o
0 ,
120
o
,
240
o
Intersect of asymptotes:
σ1 = Breakaway-point Equation: 3750 s
8-5 (o)
Asymptotes:
=
K(s s( s
K > 0:
+ 0 + 0 − 250 − 1000 − ( −5) − ( −40 5
+ 335000
− 7.288, −712.2,
Breakaway Points:
G ( s)H ( s)
6
0
0,
s
5
−2
+ 5 .247 × 10
8
s
4
0
− 1)
o
,
270
o
K < 0:
Intersect of Asymptotes:
133
o
0 ,
= − 401 . 67
+ 2 . 9375 × 10
+ 1 )( s + 2 ) 90
)
180
o
10
s
3
+ 1. 875 × 10
11
s
2
=0
σ1 =
8-6 (a)
3
−1 − 2 − 1 3
−1
−3 s −1 = 0
Breakaway-point Equation:
s
Breakaway Points;
−0.3473, −1.532,
Q(s ) = s + 5
(
= −2
1.879
)
P (s ) = s s + 3s + 2 = s (s +1)( s + 2) 2
o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
270
o
K < 0:
σ1 = Breakaway-point Equation:
s
3
−1 − 2 − ( −5) 3
−1
=1
+ 9 s + 15 s + 5 = 0 2
134
o
0 ,
180
o
−0.4475, −1.609,
Breakaway Points:
8-6 (b)
Q (s)
= s +3
Asymptotes:
P (s)
=
ε
s s
K > 0:
2
90
+ s+2 o
,
−6.9434
ϕ
270
o
K < 0:
Intersect of Asymptotes:
σ1 = 3
−1 − ( −3) 3
−1
o
0 ,
180
o
=1
+5s + 3s + 3 = 0
Breakaway-point Equation:
s
Breakaway Points:
− 4.4798
2
The other solutions are not breakaway points.
135
8-6 (c)
Q (s)
= 5s
Asymptotes:
P (s)
=
s
K > 0:
2
+ 10 180
o
K < 0: 2
− 50 = 0
Breakaway-point Equation:
5s
Breakaway Points:
− 3.162,
3.162
136
0
o
(
8-6 (d) Q(s ) = s s + s + 2 2
)
Asymptotes: K > 0:
P( s) = s + 3s + s + 5s +10 4
180
3
2
o
K < 0: 6
Breakaway-point Equation:
s
Breakaway Points:
−2,
0
o
+ 2 s + 8 s + 2 s − 33 5
4
1.784.
3
s
2
− 20 s − 20 = 0
The other solutions are not breakaway points.
137
(
) ( s + 2)
8-6 (e) Q(s ) = s −1 2
(
P( s) = s s + 2 s + 2 2
)
Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes. 3
+ 12
Breakaway-point Equation:
6s
Breakaway Points:
−1.3848
s
2
+ 8s + 4 = 0
138
8-6 (f)
Q(s ) = (s + 1)(s + 4) Asymptotes:
K > 0:
(
P (s ) = s s − 2 180
2
)
o
K < 0: 4
+ 10
Breakaway-point equations:
s
Breakaway Points:
−8.334,
s
3
+ 14
0 s
2
0.623
139
o
−8 = 0
8-6 (g) Q(s ) = s + 4s + 5 2
Asymptotes:
P( s) = s
K > 0:
90
o
,
2
(s
+ 8 s + 16
2
270
o
σ1 = 5
s
Breakaway Points:
0,
o
K < 0:
Intersect of Asymptotes:
Breakaway-point Equation:
)
−8 − ( −4 )
+ 10
4 s
− 2,
4
s
180
o
= −2
−2 + 42
0 ,
3
− 4, 140
+ 92 −2 +
s
2
+ 80
j 2 .45 ,
s
=0 −2 −
j 2 .45
(
8-6 (h) Q(s ) = s − 2 2
)( s + 4)
(
P (s ) = s s + 2s + 2 2
)
Since Q ( s ) and P ( s ) are of the same order, there are no asymptotes.
Breakaway Points:
− 2,
6.95
141
8-6 (i)
Q (s)
Asymptotes:
= ( s + 2 )( s + 0 . 5) K > 0:
180
P( s)
o
Breakaway-point Equation: Breakaway Points:
=s
−4.0205,
ε −ϕ s
2
1
K < 0: s
4
0
o
+5s + 4s −1 = 0 3
2
0.40245 The other solutions are not breakaway points.
142
8-6 (j)
Q(s ) = 2 s + 5 Asymptote s:
K > 0:
P (s ) = s 60
o
,
180
2
o
(s
2
,
300
Intersect of Asymptotes;
σ1 = Breakaway-point Equation:
6s
4
)
+2s +1 = s
0
o
2
( s + 1)
K < 0:
+ 0 − 1 − 1 − ( − 2 . 5)
+ 28
4 s
3
−1
+ 32 143
s
2
2
o
0 ,
=
0 .5 3
+ 10
s
=0
120
o
,
= 0 .167
240
o
Breakaway Points:
8-7 (a) Asymptotes:
K > 0:
−0.5316, −1, −3.135
0,
45
o
,
135
o
,
225
Intersect of Asymptotes:
σ1 = 5
+ 65
4s
Breakaway Points:
−0.6325,
ζ = 0 . 707
,
,
315
o
−2 − 2 − 5 − 6 − ( −4 )
Breakaway-point Equation:
When
o
5 s
4
−1
+ 396 −5.511
K = 13.07
144
s
3
+ 1100
= − 2 . 75 s
2
+ 1312
(on the RL)
s
+ 480 = 0
8-7 (b) Asymptotes:
K > 0:
45
o
,
135
o
,
225
Intersect of Asymptotes:
σ1 = Breakaway-point Equation: When ζ = 0 . 707 , K = 61.5
4s
3
0
o
,
315
o
− 2 − 5 − 10
= −4.25
4
+ 51 s + 160 2
145
s
+ 100 = 0
8-7 (c) Asymptotes:
K > 0:
180
o
Breakaway-point Equation: Breakaway Points:
−1.727
When
K = 9.65
ζ = 0 . 707
,
s
4
+ 4 s + 10 3
s
+ 300
2
(on the RL)
146
s
+ 500 = 0
8-7 (d)
K > 0:
90
o
,
270
o
Intersect of Asymptotes:
σ1 = When
ζ = 0 . 707
,
−2 − 2 − 5 − 6 4
−2
K = 8.4
147
= −7 . 5
8-8 (a) Asymptotes:
K > 0:
60
o
,
180
o
,
300
Intersect of Asymptotes:
σ1 =
0
o
− 10 − 20
= −10
3
Breakaway-point Equation: 3 s + 60 s + 200 = 0 2
148
Breakaway Point: (RL)
−4.2265,
K = 384.9
(b) Asymptotes:
K > 0:
45
o
,
135
o
,
225
Intersect of Asymptotes:
σ1 = Breakaway-point Equation: 4 s + 27 s Breakaway Points: (RL) −0.4258 3
2
0
o
,
315
−1 − 3 − 5
o
= −2 .25
4
+ 46 s + 15 = 0 K = 2.879 ,
149
−4.2537
K = 12.95
150
8-10
P (s)
= s + 25 3
Asymptotes:
s
2
+ 2 s + 100 K
t
> 0:
Q (s) 90
o
,
270
= 100
Intersect of Asymptotes:
σ1 = 3
−25 − 0
s
Breakaway Points:
−2.2037,
8-11 Characteristic e quation:
s
3
3
−1
= −12 . 5
+ 12 . 5 s − 50 = 0
Breakaway-point Equation: (RL)
s
o
2
−12.162
+5s + K ts + K = 0 2
151
(a) Kt = 0 :
P (s ) = s
2
( s + 5)
Q (s) = 1 o
Asymptotes: K > 0: 60 , Intersect of Asymptotes:
180
o
,
σ1 = Breakaway-point Equation:
8-11 (b)
P (s)
3s
= s + 5 s + 10 = 0 3
2
o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
2
−5 − 0
= −1. 667
3
+ 10
Q (s)
270
o
300
s
=
=0
s
o
152
Breakaway Points:
0,
−3.333
σ1 =
−5 − 0 2
−1
=0
Breakaway-point Equation: 2 s + 5 s − 10 = 0 There are no breakaway points on RL. 3
8-12
P( s)
= s + 116 2
. 84 s
Asymptotes:
+ 1843 J
L
= 0:
Q (s) 180
(RL)
2
2 . 05 s ( s
+ 5)
o
Breakaway-point Equation: − 2 . 05 s Breakaway Points:
=
4
− 479 0,
s
3
− 12532
−204.18
s
153
2
− 37782
s
=0
(
)
8-13 (a) P (s ) = s s − 1 2
Asymptotes:
Q (s ) = ( s+ 5)( s + 3) K > 0:
180 4
o
+ 16
Breakaway-point Equation:
s
Breakaway Points:
0.5239,
(RL)
s
3
+ 46
s
2
−12.254
154
− 15 = 0
(
)
8-13 (b) P (s ) = s s + 10s + 29 2
Q (s ) = 10(s + 3) o
Asymptote s: K > 0: 90 , Intersect of Asymptotes:
270
o
σ1 =
0
− 10 − ( −3) 3
−1
Breakaway-point Equation: 20 s + 190 s There are no breakaway points on the RL. 3
2
+ 600
155
= −3 . 5 s
+ 870 = 0
8-14 (a)
P( s)
= s ( s + 12 . 5)( s + 1 )
Q (s)
Asymptotes: N > 0: Intersect of Asymptotes:
= 83 .333 60
o
,
180
o
,
300
σ1 = + 27 −0.4896
Breakaway-point Equation: 3 s Breakaway Point:
(RL)
2
− 12 . 5 − 1
0
s _12 .5
o
3
=0
156
= − 4. 5
8-14 (b)
P( s)
= s + 12 . 5 s + 833 2
A > 0:
180
.333
Q (s)
= 0 .02
2
s (s
+ 12 . 5)
o
Breakaway-point Equation: 0 . 02 s Breakaway Points: (RL) 0
4
+ 0 . 5 s + 53 .125 3
157
s
2
+ 416
. 67 s
=0
8-14 (c)
P (s)
= s + 12 . 5 s + 1666 . 67 = ( s + 17 . 78 Q ( s ) = 0 . 02 s ( s + 12 . 5) 3
2
Asymptotes:
K
o
> 0:
180
)( s
− 2 . 64 +
j 9 . 3)( s
− 2 . 64 −
j 9 . 3)
o
Breakaway-point Equation: 0 . 02 s + 0 . 5 s + 3 .125 s Breakaway Point: (RL) −5.797 4
3
158
2
− 66 .67 s − 416
. 67
=0
8-15 (a)
A
= K o = 100
:
P (s)
Asymptotes: N > 0: Intersect of Asymptotes:
= s ( s + 12 . 5)( s + 1 ) 60
o
σ1 = 2
180 0
Q (s)
300
− 1 − 12 . 5
o
= − 4. 5
3
+ 27 s + 12 . 5 = 0
Breakaway-point Equation:
3s
Breakaway Points:
−0.4896
(RL)
o
159
= 41 . 67
8-15 (b)
P( s)
= s + 12 . 5 + 1666
Q (s)
2
= 0 . 02
Asymptotes:
2
s (s
. 67
= ( s + 6 .25 +
j 40 . 34 )( s
+ 6 .25 −
j 40 . 34 )
+ 12 . 5)
A > 0:
180
o
Breakaway-point Equation:
0 . 02 s
Breakaway Points:
0
(RL)
4
+ 0 . 5 s + 103 3
160
.13 s
2
+ 833
. 33 s
=0
8-15 (c)
P (s)
= s + 12 . 5 s + 833
Q (s)
3
2
. 33
= ( s + 15 . 83)(
s
− 1. 663 +
j 7 . 063)( s
− 1. 663 −
j 7 .063)
= 0 . 01 s ( s + 12 . 5)
Asymptotes:
K
o
> 0:
180
o
Breakaway-point Equation:
0 . 01 s
Breakaway Point:
−5.37
(RL)
4
+ 0 .15
s
3
+ 1. 5625
161
s
2
− 16 . 67 s − 104.
17
=0
8-16 (a)
P (s)
=s
2
(s
+ 1 )( s + 5)
Q (s)
=1
Asymptotes: K > 0: Intersect of Asymptotes:
o
45
,
135
o
,
σ1 = Breakaway-point Equation:
4s
3
+ 18
s
2
o
225 0
315
+0 −1− 5
+ 10
162
,
o
= −1. 5
4 s
=0
Breakaway point: (RL) 0, −3.851
(b)
P (s)
=s
2
(s
+ 1 )( s + 5)
Q (s)
Asymptotes: K > 0: Intersect of Asymptotes:
= 5s +1 60
o
,
σ1 =
o
180 0
,
300
o
+ 0 − 1 − 5 − ( −0 .2 ) 4
−1
=−
5. 8
= −1. 93
3
Breakaway-point Equation: 15 s + 64 s + 43 s + 10 s = 0 Breakaway Points: (RL) −3.5026 4
8-17
P( s) Q (s)
=s = 10
2
(s
+ 1 )( s + 5) + 10 = ( s + 4. 893)(
3
s
+ 1.896
2
)( s
− 0 . 394 +
j 0 .96 )( s
− 0 . 394 +
j 0 . 96 )
s
Asymptotes:
T
d
> 0:
60
o
,
180
o
,
Intersection of Asymptotes:
300
o
σ1 =
−4. 893 − 1. 896 + 0 . 3944 + 0 . 3944
There are no breakaway points on the RL.
163
4
−1
= −2
8-18 (a)
K = 1:
P (s)
=s
3
(s
+ 117
Asymptotes:
.23)( s K
L
+ 4882
90
Intersect of Asymptotes:
σ1 = Breakaway Point: (RL)
8-18 (b)
K = 1000:
P( s)
=s
Q (s)
= 1010(
3
(s
Asymptotes:
K
L
> 0:
90
+ 5000
s s
+ 4921
,
= 1010(
s
+ 1. 5948
)( s
+ 114.41
)( s
+ 4884
o
,
270
−117
.23
− 4882
.8
+ 1. 5948 + 114.41 + 4884 5
.23)( s
3
o
o
−3
= −0 .126
0
+ 117
= 1010(
Q (s)
.8 )
> 0:
270
s
+ 4882 2
. 6 )( s
.8 )
+ 5 . 6673 × 10 + 39 .18 +
5
s
+ 891089
j 423 . 7 )( s
110 )
+ 39 .18 − 423
.7 )
o
Intersect of Asymptotes:
σ1 =
−117
.23
− 4882
.8
+ 4921 5
164
−3
.6
+ 39 .18 + 39 .18
= −0 . 033
)
Breakaway-point Equation:
+ 5.279 × 10 Breakaway points: (RL) 0, −87.576 2020 s
8-19
7
+ 2 .02 × 10
7
s
6
Characteristic Equation:
s
3
10
+ 5000
s
s
2
+ 1. 5977 × 10
5
+ 572
, 400 s
P( s)
3
≅ ( s + 1. 5945)(
2
s
+ 115
Q ( s)
.6 )
Breakaway-point Equation: 1200 s
2
≅ 10 .24
+ 3775
s
s
s
s
4
+ 1. 8655 × 10
16
,000
+JL
ε
+ 115
. 6 )( s
+ 4882
+ 900
= s + 5000 s + 572 , 400 s + 900 ,000 = ( s + 1. 5945)( Since the pole at −5000 is very close to the zero at −4882.8, P ( s ) and P( s)
13
10 s
3
s
3
+ 1. 54455 × 10
+ 50 ,000 .8 )
s
2
ϕ=
Q (s)
Q ( s ) can be approximated as:
2
=0
Breakaway Points: (RL): 0, −3.146
165
18
s
2
=0
0
= 10
2
s (s
+ 5000
)
8-20 (a) α = 12 :
P ( s)
=
2
s (s
+ 12
= s +1
Q (s)
) o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
270
o
K < 0:
σ1 = Breakaway-point Equation:
8-20 (b) α = 4 :
P( s)
=s
2
(s
+4) o
Asymptotes: K > 0: 90 , Intersect of Asymptotes:
Q (s) 270
3
2s
o
+ 0 − 12 − ( −1 ) 3
−1
+ 15 s + 24 2
s
=0
180
o
= −5 . 5 Breakaway Points:
0,
−2.314, −5.186
= s +1 o
K < 0: 0 ,
σ1 = Breakaway-point Equation: 2 s + 7 s 3
0
o
0 ,
2
0
o
180
+ 0 − 4 − ( − 10 3
+8 s = 0
−1
= − 1. 5
Breakaway Points: K > 0 0. None for K < 0.
166
(c) Breakaway-point Equation:
2s
2
+ ( α + 3) s + 2 s = 0
Solutions:
s
=−
α +3 4
±
(α
+ 3) − 16 α 2
, s
4
For one nonzero breakaway point, the quantity under the square-root sign must equal zero. Thus,
α − 10 α + 9 = 0 , α = 1 or α = 9 . 2
cancellation in the equivalent G ( s ). When
8-21 (a)
P (s)
=s
2
(s
+ 3)
Q ( s)
α = 9,
The a nswer
is
α = 9.
= s +α
Breakaway-point Equation: 2 s + 3 (1 + α ) s + 6 α = 0 The roots of the breakaway-point equation are: 3
s
=
−3 ( 1 + α ) 4
±
The
α = 1 solution represents pole-zero = − 3. σ 1 = −4.
the nonzero breakaway point is at s
9( 1
+ α ) − 48 α 2
4
167
=0
For no breakaway point other than at s = 0 , set
9( 1 + α )
2
− 48 α < 0 or - 0 . 333 < α < 3
Root Locus Diagram with No Breakaway Point other than at s = 0.
8-21 (b) One breakaway point other than at s = 0: α = 0 . 333
,
168
Breaka way po int at s
= −1.
8-21 (d) Two breakaway points :
α > 3:
169
8-22
Let the angle of the vector drawn from the zero at s
=
j12
to a point s on the root locuss near the zero 1
θ. θ1 =
Let angle
of th e vect or dra wn fro m the
pole a t j 10 to s .
θ2 =
angle
of th e vect or dra wn fro
pole a t 0 to s .
θ3 =
angle
of th e vect or dra wn fro m the
pole a t
−
j 10 to s1 .
θ4 =
angle
of th e vect or dra wn fro m the
zero a t
−
j 12 to s .
be
m the
1
1
1
Then the angle conditions on the root loci are:
θ = θ 1 − θ 2 −θ 3 + θ 4 = θ 1 = θ 2 = θ 3 = θ 4 = 90
odd m ultipl es of o
Thus,
θ
180 o
= 0
The root loci shown in (b) are the correct ones.
170
o
Chapter 9
9-1 (a)
FREQUENCY DOMAIN ANALYSIS
ωn =
K=5
(b)
= 2 .24
5
= 4. 62
21 . 39
ω n = 10
K = 100
= 2 . 944
( 9 . 38 dB)
r
= 15 . 34
( 23 . 71 dB)
r
= 4.17
r
=1
r
= 1. 57
r
= ∞ ( unstab
9-2 (a)
M
(b)
M
(c)
M
(d)
M
(e)
M
(f)
M
(g)
M
(h)
M
r
1−ς
= 3 .27
2
( 9 . 8 dB)
r
= 4.12
( 12 . 3 dB)
2ζ
d / sec
Thus, minimum
Minimum BW =
2
ω n = 17 . 7 ωn
M
/ sec
r
rad / sec
( ( 1 − 2ζ
2
)+
1
= 2ζ
= 1. 618
rad / sec
rad / sec
BW = 2.44 r ad / sec
BW = 2.07 r ad / sec
/ sec
ζ
1 − 0 .416
BW = 5.16 r ad / sec
= 0.59
ζ + 2 . 917 ζ
2
= 0 .1
ωn Maxi mum
M
4ζ − 4ζ + 2 4
2
)
r
1/2
9-4
sec
= 1. 05 = 20.56 rad/sec
−πζ
Maximu
m over shoot
= 0.2
1
rad / sec
=1 −ζ
ω r = 9 .45 ra
BW = 1.12 r ad / sec
d / sec
T hus,
=
r
ωr = 0
BW = 9.18 r ad / sec
rad / sec
ω r = 3 . 5 rad
t
=1
r
BW = 0.46 r ad / sec
ω r = 0 .82
= 0.1
r
BW = 6.223
rad / sec
= 1. 05
1−ζ
M
rad / sec
ω r = 6 .25 ra
ω r = 1.25 ra
m over shoot
= 0 . 327
BW = 4.495
9-3 1
= 0 . 707
/ sec
ωr =4
ω r = 1. 5 rad
le)
= 3 . 09
r
M
2
rad / sec
6.54 =
ω r = 3 rad
( 3 . 918 dB)
=
6.54 =
20
ωr =0
r
M
ζ
rad / sec
ζ
rad / sec
( 12 .4 dB)
( 0 dB)
Maximu
= 1.46
9.24
ωr = ωn (c)
6.54 = 4.48
ωn =
K = 21.39
ζ
rad / sec
Thus,
0 .2
=e
1 −ζ
171
2
ζ = 0.456
d / sec
M
r
1
= 2ζ
= 1.232
1−ζ
Maximum M
t
2
= 1.232
r
r
=
1 − 0 .416
ζ + 2 . 917 ζ ωn
Minimum BW =
( (1 − 2ζ
9-5
2
)+
2
= 0 .2
Thus, minimum
4ζ − 4ζ + 2 4
2
)
1/2
ω n = 14. 168
= 18.7
rad/sec
rad/sec
− πζ
Maximum overshoot = 0.3
M
r
1
= 2ζ
= 1.496
1−ζ
Maximum M
r
=
t
2
1.496
=
r
1 − 0 .416
1 −ζ
=e
Thus, 0 . 3
2
ζ + 2 . 917 ζ
ζ = 0 . 358 2
ωn
Minimum BW =
( (1 − 2ζ
2
)+
= 0 .2
Thus, minimum
4ζ − 4ζ + 2 4
2
9-6
)
1/2
ω n = 6 .1246
= 1.4106
rad/sec
−πζ
M
r
= 1.4 =
ω r = 3 rad t
max
1
=
Thus, 2ζ
1−ζ
/ sec
= ωn
π ωn
1
−ζ
= 0.387
1 −ζ
Maximum overshoot = e
2
= 0.2675
1
− 2ζ
2
= 0 .8367 ω n
π 3 . 586
1
− ( 0 . 387
rad/sec
= 0 .95 )
sec
ωn =
3
= 3 . 586
At
ω = 0,
M
2
Unit-step Response:
9-7 T BW (rad/sec) Mr ________________________________________________________________ 1.14 1.17 1.26 1.63
1.54 1.09 1.00 1.09
172
r ad/sec
0 . 8367
This indicates that the steady-state value of the unit-step response is 0.9.
0 0.5 1.0 2.0
(26.75%)
2
= 2
ζ
= 0 .9 .
rad/sec
3.0 1.96 4.0 2.26 5.0 2.52 _________________________________________________________________
1.29 1.46 1.63
9-8 T BW (rad/sec) Mr _________________________________________________________________ 0 1.14 0.5 1.00 1.0 0.90 2.0 0.74 3.0 0.63 4.0 0.55 5.0 0.50 _________________________________________________________________
1.54 2.32 2.65 2.91 3.18 3.37 3.62
9-9 (a) 20
L( s )
=
When
ω = 0: ∠L ( j ω ) = −90 o
s ( 1 + 0 . 1 s )( 1
L ( jω ) =
1
+ 0 .5 s )
Pω
L( jω )
(
−0.6ω + jω 1 − 0.05ω
− 0 .05 ω = 0 2
Thus ,
ω = ±4.47
P
=0
=∞
When
2
)
=
(
0.36ω + ω 4
2
o
o
2
(1 − 0.05ω ) 2
L ( j 4.47 )
rad / sec
Φ 11 = 270 = ( Z − 0.5 Pω − P ) 180 = ( Z − 0.5 )180 o
ω = ∞: ∠L ( jω ) = −270 o
20 −0.6ω − jω 1 − 0.05ω 2
20 2
= 1,
)
L( jω )
Setting Im L ( jω )
2
= −1. 667
Thus, Z =
360 180
o o
=2
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane.
Nyquist Plot of L ( jω ):
173
=0 =0
(b) L( s )
=
10 s ( 1 + 0 . 1 s )( 1
Based on the analysis conducted in part (a), the intersect of the negative
+ 0 .5 s )
real axis by the L ( jω ) plot is at
Φ 11 = −90
o
=
δ− Z
−0.8333, and the corresponding ω
0 . 5 Pω
−P
ι
180
o
= 180
− 90
Z
is 4.47 rad/sec. o
Thus,
Z
= 0 . The closed-loop system is stable.
Nyquist Plot of L ( jω ):
(c) L( s )
=
When When
100( 1 + s ) s ( 1 + 0 . 1 s )( 1
ω = 0 : ∠L ( j 0 ) = −90 o
ω
ω = ∞: ∠L ( jω ) = −270
L( j0 ) o
= 1,
P
+ 0 .2 s )( 1 + 0 . 5 s )
=∞
L( jω )
When
=0
P
= 0.
ω = ∞: When
174
∠L ( j∞ ) = −270
o
L ( j ∞)
ω = ∞: ∠L ( jω ) = −270
o
=0 L( jω )
=0
L ( jω ) =
( 0.01ω
100(1 + jω )
Setting Im L ( jω ) Thus,
− 0.8ω
4
=0
ω = 64. 55
2
) + jω (1 − 0.17ω ) 2
0 .01 ω
ω = ± 8 . 03
2
4
=
(
) − jω (1 − 0.17ω ) ) + ω ( 1 − 0.17ω )
100(1 + jω ) 0.01ω − 0.8 ω
( 0.01ω
4
4
− 0.8ω
− 0 .8 ω 2 − 1 + 0 .17 ω 2 = 0
2
2
2
2
2
2
2
ω 4 − 63 ω 2 − 100 = 0
rad/sec
100 ( 0.01ω 4 − 0.8ω 2 ) + ω 2 (1 − 0.17ω 2 ) L ( j 8.03) = = − 10 ( 0.01ω 2 − 0.8ω 2 )2 + ω 2 (1 − 0.17ω 2 ) 2 ω=8.03 Φ 11 = 270 = ( Z − 0.5 Pω − P ) 180 = ( Z − 0.5 )180 o
o
o
Thus, Z = 2
The closed-loop system is
unstable. The characteristic equation has two roots in the right-half s-plane.
Nyquist Plot of L ( jω ):
(d) 10
L( s )
=
When
ω = 0: ∠L ( j ω ) = −180 o
P
ω
s ( 1 + 0 .2 s )( 1 + 0 . 5 s ) 2
L ( jω ) =
( 0.1ω
10 4
Setting Im L ( jω ) origin where
L ( jω )
ω = ∞.
−ω
2
) − j 0.7ω
= 0 , ω = ∞.
3
=
=2
P
=∞
(
=0
10 0.1ω − ω + j0.7ω
( 0.1ω
4
4
2
−ω
ω = ∞: ∠L ( jω ) = −360 o
When
2
)
2
+ 0.49ω
3
)
6
The Nyquist plot of L ( jω ) does not intersect the real axis except at the
175
L ( jω )
=0
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 1) 180 o
Thus, Z = 2 .
o
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( jω ):
9-9 (e) L(s ) =
When
(
3( s + 2)
s s + 3s + 1 3
ω = 0 : ∠L ( j 0 ) = −90 o
L ( jω ) =
(ω
3( jω + 2) 4
− 3ω
ω − 3ω − 2 = 0 4
Pω = 1
)
2
2
) + jω or
=
P=2
L( j0 )
=∞
(
3( jω + 2) ω − 3ω
(4
ω = 3 . 56 2
4
ω = ∞: ∠L ( j ∞) = −270 o
When 4
− 3ω
2
)
2
) − jω
2
+ω
ω = ± 1.89
o
o
Setting Im L ( jω )
2
rad/sec. L ( j 1.89 )
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 2.5 ) 180 = − 90
o
=3
Thus, Z = 2
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane.
Nyquist Plot of L ( jω ):
176
L ( j∞)
=0 = 0,
9-9 (f) L(s ) =
When
0.1
(
2
ω = 0 : ∠L ( j 0 ) = −90 o
L ( jω ) = ω =∞
Pω = 1
)
s ( s + 1) s + s + 1
or
(ω
L( j0 )
0.1
4
− 2ω
2
) + jω (1 − 2ω )
ω = 0 .5 2
2
ω = ±0 . 707
=
P=0
=∞
When
(
(ω
4
− 2ω
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 0.5 ) 180 = − 90 Nyquist Plot of L ( jω ): o
9-9 (g) L(s ) =
100
(
s ( s + 1) s + 2 2
)
Pω = 3
2
2
P=0
177
o
L ( j∞)
=0
2
2
2
L ( j 0 . 707 )
rad / sec
o
) − jω (1 − 2ω ) ) + ω ( 1 − 2ω )
0.1 ω − 2ω 4
ω = ∞: ∠L ( j ∞) = −360 o
2
2
Settiing Im L ( jω )
=0
= −0 .1333
Thus, Z = 0
The closed-loop system is stable.
When
ω = 0 : ∠L ( j 0 ) = −90 o
The phase of L ( jω ) is discontinuous at
(
Φ11 = 35.27 + 270 − 215.27 o
o
=∞
L( j0 )
o
ω = 1.414
) = 90
o
ω = ∞: ∠L ( j ∞) = −360 o
When
L ( j∞)
=0
rad/sec.
Φ11 = ( Z − 1.5) 180 = 90 o
o
Thus, P
11
=
360 180
o o
=
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of L ( jω ):
9-9 (h) L( s )
When
=
10( s s(s
+ 10
)
+ 1 )( s + 100
P
ω
)
=1
ω = 0 : ∠L ( j 0 ) = −90 o L ( jω ) =
P
=∞
L( j0 )
−101ω + jω (100 − ω )
Setting Im L ( jω )
2
= 0, ω = 0
ω = ∞: ∠L ( j ∞) = −180 o
When
=
(
10( jω + 10) − 101ω − jω 100 − ω 2
10( jω + 10) 2
=0
10201ω + ω 4
2
(100 − ω ) 2
2
)
2
is the only solution. Thus, the Nyquist plot of L ( jω ) does not intersect
the real axis, except at the origin.
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 0.5 ) 180 = − 90 o
o
The closed-loop system is stable.
Nyquist Plot of L ( jω ):
178
o
Thus, Z = 0 .
L ( j∞)
=0
2
9-10 (a) L( s )
K
= s(s
+ 2 )( s + 10
For stability, Z = 0.
Pω
)
=1
P
o
Φ11 = −0 . 5 Pω × 180
=0
= −90
o
This means that the (−1, j0) point must not be
enclosed by the Nyquist plot, or
0 < 0 .004167 K Nyquist Plot of L ( jω ):
9-10 (b) L( s )
K(s
= s(s
< 1.
+ 1)
Pω
+ 2 )( s + 5)( s + 15)
For stability, Z = 0.
Φ11 = −0 . 5 × 180
by the Nyquist plot, or 0
o
< 0 . 000517
Thus,
0
< K < 240
=1
P
=0
= −90 9K
o
This means that the (−1, j0) must not be enclosed
0
Φ 11 = 180
Thus, the system is unstable for all K > 0. For K < 0, the critical point is (1, j0), all K < 0. Thus, the system is unstable for all values of K.
Nyquist Plot of L ( jω ):
9-11 (a) G (s)
=
K (s
+ 5)
2
P
ω
=0
P
=0
180
o
− 180
, not
Φ11 = 0
o
for
o
.
∠G ( j 0 ) = 0
o
(K
> 0)
∠G ( 0 ) = 180
o
∠G ( j ∞ ) = 0
o
(K
0)
(K < 0)
G ( j ∞)
=0
For stability, Z = 0.
Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o
0 K
0)
∠G ( 0 ) = 180
o
(K
< 0)
G ( j0 )
K
=
625 G ( j∞ )
=0
o
(K > 0)
∠G ( j ∞ ) = 180
181
o
(K < 0)
G ( j ∞)
=0
For stability, Z = 0.
Φ 11 = − ( 0.5 Pω + P ) 180 = 0 o
0
2500
K K
< − 625
−625 < K < 0
o
o
Stable
Φ 11 = 360
o
Φ11 = 180
o
Unstable Unstable
o
Φ11 = 0
Stable
The system is stable for − 625 < K < 2500.
9-12
(
)
(
)
s s + 2s + s +1 + K s + s + 1 = 0 3
2
Leq ( s) =
(
2
K s + s +1
(
2
)
s s + 2s + s + 1
Leq ( jω ) =
3
2
K
(ω
Setting Im L
eq
4
(1 − ω
−ω
( jω )
2
Pω = 1
)
P =0
) + jω = K − ( ω ) + jω (1 − 2ω ) (ω 2
2
L eq ( j 0 )
+ω
4
4
−ω
2
ω 4 − 2ω 2 + 1 = 0 Thus,
ω = ±1
L eq ( j 1)
rad/sec are the real solutions.
= −K
For stability,
Φ11 = − ( 0 . 5 Pω + P )180
o
= − 90
o
When K = 1 the system is marginally stable. K K
>0 −1
2 s
K
1
− 2K +1
2
+1
K s
0
=
− 1)
(K
2
+1
K
K
K>0
When K = 1 the coefficients of the s row are all zero. The auxiliary equation is s + 1 = 0 The solutions are ω = ± 1 rad/sec. Thus the Nyquist plot of L eq ( j ω ) intersects the −1 point when K = 1, when ω = ± 1 1
2
∞, except at K = 1.
rad/sec. The system is stable for 0 < K
0
Thus
K
P
>9
0
10,0 00
Root Contours: G eq ( s )
(c)
K
I
=
100 K s
3
+ 10
s
2
P
+ 100
s s
+ 10 ,000
=
100 K P s (s
+ 23 . 65)(
s
− 6 .825 +
j 19 .4 )( s
100 ( K P s + 100 )
(
s s + 10 s + 100 2
)
The following maximum overshoots of the system are computed for various values of K
ymax
j19 .4 )
= 100 G ( s) =
KP
− 6 . 825 −
15
20
1.794
1.779
When K P = 25,
22 1.7788
24 1.7785
25
26
30
1.7756
1.779
1.782
minimum ymax = 1.7756
10-11 (a) Forward-path Transfer Function:
220
P
. 40 1.795
100
1000
1.844
1.859
G ( s) =
100 ( K P s + K I )
(
s s + 10 s + 100 2
For K
)
v
(
=
100 K
= 10 ,
I
100
K
I
= 10
)
s3 + 10 s2 + 100 K + 1 s + 1000 = 0 P
(b) Characteristic Equation: Routh Tabulation: s
3
s
2
s
1
s
0
1
100 + 100 K
10
1000
100 K
P
For stability,
KP > 0
0
P
1000
Root Contours: G
(c)
eq
(s)
=
100 K s
3
+ 10
s
2
P
s
+ 100
s
The maximum overshoots of the system for different values of K
+ 1000
P
ranging from 0.5 to 20 are
computed and tabulated below. KP ymax
0.5
1.0
1.393
1.275
1.6
1.7
1.8
1.9
1.2317
1.2416
1.2424
1.2441
When K P = 1.7,
2.0
3.0
5.0
1.246
1.28
1.372
maximum ymax = 1.2416
10-12 Gc (s ) = K P + K D s + where K
P
KI
=
s K
K Ds + K P s + K I 2
=
P2
s + K D1K I 2
= (1 + K D1 s ) K P2 + K
D
= K D1K P 2
Forward-path Transfer Function:
221
s
K I2
K
I
=
K
I2
10 1.514
20 1.642
G ( s) = G c ( s ) G p ( s) = Thus K
I
100 ( 1 + K D1 s ) ( K P2 s + K I 2 )
(
s s + 10 s + 100 2
For stability, K
s s + 10 s + 100 2
> 9.
Select K
s + 10 s + (100 + 100 KP 2 ) s + 10,000 = 0 3
)
P2
= 10
K
P
=
K
P2
D1
= 0 .2 ,
2
for fast rise time.
1000 ( 1 + K D 1s ) ( s + 10 )
G ( s) = When K
= K I 2 = 100
Characteristic Equation:
100 ( K P 2 s + 100 )
P2
sG ( s )
=0)
D1
Forward-path Transfer Function:
(
s→0
= K I 2 = 100
Consider only the PI controller, (with K
G ( s) =
= lim
Kv
)
(
s s + 10 s + 100 2
)
K
the rise time and overshoot requirements are satisfied.
D
=
K
D1
K
P2
= 0 .2 × 10 = 2
+ K D 1 K I 2 = 10 + 0 .2 × 100 = 30 G (s) c
= 30 + 2 s +
100 s
Unit-step Response
10-13 Process Transfer Function: Gp ( s) =
Y ( s) U (s )
=
e
−0.2s
1 + 0.25 s
1
≅
(1 + 0.25s ) (1 + 0.2s + 0.02s
2
)
(a) PI Controller: KP + G ( s) = G c ( s ) G p ( s) ≅
For K
v
= 2,
K
G ( s) =
Thus
v
KI
s 2 1 + 0.2 s + 0.02 s
(1 + 0.25s ) ( = lim
s→0
sG ( s )
=
4
200 ( 2 + K P s )
(
200 K
× 50
s ( s + 4 ) s + 10s + 50 2
I
=
K
I
)
=
200 ( K P s + K I )
(
s ( s + 4 ) s + 10s + 50 2
=2
Thus
)
The following values of the attributes of the unit-step response are computed for the system with various values for K . P
KP
ts
Max overshoot (%)
(sec)
222
ts
(sec)
) K
I
=2
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0
19.5 13.8 8.8 4.6 1.0 0 0 0 0
0.61 0.617 0.615 0.606 0.5905 0.568 0.541 0.5078 0.44
223
2.08 1.966 1.733 0.898 0.878 0.851 1.464 1.603 1.618
224
225
Unit-step Response.
226
The unit-step response shows a maximum overshoot of 26%. Although the relative damping ratio of the complex roots is 0.707, the real pole of the third-order system transfer function is at −0.667 which adds to the overshoot.
(c) G ( s) = G c ( s ) G p ( s) =
For K
v
= 100
, K
I2
=
K
For a small overshoot, K
= 100
I
G ( s) =
D1
0.00667 (1 + K D1 s ) ( K P2 s + K I 2 )
(
s s + 0.00667 2
. Let us select K
P2
)
= 50 .
Then
0.00667 (1 + K D1 s ) ( 50 s + 100 )
(
s s + 0.00667 2
)
must be relatively large. When K
D1
= 100
, the maximum overshoot is
approximately 4.5%. Thus, KP
=
K
K
D
=
K
I
= 100
K
P2 D1
+ K D 1 K I 2 = 50 + 100 × 100 = 10050 K
P2
= 100 × 50 = 5000
System Characteristic Equation: s + 33 . 35 s + 67 . 04 s + 0 .667 = 0 3
Roots:
2
− 0.01,
Unit-step Response.
227
− 2.138,
− 31.2
10-16 (a) G (s) p
=
Z (s)
=
F (s)
1 2
Ms
=
+Ks
1 150 s
2
=
+1
0 . 00667 s
+ 0 . 00667
2
The transfer function G p ( s ) has poles on the j ω axis. The natural undamped frequency is
ω n = 0 . 0816
rad/sec.
(b) PID Controller:
= lim
Kv
s →0
sG ( s )
=
ζ = 1 and ω n = 1
(
2
s s + 0.00667 2
= 100
KI
T hus
KI
)
)
= 100
s + 0.00667 K D s + 0.00667 (1 + K P ) s + 0.00667 K I = 0 3
Characteristic Equation: For
(
0.00667 K D s + K P s + K I
G ( s) = G c ( s ) G p ( s) =
2
rad/sec, the second-order term of the characteristic equation is s
2
+ 2 s + 1.
Dividing the characteristic equation by the seond-order term.
s + ( 0.00667 K D − 2 ) s + 2s + 1 s + 0.00667 K D s + ( 0.00667 + 0.00667 K P ) s + 0.00667 K I 2
3
2
s + 2s + s 3
2
( 0.00667 K ( 0.00667 K ( 0.00667 K
D
− 2 ) s + ( 0.00667 KP − 0.99333 ) s + 0.00667 K I
D
− 2 ) s + ( 0.01334 K D − 4 ) s + 0.00667 K D − 2
P
− 0.01334 K D + 3.00667 ) s + 0.00667 K I − 0.00667 K D + 2
2
2
For zero remainder, 0 . 00667 K
−0 . 00667
K
P D
− 0 . 01334
K
+ 0 . 00667
K
D I
+ 3. 00667 = 0 +2 =0
From Eq. (2),
228
(1) (2)
0 . 00667 K
D
= 0 . 00667
K
P
= 0 . 01334
K
From Eq. (1), 0 . 00667 K
I
+ 2 = 2 . 667
D
− 3 . 00667 = 2 .3273
K
Thus
D
= 399 K
Th us
. 85
P
= 348
. 93
Forward-path Transfer Function:
(
0.00667 399.85 s + 348.93 s + 100
G ( s) =
2
(
s s + 0.00667 2
)
)
Characteristic Equation:
s 3 + 2.667 s 2 + 2.334 s + 0.667 = ( s + 1) Roots:
− 1,
− 1,
2
( s + 0.667 ) = 0
− 0.667
Unit-step Response.
The maximum overshoot is 20%.
10-17 (a) Process Transfer Function: G
p
( s)
=
4
G ( s) = G c ( s ) G p ( s) =
2
s
Characteristic Equation: K
P
= 0 .25
a nd
Forward-path Transfer Function
K
D
s
2
+ 4K Ds + 4 K P =
= 0 . 3535
Unit-step Response.
229
s
2
+ 1.414
s
4 ( KP + KD s) s +1 = 0
2
for
ζ = 0 .707
,
ωn =1
rad/sec
Maximum overshoot = 20.8%
(b)
Select a relatively large value for K The closed-loop zero at s
D
= −K P
and a small value for K / K
D
P
dynamics are governed by the other closed-loop poles. Let K The following results show that the value of K Kp
For BW
≤
40 rad/sec and M
P
= 10
D
and use small values of K
tr
.
= 1, we can again select
ts
(sec)
0 0 0
r
P
is not critical as long as it is small.
Max overshoot (%)
0.1 0.05 0.2
(c)
so that the closed-loop poles are real.
is very close to one of the closed-loop poles, and the system
(sec)
0.0549 0.0549 0.0549
K
D
= 10
and a small value for K
0.0707 0.0707 0.0707
P
.
The following frequency-domain results substantiate the design. Kp
PM (deg)
Mr
0.1 0.05 0.2
89.99 89.99 89.99
1 1 1
10-18 (a) Forward-path Transfer Function: G ( s) = G c ( s ) G p ( s) =
2
40 40 40
Characteristic Equation:
10,000 ( K P + K D s ) s
BW (rad/sec)
s
( s + 10)
Routh Tabulation:
230
3
+ 10
s
2
+ 10 ,000
K
D
s
+ 10 ,000
K
P
=0
s
3
s
2
s
1
s
0
1
10 ,000 K
10
10, 000 K
10,000 K
D
10,000
− 1000 K
K
D
P
The system is stable for K
0
P
P
(b) Root Locus Diagram: 10,000 K P
G ( s) =
s
Root Contours: 0 ≤ K
G
eq
D
< ∞,
(s)
=
2
( s + 10 )
K
= 0 . 001 ,
P
10 ,000 K s
3
+ 10
s
2
D
0 . 002 , 0 .005 , 0 . 01 . s
+ 10 ,000
231
K
P
P
>0
and
K
D
> 0 .1 K P
(c)
The root contours show that for small values of K means that if we choose K
P
P
the design is insensitive to the variation ofK
to be between 0.001 and 0.005, the value of K
for a relative damping ratio of 0.707. Let K
P
= 0 . 001
and K
D
= 0 . 005
D
P
. This
can be chosen to be 0.005 . G (s) c
= 0 . 001 + 0 . 005
The forward-path transfer function becomes
10 (1 + 5 s )
G ( s) =
s
( s + 10 )
−0.2, which is very close to s = 0, G(s) can be approximated as:
Since the zero of G ( s ) is at s =
G (s) ≅ For the second-order system,
2
ζ = 0 . 707 t
r
=
50 s ( s + 10 ) . Using Eq. (7-104), the rise time is obtained as
1 − 0 .4167
ζ + 2 . 917 ζ ωn
Unit-step Response:
232
2
= 0 . 306
sec
s.
(d) Frequency-domain Characteristics: 10 (1 + 5 s )
G ( s) =
s PM (deg)
2
( s + 10 )
GM (dB)
∞
63
Mr
BW (rad/sec)
1.041
7.156
10-19
0 25.92 ∗ A = 0 −2.36
1
0
0
−1 0 0 s −25.92 s 0 0 ∗ sI − A = 0 0 s − 1 2.36 0 0 s
0 0 0 0 0 1 0 0 0
s
0
0
∆ = sI − A = s 0 s − 1 + ∗
0
0
s3 2 1 25.92 s ∗ −1 ( sI − A ) = ∆ − 2.36 s 2 −2.36 s
s
−25.92
0
0
s
−1 = s
2.36
0
s
0 2
(s
2
− 25.92
s 0 0 3 2 −2.36 s − 25.92 s s − 25.92 3 −2.36 s 0 s − 25.92s 2
s
0
3
233
0
)
( sI − A )
−1
∗
−0.0732 s 2 0 −0.0732 1 −0.0732 s 3 ∗ −1 = B = ( sI − A ) 0 ∆ 0.0976 s 2 − 2.357 0.0976 3 0.0976 s − 2.357 s
(
Y ( s) = D sI − A
∗
)
−1
B = [0
0
(
0 ] sI − A
1
∗
)
−1
B=
s
2
(s
3
3
10-20
1
c
=
K
P
+
)
D
which
.
Let us first attempt to compensate the system with a PI controller. G (s)
− 25.92
)
2
The system cannot be stabilized by the PD controller, since the s and the s terms involve K D
2
2
s + 0.0976 s + ( 0.0976 KP − 25.92 ) s − 2.357 K D s − 2.357 KP = 0 4
Characteristic Equation:
require opposite signs for K
(
0.0976 s − 24.15
K
I
G ( s) = Gc ( s ) Gp ( s) =
Then
s
100 ( K P s + K I )
(
s s + 10 s + 100 2
)
Since the system with the PI controller is now a type 1 system, the steady-state error of the system due to a step input will be zero as long as the values of K and K are chosen so that the system is stable. Let us choose the ramp-error constant K
= 100
v
performance characteristics are obtained with K KP 10 20 30 40 50 100
I
P
I
. Then, K
= 100
I
= 100
. The following frequency-domain
and various value of K
PM (deg)
P
ranging from 10 to 100.
GM (dB)
∞ ∞ ∞ ∞ ∞ ∞
1.60 6.76 7.15 6.90 6.56 5.18
Mr
BW (rad/sec)
29.70 7.62 7.41 8.28 8.45 11.04
50.13 69.90 85.40 98.50 106.56 160.00
The maximum phase margin that can be achieved with the PI controller is only 7.15 deg when K
Thus, the overshoot requirement cannot be satisfied with the PI controller alone. Next, we try a PID controller.
Gc (s ) = K P + K Ds +
KI
=
(1 + K s ) (K D1
s
Based on the PI-controller design, let us select K becomes
G ( s) =
P2
= 30 .
P2
s + KI2 )
=
P
= 30 .
(1 + K s ) (K D1
s
P2
s + 100 )
s
Then the forward-path transfer function
100 ( 30 s + 100 )( 1 + K D1s )
(
s s + 10 s + 100 2
)
The following attributes of the frequency-domain performance of the system with the PID controller are obtained for various values of K ranging from 0.05 to 0.4. D1
K D1 0.05 0.10 0.20 0.30
PM (deg) 85.0 89.4 90.2 90.2
GM (dB)
∞ ∞ ∞ ∞
234
Mr
BW (rad/sec)
1.04 1.00 1.00 1.00
164.3 303.8 598.6 897.0
0.40
∞
90.2
We see that for values of K
D1
K
D1
P
=
= 0 .2, K
1201.0
greater than 0.2, the phase margin no longer increases, but the
bandwidth increases with the increase in K K
1.00
P2
K
I
=
D1
K
. Thus we choose
I2
= 100
,
K
D
=
K
D1
K
P2
= 0 .2 × 30 = 6 ,
+ K D 1 K I 2 = 30 + 0 .2 × 100 = 50
The transfer function of the PID controller is
G (s) c
= 50 + 6 s +
100
s The unit-step response is show below. The maximum overshoot is zero, and the rise time is 0.0172 sec.
10-21 (a) Gp (s ) =
4 s
2
G ( s) = Gc ( s ) G p ( s ) =
4 (1 + aTs ) s (1 + Ts )
Root Contours: ( T is fixed and a varies)
235
2
G
eq
(s)
=
4 aTs Ts
3
+ s +4 2
small value for T and a large and a = 100.
Select a value for a. Let T = 0.02
G (s) c
=
1+ 2 s
G ( s) =
1 + 0 . 02 s
400 ( s + 0.5) s
The characteristic equation is The roots are: The system transfer function is
s
3
+ 50
s
2
+ 400
s
2
( s + 50)
+ 200 = 0
−0.5355, −9.3, −40.17 Y (s)
=
R (s )
400 ( s + 0.5 )
( s + 0.5355) ( s + 9.3) ( s + 40.17 )
Since the zero at −0.5 is very close to the pole at −0.5355, the system can be approximated by a secondorder system,
Y (s)
=
R (s )
373.48
( s + 9.3)( s + 40.17 )
The unit-step response is shown below. The attributes of the response are: Maximum overshoot = 5%
t
s
= 0 . 6225
Unit-step Response.
236
sec
t
r
= 0 .2173
sec
The following attributes of the frequency-domain performance are obtained for the system with the phase-lead controller. PM = 77.4 deg GM = infinite Mr = 1.05 BW = 9.976 rad/sec
10-21 (b)
The Bode plot of the uncompensated forward-path transfer function is shown below. The diagram shows that the uncompensated system is marginally stable. The phase of G ( jω ) is
ωm
frequencies. For the phase-lead controller we need to place
−180 deg at all
at the new gain crossover frequency
to realize the desired phase margin which has a theoretical maximum of 90 deg. For a desired phase margin of 80 deg,
The gain of the controller is 20 log
10
1 + sin 80
o
1 − sin 80
o
=
a
= 42
a
= 130
dB. The new gain crossover frequency is at
G ( jω )
42
=−
= −21
dB
2 Or
G ( jω )
1 T 1 aT
=
=
aω
4
ω m
2
= 0 . 0877
=
Gc (s ) =
1 + aTs 1 + Ts
× 6 . 75 = 77
130
= 0 . 592
Thus
=
Thus
aT
ω = 2
Thus
ω = 6 . 75
45 . 61
T
rad/sec
= 0 . 013
= 1. 69
1 + 1.702 s
4 (1 + 1.702s )
G( s) =
1 + 0.0131s
s
Bode Plot.
237
2
(1 + 0.0131s )
10-22 (a) Forward-path Transfer Functi on: 1000 (1 + aTs )
1000a s +
1
aT G ( s) = G c ( s ) GP ( s) = = s ( s + 10 ) (1 + Ts ) 1 s ( s + 10 ) s + T
238
Set 1/aT = 10 so that the pole of G ( s ) at s = becomes s
ωn =
2 ζω
1000 a
n
=
1
2
+
=2
T
−10 is cancelled. 1
s
T
+ 1000
a
c
=
=0
1000 a
a = 40
Thus
Controller Transfer Function: G (s)
The characteristic equation of the system
and
T = 0.0025
Forward-path Transfer Function:
1 + 0 . 01 s
G ( s) =
1 + 0 . 0025 s
40,000 s ( s + 400 )
The attributes of the unit-step response of the compensated system are: t
Maximum overshoot = 0
= 0 . 0168
r
t
sec
s
= 0 . 02367
sec
(b) Frequency-domain Design The Bode plot of the uncompensated forward-path transfer function is made below.
G (s) =
1000 The attributes of the system are PM = 17.96 deg, GM = infinite.
s ( s + 10 )
Mr = 3.117, and BW = 48.53 rad/sec. To realize a phase margin of 75 deg, we need more than 57 deg of additional phase. Let us add an additional 10 deg for safety. Thus, the value of φ for the phase-lead controller is chosen to be m
67 deg. The value of a is calculated from
The gain of the controller is 20 log
10
1 + sin 67
o
1 − sin 67
o
a
=
a
= 20
log
10
= 24. 16
24. 16
= 27 . 66
dB. The new gain crossover frequency
is at G ( jω
From the Bode plot 1 T
=
aT
=
ωm '
'
m
)
=−
27 . 66
= − 13 . 83
dB
2
is found to be 70 rad/sec. Thus,
24. 16
× 70 = 344 =
1 + aTs
Thus
G (s)
PM = 75.19 deg
GM = infinite
c
T
or
=
1
= 0 . 0029
+ 0 . 0702
aT
= 0 . 0702
s
1 + Ts 1 + 0 . 0029 s The compensated system has the following frequency-domain attributes: Mr = 1.024 BW = 91.85 rad/sec
The attributes of the unit-step response are: Rise time tr = 0.02278 sec Settling time ts = 0.02828 sec
239
Maximum overshoot = 3.3%
10-23 (a) Forward-path Transfer Function: ( N = 10) G ( s) = G c ( s ) G p ( s) =
200 ( 1 + aTs )
s ( s + 1) ( s + 10 ) ( 1 + Ts )
Starting with a = 1000, we vary T first to stabilize the system. The following time-domain attributes are obtained by varying the value of T.
240
T
tr
Max Overshoot (%)
ts
0.0001
59.4
0.370
5.205
0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010
41.5 29.9 22.7 18.5 16.3 15.4 15.4 15.5 16.7
0.293 0.315 0.282 0.254 0.230 0.210 0.192 0.182 0.163
2.911 1.83 1.178 1.013 0.844 0.699 0.620 0.533 0.525
The maximum overshoot is at a minimum when T = 0.0007 or T = 0.0008. The maximum overshoot is 15.4%.
Unit-step Response. ( T = 0.0008 sec a = 1000)
10-23 (b) Frequency-domain Design. Similar to the design in part (a), we set a = 1000, and vary the value of T between 0.0001 and 0.001. The attributes of the frequency-domain characteristics are given below. T 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007
PM (deg)
GM
17.95 31.99 42.77 49.78 53.39 54.69 54.62
60.00 63.53 58.62 54.53 51.16 48.32 45.87
Mr
(dB)
241
3.194 1.854 1.448 1.272 1.183 1.138 1.121
BW (rad/sec) 4.849 5.285 5.941 6.821 7.817 8.869 9.913
0.0008 0.0009 0.0010
53.83 52.68 51.38
43.72 41.81 40.09
1.125 1.140 1.162
The phase margin is at a maximum of 54.69 deg when T = 0.0006. The performance worsens if the value of a is less than 1000.
10-24 (a) Bode Plot.
The attributes of the frequency response are: PM = 4.07 deg
Mr = 23.24
GM = 1.34 dB
BW = 4.4 rad/sec
10-24 (b) Single-stage Phase-lead Controller. G ( s) =
6 (1 + aTs )
s (1 + 0.2s ) (1 + 0.5s ) (1 + Ts )
We first set a = 1000, and vary T. The following attributes of the frequency-domain characteristics are obtained. T 0.0050 0.0010 0.0007 0.0006
PM (deg) 17.77 43.70 47.53 48.27
242
Mr 3.21 1.34 1.24 1.22
10.92 11.88 12.79
0.0005 0.0004 0.0002 0.0001
48.06 46.01 32.08 19.57
1.23 1.29 1.81 2.97
The phase margin is maximum at 48.27 deg when T = 0.0006. Next, we set T = 0.0006 and reduce a from 1000. We can show that the phase margin is not very sensitive to the variation of a when a is near 1000. The optimal value of a is around 980, and the corresponding phase margin is 48.34 deg. With a = 980 and T = 0.0006, the attributes of the unit-step response are: Maximum overshoot = 18.8%
tr = 0.262 sec
(c) Two-stage Phase-lead Controller. ( a = 980,
G( s) =
ts = 0.851 sec
T = 0.0006)
(
)
6 ( 1 + 0.588s ) 1 + bT s 2 s (1 + 0.2s )(1 + 0.5 s ) (1 + 0.0006 s ) 1 + T s 2
(
)
Again, let b = 1000, and vary T . The following results are obtained in the frequency domain. 2
T2
Mr
PM (deg)
0.0010 93.81 1.00 0.0009 94.89 1.00 0.0008 96.02 1.00 0.0007 97.21 1.00 0.0006 98.43 1.00 0.0005 99.61 1.00 0.0004 100.40 1.00 0.0003 99.34 1.00 0.0002 91.98 1.00 0.0001 73.86 1.00 Reducing the value of b from 1000 reduces the phase margin. Thus, the maximum phase margin of 100.4 deg is obtained with b = 1000 and T 2 = 0.0004. The transfer function of the two-stage phaselead controller is
Gc ( s) =
( 1 + 0.588s ) (1 + 0.4s ) (1 + 0.0006s ) (1 + 0.0004s )
(c) Unit-step Responses.
243
10-25 (a)
The loop transfer function of the system is
G ( s) H( s) =
1 + R2C s
10 K pK aK e Ns (1 + 0.05 s ) s
The characteristic equation is For root locus plot with R
2
3
+ 20
RCs 1
s
2
=
68.76 s (1 + 0.05s )
+ 6 .876 × 10
−4
6.876 × 10 R2 s s + 20 s + 687.6 3
2
+ 687
2s .6
=0
as the variable parameter, we have −4
G eq ( s ) =
R s
−6
1 + R2 × 10 s
2
−4
=
6.876 × 10 R2 s
( s + 21.5) ( s − 0.745 +
Root Locus Plot.
244
j 5.61) ( s − 0.745 − j 5.61 )
When R
2
= 2 . 65 × 10
5
, the roots are at
−6 . 02 ±
j 7 . 08 , and the relative damping ratio is 0.65 which is
maximum. The unit-step response is plotted at the end together with those of parts (b) and (c).
(b) Phase-lead Controller. G ( s) H( s) =
68.76 (1 + aTs )
s (1 + 0.05 s ) (1 + Ts )
(
)
(
)
Characteristic Equation: Ts + 1 + 20 T s + 20 + 1375.2 aT s + 1375.2 = 0 3
2
With T = 0.01, the characteristic equation becomes
3 2 s + 120 s + ( 2000 + 1375.2 a ) s + 137520 = 0 The last equation is conditioned for a root contour plot with a as the variable parameter. Thus 1375 .2 as G eq ( s ) = 3 2 s + 120 s + 2000 s + 137 ,520 From the root contour plot on the next page we see that when a = 3.4 the characteristic equation roots are at − 39 .2 , − 40 .4 + j 43 . 3, and − 40 .4 − j 43 . 3 , and the relative damping ratio is maximum and is 0.682.
Root Contour Plot ( a varies).
245
Unit-step Responses.
10-25 (c) Frequency-domain Design of Phase-lead Controller. For a phase margin of 60 deg, a = 4.373 and T = 0.00923. The transfer function of the controller is G (s) c
=
1 + aTs 1
+ Ts
=
1
+ 0 . 04036
s
1 + 0 . 00923 s
246
10-26 (a) Time-domain Design of Phase-lag Controller. Process Transfer Function:
Gp ( s) =
200 s ( s + 1 )( s + 10 )
For the uncompensated system, the two complex characteristic equation roots are at s
= − 0 .475 +
j 0 .471
and −0 .475 − j 0 .471 which correspond to a relative damping ratio of 0.707, when the forward path gain is 4.5 (as against 200). Thus, the value of a of the phase-lag controller is chosen to be a
=
4. 5
= 0 . 0225
Select T = 1000
which is a large number.
200 Then G (s) c
=
1 + aTs 1 + Ts
=
1
+ 22 . 5 s
G ( s) = Gc ( s ) Gp ( s) =
1 + 1000 s
Unit-step Response.
Maximum overshoot = 13.6
tr = 3.238 sec
ts = 18.86 sec
Bode Plot (with phase-lag controller, a = 0.0225, T = 1000)
247
4.5 ( s + 0.0889) )
s ( s + 1 )( s + 10 )( s + 0.001)
PM = 59 deg.
Mr = 1.1
GM = 27.34 dB
BW = 0.6414 rad/sec
10-26 (b) Frequency-domain Design of Phase-lag Controller. For PM = 60 deg, we choose a = 0.02178 and T = 1130.55. The transfer function of the phase-lag controller is 1 + 24. 62 s G (s) = GM = 27.66 dB Mr = 1.093 BW = 0.619 rad/sec c 1 + 1130 . 55 s
Unit-step Response.
Max overshoot = 12.6%, tr = 3.297 sec
ts = 18.18 sec
10-27 (a) Time-domain Design of Phase-lead Controller Forward-path Transfer Function.
G ( s) = Gc ( s ) Gp ( s) =
K (1 + aTs ) s ( s + 5)
2
(1 + Ts )
248
K
v
= lim
s →0
sG ( s )
=
K 25
= 10
Thus
K = 250
With K = 250, the system without compensation is marginally stable. For a > 1, select a small value for T and a large value for a. Let a = 1000. The following results are obtained for various values of T ranging from 0.0001 to 0.001. When T = 0.0004, the maximum overshoot is near minimum at 23%. T 0.0010 0.0005 0.0004 0.0003 0.0002 0.0001
tr
ts
Max Overshoot (%)
(sec)
(sec)
33.5 23.8 23.0 24.4 30.6 47.8
0.0905 0.1295 0.1471 0.1689 0.1981 0.2326
0.808 0.6869 0.7711 0.8765 1.096 2.399
As it turns out a = 1000 is near optimal. A higher or lower value for a will give larger overshoot.
Unit-step Response.
(b) Frequency-domain Design of Phase-lead Controller 250 ( 1 + aTs )
G ( s) = s
2
( s + 5 ) (1 + Ts ) 2
Setting a = 1000, and varying T, the following attributes are obtained. T 0.00050 0.00040 0.00035 0.00030 0.00020
Mr
PM (deg) 41.15 42.85 43.30 43.10 38.60
1.418 1.369 1.355 1.361 1.513
BW (rad/sec) 16.05 14.15 13.16 12.12 10.04
When a = 1000, the best value of T for a maximum phase margin is 0.00035, and PM = 43.3 deg. As it turns out varying the value of a from 1000 does not improve the phase margin. Thus the transfer function of the controller is
249
G (s) c
=
1 + aTs 1
1
=
+ Ts
+ 0 . 35
s
+ 0 . 00035
1
G ( s) =
and s
250 ( 1+ 0.35s ) s ( s + 5)
2
(1 + 0.00035s )
10-27 (c) Time-domain Design of Phase-lag Controller Without compensation, the relative damping is critical when K = 18.5. Then, the value of a is chosen to be 18 . 5 a = = 0 . 074 250 We can use this value of a as a reference, and conduct the design around this point. The value of T is preferrably to be large. However, if T is too large, rise and settling times will suffer. The following performance attributes of the unit-step response are obtained for various values of a and T. a
T
0.105 0.100 0.095 0.090 0.090 0.090 0.090 0.090 0.090 0.090
Max Overshoot (%)
tr
ts
2.6 2.9 2.6 2.5 2.1 1.9 1.7 1.4 0.9 0.7
1.272 1.348 1.422 1.522 1.532 1.538 1.543 1.550 1.560 1.566
1.82 1.82 1.82 2.02 2.02 2.02 2.02 2.22 2.22 2.22
500 500 500 500 600 700 800 1000 2000 3000
As seen from the results, when a = 0.09 and for T ≥ 2000, the maximum overshoot is less than 1% and the settling time is less than 2.5 sec. We choose T = 2000 and a = 0.09. The corresponding frequency-domain characteristics are: PM = 69.84 deg
Mr = 1.004
GM = 20.9 dB
BW = 1.363 rad/sec
(d) Frequency-domain Design of Phase-lag Controller G ( s) =
250 ( 1 + aTs ) s ( s + 5)
2
a 1)
1 + Ts
By selecting a small value for T, the value of a becomes the critical design parameter in this case. If a is too small, the overshoot will be excessive. If the value of a is too large, the oscillation in the step response will be objectionable. By trial and error, the best value of a is selected to be 6, and T = 0.001. The following performance attributes are obtained for the unit-step response. tr = 0.01262 sec
Maximum overshoot = 0%
ts = 0.1818 sec
However, the step response still has oscillations due to the compliance in the motor shaft. The unitstep response of the phase-lead compensated system is shown below, together with that of the uncompensated system.
(b) Phase-lead and Second-order Controller The poles of the process G p ( s ) are at −161.3, order term is
+ 262
2
s
s
+ 2 ,624, G
The value of
ωn
is set to
Let the two poles of G
c1
c1
( s)
=
s
2
=
+ 262 s
2
s
2
2
417 .1
+ 2ζ pω n s + ω n
and
s
+ 2 ,624,
s
2
= 1620
= − 1620
G c1 ( s )
The second-
417 .1 . Let the second-order controller transfer function be
2 ,624, 417 .1
( s ) be at s
−131+ j1614.7 and −131 − j1614.7.
rad/sec, so that the steady-state error is not affected.
− 1620
. Then,
ζ p = 405
+ 262
s
+ 2 ,624,
417 .1
+ 3240
s
+ 2 ,624,
417 .1
6.087 × 10
10
G ( s) = G c ( s ) G c1 ( s ) G p ( s) =
(
.
(1 + 0.006 s )
)
s ( s + 161.3 ) s + 3240 s + 2,624,417.1 ( 1 + 0.001s )
The unit-step response is shown below, and the attributes are:
252
2
tr = 0.01012 sec
Maximum overshoot = 0.2
ts = 0.01414 sec
The step response does not have any ripples.
Unit-step Responses
10-29 (a) System Equations. e
a
= R a i a + K bω m
T
= K i ia
m
Tm = Jm
dω m dt
+ Bmωm + KL ( θ m − θL ) + BL ( ωm − ω L )
dω L
K L ( θ m − θ L ) + BL ( ω m − ω L ) = J L
dt
State Equations in Vector-matrix Form:
dθ L dt 0 K dω L − L dt J L dθ = 0 m dt K L dω m J m dt
−
1
0
BL
KL
JL
JL
0
0
BL Jm
−
KL Jm
θ 0 BL L 0 JL ω L + 0 ea θ 1 m Ka B + Bl Ki K b ω m − m − R J a m Jm J m Ra 0
253
State Diagram:
Transfer Functions:
Ωm ( s)
=
Ea ( s ) Ω L ( s)
=
E a (s ) Ωm ( s)
=
Ea ( s ) Ω L ( s) E a (s )
=
(
)
Ki s + BL s + KL / Ra 2
J m J L s + ( K e J L + BL J L + BL J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e 2
2
K i ( B Ls + K L ) / R a
J m J L s + ( K e J L + BL J L + B L J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e 3
2
(
133.33 s + 10 s + 3000 2
)
s + 318.15 s + 60694.13 s + 58240 3
2
=
(
133.33 s + 10s + 3000 2
( s + 0.9644) ( s + 158.59 +
j187.71) ( s + 158.59 − j187.71)
1333.33 ( s + 300 )
( s + 0.9644 ) ( s + 158.59 +
j187.71 ) ( s + 158.59 − j187.71)
(b) Design of PI Controller.
G ( s) =
Kv
Ω L (s )
= lim
E (s )
s→0
With K I
1333.33 K P s + =
sG ( s )
KP 20 18
KI
s ( s + 0.9644 ) s + 317.186s + 60388.23
=
= 14 .56 ,
(
( s + 300 ) KP
1333 .33 0 . 9644
× 300
× 60388
2
K
I
. 23
= 6 . 87
KI
= 100
) Thus
KI
= 14 .56
we study the effects of varying K P . The following results are obtained. tr
ts
(sec)
(sec)
0.00932 0.01041
0.02778 0.01263
254
Max Overshoot (%) 4.2 0.7
)
17 16 15 10 With K I
= 14 .56
0.01113 0.01184 0.01303 0.02756
0.01515 0.01515 0.01768 0.04040
0 0 0 0.6
and K P ranging from 15 to 17, the design specifications are satisfied.
Unit-step Response:
(c) Frequency-domain Design of PI Controller ( KI = 14.56) G ( s) =
(
1333.33 ( K P s + 14.56 )( s + 300 )
s s + 318.15 s + 60694.13s + 58240 3
2
The following results are obtained by setting K I KP
PM (deg)
20 18 17 16 15 10 8 7 6 5
65.93 69.76 71.54 73.26 74.89 81.11 82.66 83.14 83.29 82.88
GM (dB)
∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞
= 14 .56
)
and varying the value of K P .
Mr
BW (rad/sec)
Max Overshoot (%)
1.000 1.000 1.000 1.000 1.000 1.005 1.012 1.017 1.025 1.038
266.1 243 229 211.6 190.3 84.92 63.33 54.19 45.81 38.12
4.2 0.7 0 0 0 0.6 1.3 1.9 2.7 4.1
255
tr
ts
(sec)
(sec)
0.00932 0.01041 0.01113 0.01184 0.01313 0.0294 0.04848 0.03952 0.04697 0.05457
0.02778 0.01263 0.01515 0.01515 0.01768 0.0404 0.03492 0.05253 0.0606 0.0606
From these results we see that the phase margin is at a maximum of 83.29 degrees when K P = 6 . However, the maximum overshoot of the unit-step response is 2.7%, and M r is slightly greater than one. In part (b), the optimal value of K P from the standpoint of minimum value of the maximum overshoot is between 15 and 17. Thus, the phase margin criterion is not a good indicator in the present case.
10-30 (a) Forward-path Transfer Function Gp ( s) =
K Θ m ( s) Tm ( s )
=
(
(
( = + 2100 s + 10,000 ) s ( s + 4.937 ) ( s
100 K s + 10 s + 100
s s + 20 s 3
2
2
)
10,000 s + 10 s + 100 2
2
)
+ 15.06 s + 2025.6
)
The unit-step response is plotted as shown below. The attributes of the response are: tr = 0.01345 sec
Maximum overshoot = 57%
ts = 0.4949 sec
(b) Design of the Second-order Notch Controller The complex zeros of the notch controller are to cancel the complex poles of the process transfer function. Thus
G (s) c
=
s
2
s
2
+ 15 . 06 s + 2025 + 90 ζ p s + 2025
.6 .6
( G ( s) = G ( s ) G ( s) = s ( s + 4.937 ) ( s
10,000 s + 10 s + 100
and
c
p
The following results are obtained for the unit-step response when various values of The maximum overshoot is at a minimum of 4.1% when ζ
p
= 1.222
ζp
are used.
. The unit-step response is
plotted below, along with that of the uncompensated system.
ζp
2 ζω
2.444 2.333 2.222 1.667 1.333 1.222 1.111 1.000
200 210 200 150 120 110 100 90
Max Overshoot (%)
n
7.3 6.9 6.5 4.9 4.3 4.1 5.8 9.8
Unit-step Response
256
2
2
)
+ 90ζ p s + 2025.6
)
10-30 (c) Frequency-domain Design of the Notch Controller The forward-path transfer function of the uncompensated system is
( G ( s) = s ( s + 4.937 ) ( s
10000 s + 10 s + 100 2
2
)
+ 15.06 s + 2025.6
)
The Bode plot of G ( jω ) is constructed in the following. We see that the peak value of G ( jω ) is approximately 22 dB. Thus, the notch controller should provide an attentuation of at the resonant frequency of 45 rad/sec. Using Eq. (10-155), we have
G ( j 45) c
ζz
=
ζp
=
0 .167
ζp
Notch Controller Transfer Function
G (s) c
=
s
+ 15 .06 s + 2025
s
2
2
+ 189
.216 s
= 0 . 0794
Thus
ζ p = 2 .1024
Forward-path Transfer Function
.6
+ 2025
−22 dB or 0.0794
G (s ) =
.6
Bode Plots
257
( s ( s + 4.937 ) (s
10,000 s + 10 s + 100 2
2
)
+ 189.22s + 2025.6
)
Attributes of the frequency response: PM = 80.37 deg
M
GM = infinite
r
= 1. 097
BW = 66.4 rad/sec
Attributes of the frequency response of the system designed in part (b): PM = 59.64 deg
GM = infinite
M
r
= 1. 048
10-31 (a) Process Transfer Function Gp ( s) =
(
500 ( s + 10 )
s s + 10 s + 1000 2
)
The Bode plot is constructed below. The frequency-domain attributes of the uncompensated system are: PM = 30 deg
GM = infinite
M
r
= 1.86
The unit-step response is oscillatory.
(b) Design of the Notch Controller
258
and
BW = 3.95 rad/sec
BW = 126.5 rad/sec
For the uncompensated process, the complex poles have the following constants:
ωn =
1000
= 31 .6
2 ζω
rad / sec
n
= 10
Thus
ζ
= 0.158
The transfer function of the notch controller is
G (s) c
=
s
2
+ 2ζ zω n s + ω n
s
2
2
+ 2ζ p s + ω n 2
For the zeros of G ( s ) to cancel the complex poles of G p ( s ) , c
ζ z = ζ = 0 .158
.
From the Bode plot, we see that to bring down the peak resonance of G ( j ω n ) in order to smooth out the magnitude curve, the notch controller should provide approximately −26 dB of attenuation. Thus, using Eq. (10-155),
ζz ζp
−26
= 10
= 0 . 05
20
ζp =
Thu s
0 .158
= 3 .1525
0 . 05
The transfer function of the notch controller is
G (s) c
=
s s
2
2
+ 10 s + 1000
+ 199
. 08 s
G ( s) = G c ( s ) G p ( s) =
+ 1000
(
500 ( s + 10 )
s s + 199.08 s + 1000 2
The attributes of the compensated system are: PM = 72.38 deg
GM = infinite
Maximum overshoot = 3.4%
t
r
= 0 . 3868
M
r
=1
sec
Bode Plots
259
BW = 5.44 rad/sec t
s
= 0 .4848
sec
)
Step Responses
10-31 (c) Time-domain design of the Notch Controller
260
With
ζ z = 0 .158
ω n = 31 . 6 ,
and
the forward-path transfer function of the compensated system is
G ( s) = G c ( s ) G p ( s) =
500 ( s + 10 )
(
s s + 63.2ζ p s + 1000 2
)
The following attributes of the unit-step response are obtained by varying the value of
ζp
2 ζω
1.582 1.741 1.899 2.057 2.215 2.500 3.318 When
ζ p = 2 .5
tr
Max Overshoot (%)
n
100 110 120 130 140 158.25 209.7
ζ p. ts
(sec)
0 0 0 0 0.2 0.9 4.1
(sec)
0.4292 0.4172 0.4074 0.3998 0.3941 0.3879 0.3884
0.5859 0.5657 0.5455 0.5253 0.5152 0.4840 0.4848
the maximum overshoot is 0.9%, the rise time is 0.3879 sec and the setting
time is 0.4840 sec. These performance attributes are within the required specifications.
10-32
Let the transfer function of the controller be
(
20,000 s + 10 s + 50
Gc ( s) =
2
( s + 1000 )
)
2
Then, the forward-path transfer function becomes
For G
For
cf
(s)
= 1,
K
± 20% variation in K,
= lim
v
sG ( s )
s→0
K
we let
G cf ( s ) =
10
(
and
2
s s + 10 s + 100 6
2
K
= 50
8
10
= 4000
min
=
(
20,000 K s + 10 s + 50
G ( s) = Gc ( s ) Gp ( s) =
K
max
50 ( s + 1 )
)
) ( s + 1000)
2
Thus t he nom inal K
= 6000
= 5000
. To cancel the complex closed-loop poles,
where the (s + 1) term is added to reduce the rise time.
s + 10 s + 50 2
Closed-loop Transfer Function:
Y (s) R (s )
10 K ( s + 1) 6
=
(
s s + 10 s + 100 2
)( s + 1000 ) + 20,000K ( s 2
2
+ 10 s + 50
)
Characteristic Equation: K = 4000:
Roots:
s
5
+ 2010
− 97 . 7 ,
s + 1,020 , 100 s + 9 . 02 × 10 s + 9 × 10 − 648 . 9 , − 1252 . 7 , − 5 . 35 + j 4. 6635 , 4
3
Max overshoot K = 5000:
Roots:
s
5
+ 2010
−132
.46 ,
s
4
≅ 6.7%
+ 1,02010
587 .44,
Max overshoot
7
≅ 4%
2
+ 4 × 10 = 0 − 5 . 35 − j 4. 6635
8
9
Rise time < 0.04 sec
0 s + 1.1 × 10 s + 1.1 × 10 s + 5 × 10 − 1279 . 6 , − 5 .272 + j 4. 7353 , − 5 .272 3
8
2
9
Rise time < 0.04 sec
261
9
=0 − j 4. 7353
K = 6000
Roots:
s
5
+ 2010
−176
s
. 77 ,
, 100 s + 1. 3 × 10 s + 1.3 × 10 s + 6 × 10 = 0 − 519 . 37 , − 1303 .4, − 5 .223 + j 4. 7818 , − 5 .223 − j 4. 7818 4
+ 1,020
Max overshoot
3
8
≅ 2.5%
2
9
9
Rise time < 0.04 sec
Thus all the required specifications stay within the required tolerances when the value of K varies by plus and minus 20%.
Unit-step Responses
10-33
Let the transfer function of the controller be
Gc ( s) =
(
200 s + 10 s + 50 2
( s + 100 )
)
2
The forward-path transfer function becomes
G ( s) = G c ( s ) G p ( s) =
(
200,000 K s + 10 s + 50 2
s ( s + a ) ( s + 100 )
)
2
For a = 10, K
v
= lim
sG ( s )
s →0
=
10
7
10
K 5
= 100
K
= 100
K
Thus
=1
Characteristic Equations: ( K = 1) a = 10:
Roots:
+ 210 s + 2 .12 × 10 s + 2 .1 × 10 s + 10 = 0 −4. 978 + j 4. 78 , − 4. 978 − j 4. 78 , − 100 + j 447 s
4
3
5
2
6
7
.16
− 100 −
s + 208 s + 2 .116 × 10 s + 2 . 08 × 10 s + 10 = 0 Roots: − 4. 939 + j 4. 828 , − 4. 939 − j 4. 828 , − 99 . 06 + j 446 .97 , − 99 . 06 4
a = 8:
a = 12:
s
4
3
+ 212
s
3
5
2
6
+ 2 .124 × 10 + 2 .12 × 10 5
262
6
s
+ 10
j 447 .16
7
7
=0
−
j 446 . 97
Roots: − 5. 017 + j 4. 73 ,
− 5 . 017 −
Unit-step Responses:
j 4. 73 ,
− 100
. 98
+
j 447 . 36 ,
− 100
. 98
−
j 447 . 36
All three responses for a = 8, a = 10, and 12 are similar.
10-34 Forward-path Transfer Function: G ( s) =
Y ( s)
=
E (s )
K s ( s + 1 )( s + 10 ) + K Kt s
Kv = lim sG( s ) = s →0
K =1 10 + KKt
Characteristic Equation:
s + 11s + (10 + KKt ) s + K = s + 11 s + Ks+ K = 0
For root loci,
G eq ( s ) =
3
2
3
K ( s + 1) s
2
( s + 11)
263
2
Root Locus Plot ( K varies)
The root loci show that a relative damping ratio of 0.707 can be realized by two values of K. K = 22 and 59.3. As stipulated by the problem, we select K = 59.3.
10-35 Forward-path Transfer Function: G ( s) =
10 K s ( s + 1) ( s + 10 ) + 10 K t s
K
v
= lim
s →0
sG ( s )
264
10 K
= 10
+ 10
K
= t
K 1+ K
=1 t
Thus
K
t
=
K
−1
s ( s + 1 )( s + 10 ) + 10 Kt + 10 K = s + 11 s +10 Ks + 10 K = 0 3
Characteristic Equation: When K = 5.93 and K s
3
The roots are:
t
=
K
− 1 = 4. 93
+ 11 s + 10 . 046 2
− 10 . 046
s
, the characteristic equation becomes
+ 4. 6 = 0
− 0 .47723 +
,
2
− 0 .47723 −
j 0 .47976 ,
j 0 .47976
10-36 Forward-path Transfer Function: G ( s) =
K (1 + aTs )
(
s ( (1 + Ts ) s + 10 s + KK t 2
K
)
v
= lim
s →0
=
sG ( s )
1 K
= 100
Thu s
t
Let T = 0.01 and a = 100. The characteristic equation of the system is written:
(
)
s + 110 s + 1000s + K 0.001s + 101s +100 = 0 4
3
2
2
To construct the root contours as K varies, we form the following equivalent forward-path transfer function:
G eq ( s ) =
(
0.001 K s + 101,000 s + 100,000 s
2
2
( s + 10) ( s + 100 )
) = 0.001K ( s + 1)( s + 50499 ) s
2
( s + 10) ( s + 100)
From the root contour diagram we see that two sets of solutions exist for a damping ratio of 0.707. These are: K = 20: Complex roots: − 1.158 + j 1.155 , − 1.158 − j1 .155 K = 44.6:
Complex roots: − 4. 0957 + j 4. 0957 ,
The unit-step responses of the system for K = 20 and 44.6 are shown below.
Unit-step Responses:
265
− 4. 0957 −
j 4. 0957
K
t
= 0 . 01
10-37 Forward-path Transfer Function: G ( s) =
K s K K1 Ki N s J t L a s + ( Ra J t + L a Bt + K 1 K 2 J t ) s + R a Bt + K 1 K 2 Bt + K b K i + K K1 Ki Kt 2
1.5 × 10 K 7
G ( s) =
Ramp Error Constant:
Thus
(
s s + 3408.33s + 1,204,000 + 1.5 × 10 KK t K
2
v
8
= lim
s →0
sG ( s )
1.204
+ 150
1.204
15 K
=
KK
t
= 0 .15
+ 150
KK
)
= 100 t
K
The forward-path transfer function becomes
1.5 × 10 K 7
G ( s) =
(
s s + 3408.33s + 150,000 K
Characteristic Equation:
2
s
3
+ 3408
. 33 s
+ 150
)
,000 Ks
+ 1. 5 × 10
7
K
=0
When K = 38.667 the roots of the characteristic equation are at
− 0 .1065
,
− 1. 651 +
− 1.651 −
j 1. 65 ,
(ζ
j1. 65
The forward-path transfer function becomes
G ( s) =
(
5.8 × 10
8
s s + 3408.33 s + 5.8 × 10 2
266
6
)
≅ 0 . 707
for the complex roots)
Unit-step Response
Unit-step response attributes:
10-38
Maximum overshoot = 0
Rise time = 0.0208 sec
Settling time = 0.0283 sec
(a) Disturburnce -to-Output Transfer Function Y (s)
==
TL ( s) For T ( s ) L
r =0
= 1/
2 (1 + 0.1s )
c
=1
s
t →∞
= lim
sY ( s )
s →0
=
1
≤ 0 . 01 Thus 10 K Performance of Uncompensated System. K = 10, G ( s ) = 1 lim y ( t )
10-38 (b)
G (s)
s (1 + 0.01 s)(1 + 0.1 s ) + 20 K
K
≥ 10
c
G ( s) =
200 s (1 + 0.01s ) (1 + 0.1s )
The Bode diagram of G ( jω ) is shown below. The system is unstable. The attributes of the frequency response are:
PM = −9.65 deg
GM =
−5.19 dB.
(c) Single-stage Phase-lead Controller Design To realize a phase margin of 30 degrees, a = 14 and T = 0.00348. G (s) c
=
1 + aTs 1
+ Ts
=
1 + 0 . 0487 s 1
+ 0 . 00348
267
s
The Bode diagram of the phase-lead compensated system is shown below. The performance attributes are: PM = 30 deg GM = 10.66 dB M = 1. 95 BW = 131.6 rad/sec. r
(d) Two-stage Phase-lead Controller Design Starting with the forward-path transfer function
G ( s) =
200 ( 1+ 0.0487 s )
s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348 s )
The problem becomes that of designing a single-stage phase-lead controller. For a phase margin or 55 degrees, a = 7.385 and T = 0.00263. The transfer function of the second-stage controller is G
Thus
G ( s) =
c1
( s)
=
1
+ aTs
1
+ Ts
=
1
+ 0 . 01845
s
1
+ 0 . 00263
s
200 ( 1+ 0.0487 s ) (1 + 0.01845 s )
s (1 + 0.1s ) (1 + 0.01s ) (1 + 0.00348s )( 1 + 0.00263s )
The Bode diagram is shown on the following page. The following frequency-response attributes are obtained: PM = 55 deg
GM = 12.94 dB
M
r
= 1.11
Bode Plot [parts (b), (c), and (d)]
268
BW = 256.57 rad/sec
10-39 (a) Two-stage Phase-lead Controller Design. The uncompensated system is unstable. PM = −43.25 deg and GM = −18.66 dB. With a single-stage phase-lead controller, the maximum phase margin that can be realized affectively is 12 degrees. Setting the desired PM at 11 deg, we have the parameters of the single-stage phaselead controller as a = 128.2 and T = 0 . 00472 . The transfer function of the single-stage controller 1
G
is
c1
( s)
=
1
+ aT 1 s
1
+ T1 s
1
=
+ 0 . 6057
s
1 + 0 . 00472
s
Starting with the single-stage-controller compensated system, the second stage of the phase-lead controller is designed to realize a phase margin of 60 degrees. The parameters of the second-stage controller are: b = 16.1 and T = 0 . 0066 . Thus, 2
G
c2
(s)
=
1 + bT s 2
1+T s
=
2
G (s) c
= G c 1 ( s )G c 2 ( s ) =
1 + 0 .106 s 1 + 0 . 0066 s
1 + 0 . 6057 s
1
+ 0 .106
s
1 + 0 . 00472 s 1 + 0 . 0066 s
Forward-path Transfer Function:
269
1,236,598.6 ( s + 1.651 )( s + 9.39 )
G ( s) = G c1 ( s ) G c2 ( s ) G p ( s) =
s ( s + 2)( s + 5 ) ( s + 211.86 ) ( s + 151.5 )
Attributes of the frequency response of the compensated system are: GM = 19.1 dB
M
PM = 60 deg
= 1. 08
r
BW = 65.11 rad/sec
The unit-step response is plotted below. The time-response attributes are: t
Maximum overshoot = 10.2%
s
= 0 .1212
t
sec
r
= 0 . 037
sec
(b) Single-stage Phase-lag Controller Design. With a single-stage phase-lag controller, for a phase margin of 60 degrees, a = 0.0108 and T = 1483.8. The controller transfer function is 1 + 16 . 08 s G (s) = c 1 + 1483 . 8 s The forward-path transfer function is G (s)
= G c ( s )G p ( s ) =
β+ γ β+ γβ+ γβ+ γ 6 .5 s
s s
s
2
0 . 0662
5
s
0 . 000674
From the Bode plot, the following frequency-response attributes are obtained: PM = 60 deg
M
GM = 20.27 dB
r
= 1. 09
BW = 1.07 rad/sec
The unit-step response has a long rise time and settling time. The attributes are: Maximum overshoot = 12.5%
t
s
= 12 . 6
t
sec
r
=
2 .126 sec
(c) Lead-lag Controller Design. For the lead-lag controller, we first design the phase-lag portion for a 40-degree phase margin. The result is a = 0.0238 and T = 350 . The transfer function of the controller is 1
G
c1
( s)
=
(s)
=
1
+ 8 . 333
s
1 + 350 s The phase-lead portion is designed to yield a total phase margin of 60 degrees. The result is b = 4.8 and T = 0 .2245 . The transfer function of the phase-lead controller is 2
G
c2
1
+ 1.076
s
1 + 0 .2245 s The forward-path transfer function of the lead-lag compensated system is
G ( s) = Frequency-response attributes: Unit-step response attributes:
68.63 ( s + 0.12 )( s + 0.929 )
s ( s + 2 ) ( s + 5 ) ( s + 0.00286 ) ( s + 4.454 ) PM = 60 deg GM = 13.07 dB M = 1. 05 BW = 3.83 rad/sec r Maximum overshoot = 5.9%
270
t
s
= 1. 512
sec
t
r
= 0 . 7882
sec
Unit-step Responses.
10-40 (a)
The uncompensated system has the following frequency-domain attributes: PM = 3.87 deg The Bode plot of G
p
M
GM = 1 dB
r
= 7 .73
BW = 4.35 rad/sec
( j ω ) shows that the phase curve drops off sharply, so that the phase-lead
controller would not be very effective. Consider a single-stage phase-lag controller. The phase margin of 60 degrees is realized if the gain crossover is moved from 2.8 rad/sec to 0.8 rad/sec. The attenuation of the phase-lag controller at high frequencies is approximately −15 dB. Choosing an attenuation of
−17.5 dB, 20 log
we calculate the value of a from a = −17 . 5 dB Thus a = 0.1334
10
The upper corner frequency of the phase-lag controller is chosen to be at 1/aT = 0.064 rad/sec. Thus, 1/T = 0.00854 or T = 117.13. The transfer function of the phase-lag controller is 1 + 15 .63 s G (s) = c 1 + 117 .13 The forward-path transfer function is
G ( s) = G c ( s ) G p ( s) =
5 ( 1 + 15.63 s ) (1 − 0.05 s )
s (1 + 0.1 s ) (1 + 0.5 s ) (1 + 117.13 s ) (1 + 0.05 s )
From the Bode plot of G ( jω ) , the following frequency-domain attributes are obtained: PM = 60 deg
GM = 18.2 dB
M
r
= 1. 08
BW = 1.13 rad/sec
The unit-step response attributes are: maximum overshoot = 10.7%
t
s
= 10 .1
sec
t
Bode Plots
271
r
= 2 .186
sec
10-40 (b)
Using the exact expression of the time delay, the same design holds. The time and frequency domain attributes are not much affected.
272
10-41 (a) Uncompensated System. Forward-path Transfer Function: G ( s) = The Bode plot of G ( jω ) is shown below.
10
(1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s )
The performance attributes are: PM = −10.64 deg The uncompensated system is unstable.
GM =
−2.26 dB
(b) PI Controller Design.
(
10 K p s + K I
Forward-path Transfer Function: G ( s) = Ramp-error Constant: K v = lim sG s →0
G ( s) =
)
s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s ) ( s ) = 10 K I = 0 .1 Thu s K I = 0 . 01 0.1 (1 + 100K P s )
s (1 + s ) (1 + 10 s )( 1 + 2s ) (1 + 5s )
The following frequency-domain attributes are obtained for various values of K
P
.
KP
PM (deg)
GM (dB)
Mr
0.01 0.02 0.05 0.10 0.12 0.15 0.16 0.17 0.20
24.5 28.24 38.84 50.63 52.87 53.28 52.83 51.75 49.08
5.92 7.43 11.76 12.80 12.23 11.22 10.88 10.38 9.58
2.54 2.15 1.52 1.17 1.13 1.14 1.16 1.18 1.29
The phase margin is maximum at 53.28 degrees when K
P
= 0 .15 .
The forward-path transfer function of the compensated system is
G ( s) =
0.1 (1 + 15 s )
s (1 + s ) (1 + 10 s )( 1 + 5 s ) ( 1 + 2 s )
The attributes of the frequency response are: PM = 53.28 deg
M
GM = 11.22 dB
= 1.14
r
BW = 0.21 rad/sec
The attributes of the unit-step response are: Maximum overshoot = 14.1%
t
r
= 10 . 68
sec
Bode Plots
273
t
s
= 48
sec
BW (rad/sec) 0.13 0.13 0.14 0.17 0.18 0.21 0.22 0.23 0.25
10-41 (c) Time-domain Design of PI Controller.
274
By setting K
= 0 . 01
I
and varying K
we found that the value of K
P
P
that minimizes the maximum
overshoot of the unit-step response is also 0.15. Thus, the unit-step response obtained in part (b) is still applicable for this case.
10-42 Closed-loop System Transfer Function. Y (s)
1
=
s + ( 4 + k 3 ) s + ( 3 + k 2 + k3 ) s + k1 3
R (s )
2
For zero steady-state error to a step input, k
=1.
1 2
we divide the characteristic polynomial by s
s + ( 2 + k2 )
For the complex roots to be located at −1 +j and −1
+2s+ 2
− j,
and solve for zero remainder.
s + 2 s + 2 s + ( 4 + k3 ) s + ( 3 + k2 + k3 ) s + 1 2
3
2
s + 3
2s
+ 2s
2
( 2 + k ) s + (1 + k + k ) s + 1 ( 2 + k ) s + ( 4 + 2k ) s + 4 + 2k 2
3
2
3
2
3
3
( -3+k
2
− k 3 ) s − 3 − 2 k3
− 3 − 2k3 = 0
For zero remainder,
−0.5.
k
Thus
−3 + k 2 − k 3 = 0
The third root is at steady-state error.
3
3
= −1. 5 k
Thus
2
= 1. 5
Not all the roots can be arbitrarily assigned, due to the requirement on the
10-43 (a) Open-loop Transfer Function. G ( s) =
X 1 (s )
=
E (s )
k3 s s + ( 4 + k 2 ) s + 3 + k 1 + k 2 2
Since the system is type 1, the steady-state error due to a step input is zero for all values of k , k , and k 1
2
3
that correspond to a stable system. The characteristic equation of the closed-loop system is
s + ( 4 + k2 ) s + ( 3 + k1 +k 2 )s + k 3 = 0 3
For the roots to be at
−1 + j, −1 − j,
2
and
−10, the equation should be: 3 2 s + 12 s + 22 s + 20 = 0
Equating like coefficients of the last two equations, we have 4 + k = 12 2
3
+ k 1 + k 2 = 22 k
3
= 20
k
Thus
k
Thus
(b) Open-loop Transfer Function. Y (s ) Gc ( s) 20 = = 2 E ( s ) ( s + 1) ( s + 3 ) s s + 12 s + 22
(
Thus
)
10-44 (a)
275
Thus
2 1
=8 = 11
k3 = 20
Gc ( s) =
20 ( s + 1 ) ( s + 3 )
(
s s + 12 s + 22 2
)
0 25.92 ∗ A = 0 −2.36
0
1
0
0
0
0
0
0
1
0
0
0
0 −0.0732 ∗ B = 0 0.0976
The feedback gains, from k 1 to k 4 :
−2.4071E+03
−4.3631E+02
The A 0.0000E+00
−1.5028E+02
∗
− B ∗K
0.0000E+00 2/3258E+02
−1.0182E+02
matrix of the closed-loop system
1.0000E+00
−3.1938E+01
−8.4852E+01
0.0000E+00 0.0000E+00 −7.4534E+00 0.0000E+00 1.0000E+00 8.2816E+00 9.9379E+00
−6.2112E+00
0.0000E+00 4.2584E+01
The B vector 0.0000E+00
−7.3200E−02
0.0000E+00 9.7600E−02
Time Responses:
10-44 (b)
The feedback gains, from k 1 to k 2 :
−9.9238E+03 The A
−1.6872E+03 ∗
− B ∗K
−1.3576E+03
matrix of the closed-loop system
276
−8.1458E+02
0.0000E+00
−7.0051E+02
1.0000E+00
−1.2350E+02
0.0000E+00 9.6621E+02
0.0000E+00 0.0000E+00 −5.9627E+01 0.0000E+00 1.0000E+00 1.3251E+02 7.9503E+01
−9.9379E+01
0.0000E+00 1.6467E+02
The B vector 0.0000E+00
−7.3200E−02
0.0000E+00 9.7600E−02
Time Responses:
10-45
The solutions are obtained by using the pole-placement design option of the linsys program in the ACSP software package.
(a)
The feedback gains, from k 1 to k 2 :
−6.4840E+01
−5.6067E+00
The A 0.0000E+00
−3.0950E+02
0.0000E+00
−4.6190E+02
∗
− B ∗K
2.0341E+01
2.2708E+00
matrix of the closed-loop system
1.0000E+00 0.0000E+00 0.0000E+00 1.1463E+02 1.4874E+01 0.0000E+00 0.0000E+00 1.0000E+00 −3.6724E+01 1.7043E+02 1.477eE+01
−3.6774E+01
The B vector 0.0000E+00
−6.5500E+00
0.0000E+00
277
−6.5500E+00 (b) Time Responses: ∆x ( 0 ) =
0 .1
0
0
0
∆x ( 0 ) =
0 .1
0
0
0 , the initial position of
With the initial states
'
'
∆x 1 or ∆y1 is preturbed downward ∆x 3 = ∆y 2 is
from its stable equilibrium position. The steel ball is initially pulled toward the magnet, so
negative at first. Finally, the feedback control pulls both bodies back to the equilibrium position. With the initial states
∆x ( 0 ) =
0
0
0 .1
'
0 , the initial position of
∆x 3 or ∆y 2
is preturbed
downward from its stable equilibrium. For t > 0, the ball is going to be attracted up by the magnet toward the equilibrium position. The magnet will initially be attracted toward the fixed iron plate, and then settles to the stable equilibrium position. Since the steel ball has a small mass, it will move more actively.
10-46 (a) Block Diagram of System.
278
u = −k1 x1 + k 2 ∫ ( − x1 + w1 ) dt State Equations of Closed-loop System:
dx1 dt −2 − k1 1 x1 0 = x + k dx − k 0 2 2 2 2 dt
1 w1 0 w2
Characteristic Equation:
sI − A = For s =
−10, −10, X (s)
sI
=
X ( s) =
−1
k2
s
−A =
X 1 (s)
W1 ( s )
s + 2 + k1
=
(
=
s2
1+ 2 s
s
−1
−2
+ 18
W2 (s)
=
200 + W2s
s s + 20 s + 200 2
2
+ 20 s + 200 = 0
200 W 1 ( s ) s
1
= s + ( 2 + k1 ) s + k2 = 0
)
s
+ s −1W 2 ( s ) −1
W2 s
Thus
+ 200
s
k1
= 18
200 W 1 ( s )
−2
=
=
const ant
W2
s
2
k2
and
+ sW 2 ( s )
+ 20 s + 200
lim x( t) = lim sX ( s) = 1 t →∞
10-46 (b) With PI C ontroller: Block Diagram of System:
279
s →0
= 200
Set K P
=
2 and
KI
= 200 .
X (s ) =
(K
P
s + K I ) W1 ( s ) + sW2 ( s ) s + 20 s + 200 2
Time Responses:
280
=
( 2s + 200 ) W ( s ) + sW ( s ) 1
s + 20 s + 200 2
2
Chapter 11
THE VIRTUAL LAB
Part 1) Solution to Lab questions within Chapter 11 11-5-1 Open Loop Speed 1. Open loop speed response using SIMLab: a. +5 V input:
The form of response is like the one that we expected; a second order system response with overshoot and oscilla tion. Considering an amplifier gain of 2 and K b = 0.1 , the desired set point should be set to 2.5 and as seen in the figure, the final value is approximately 50 rad/sec which is armature voltage divided by K b . To find the above response the systems parameters are extracted from 11-3-1 of the text and B is calculated from 11-3 by having τ m as:
τm =
Ra J m , Ra B + k b k m
B=
Ra J m − k b k m τ m = 0.000792kg ⋅ m 2 / sec Ra τ m
b. +15 V input:
282
c. –10 V input:
2. Study of the effect of viscous friction:
283
The above figure is plotted for three different friction coefficients (0, 0.001, 0.005) for 5 V armature input. As seen in figure, two important effects are observed as the viscous coefficient is increased. First, the final steady state velocity is decreased and second the response has less oscillation. Both of these effects could be predicted from Eq. (11.1) by increasing damping ratio ζ.
284
3. Additional load inertia effect:
As the overall inertia of the system is increased by 0.005 / 5.2 2 and becomes 1.8493 × 10 −3 kg.m2 , the mechanical time constant is substantially increased and we can assume the first order model for the motor (ignoring the electrical sub-system) and as a result of this the response is more like an exponential form. The above results are plotted for 5 V armature input.
285
4. Reduce the speed by 50% by increasing viscous friction:
As seen in above figure, if we set B=0.0075 N.s/m the output speed drop by half comparing with the case that B=0 N.s/m. The above results are plotted for 5 V armature input.
286
5. Study of the effect of disturbance:
Repeating experiment 3 for B=0.001 N.s/m and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of open loop system is like the effect of higher viscous friction and caused to decrease the steady state value of speed.
287
6. Using speed response to estimate motor and load inertia:
Using first order model we are able to identify system parameters based on unit step response of the system. In above plot we repeated the experiments 3 with B=0.001 and set point voltage equal to 1 V. The final value of the speed can be read from the curve and it is 8.8, using the definition of system time constant and the cursor we read 63.2% of speed final value 5.57 occurs at 0.22 sec, which is the system time constant. Considering Eq. (11-3), and using the given value for the rest of parameters, the inertia of the motor and load can be calculated as:
J=
τ m ( Ra B + K m K b ) 0.22(1.35 × 0.001 + 0.01) = = 1.8496 × 10 − 3 kg.m2 Ra 1.35
We also can use the open loop speed response to estimate B by letting the speed to coast down when it gets to the steady state situation and then measuring the required time to get to zero speed. Based on this time and energy conservation principle and knowing the rest of parameters we are able to calculate B. However, this method of identification gives us limited information about the system parameters and we need to measure some parameters directly from motor such as Ra , K m , K b and so on. So far, no current or voltage saturation limit is considered for all simulations using SIMLab software.
288
7. Open loop speed response using Virtual Lab: a. +5 V:
289
b. +15 V:
c. –10 V:
290
Comparing these results with the part 1, the final values are approximately the same but the shape of responses is closed to the first order system behavior. Then the system time constant is obviously different and it can be identified from open loop response. The effect of nonlinearities such as saturation can be seen in +15 V input with appearing a straight line at the beginning of the response and also the effects of noise and friction on the response can be observed in above curves by reducing input voltage for example, the following response is plotted for a 0.1 V step input:
291
8. Identifying the system based on open loop response:
Open loop response of the motor to a unit step input voltage is plotted in above figure. Using the definition of time constant and final value of the system, a first order model can be found as:
G( s ) =
9 , 0.23s + 1
where the time constant (0.23) is found at 5.68 rad/sec (63.2% of the final value).
292
11-5-2 Open Loop Sine Input 9. Sine input to SIMLab and Virtual Lab (1 V. amplitude, and 0.5, 5, and 50 rad/sec frequencies) a. 0.5 rad/sec (SIMLab):
293
b. 5 rad/sec (SIMLab):
c. 50 rad/sec (SIMLab):
294
d. 0.5 rad/sec (Virtual Lab):
e. 5 rad/sec (Virtual Lab):
295
f. 50 rad/sec (Virtual Lab):
10. Sine input to SIMLab and Virtual Lab (20 V. amplitude) a. 0.5 rad/sec (SIMLab):
296
b. 5 rad/sec (SIMLab):
c. 50 rad/sec (SIMLab):
297
d. 0.5 rad/sec (Virtual Lab):
e. 5 rad/sec (Virtual Lab):
298
f. 5 rad/sec (Virtual Lab):
In both experiments 9 and 10, no saturation considered for voltage and current in SIMLab software. If we use the calculation of phase and magnitude in both SIMLab and Virtual Lab we will find that as input frequency increases the magnitude of the output decreases and phase lag increases. Because of existing saturations this phenomenon is more sever in the Virtual Lab experiment (10.f). In this experiments we observe that M = 0.288 and ϕ = −93.82 o for ω = 50.
299
11-5-3 Speed Control 11. Apply step inputs (SIMLab) In this section no saturation is considered either for current or for voltage.
a. +5 V:
300
b. +15 V:
c. -10 V:
301
12. Additional load inertia effect: a. +5 V:
b. +15 V:
302
c. -10 V:
13. Study of the effect of viscous friction:
303
As seen in above figure, two different values for B are selected, zero and 0.0075. We could change the final speed by 50% in open loop system. The same values selected for closed loop speed control but as seen in the figure the final value of speeds stayed the same for both cases. It means that closed loop system is robust against changing in system’s parameters. For this case, the gain of proportional controller and speed set point are 10 and 5 rad/sec, respectively.
14. Study of the effect of disturbance:
Repeating part 5 in section 11-5-1 for B=0.001 and T L =0.05 N.m result in above figure. As seen, the effect of disturbance on the speed of closed loop system is not substantial like the one on the open loop system in part 5, and again it is shown the robustness of closed loop system against disturbance. Also, to study the effects of conversion factor see below figure, which is plotted for two different C.F. and the set point is 5 V.
304
By decreasing the C.F. from 1 to 0.2, the final va lue of the speed increases by a factor of 5.
15. Apply step inputs (Virtual Lab) a. +5 V:
305
b. +15 V:
c. –10 V:
306
As seen the responses of Virtual Lab software, they are clearly different from the same results of SIMLab software. The nonlinearities such as friction and saturation cause these differences. For example, the chattering phenomenon and flatness of the response at the beginning can be considered as some results of nonlinear elements in Virtual Lab software.
11-5-4 Position Control 16. 160 o step input (SIMLab)
307
17. –0.1 N.m step disturbance
18. Examine the effect of integral control
308
In above figure, an integral gain of 1 is considered for all curves. Comparing this plot with the previous one without integral gain, results in less steady state error for the case of controller with integral part.
19. Additional load inertia effect (J=0.0019, B=0.004):
309
20. Set B=0:
21. Study the effect of saturation
310
The above figure is obtained in the same conditions of part 20 but in this case we considered ± 10 V. and ± 4 A. as the saturation values for voltage and current, respectively. As seen in the figure, for higher proportional gains the effect of saturations appears by reducing the frequency and damping property of the system.
22. Comments on Eq. 11-13 After neglecting of electrical time constant, the second order closed loop transfer function of position control obtained in Eq. 11-13. In experiments 19 through 21 we observe an under damp response of a second order system. According to the equation, as the proportional gain increases, the damped frequency must be increased and this fact is verified in experiments 19 through 21. Experiments16 through 18 exhibits an over damped second order system responses.
23. In following, we repeat parts 16 and 18 using Virtual Lab:
Study the effect of integral gain of 5:
311
312
Ch. 11 Problem Solutions Part 2) Solution to Problems in Chapter 11 11-1. In order to find the current of the motor, the motor constant has to be separated from the electrical component of the motor.
The response of the motor when 5V of step input is applied is:
a) The steady state speed: 41.67rad/sec b) It takes 0.0678 second to reach 63% of the steady state speed (26.25rad/sec). This is the time constant of the motor. c) The maximum current: 2.228A 313
11.2
The steady state speed at 5V step input is 50rad/sec. a) It takes 0.0797 seconds to reach 63% of the steady state speed (31.5rad/sec). b) The maximum current: 2.226A c) 100rad/sec
314
11-3
a) b) c) d)
50rad/sec 0.0795 seconds 2.5A. The current When Jm is increased by a factor of 2, it takes 0.159 seconds to reach 63% of its steady state speed, which is exactly twice the original time period . This means that the time constant has been doubled.
315
11-4 Part 1: Repeat problem 11-1 with TL = -0.1Nm
a) It changes from 41.67 rad/sec to 25 rad/sec. b) First, the speed of 63% of the steady state has to be calculated. 41.67 - (41.67 - 25) × 0.63 = 31.17 rad/sec. The motor achieves this speed 0.0629 seconds after the load torque is applied c) 2.228A. It does not change Part 2: Repeat problem 11-2 with TL = -0.1Nm a) It changes from 50 rad/sec to 30 rad/sec. b) The speed of 63% of the steady state becomes 50 - (50 - 30) × 0.63 = 37.4 rad/sec. The motor achieves this speed 0.0756 seconds after the load torque is applied c) 2.226A. It does not change.
316
Part 3: Repeat problem 11-3 with TL = -0.1Nm
a) It changes from 50 rad/sec to 30 rad/sec. b) 50 - (50 - 30) × 0.63 = 37.4 rad/s The motor achieves this speed 0.0795 seconds after the load torque is applied. This is the same as problem 11-3. c) 2.5A. It does not change d) As TL increases in magnitude, the steady state velocity decreases and steady state current increases; however, the time constant does not change in all three cases.
317
11-5 The steady state speed is 4.716 rad/sec when the amplifier input voltage is 5V:
11-6
a) 6.25 rad/sec. b) 63% of the steady state speed: 6.25 × 0.63 = 3.938 rad/sec It takes 0.0249 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 1 Ampere.
318
11-7 a) 9.434 rad/sec. b) 63% of the steady state speed: 9.434 × 0.63 = 5.943 rad/sec It takes 0.00375 seconds to reach 63% of its steady state speed. c) The maximum current drawn by the motor is 10 Amperes. d) When there is no saturation, higher Kp value reduces the steady state error and decreases the rise time. If there is saturation, the rise time does not decrease as much as it without saturation. Also, if there is saturation and Kp value is too high, chattering phenomenon may appear. 11-8
a) The steady state becomes zero. The torque generated by the motor is 0.1 Nm. b) 6.25 - (6.25 - 0) × 0.63 = 2.31 rad/sec. It takes 0.0249 seconds to reach 63% of its new steady state speed. It is the same time period to reach 63% of its steady state speed without the load torque (compare with the answer for the Problem 11-6 b). 11-9 The SIMLab model becomes
The sensor gain and the speed input are reduced by a factor of 5. In order to get the same result as Proble m 11-6, the Kp value has to increase by a factor of 5. Therefore, Kp = 0.5. The following graphs illustrate the speed and current when the input is 2 rad/sec and Kp = 0.5. 319
11-10
320
a) 1 radian. b) 1.203 radians. c) 0.2215 seconds.
11-11
a) The steady state position is very close to 1 radian. b) 1.377 radians. c) 0.148 seconds. It has less steady state error and a faster rise time than Problem 11-10, but has larger overshoot.
321
11-12 Different proportional gains and the ir corresponding responses are shown on the following graph.
As the proportional gain gets higher, the motor has a faster response time and lower steady state error, but if it the gain is too high, the motor overshoot increases. If the system requires that there be no overshoot, Kp = 0.2 is the best value. If the system allows for overshoot, the best proportional gain is dependant on how much overshoot the system can have. For instance, if the system allows for a 30% overshoot, Kp = 1 is the best value.
322
11-13 Let Kp = 1 is the best value.
As the derivative gain increases, overshoot decreases, but rise time increases.
323
11-14
324
11-15 There could be many possible answers for this problem. One possible answer would be Kp = 100 Ki= 10 Kd= 1.4
The Percent Overshoot in this case is 3.8%.
325
11-16 0.1 Hz
0.2 Hz
326
0.5 Hz
1 Hz
327
2Hz
5Hz
328
10Hz
50Hz
329
As frequency increases, the phase shift of the input and output also increase. Also, the amplitude of the output starts to decrease when the frequency increases above 0.5Hz. 11-17
As proportional gain increases, the steady state error decreases.
330
11-18
Considering fast response time and low overshoot, Kp =1 is considered to be the best value.
331
11-19 It was found that the best Kp = 1
As Kd value increases, the overshoot decreases and the rise time increases.
332
Appendix H
H-1 (a) L( s )
5( s
=
s(s
GENERAL NYQUIST CRITERION
− 2)
Pω
+ 1 )( s − 1 )
When ω = 0 : ∠L ( j 0 ) = −90
o
L ( jω ) =
5( jω − 2)
(
− jω 1 + ω
The Nyquist plot of
2
)
=
P
L( j0 ) o
Whenω = ∞: ∠L ( j ∞) = −180
=1
=∞
L ( j∞)
− 5(ω + 2 j )
(
ω 1+ω
L ( jω )
2
=1
)
=0
L ( jω )
When
= 0 , ω = ∞.
does not intersect the real axis except at the origin.
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 1.5 ) 180 = − 90 o
o
o
Thus, Z=1.
The closed-loop system is unstable. The characteristic equation has 1 root in the right-half s-plane. Nyquist Plot of
(b)
L ( jω )
50
= s( s
+ 5)( s − 1 )
L ( jω ):
Pω
When ω = 0 : ∠L ( j 0 ) = 90
=1 o
When ω = ∞: ∠L ( j ∞) = −270
P
L( j0) o
=1
=∞
L ( j∞)
333
=0
50
L ( jω ) =
2
(
−4ω − j ω 5 + ω 2
(
50 −4ω + jω 5 + ω
2
)
=
16ω + ω 4
2
2
(5+ ω ) 2
)
2
For Im L ( jω ) = 0 , ω = ∞. Thus, the Nyquist plot of L(s) intersects the real axis only at the origin where ω = ∞. Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 1.5 ) 180 = 90 o
o
o
Thus, Z = 2
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. The Nyquist plot of
(c) L ( s ) =
(
3( s + 2)
s s + 3s + 1 3
L ( jω ):
Pω = 1
)
When ω = 0 : ∠L ( j 0 ) = −90
o
When ω = ∞: ∠L ( j ∞) = −270
L ( jω ) =
(ω
3( jω + 2) 4
− 3ω
ω − 3ω − 2 = 0 4
2
2
) + jω or
=
P=2
L( j0 ) o
=∞
L ( j∞)
=0
(
3( jω + 2) ω − 3ω
(4
ω = 3 . 56 2
4
4
− 3ω
2
)
2
2
) − jω
+ω
ω = ± 1.89 rad/sec.
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 2.5 ) 180 = − 90 o
o
Setting
2
o
L ( j 1.89 )
Im L ( jω )
= 0,
=3
Thus, Z = 2
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane.
334
Nyquist Plot of
(d) L ( s ) =
100
(
s ( s + 1) s + 2 2
L ( jω ):
Pω = 3
)
When ω = 0 : ∠L ( j 0 ) = −90
o
P=0 L( j0 )
=∞
o
When ω = ∞: ∠L ( j ∞) = −360 L ( j∞) = 0 The phase of L ( jω ) is discontinuous at ω = 1.414 rad/sec.
(
Φ11 = 35.27 + 270 − 215.27 o
o
o
) = 90
o
Φ11 = ( Z − 1.5) 180 = 90 o
o
Thus,
P
11
=
360 180
o o
=
2
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. Nyquist Plot of
L ( jω ):
335
s − 5s + 2 2
(e)
L(s ) =
(
s s + 2 s + 2 s + 10 3
2
When ω = 0 : ∠L ( j 0 ) = −90
o
When ω = ∞: ∠L ( j ∞) = −180
)
Pω = 1
L( j0 )
=∞
o
L ( j∞)
P=2
=0
( 2 − ω ) − j5ω ( ω − 2 ω ) − jω (10 − 2ω ) ( 2 − ω ) − j5ω L ( jω ) = = ( ω − 2ω ) + jω (10 − 2ω ) (ω − 2ω ) + ω (10 − 2ω ) 2
4
2
2
2
4
( 2 − ω )(10 − 2ω ) + 5( ω
Setting L( jω) = 0,
4
2
2
4
− 2ω
or ω − 3 .43ω + 2 .86 = 0 2 Thus, ω = 1.43 o r 2.0 1. ω = ±1.2 rad / sec or L ( j1.2 ) = − 0 . 7 L ( j1.42 ) = −0 . 34 4
2
2
2
2
2
2
2
2
)= 0
2
± 1.42 r ad
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 2.5 )180 = −90 o
o
o
/ sec
Thus, Z = 2 .
The closed-loop system is unstable. The characteristic equation has two roots in the right-half s-plane. L ( jω ):
Nyquist Plot of
(f)
L(s ) =
(
) ( s + 2 ) = −0.1s − 0.2 s + 0.1s + 0.2 s ( s + s + 1) s ( s + s + 1)
− s −1 2
3
2
2
When ω = 0 : ∠L ( j 0 ) = −90
2
o
When ω = ∞: ∠L ( j∞ ) = 180
L( j0 ) o
=∞
L( j∞)
336
= 0 .1
Pω = 1
P=0
( 0.2 + 0.2ω ) + 0.1 jω ( 1 + ω ) = (1+ ω ) ( 0.2 + j 0.1ω ) − ω − jω ( 1− ω ) L ( jω ) = −ω + jω (1 − ω ) ω + ω ( 1− ω ) 2
2
2
2
2
2
4
Setting Im L ( jω ) = 0 , ω = 2 . 2
2
ω = ±1.414 rad/sec
Thus,
Φ 11 = ( Z − 0.5 Pω − P )180 = ( Z − 0.5 ) 180 = − 90 o
2
o
H-2 (a)
L(s ) =
L ( jω ):
K ( s − 2)
(
s s −1 2
)
Pω = 1
P =1
For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P ) 180 = −270 o
o
For K > 0, Φ 11 = −90 . For K < 0, Φ 11 = +90
o
≠ −270
o
The system is unstable. o
The system is unstable.
Thus the system is unstable for all K.
337
2
L ( j 1.414 )
Thus, Z = 0
o
The closed-loop system is stable. Nyquist Plot of
2
= − 0 .3
Nyquist plot
(b)
L( s )
K
= s(s
+ 10
)( s
P
ω
− 2)
=1
P
=1
For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P ) 180 = −1.5 × 180 = − 270 o
For K > 0, Φ 11 = 90
o
For K < 0, Φ 11 = −90 Nyquist Plot of
o
The system is unstable.
. o
o
≠ −270
o
.
The system is unstable for all values of K .
L ( jω ):
338
(c) L ( s ) =
(
K ( s + 1)
s s + 3s + 1 3
P
)
ω
=1
P
=
2
For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P− 1 ) 180 = − 450 o
For K > 0, Φ 11 = −90
o
For K < 0, Φ 11 = +90
o
o
The system is unstable.
.
when K > −1.
Φ11 = −270
o
when K < −1.
The Nyquist plot of L ( jω ) crosses the real axis at K, and the phase crossover frequency is 1.817 rad/sec. The system is unstable for all values of K. Nyquist Plot of
(d) L ( s ) =
(
L ( jω ):
K s − 5s + 2
(
2
)
s s + 2 s + 2 s + 10 3
2
P
)
ω
=1
P
=
2
For stability, Z = 0. Φ 11 = − ( 0.5 Pω + P ) 180 = −2.5 × 180 = − 450 o
The Nyquist plot of
L ( jω )
o
o
intersects the real axis at the following points:
ω = ∞: L ( j 0 ) = 0 ω = 1.42 rad / sec: L ( j 1.42 ) = − 0 . 34 ω = 1.2 rad / sec: L ( j 1.4 ) = −0 . 7 K
K
339
0
0 or K < 15971.67 Thus, for stability, 0 < K < 30.77 T = 0.5 sec. Forward-path transfer function: K ( 0 . 09883 z + 0 . 07705) G G (z ) = ho p 2 z − 1.4724 z + 0 .47237 w
2
2
+ ( 0 . 09883
0 . 009287
(b)
Characteristic equation: z
Let
z
=
w
+1
w
−1
.
K
− 1.4724
)z
+ ( 0 . 07705
K
+ 0 .47237
)
=0
The characteristic equation becomes + ( 1. 05526 − 0 .1541
2
0 .17588 Kw
+ 2 .94477 − 0 . 02178
K )w
K
=0
For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are:
− 0 . 1541 − 0 . 02178
1. 05526 2 . 9447
Stability condition:
(c) T = 1 sec.
0
0 or > 0 or
K K
< 6 . 8479 < 135 .2
< 6 . 8479
K
Forward-path transfer function: G
ho
G
p
(z )
=
+ 0 .19652
K ( 0 .3214 z z
2
− 1.2231
)
+ 0 .22313
z
Characteristic equation: z
Let
z
=
w
+1
w
−1
.
2
+ ( 0 . 3214
K
− 1.2231
)z
+ ( 0 .22313 + 0 .19652
=0
K )
The characteristic equation becomes
0 . 5179 Kw
2
+ ( 1. 55374 − 0 . 393
K )w
+ 2 .4462 − 0 .12488
K
=0
For stability, all the coefficients of the characteristic equation must be of the same sign. Thus, the conditions for stability are :
1. 55374 2 .4462
Stability condition:
I-23
0
0 − 0 .12488 K > 0
K
or or
< 3 . 9535
(a)
Roots: -1.397 -0.3136 + j0.5167 Unstable System.
-0.3136 – j0.5167
(b)
Roots: 0.3425 -0.6712 + j1.0046 Unstable System.
-0.06712 – j1.0046
361
< 3 . 9535 < 19 . 588
K
I-24
(c)
Roots: -0.4302 Unstable System.
-0.7849 + j1.307
(d)
Roots: 0.5 -0.8115 Stable System.
-0.7849 – j1.307
-0.0992 + j0.7708
-0.0992 – j0.7708
(a) Forward-path Transfer Function: T = 0.1 sec. 5 = 1 − z −1 Z 2.5 − 1.25 + 1.25 = 0.02341z + 0.021904 −1 G ( z ) = (1 − z ) Z 2 ) s 2 s s + 2.5 z 2 − 1.8187z + 0.8187 ( s ( s + 2) Closed-loop Transfer Function: Y (z) R( z )
=
G (z) 1+ G ( z)
=
0 . 0234 z z
2
+ 0 . 021904
− 1. 7953
z
+ 0 .8406
(b) Unit-Step Response y(kT)
(c) Forward-path Transfer Function: T = 0.05 sec. 5 = 0.0060468 z + 0.0058484 −1 G ( z ) = (1 − z ) Z 2 s ( s + 2) z 2 − 1.9048z + 0.9048 Closed-loop Transfer Function: Y (z) R( z )
=
0 . 006046 z
2
z
− 1. 8988
362
+ 0 . 005848 z
+ 0 . 91069
4
Unit-step Response y(kT)
(a)
I-25
(
Y ( z) = 1 − z
−1
(
= 10
Y ( z)
G( z) =
−1
∗ p
= lim
z →1
K
∗ a
G (z)
=
1 2
T
)T
2
K T
−
E (z)
t
z
−1
(
U ( z ) = 10 E( z) − Kt 1 − z U (z)
z ( z + 1) 10( z −1) z − 1 + K tT
3
10( z z
) Z s
1
− 1)
−1+ K tT
2
U ( z)
E (z)
E ( z)
5T ( z + 1)
( z − 1) ( z − 1 + K T ) t
lim ( z
K
− 1)
2
∗
=
v
G (z)
1 T
lim ( z z →1
−1 )G ( z ) =
10 T
2
K T
2
t
=
10 K
t
=0
(b) Forward-path Transfer Function:
Closed-loop Transfer Function:
5T ( z + 1) 2
G( z) =
=
U (z)
Th us
−1
2
=
=∞
z →1
3
2( z − 1)
E( z) Error Constants: K
U ( z)
1
U (z)
Y ( z) = 1 − z
) Z s
Y ( z)
( z − 1) ( z − 1 + K T )
R (z )
t
5T ( z + 1) 2
=
(
) (
)
z + K tT + 5T − 2 z + 5T − K tT + 1 2
2
2
(c) Characteristic Equation: T = 0.1 sec.
(
) (
)
F (z ) = z + Kt T + 5T − 2 z + 5T − K tT + 1 = z + ( 0.1Kt − 1.95) z + ( 1.05 − 0.1Kt ) = 0 2
2
2
363
2
For stability, from Eq. (6-62), where Thus,
a
0
= 1. 05 − 0 .1 K t F (1 ) a0
= 0 .1 > 0
=
> 0,
F (1 ) a
1
F ( − 1)
0 .5 K
t
> 0,
a0
= 4 − 0 .2 K t > 0
− 0 .1 K t < a 1 = 1
1. 05
Stability Condition:
(d) Unit-step Response:
F ( −1 )
or
0.5 < K t
< K t < 20
=5 Y (z)
=
R( z)
(e) Unit-step Response
< a2
= 0 .1 K t − 1. 95
K
t
0 . 05 ( z z
2
z
2
+ 1)
− 1.45 z + 0 . 55
=1
Y (z) R( z )
=
0 . 05 ( z
+1)
364
− 1. 85 z + 0 . 95
or
< 20 . 5
K
t
< 20
I-26 (a) T = 0.1 sec.
s + 37.06 s + 141.2 E (z ) ( s + 37.06 z + 141.2 ) Ωm ( z) −1 − 30 + 14.2 + 29.7 + 0.418 = (1 − z ) Z 2 E( z) s s + 4.31 s + 32.7 s 1000( z + 20)
Gp (s ) =
Ω m ( z)
(
= 1− z
2
(
= 1− z
=
−1
)
z
+
z −1
3 . 368 z 3
−30 z
− 1. 6876
( z −1)
+ 1. 7117
2
z
14.2 z
2
z
+ 0 . 71215
29.7 z
+
2
z
+
z − 0.65
− 0 . 3064
=
− 0 . 024576
−1
100( s + 20)
)Z s
2
z − 0.038 0.418 z
+ 1. 7117
z
− 0 . 3064
− 1 )( z − 0 . 65)(
z
− 0 . 038
3 . 368 z (z
2
(b) Closed-loop Transfer Function: T = 0.1 sec. Ωm ( z ) R (z )
=
3. 368 z
+ 1. 7117
2
+ 1. 6805
z
3
3
+ 1. 6805
2
z
z
− 0 . 3074
+ 2 .424
z
− 0 . 331
Characteristic Equation: z
z
2
+ 2 .424
z
− 0 . 331 = 0
Roots of Characteristic Equation: z
= 0 .125
,
− 0 . 903 +
The complex roots are outside the unit circle
365
j1. 3545 , z
− 0 . 903 −
j 1. 3545
= 1, so the system is unstable.
)
(c) T = 0.01 sec.
Forward-path Transfer Function
Ωm ( z ) E (z)
Ωm ( z )
Closed-loop Transfer Function
R (z )
= =
− 2 . 6785
3
z
z 2
z
0 . 047354
− 2 . 6312
3
z
+ 0 . 005974
2
0 . 04735 z
z
− 0 . 69032
z
+ 0 . 005875
− 0 . 03663
3z
+ 2 . 3748
2
− 0 . 03663
3z
+ 2 . 3689
2
z
− 0 . 72695
Characteristic Equation: − 2 . 6312 Characteristic Equation Roots: z
z
3
= 0 . 789
,
+ 2 . 3748
2
z
+
0 . 921
T = 0.001 sec. Forward-path Transfer Function Ω m ( z ) −1.43 × 10 =
−6
z
−6
z
j 0 .27 ,
+ 4. 98 × 10
−4
z
3
− 2 . 9635
2
z
−4
+ 4. 98 × 10
3
R (z )
− 2 . 963
3
z
z
+ 2 .9271 z
2
j 0 .27
− 2 . 384 × 10
2
z
−7
z
− 4. 89 × 10
z
− 4. 89 × 10
−4
− 0 .9636
z
− 2 . 384 × 10
+ 2 . 9271
2
z
−
0 . 921
3
E (z)
Closed-loop Transfer Function Ω m ( z ) −1.43 × 10 =
− 0 . 72695 = 0
z
−7
−4
− 0 . 9641
Characteristic Equation: z
3
− 2 .963
+ 2 . 9271
2
z
z
− 0 . 9641 = 0
Characteristic Equation Roots: z
= 0 . 99213
,
+
0 . 98543
j 0 . 02625 ,
−
0 . 98543
j 0 . 02625
(d) Error Constants: K
∗ p
= lim
∗
1
Kv
∗
z →1
=
Ka =
lim ( z z →1
T
1 T
G ho G p ( z )
2
− 1)
2
= lim
z →1
+ 1.7117
z
− 0 . 3064
− 1)( z − 0 .65)(
z
− 0 . 038
3 . 368 z (z
G ho G p ( z )
2
3. 368 z
= − lim
z →1
lim ( z − 1) Gho Gp ( z) = z →1
T
− 0 . 65)(
(z
2
z →1
Steady-state Errors: Step Input: Ramp Input:
e
e
∗ ss
∗ ss
= =
1 1+ K 1 K
∗
+ 1. 7117
( 3.368 z lim
1
2
2
∗
=0
p
=
v
366
T 14. 177
=∞ )
2
z
z
− 0 . 3064
− 0 . 038
=
14. 177
)
T
+ 1.7117 z − 0.3064 ( z − 1)( z − 0.038)
) ( z − 1) = 0
e
Parabolic Input:
I-27
∗ ss
=
1 K
=∞
∗ a
(a) Forward-path Transfer Function: (no zero-or der hold) T = 0.5 sec. G (z)
0 .1836 zK
= z
2
z
2
− 1. 0821
z
+ 0 . 0821
0 .1836 zK
= (z
− 1 )( z − 0 . 0821
T = 0.1 sec. G (z)
=
0 . 0787 zK
− 1. 6065
z
+ 0 . 6065
367
=
0 . 0787 zK (z
− 1)( z − 0 .6065)
)
(b)
Open-loop Transfer Function: (with zero-order hold) T = 0.5 sec. G (z)
=
K ( 0 . 06328 z
− 1. 0821
+ 0 . 02851
z
+ 0 . 08021
K ( 0 . 00426 z
+ 0 .003608
z
2
)
=
0 . 06328 K ( z (z
+ 0 .4505)
− 1 )( z − 0 . 0821
)
T = 0.1 sec. G (z)
=
z
I-28
2
− 1. 6065
Forward-path Transfer Function: G( z) =
z
+ 0 . 6065
(
)
=
0 . 00426 K ( z (z
− 1 )( z − 0 .6065)
) + 2.4644 z − 0.7408 )
0.0001546 K z + 3.7154 z + 0.8622
(
z z − 2.7236 z 3
2
2
368
+ 0 .8468
)
I-29 (a)
P (z )
=
z
3
−1
Q (z)
=
z
2
+ 1. 5 z − 1 = ( z − 0 . 5)(
The system is unstable for all values of K .
369
z
+ 2)
I-30
(a) Bode Plot:
The system is stable.
(b) Apply w-transformation, z
Then
G
=
ho
2
+ wT
2
− wT
G(w)
The Bode diagram of
G
ho
= G ho G ( z ) G (w )
z=
2 + wT
=
10( 1
2 − wT
− 0 .0025 w (1 + w )
is plotted as shown below.
The gain and phase margins are determined as follows: GM = 32 dB
PM = 17.7 deg.
370
2
w )
Bode plot of
I-31
G
ho
G (w ) :
2 16.67 N 1 − e −Ts 0.000295 ( z + 3.39 z + 0.714) G hoG ( z ) = Z = s s ( s + 1)( s + 12.5) ( z − 1) ( z − 0.9486 )( z − 0.5354 )
The Bode plot of
G
ho
G(z)
is plotted as follows. The gain margin is 17.62 dB, or 7.6.
Thus selecting an integral value for N, the maximum number for N for a stable system is 7. Bode Plot of G ho G ( z )
371
I-32
(a)
G (s) c
=2+
200 s
Backward-rectangular Integration Rule: Gc ( z ) = 2 +
200T z −1
=
2 z − 2 + 200T z −1
=
2 + ( 200T − 2 ) z 1− z
−1
−1
Forward-rectangular Integration Rule: Gc ( z ) = 2 +
200Tz
=
z −1
( 2 + 200T ) z − 2 ( 2 + 200T ) − 2 z =
z −1
1−z
−1
−1
Trapazoidal Integration Rule: 200T ( z + 1)
Gc ( z ) = 2 +
(b)
G (s) c
2 ( z − 1)
=
( 4 + 200T ) z + 200T 2 ( z − 1)
−2
=
( 4 + 200T ) + ( 200T − 2 ) z
(
2 1−z
−1
)
= 10 + 0 .1 s
The controller transfer function does not have any integration term. The differentiator is realized by backward difference rule. G c ( z ) = 10 +
(c)
G (s) c
= 1 + 0 .2 s +
0.1 ( z − 1)
=
(10T + 0.1) z − 0.1
Tz
z
= (10T + 0.1) − 0.1z
5 s
Backward-rectangular Integration Rule: Gc ( z ) = 1 +
0.2 ( z − 1)
+
Tz
5T
( z − 1)
=
( T + 0.2 ) − 0.2 z
−1
+
T
5Tz
−1
1− z
−1
Forward-rectangular Integration Rule: Gc ( z ) = 1 +
0.2 ( z − 1)
+
Tz
5Tz z −1
=
( T + 0.2 ) − 0.2 z
−1
T
+
5T 1− z
−1
Trapezoidal Integration Rule:
Gc ( z ) = 1 +
0.2 ( z − 1) Tz
+
5T ( z + 1) 2 ( z − 1)
=
372
( T + 0.2 ) − 0.2z T
−1
+
( ) 2 (1 − z )
5T 1 + z
−1
−1
−1
−1
I-33
(a)
G (s) c
=
10 s
T = 0.1 sec
+ 12
(
Gc (z ) = 1 − z
−1
(
10 ( s + 1.5 ) s + 10
(
Gc (z ) = 1 − z
(
(c)
s
=
G (s) c
(
G (z) c
=
I-35
−1
) z − 1 − z − e z
z
−1.2
= 0.5825 z − 0.301
T = 1 sec
10 ( s + 1.5 )
) Z s ( s + 10) = ( 1 − z ) Z ) z − 1 + z − e 1.5 z
8.5
−1
−1
1.5 s
+
s + 10 8.5
= 10 ( z − 0.9052) z − 0.368
−1
) Z s + 1.55 = ( 1 − z ) z − e 1
z
−1
−0.155
= z −1 z − 0.8564
+ 0 . 01 s
Gc (z ) = 1 − z
I-34
1
1 + 0 .4 s 1
(
1
−1
T = 0.1 sec
+ 1. 55
s
Gc (z ) = 1 − z
(d)
−1
−1
= 1− z
10
= 0.8333 1 − z
(b) G c ( s ) =
) Z s ( s + 12) = 0.8333( 1 − z ) Z s − s + 12
−1
) Z
−1 0.025 z + 0.975z = 40 z − 0.975 = 40 ( 1 − z ) Z z − 1 z − e−10 z − 0.0000454 s (1 + 0.01 s ) 1 + 0.4 s
(a) Not physically realizable, since according to the form of Eq. (11-18), (b) Physically realizable. (c) Physically realizable. (d) Physically realizable. (e) Not physically realizable, since the leading term is 0.1z. (f) Physically realizable. (a)
G (s) c
= 1 + 10
s
K
P
=1
K
D
= 10
373
Thus
Gc ( z ) =
b
0
≠0
but a
0
= 0.
( T + 10) z − 10 Tz
(
) Z s4 = 2T ( z + 1) 2
G ho Gp ( z ) = 1 − z
−1
G ( z ) = G c ( z ) Gho G p ( z ) =
( z − 1) 2T ( z + 1) [(T + 10 ) z − 10 ] z ( z − 1) 3
2
2
By trial and error, when T = 0.01 sec, the maximum overshoot of When T = 0.01 sec, G( z) =
0.02 ( z + 1) (10.01 z − 10 ) z ( z − 1)
Y ( z)
2
R (z )
=
y ( kT )
is less than 1 percent.
0.02 ( z + 1)( 10.01 z − 10 ) z − 1.7998 z + 1.0002 z − 0.2 3
2
When the input is a unit-step function, the output response y ( kT ) is computed and tabulated in the following for 40 sampling periods. The maximum overshoot is 0.68%, and the final value is 1. Sampling Periods k y ( kT ) ------------------------------
I-36
1 s3
(a) G ho Gp ( z ) = ( 1 − z −1 ) Z
2 = 2T ( z + 1) ( z − 1 )2
G ( z ) = G c ( z ) Gho G p ( z ) =
=
Characteristic Equation:
(
K PTz + KD ( z − 1 ) 2T
( z + 1) ( z − 1) 2
2
Tz
( K PT + KD ) z2 + K P Tz − K D
2T
z ( z − 1)
)
2
(
)
z + 2 K PT + K DT − 1 z + 2K PT + 1 z − 2 K DT = 0 3
2
2
2
For two roots to be at z = 0.5 and 0.5, the characteristic equation should have z − z + 0 .25 as a 2 factor. Dividing the characteristic equation by z − z + 0 .25 and solving for zero remainder, we get 2
374
4K T
2
P
Solving for
K
P
and K
+ 2 K D T − 0 .25 = 0
D
K
The third root is at
z
− 0 . 5 K P T − 2 .5 K D T + 0 .25 = 0 2
a nd
from these two equations, we have P
=
0 . 0139 T
K
2
0 . 0972
=
D
T
= 1 − 2 K P T − 2 K D T = 0 . 7778 2
T
f or
= 0 . 01
sec
The forward-path transfer function is G( z) =
Y (z) R( z )
=
0.2222 ( z + 1) ( z − 0.8749) z ( z − 1)
z
3
+ 0 . 0278
2
0 .2222 z
− 1. 7778
2
z
2
z
+ 1. 0278
− 0 .1944 z
− 0 .1944
Unit-step Response:
(b)
(b) KP = 1, T = 0.01 sec G (z)
= G c ( z ) G ho G p ( z ) =
=
2T
δ+ ι + − β− γ δ + ι + β− γ T
K
z
D
z z
0 . 02
0 . 01
2
1
Tz
K
D
2
KD z
2
z z
1
0 . 01 z
− KD
2
The unit-step response of the system is computed for various values of
K
D
. The results are
tabulated below to show the values of the maximum overshoot. KD Max overshoot (%)
1.0 14
5.0 0.9
6.0 0.67
7.0 0.5
375
8.0 0.38
9.0 0.31
9.1 0.31
9.3 0.32
9.5 0.37
10.0 0.68
I-37
(a) Phase-lead Controller Design:
(
G ( z ) = G ho Gp ( z ) = 1 − z
−1
) Z s
= 0.02 ( z + 1) z ( z − 1) 2
4
3
T = 0.1 sec
Closed-loop Transfer Function: Y (z ) R (z )
=
Gho Gp ( z)
0.02 ( z + 1)
=
1 + G ho Gp ( z )
2
=
z
With the w-transformation,
T
G ( jω
w
)
+w =
2 T
From the Bode plot of
The system is unstable.
z − 1.98 z + 1.02 2
20
+w
20
−w
G ( w) =
4 (1 − 0.05 w ) w
2
−w
the phase margin is found to be −5.73 degrees.
For a phase margin of 60 degrees, the phase-lead controller is G (w ) c
=
1 + aTw 1
+ Tw
=
1
+ 1.4286
w
1
+ 0 . 0197
w
The Bode plot is show below. The frequency-domain characteristics are: PM = 60 deg
M
GM = 10.76 dB
r
= 1.114
The transfer function of the controller in the z-domain is Gc ( z ) =
21.21 ( z − 0.9222 )
( z + 0.4344)
(b) Phase-lag Controller Design: Since the phase curve of the Bode plot of
G ( jω
w
)
is always below −180 degrees, we cannot
design a phase-lag controller for this system in the usual manner.
376
Bode Plots for Part (a):
G ho G p ( z )
=
Λ ε− ϕΜ Μ Ν β+ 1
z
−1
4500 K
Z
s
2
s
Ο β + Π = γΠ Θ β− γβ− K 0 . 002008
361 .2
377
z
1
z
z
0 . 001775
0 . 697
γ
γ
I-38
(a) Forward-path Transfer Function: ∗
Kv =
1 T
lim z →1
[ ( z − 1) G
]
G p ( z ) = lim
ho
K ( 2.008 z + 1.775)
= 1000
z − 0.697
z →1
Thus
K
∗ v
= 80 .1
(b) Unit-step Response:
Maximum overshoot = 60 percent.
(c) Deadbeat-response Controller Design: (K = 80.1) G ho Gp ( z ) =
G ho G p ( z
−1
)
=
Q(z P(z
−1 −1
)
=
0.16034 z + 0.14217
( z − 1) ( z − 0.697 )
0 . 16034 z
−1
1 − 1. 697 z
)
+ 0 .14217
−1
+ 0 . 697
z z
−2
Q (1 )
−2
= 0 . 3025
Digital Controller: −1
Gc ( z ) =
−1
P (z ) −1
Q (1) − Q (z )
= =
1 − 1.697 z + 0.697 z −1
0.3025 − 0.16034 z − 0.14217z z − 0.53 z − 0.47 2
M (z)
Closed-loop system transfer function: Unit-step response:
Y (z)
−2
3.3057 ( z − 1 ) ( z − 0.697 )
G ( z ) = G c ( z ) Gho G p ( z ) =
Forward-path transfer function:
−2
= 0 . 53
z
378
−1
+z
−2
+z
=
−3
3.3057 ( z − 1 ) ( z − 0.697 ) z − 0.53 z − 0.47 2
0 . 53 z
+L
z
+ 0 .47 2
Deadbeat Response:
I-39
G p (s)
2500
=
(
G hoG ( z ) = 1 − z
G ho
T = 0.05 sec
+ 25)
s(s
−1
) Z
2.146 z + 1.4215 = s ( s + 25 ) ( z − 1 ) ( z − 0.2865 ) 2500
2
( ) = 2.146 z + 1.4215 z G( z ) = P ( z ) 1 − 1.2865 z + 0.2865z Q z
−1
−1
−1
−1
−2
−1
Q (1 )
−2
= 3 . 5675
Deadbeat Response Controller Transfer Function:
( )=Q
Gc z
−1
( ) = 1 − 2.865z + 0.2865z (1) − Q ( z ) 3.5675 − 2.146 z − 1.4215 z P z
−1
−1
−2
−1
−1
−2
G (z) c
Forward-path Transfer Function: G (z)
= G c ( z ) G ho G p ( z ) =
2 .146 z 3 . 5675 z
+ 1.4215
− 2 . 146
z
0 . 6015 z
+ 0 . 3985
2
− 1.4215
Closed-loop System Transfer Function: M (z)
Unit-step response:
Y (z)
=
= 1 + 0 . 6015
z
379
−1
z
+z
2
−2
+z
−3
+L
=
(z
− 1 )( z − 0 .285)
3 . 5675 z
2
− 2 .146
z
− 1.4215
I-40
The characteristic equation is z + ( − 1.7788 + 0.1152 k1 + 22.12 k 2 ) z + 0.7788 + 4.8032 k1 − 22.12 k 2 = 0 2
For the characteristic equation roots to be at 0.5 and 0.5, the equation should be z
2
− z + 0 .25 = 0
Equating like coefficients in the last two equations, we have − 1.7788 + 0 .1152 0 . 7788
Solving for the value of
k and k 1
2
+ 4. 8032
k k
1
1
+ 22 .12 k 2 = − 1
− 22 . 12 k 2 = 0 .25
from the last two equations, we have
380
k
1
= 0 . 058
and k
2
= 0 . 035
.
