FIGURE 5.8. Percent overshoot and normalized peak time versus damping ratio z
for a second-order system (Eq. 5.8). Dorf/Bishop. Modern Control Systems 9/E.
Dorf/Bishop Modern Control Systems 9/E 2.0 1.8
z 0.1
1.6
0.2
1.4 0.4 1.2 0.7 y(t) 1.0
1.0
0.8 2.0
0.6 0.4 0.2 0.0
0
2
4
6
8
10
12
14
vn t (a)
vn t 4.0
2.0
0.0 2.0
6.0
8.0
10.0
12.0
14.0
Output
1.5
1.0
0.5
0.0 12.0
Da
mp
ing
14.0
10.0
0.4
8.0
0.8
6.0
1.2
fac
tor
z
4.0 1.6
vn t
2.0
2.0
(b)
FIGURE 5.5 (a) Transient response of a second-order system (Eq. 5.9) for a step input. (b) The transient response of a second-order system (Eq. 5.9) for a step input as a function of and vnt. (Courtesy of Professor R. Jacquot, University of Wyoming.) © 2001 by Prentice Hall, Upper Saddle River, NJ.
Dorf/Bishop Modern Control Systems 9/E
1.0
z 0.10
0.8
0.25
0.6
0.50
0.4 y(t) vn
1.0
0.2 0.0 0.2 0.4 0.6 0.8
0
1
2
3
4
5
6
7
8
9
10
vn t
FIGURE 5.6 Response of a second-order system for an impulse function input.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Dorf/Bishop Modern Control Systems 9/E
Mpt Overshoot
ess
1.0 d 1.0 0.9 y(t)
1.0 d
0.1 0
Tp Peak time
Tr1
Ts Settling time
Tr Rise time
FIGURE 5.7 Step response of a control system (Eq. 5.9).
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Time
Percent maximum overshoot
Dorf/Bishop Modern Control Systems 9/E
100
5.00
90
4.80
80 70
Percent overshoot
4.60 4.40 4.20
60 50
v n Tp
4.00 v n Tp
40
3.80
30
3.60
20
3.40
10
3.20
0 3.00 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Damping ratio, z
FIGURE 5.8 Percent overshoot and normalized peak time versus damping ratio z for a second-order system (Eq. 5.8).
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Dorf/Bishop Modern Control Systems 9/E
3.5
3.0
Actual rise time 2.5 v nTr
Linear approximation 2.16z 0.60 Tr1 vn
1
2.0
1.5
1.0
0
0.1 0.2
0.3 0.4 0.5 0.6 z
0.7
0.8 0.9
1
FIGURE 5.9 Normalized rise time Tr versus z for a second-order system. 1
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Dorf/Bishop Modern Control Systems 9/E
1.6
v n 10 rad/s
1.4 v n 1 rad/s
Amplitude
1.2 1.0 0.8 0.6 0.4 0.2 0
0
1
2
3
5 4 6 Time (seconds)
7
8
9
FIGURE 5.10 The step response for z 0.2 for vn 1 and vn 10.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
10
Dorf/Bishop Modern Control Systems 9/E
jv roots of the closed-loop system
1 g
1
z
FIGURE 5.12 An s-plane diagram of a third-order system.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
s
Dorf/Bishop Modern Control Systems 9/E 10 z
2 0.
4
2.0 a zvn
0.3
0.
1 0.
5.0
0.7
1.0
0.6
0.5
0.8 1.0
0.5
0.2
0.1 1
2
5
10
20 50 Percent overshoot
100
200
500
1000
(a) 3.5 A: 5 B: 2 C: 1 D: 0.5
D 3.0
Output
2.5
2.0
C
1.5
B A
1.0
0.5
0
0
2
4
6
8
10
12
Time (v n t) (b)
FIGURE 5.13 (a) Percent overshoot as a function of z and vn when a second-order transfer function contains a zero. (From R. N. Clark, Introduction to Automatic Control Systems, New York, Wiley, 1962, redrawn with permission.) (b) The response for the second-order transfer function with a zero for four values of the ratio (a/zvn ): A 5, B 2, C 1, and D 0.5 when z 0.45.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Dorf/Bishop Modern Control Systems 9/E
z 0.707
jv
r1 z 0.707 and zv n 1
45 zv n 1
rˆ1
FIGURE 5.15 Specifications and root locations on the s-plane.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
s
Dorf/Bishop Modern Control Systems 9/E
jv
1
1
1
s 1
1
0
0
1 t
FIGURE 5.17 Impulse response for various root locations in the s-plane. (The conjugate root is not shown.)
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Dorf/Bishop Modern Control Systems 9/E
y(t)
Input Output e ss e ss
FIGURE 5.20 Triangular wave response.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
t
Dorf/Bishop Modern Control Systems 9/E
Amplifier R(s)
Ka
Motor constant
D(s)
5
Load 1 s(s + 20)
FIGURE 5.51 Control system model with a second-order model of the motor and load.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
Y(s)
Dorf/Bishop Modern Control Systems 9/E
Select Ka. Ka=30; t=[0:0.01:1]; nc=[Ka*5];dc=[1]; sysc=tf(nc,dc); ng=[1];dg=[1 20 0]; sysg=tf(ng,dg); sys1=series(sysc,sysg); sys=feedback(sys1, [1]); [y,T]=step(sys,t); plot(T,y), grid xlabel('Time (sec)') ylabel('y(t)')
Compute the closed-loop transfer function.
(a) 1.2 Ka = 60. 1
y(t)
0.8 Ka = 30. 0.6 0.4 0.2 0 0
0.1
0.2
0.3
0.4 0.5 0.6 Time (sec)
0.7
0.8
0.9
(b)
FIGURE 5.52 Response of the system to a unit step input, r(t) 1, t 0. (a) MATLAB script. (b) Response for Ka 30 and 60.
© 2001 by Prentice Hall, Upper Saddle River, NJ.
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