自动控制原理(中英文对照 李道根)习题3.题解

P3.2 Consider the system described by the block diagram shown in Fig. P3.2(a). Determine the polarities of two feedbacks for each of the following step responses shown in Fig. P3.2(b), where “0” indicates that the feedback is open.
t 0
(d) (s )
12.5
2

12.5
2
c(t ) 0 .5 t 0
system. The unit step response, with a closed-loop gain k 0.5 , is sketched as shown.
G ( s) 1 s( s 1)
P3.4 The open-loop transfer function of a unity negative feedback system is
6 ( s 1 )( s 2)(s 3) s 6s 11s 6 By inspection, the characteristic roots are 1 , 2 , 3 .
(b) (s )
6
3
2

ቤተ መጻሕፍቲ ባይዱ
c (t ) 1.0 t 0
Obviously, all three transient components are decayed exponential terms. Therefore, its unit step response, with a closed-loop gain k 1 , is sketched as shown.
2 ln 2 p
Since t p 1 sec . , solving the formula for calculating the peak time, t p
Determine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion). Solution: Writing he closed-loop transfer function 2 n 1 (s ) 2 2 2 s s 1 s 2 n s n we get n 1 , 0.5 . Since this is an underdamped second-order system with 0.5 , the system performance can be estimated as follows. Rising time t r Peak time t p
0 0
where the sign of k 2 s is depended on the outer feedback and the sign of k1 k 2 is depended on the inter feedback. Case (1). The response presents a sinusoidal. It means that the system has a pair of pure imaginary roots, i.e. the characteristic polynomial is in the form of ( s ) s 2 k1 k 2 . Obviously, the outlet feedback is “–”and the inner feedback is “0”. Case (2). The response presents a diverged oscillation. The system has a pair of complex conjugate roots with positive real parts, i.e. the characteristic polynomial is in the form of ( s ) s 2 k 2 s k1 k 2 . Obviously, the outlet feedback is “+” and the inner feedback is “–”. Case (3). The response presents a converged oscillation. It means that the system has a pair of complex conjugate roots with negative real parts, i.e. the characteristic polynomial is in the form of ( s ) s 2 k 2 s k1 k 2 . Obviously, both the outlet and inner feedbacks are “–”. Case (4). In fact this is a ramp response of a first-order system. Hence, the outlet feedback is “0” to produce a ramp signal and the inner feedback is “–”. Case (5). Considering that a parabolic function is the integral of a ramp function, both the outlet and inner feedbacks are “0”. P3.3 Consider each of the following closed-loop transfer function. By considering the location of the poles on the complex plane, sketch the unit step response, explaining the results obtained. (a) ( s) (c) ( s)
k (t ) dc(t ) (t ) e t 2e 2t dt
As we know that the transfer function is the Laplace transform of corresponding impulse response, i.e.
C ( s) 1 2 s 2 4s 2 L[ (t ) e t 2e 2t ] 1 2 R (s ) s 1 s 2 s 3s 2
p
1.3 1 100 % 30 % 1
t (s) 0 0. 1
Solving the formula for calculating the overshoot,
15
Figure P3.5
■Solutions
p e
1 2
0.3 , we have ln p 0.362
20 s 12s 20
By inspection, the characteristic roots are 2 , 10 . This is an overdamped second-order system. Therefore, considering that the closed-loop gain is k 1 , its unit step response can be sketched as shown.
[(s＋1 ） 2 2 ]( s 5) By inspection, the characteristic roots are 1 j 2 , 5 . Since (s 2s 5)(s 5) 0.1 5 , there is a pair of dominant poles, 1 j 2 , for this
arccos n 1 2 n 1
2

arccos 0.5
1 1 0.5 2
2.42 sec .

1 2

1 1 0.5 2
3.62 sec .
1 0.52
Percent overshoot p e Setting time t s
R (s) k1 k2

C (s )

0
s
0
s
(a) Block diagram c (t ) c (t ) c (t )
1.0 t 0 (1)
1.0 t 0 (2) c (t ) c (t ) 1.0 0
(4 )
1.0 t 0 (3)
A sy mp totic line t
Parabolic 1 .0 t
14
■Solutions
(c) (s )
4 s 2s 2
2

4 ( s 1) 2 1
2.0
c (t )
This is an underdamped second-order system, because its characteristic roots are 1 j . Hence, transient component is a decayed sinusoid. Noting that the closed-loop gain is k 2 , the unit step response can be sketched as shown.
0
(5 )
(b) U nit-step resp onses
Figure P3.2
Solution: In general we have
13
■Solutions
k1 k 2 C ( s) 2 R( s ) s k 2 s k1 k 2
0 0
Note that the characteristic polynomial is ( s ) s 2 k 2 s k1 k 2
3
100 % e 0.5
100 % 16.3 %
n

3 6 sec . (using a 5% setting criterion) 0.5 1
c (t ) 1 .3 1 .0
P3.5 A second-order system gives a unit step response shown in Fig. P3.5. Find the open-loop transfer function if the system is a unit negative-feedback system. Solution: By inspection we have
20 , s 12s 20 4
2
(b) ( s) (d) ( s)
20 (s 2)( s 10)
s 2 2s 2
2
,
6 s 6s 11s 6 12.5
3 2
( s 2 2s 5)( s 5)
c( t ) 1.0 t 0
Solution: (a) (s )
■Solutions
■Solutions
P3.1 The unit step response of a certain system is given by c (t ) 1 e t e 2t , t 0 (a) Determine the impulse response of the system. (b) Determine the transfer function C ( s) R ( s) of the system. Solution: The impulse response is the differential of corresponding step response, i.e.