自动控制原理(中英文对照李道根)习题3题解

(d) (s)
12.5
(s 2 2s 5)(s 5)
Solution: (a) (s)
■Solutions
■Solutions
P3.1 The unit step response of a certain system is given by c(t) 1 e t e2t , t 0
(a) Determine the impulse response of the system. (b) Determine the transfer function C(s) R(s) of the system.
R (s)
k1
s
0
k2
s
0
C (s )
(a) Block diagram
c (t)
c (t)
c(t)
1.0
1.0
1.0
t
t
t
0
0
0
(1)
(2)
(3)
c (t)
c(t) Asy mp totic
Parabolic
line
1.0
1.0
t 0
(4 )
t
0
(5 )
(b) Unit-step resp onses
response, i.e.
C(s) L[ (t) e t 2e2t ] 1 1 2 s 2 4s 2
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.
location of the poles on the complex plane, sketch the unit step response, explaining the results
obtained.
(a)
(s)
s2
20 12s
20
,
(b)
(s)
s3
6 6s2 11s
6
(c) (s) 4 , s2 2s 2
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 k2 s k1k2 . Obviously, the outlet feedback is “+” and the inner feedback is “–”.
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 (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 k2 s k1k2 . Obviously, both the outlet and inner feedbacks are “–”.
Figure P3.2
Solution: In general we have
13
■Solutions
C(s)
k1 k 2
R(s)
s2
0
k
2
s
0
k1 k 2
Note that the characteristic polynomial is
(s)
s2
0
k2
s
0
k1 k 2
where the sign of k2 s is depended on the outer feedback and the sign of k1k2 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 k1k2 . Obviously, the outlet feedback is “–”and the inner feedback is “0”.
Solution: The impulse response is the differential of corresponding step response, i.e.
k(t) dc(t) (t) et 2e 2t dt
As we know that the transfer function is the Laplace transform of corresponding impulse
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