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.
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
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 )
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)
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
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
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
■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.
2 ln 2 p
Since t p 1 sec . , solving the formula for calculating the peak time, t p
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
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