Modern Control Engineering

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248 Chapter 5 / Transient and Steady-State Response Analyses

Solution.The closed-loop transfer function is

MATLAB Program 5–24 produces the unit-acceleration response. The resulting response, together

with the unit-acceleration input, is shown in Figure 5–63.

C(s)

R(s)

=

2

s^2 +s+ 2

MATLAB Program 5–24

num = [2];

den = [1 1 2];

t = 0:0.2:10;

r = 0.5*t.^2;

y = lsim(num,den,r,t);

plot(t,r,'-',t,y,'o',t,y,'-')

grid

title('Unit-Acceleration Response')

xlabel('t Sec')

ylabel('Input and Output')

text(2.1,27.5,'Unit-Acceleration Input')

text(7.2,7.5,'Output')

Unit-Acceleration Response

t Sec

012345678910

Input and Output

50

0

10

5

15

20

25

30

35

40

45

Unit-Acceleration Input

Ouput

Figure 5–63

Response to unit-

acceleration input.

A–5–15. Consider the system defined by

C(s)

R(s)

=

1

s^2 + 2 zs+ 1

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