Modern Control Engineering

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186 Chapter 5 / Transient and Steady-State Response Analyses

To: Y2

1.5

2

1

0.5

0

04812

Time (sec)

1.5

2

1

0.5

0

04812

To: Y1

0.4

0.6

0.2

−0.2

00

−0.4

0.4

0.6

0.2

−0.2

−0.4

04812 04812

From: U1 From: U2

Step Response

Amplitude

Figure 5–18

Unit-step response

curves.

MATLAB Program 5–1

A = [–1 –1;6.5 0];

B = [1 1;1 0];

C = [1 0;0 1];

D = [0 0;0 0];

step(A,B,C,D)

signalu 1 as the input, we assume that signal u 2 is zero, and vice versa. The four transfer functions

are

Assume that u 1 andu 2 are unit-step functions. The four individual step-response curves can then

be plotted by use of the command

step(A,B,C,D)

MATLAB Program 5–1 produces four such step-response curves. The curves are shown in Figure 5–18.

(Note that the time vector tis automatically determined, since the command does not include t.)

Y 2 (s)

U 2 (s)

=

6.5

s^2 +s+6.5

Y 2 (s)

U 1 (s)

=

s+7.5

s^2 +s+6.5

,

Y 1 (s)

U 2 (s)

=

s

s^2 +s+6.5

Y 1 (s)

U 1 (s)

=

s- 1

s^2 +s+6.5

,

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