Modern Control Engineering

(Chris Devlin) #1
MATLAB Program 7–29 produces the unit-ramp response curves. [Note that the unit-ramp
response is obtained as the unit-step response of C(s)/sR(s).] The resulting curves are shown in
Figure 7–150. The compensated system has a steady-state error equal to one-half that of the orig-
inal uncompensated system.

554 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method

Outputs

1.8

0.8

0

1.2

1.6

0.6

0.2

1

1.4

0.4

t Sec

012 3 4 5 6

Unit-Step Responses of Uncompensated System and Compensated System

Uncompensated system

Compensated system

Figure 7–149
Unit-step responses
of the uncompensated
and compensated
systems.

MATLAB Program 7–28


%Unit-step responses


num1 = [10];


den1 = [1 1 10];


num2 = [95.238 286.6759];


den2 = [1 15.3339 110.5719 286.6759];


t = 0:0.01:6;


[c1,x1,t] = step(num1,den1,t);


[c2,x2,t] = step(num2,den2,t);


plot(t,c1,'.',t,c2,'-')


grid;


title('Unit-Step Responses of Uncompensated System and Compensated System')


xlabel('t Sec');


ylabel('Outputs')


text(1.70,1.45,'Uncompensated System')


text(1.1,0.5,'Compensated System')


1/Kv=0.05. The static velocity error constant of the uncompensated system is 10 sec–1. Hence,
the original uncompensated system will have twice as large a steady-state error in following the
unit-ramp input.

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