control system

1. LAB#05
Time Response of a Control System:-
>> %Example of Time Response of a Control System:-
>> 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) %This command is To get the graph of unit step;
Graph:-
>> a=step(A,B,C,D) %This command gives the value of step response;
>> impulse(A,B,C,D) %This command is to get the graph of impulse response;

2. >> step(A,B,C,D,1) %This command gives the Graph of U1 i/p and o/p;
>> step(A,B,C,D,2) %This command gives the Graph of U2 i/p and o/p;

3. Exercise#1: - Find transient and steady state characteristics of the following transfer
functions.
num=[1 2];
den=[1 7 12];
a=step(num,den);
subplot(3,1,1);
plot(a)
grid on
title('step response')
%Now impulse response;
b=impulse(num,den);
subplot(3,1,2);
plot(b)
grid off
title('impulse response');
%Now ramp response;
t=0:0.1:1;
u=t;
c=lsim(num,den,u,t);
subplot(3,1,3);
plot(c)
title('ramp response')

4. 2nd
part:-
>> numA=[0 2 3 1];
>> denA=[1 3 1 0];
>> a=step(numA,denA);
>> subplot(3,1,1);
>> plot(a)
>> title('step response')
>> % for impulse response;
>> b=impulse(numA,denA);
>> subplot(3,1,2);
>> plot(b)
>> title('impulse response');
>> % for ramp response;
>> t=0:0.01:2;

5. >> u=t;
>> c=lsim(numA,denA,u,t);
>> subplot(3,1,3)
>> plot(c)
>> title('ramp response')
Exercise#2: - Reduce the following open loop systems into a single block and find time
response of the system.
(i)
num1=[1];
den1=[9 17];
num2=9*[1 3];
den2=[2 9 27];
S
9S + 17
9(S+3)
2S2 + 9s + 27

6. [numA,denA]=series(num1,den1,num2,den2)
a=step(numA,denA);
subplot(3,1,1);
plot(a)
title('step response')
b=impulse(denA,denA);
subplot(3,1,2)
plot(b)
title('impulse response')
%Now ramp response;
t=0:0.01:3;
u=t;
c=lsim(numA,denA,u,t)
subplot(3,1,3);
plot(c)
title('ramp response')
(ii)
4/5
S + 3
2S + 4
S3
S2
S - 6

7. num1=[2 4];
den1=[1 0 0 0];
num2=[4/5];
den2=[1 3];
[numa,dena]=series(num1,den1,num2,den2);
num3=[1 0 0];
den3=[0 1 -6];
[numA,denA]=series(numa,dena,num3,den3)
%step response;
a=step(numA,denA);
subplot(3,1,1);
plot(a)
title('step response')
%impulse response;
b=impulse(denA,denA);
subplot(3,1,2)
plot(b)
title('impulse response')
%Now ramp response;
t=0:0.01:3;
u=t;
c=lsim(numA,denA,u,t)
subplot(3,1,3);
plot(c)
title('ramp response')
Exercise#3: - Find time response of following closed loop systems

8. (i)
>> numA=[0 9];
>> denA=[1 5];
>> numB=[0 1];
>> denB=[0 1];
>> [num1,den1]=feedback(numA,denA,numB,denB)
num1 =
0 0 9
den1 =
0 1 14
>> a=step(num1,den1);
> > subplot(3,1,1);
>> plot(a)
>> title('step response')
>> % for Impulse response;
C(S)R(S)
-
9
S + 5

9. >> b=impulse(num1,den1);
>> subplot(3,1,2);
>> plot(b)
>> title('impulse response')
>> %for ramp response;
>> t=0:0.001:2;
>> u=t;
>> c=lsim(num1,den1,u,t);
>> subplot(3,1,3);
>> plot(c)
>> title('ramp response')

10. (ii)
>> numA=[0 1];
>> denA=[1 1];
>> numB=[0 2];
>> denB=[1 0];
>> [num1,den1]=feedback(numA,denA,numB,denB)
num1 =
0 1 0
den1 =
1 1 2
>> a=step(num1,den1);
>> subplot(3,1,1);
>> plot(a)
>> title('step response')
C(S)R(S)
-
1
S +1
2
S

11. >> % for impulse response;
>> b=impulse(num1,den1);
>> subplot(3,1,2);
>> plot(b)
>> title('impulse response')
>> % for ramp response;
>> t=0:0.01:1;
>> u=t;
>> c=lsim(num1,den1,u,t);
>> subplot(3,1,3);
>> plot(c)
>> title('ramp response')

12. C(S)R(S)
-
9
S + 5