Control Theory-Lab
CASIAN INSTITUTE OF TECHNOLOGY
CONTROL THEORY
Name: Dilesha Herath
ID.No. st20000297
Date: 23/04/2013
Exercise
Consider the following plant.
G( s)
(a) Determine the stability of this plant.
1 ( s 1) 2
>> numerator = [0 0 1]; denominator = [1 -2 1]; >> [z p k] = tf2zp(numerator,denominator) z = Empty matrix: 0-by-1
p = 1 1
k = 1 >> x1=1; x2=-2; x3=1; x4=0; >> alpha1=(x1/x2); y1=(x3-alpha1*x4); alpha2=(x2/y1); >> if alpha1>0 else if alpha2>0 disp('the system is stable') else disp('the system is not stable') alpha1, alpha2 end 1
ASIAN INSTITUTE OF TECHNOLOGY
CONTROL THEORY
end the system is not stable alpha1 = -0.5000
alpha2 = -2
(b) Assume that PID controller, K (s) K P K I / s K D s , is selected to compensate the requirement of system stability and zero steady state error to unit step input, use the Routh-Hurwitz test to show that a cascade of PID controller with the original plant and unity negative feedback configuration can meet the requirements. Remember that the system must be stabilized first before the consideration of steady state error.
>> Kd=1; Kp=1; Kc=1; >> alpha1=(x1/x2); y1=(x3-alpha1*x4); y2=0; alpha2=(x2/y1); z1=(x4-alpha2*y2); alpha3=y1/z1; >> if alpha1> Kd=25; Kp=25; Kc=25; >> alpha1=(x1/x2); y1=(x3-alpha1*x4); y2=0; alpha2=(x2/y1); z1=(x4-alpha2*y2); alpha3=y1/z1; >> if alpha1>numerator = [Kd Kp Ki]; denominator = [1 (Kd-2) (Kp+1) Ki]; >> t=[0:0.1:10]; y=step(numerator, denominator, t); plot(t,y)
>> roots(num) ans = -0.5000 + 0.8660i -0.5000 - 0.8660i The system is unstable. >> Kd=25; Kp=25; Kc=25; >>numerator = [Kd Kp Ki]; denominator = [1 (Kd-2) (Kp+1) Ki]; 4
ASIAN INSTITUTE OF TECHNOLOGY
CONTROL THEORY
>> t=[0:0.1:10]; y=step(numerator, denominator, t); plot(t,y)
>> roots(num) ans = -0.5000 + 0.8660i -0.5000 - 0.8660i
The system is perfect but the output is not perfect.
>> Kd=100; Kp=50; Kc=50; >>numerator = [Kd Kp Ki]; denominator = [1 (Kd-2) (Kp+1) Ki]; >>
CONTROL THEORY
Name: Dilesha Herath
ID.No. st20000297
Date: 23/04/2013
Exercise
Consider the following plant.
G( s)
(a) Determine the stability of this plant.
1 ( s 1) 2
>> numerator = [0 0 1]; denominator = [1 -2 1]; >> [z p k] = tf2zp(numerator,denominator) z = Empty matrix: 0-by-1
p = 1 1
k = 1 >> x1=1; x2=-2; x3=1; x4=0; >> alpha1=(x1/x2); y1=(x3-alpha1*x4); alpha2=(x2/y1); >> if alpha1>0 else if alpha2>0 disp('the system is stable') else disp('the system is not stable') alpha1, alpha2 end 1
ASIAN INSTITUTE OF TECHNOLOGY
CONTROL THEORY
end the system is not stable alpha1 = -0.5000
alpha2 = -2
(b) Assume that PID controller, K (s) K P K I / s K D s , is selected to compensate the requirement of system stability and zero steady state error to unit step input, use the Routh-Hurwitz test to show that a cascade of PID controller with the original plant and unity negative feedback configuration can meet the requirements. Remember that the system must be stabilized first before the consideration of steady state error.
>> Kd=1; Kp=1; Kc=1; >> alpha1=(x1/x2); y1=(x3-alpha1*x4); y2=0; alpha2=(x2/y1); z1=(x4-alpha2*y2); alpha3=y1/z1; >> if alpha1> Kd=25; Kp=25; Kc=25; >> alpha1=(x1/x2); y1=(x3-alpha1*x4); y2=0; alpha2=(x2/y1); z1=(x4-alpha2*y2); alpha3=y1/z1; >> if alpha1>numerator = [Kd Kp Ki]; denominator = [1 (Kd-2) (Kp+1) Ki]; >> t=[0:0.1:10]; y=step(numerator, denominator, t); plot(t,y)
>> roots(num) ans = -0.5000 + 0.8660i -0.5000 - 0.8660i The system is unstable. >> Kd=25; Kp=25; Kc=25; >>numerator = [Kd Kp Ki]; denominator = [1 (Kd-2) (Kp+1) Ki]; 4
ASIAN INSTITUTE OF TECHNOLOGY
CONTROL THEORY
>> t=[0:0.1:10]; y=step(numerator, denominator, t); plot(t,y)
>> roots(num) ans = -0.5000 + 0.8660i -0.5000 - 0.8660i
The system is perfect but the output is not perfect.
>> Kd=100; Kp=50; Kc=50; >>numerator = [Kd Kp Ki]; denominator = [1 (Kd-2) (Kp+1) Ki]; >>