Applications of MATLAB in Control Part II.pdf

EE549: Computer Applications in Electrical Engineering
Lecture 10
Applications of MATLAB in Control II
Lecturer
Dr. Karim Ibrahim
1

Lecture (10) Contents
 Control System Response
• The Step Response
• The Impulse Response
• Arbitrary Input Response
• Steady State Response
 Control Design
• Root Locus Plot
• Bode Diagram Plot
• Nyquist Plot of Frequency Response
• Nichols Chart of Frequency Response
• PID Controllers
2

What is System Response?
 The response of a system is the shape and characteristics of the output of this system for a
certain input.
 In the analysis of systems, some basic test input signals are used such as the unit step, the
impulse, and the ramp function.
 To get the step response of a certain system using MATLAB, the system should first be
defined using either block diagrams, transfer functions, or zeros-poles-gain, then use the
proper function syntax.
3

The Step Response
 The response of the system to a step input can be acquired using the next command:
 The response of the system to a unit impulse be acquired using the next command:
4
The Impulse Response

Arbitrary Input Response
 If it is required to check the system response to a sinusoidal input or in general an arbitrary
input signal. In this case, you may use the following command:
5

Example (10-1)
 For the next system transfer function:
Find:
i. The Step Response of the system for 15secs.
ii.The value of the rise time, overshoot and overshoot
time, settling time, and the steady state error.
6
>> sys=tf([1 3],[1 1 4]);
>> t=0:0.01:15;
>> step(sys,t)
% For Manual calculations:
>> y=step(sys,t);
>> Y_analysis = stepinfo(y,t)
Y_analysis =
RiseTime: 0.4688
SettlingTime: 7.9585
SettlingMin: 0.5650
SettlingMax: 1.1664
Overshoot: 55.4905
Undershoot: 0
Peak: 1.1664
PeakTime: 1.2800
>> SSV=y(end) % Steady state value
>> SSE=1-SSV % Steady state error
SSV =
0.7502
SSE =
0.2498

7
Right click your mouse and
then activate the transient
and steady state
characteristics as shown:

Example (10-2)
Find:
i. The impulse Response of the system for 15secs.
ii.The value of the peak response and the settling time
9
>> sys=tf([1 3],[1 1 4]);
>> t=0:0.01:15;
>> impulse(sys,t)
>> Y=impulse(sys,t);
>> Y_analysis = lsiminfo(Y,t)
Y_analysis =
SettlingTime: 7.4109
Min: -0.5926
MinTime: 1.9600
Max: 1.3337
MaxTime: 0.3400
>> SSV=Y(end) % Steady state value
>> SSE=0-SSV % Steady state error
SSV =
-8.9451e-04
SSE =
8.9451e-04

Example (10-3)
 For the next system transfer function
Find the response of the system for an input signal u(t),
where:
i. U(t) is a unit ramp function, where U(t)=t.
ii.U(t) is a cosine function, where U(t)=cos(t).
11
>> sys=tf([2],[1 2 6]);
>> t=0:0.01:10;
>> u1=t; % Unit Ramp input
>> u2=cos(t); % Cos input
>> figure(1);
>> lsim(sys,u1,t);
>> grid;
>> figure(2);
>> lsim(sys,u2,t);
>> grid;
>> Y1=lsim(sys,u1,t);
>> Y1_analysis = lsiminfo(Y1,t)
>> SSV1=Y1(end) % Steady state value
>> SSE1=u1(end)-SSV1 % Steady state error
>> Y2=lsim(sys,u2,t);
>> Y2_analysis = lsiminfo(Y2,t)
>> SSV2=Y2(end) % Steady state value
>> SSE2=u2(end)-SSV2 % Steady state error

Example (10-3) – Solution tips
 The time period of analysis should be specified first then the input function is defined and
finally the LINEAR SIMULATION function should be applied on the defined system.
 In the code, “u1” is a ramp function as it increases with time in the same rate as the time
vector “t”, while “u2” is a cosine function.
 Both signals are applied to the same system in a time interval specified by “t” and the output
of each input signal is displayed in a separate figure using the “figure” command.
 Note that the linear simulation function can also display the peak response of the system
output using the right mouse key and then the characteristics option.
 The plots displayed in each figure represent the input signal (in gray) and the output (in blue).
12

Root Locus Plot
18
Bode Diagram Plot
 The root locus plot of the system can be obtained using the next command:
>> rlocus(sys) Calculates and plots the root locus of the open-loop model sys
 The bode diagram plot of the system can be obtained using the next command:
>> bodeplot(sys) Calculates and plots the root locus of the open-loop model sys

Nyquist Plot of Frequency Response
19
Nichols Chart of Frequency Response
 The Nyquist Plot of the system can be obtained using the next command:
>> nyquist(sys) Creates a Nyquist plot of a dynamic system
 The Nichols Chart of the system can be obtained using the next command:
>> nichols(sys) Creates a Nichols Chart of a dynamic system

Example (10-4)
 For the next feedback system
Plot the Root Locus, Bode Diagram, Nyquist Plot and
Nichols Chart.
20
>> G=tf([2],[1 2 0])
>> H=tf([4],[1 5])
>> Sys=feedback(G,H)
>> figure(1)
>> rlocus(Sys), grid on
>> figure(2)
>> bodeplot(Sys), grid on
>> figure(3)
>> nyquist(Sys), grid on
>> figure(4)
>> nichols(Sys), grid on

Simulink - Example (10-5)
Using MATLAB Simulink Find:
i. The Step Response of the system for 10secs.
ii.The value of the rise time, overshoot and overshoot time, settling time, and the steady state
error.
25

26
Example (10-5) – Solution (A)

27
Example (10-5) – Solution (B)
Step A
Step B Step C

28
Step D Step E

29
Step F

PID Controller - Example (10-6)
 For the next feedback system
 Analyze the response of the system with step input when adding: Proportional (P), Differential
(D), Integral Controllers (I).
 Finally study the effect of the PID controller on the system
 Tune the PID Controller Manually to enhance the system response.
 Compare between the original and tuned system responses.
30

31
Example (10-6) – Original Response
G=tf([2],[1 2 0])
H=tf([4],[1 5])
figure(1)
Sys=feedback(G,H)
step(Sys)
grid on
title('Original Step response’)
legend('Original’)
struct2table([stepinfo(Sys)])

Proportional Control (P-control)
Proportional Gain: Kp*e(t)
Consider a system, G(s) with 2 real poles (-z1, -z2)
KP
As Kp increases:
• Damping decreases
• Overshoot occurs
• Rise time decreases
• Too much gain can cause instability

33
Example (10-6) – (P-Control)
G=tf([2],[1 2 0])
H=tf([4],[1 5])
Sys=feedback(G,H)
figure(2)
GS=tf([5],[1]) % Static Gain
GS_Sys=feedback(G*GS,H)
step(GS_Sys)
grid on
hold on
step(Sys)
title('Step response with adding Gain
"P-control"')
legend('With P-control','Original')
struct2table([stepinfo(Sys)
stepinfo(GS_Sys)])
Static Gain =5

Differential Control (D-control)
Differential Gain:
Kd*S
• Adds a zero to system
• Decrease overshoot
• Increases damping
• Decreases settling time

35
Example (10-6) – (D-Control)
G=tf([2],[1 2 0])
H=tf([4],[1 5])
Sys=feedback(G,H)
figure(3)
DS=tf([1 0],[1]); % Differentiator
DS_Sys=feedback(G*DS,H)
step(DS_Sys)
grid on
hold on
step(Sys)
title('Step response with adding
Diffrentiator "D-control"')
legend('With D-control','Original')
stepinfo(DS_Sys)])
Differentiator = S

Integral Control (I-control)
Differential Gain:
KI
𝑺
• Adds a pole at the origin
• Increases overshoot
• Decreases rise time
• Eliminates steady state error

37
Example (10-6) – (I-Control)
G=tf([2],[1 2 0])
H=tf([4],[1 5])
Sys=feedback(G,H)
figure(4)
IS=tf([1],[1 0]); % Integrator
IS_Sys=feedback(G*IS,H)
step(IS_Sys)
grid on
hold on
step(Sys)
Integratortor "I-control"')
legend('With I-control','Original')
stepinfo(IS_Sys)])
Integrator = 1/S

39
Example (10-6) – (PID-Control) G=tf([2],[1 2 0])
H=tf([4],[1 5])
Sys=feedback(G,H)
figure(5)
GS=tf([5],[1]); % Static Gain
IS=tf([1],[1 0]); % Integrator
All_Sys=feedback(G*(GS+DS+IS),H);
step(All_Sys)
grid on
hold on
step(Sys)
Integratortor "PID-control"')
legend('With PID','Original')
stepinfo(All_Sys)])
PID = 5 + S + 1/S

40
Example (10-6)-(PID-Control)-Manual Tuning
G=tf([2],[1 2 0])
H=tf([4],[1 5])
Sys=feedback(G,H)
figure(6) % Manually Tuned
GS=tf([2.5],[1]); % Static Gain
IS=tf([0.01],[1 0]); % Integrator
All_Sys=feedback(G*(GS+DS+IS),H);
step(All_Sys)
grid on, hold on
step(Sys)
Integratortor "PID-control"')
legend('With PID','Original')
stepinfo(All_Sys)])
PID = 2.5 + 2*S + 0.01/S

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