How to plot the phase margin?

Hello I have one tricky question where I am struck.I know how to plot the bode diagtam using the margin command but i am not sure how can i implement more than 30% margin part. my system and requirement is below. Please help me to understand how can i do this.

Answer 1

[Bora Eryilmaz]

The transfer function 20/(s+1) is your controller. As it is, it does not give you a 50-degree phase margin. You need to modify this controller to achieve the desired phase margin. There are couple ways of doing this:

• Either reduce the gain of the controller to achieve a larger phase margin, but at a lower frequency.
• Or add additional phase using controller terms such as lead-lags.

% Initial compensator
C = zpk([], -1, 20);

% The plant
G = zpk(-20, [-2 -5], 1);

% Phase and gain margin of the open-loop transfer function with initial
% compensator
L = C*G;
margin(L)

% Modify the compensator C to obtain the desired phase margin (this part is
% up to you, since this seems like a homework question).
C = zpk([], -2, 10)  % This is my trivial change, will not give 50 degrees phase margin.

C =
 
   10
  -----
  (s+2)
 
Continuous-time zero/pole/gain model.

L = C*G;
margin(L)

% Bode plot of the compentator
bode(C)

% Closed-loop system
T = L / (1 + L);
bode(T)

step(T)

[Hardik]

Hello thank you sir for your explanation i have one question that can i use the following code to get moe than 50% phase margin. I have set the phase margin to 51%.

clc;
close all;
s=tf('s'); % To enter the transfer function directly
V= 51/(s+1); % This will give 51 degrees phase margin
G= (s+20)/((s+2)*(s+5));
margin(V)
M = C*G;
margin(M)
bode(C)
S = M / (1 + M);
bode(S)
step(S)

Did I do this correctly?

[Bora Eryilmaz]

Please run your code and see yourself if

margin(M)

gives you the desired phase margin.

C= 51/(s+1); % This will give 51 degrees phase margin

How do you know that the controller C above will give you a 51 degree phase margin? Its gain is 51, but that doesn't necessarily give you a 51 degree phase margin.

[Hardik]

SIr i run the code and I understand that I had to reduce the gain value I have use the following code and in bode plot it says Pm = 61.3 deg.

clc;

close all;

clear all;

s=tf('s'); % To enter the transfer function directly

V= 1.5/(s+1); % This will give 61.3 degrees phase margin

G= (s+20)/((s+2)*(s+5));

M=V*G

margin(M)

Did i do it correctly this time sir?

[Bora Eryilmaz]

Looks like it. If the phase margin is better than 50 degrees, then your controller satisties the requirements of your problem. Also take a look at the step response of the closed loop system to make sure it is well behaved:

s = tf('s'); % To enter the transfer function directly
V = 1.5/(s+1); % This will give 61.3 degrees phase margin
G = (s+20)/((s+2)*(s+5));
M = V*G;
margin(M)

step(M/(1+M))

You will see that the steady-state step response value of your closed-loop system is not 1. This was not part of your question, but normally this would not be an acceptable step response... As a small hint, if your controller had an integrator, the steady-state response would have converged to 1, which is more desirable.

[Hardik]

Okay Sir. I understand. thank you for all your help.

[Hardik]

Hello Sir, in my next part of problem I have to use the Feedback command in Matlab to find the closed loop transfer function of the compensated system. Provide the overall closed loop transfer function.

I have done the coding part but I am bit confused here that Do i have to use feedback command on original given system or the output TF i get from this part i have to use?

And if i have to use new obtanined TF then do i also have to use the plant value (s+20) and multiply them and then use the feedback command ?