How can I fix this error?

132 views (last 30 days)
sadel
sadel on 20 Jun 2011
Hi all!
How can I fix this error?
sys=tf([1 1 1 1],[1 1]);
t=0:0.1:10;
step(sys,t)
??? Error using ==> DynamicSystem.step at 84 Cannot simulate the time response of models with more zeros than poles.
I know that my system is unstable but how can I receive an "unstable" graph for this system on my axes?
  7 Comments
Arnaud Miege
Arnaud Miege on 20 Jun 2011
You can't have systems that are not proper (order of numerator -> order of denumerator) in the control system toolbox. Paulo's suggestion with the symbolic math toolbox is about as good as it gets.
Fangjun Jiang
Fangjun Jiang on 20 Jun 2011
@Arnaud, Can the Symbolic Math Toolbox be used to define a proper transfer function and plot out its step response? I used Maple long time ago and remember it can be done.

Sign in to comment.

Accepted Answer

Paulo Silva
Paulo Silva on 20 Jun 2011
Just FYI the following code won't provide the same graph has the step function, it just plots the function having s as the variable, the step fuction just works for proper systems (n poles >= n zeros).
"Impulse response":
syms s
TFC=evalc('tf([1 1 1 1],[1 1])');
[a b] = strread(TFC, '%s %s', 'delimiter',char(10));
num=sym(char(a(2)));
den=sym(char(a(3)));
ezplot(num/den,[0 100])
"Step response":
syms s
TFC=evalc('tf([1 1 1 1],[1 1 0])');
[a b] = strread(TFC, '%s %s', 'delimiter',char(10));
num=sym(char(a(2)));
den=sym(char(a(3)));
ezplot(num/den,[0 100])
The scales aren't the same like you get with the step function.
  9 Comments
Fangjun Jiang
Fangjun Jiang on 21 Jun 2011
Now that is the step response! You are taking the transfer function symbolic equation, doing an inverse Laplace transform to get the transfer function in time domain and then get the response in time domain.
Don't take my question and comment as offense. I am just trying to make sure that we are speaking the same language. You got to admit though, that the code in your answer (not the one in the comment) is not plotting the step response or impulse response, whether or not it is a proper or non-proper system.
Paulo Silva
Paulo Silva on 21 Jun 2011
Yes I admit my mistake, it wasn't the first time I made mistakes because sadel or other similar person wants to do strange things, fixing MATLAB "errors" :)

Sign in to comment.

More Answers (4)

Fangjun Jiang
Fangjun Jiang on 20 Jun 2011
Your transfer function can be simplified as s^2+1. If your transfer function is 1, then its step response is the step input itself. If your transfer function is s, it means the derivative of the input. The step response would be infinite (impulse) at the time of the step and zero at the rest. If your transfer function is s^2, it means the derivative of the derivative of the input. The step responsive would be impulse and negative impulse at the time of the step and zero at the rest.

sadel
sadel on 20 Jun 2011
Let's say that we have this system in open loop.
sys=tf([1 1 1 1 1 1],[1]);
I receive the same error because it is unstable. more zeros than poles etc.
If I use the closed loop of this system, I can receive a graph. But why? It's unstable, too.
sys=tf([1 1 1 1 1 1],[1 1 1 1 1 2]);
t=0:0.01:30;
step(sys,t)
I wanna say that I want a graph for the open loop system but only for 30 seconds. I don't care if this goes to infinity. How can I fool the matlab about the number of the zeros and poles? p.s. I agree with Fangjun. Paulo, 1/(s+a)=exp(-a*t). It's not the same graph.
  1 Comment
Paulo Silva
Paulo Silva on 20 Jun 2011
I never said it was the same graph, I tried to help the best I could, please read what I said in the first answer:
The scales aren't the same like you get with the step function.
In your code just rethrow the error to an errordlg or to the command line when the step function returns errors.
Your example:
sys=tf([1 1 1 1 1 1],[1 1 1 1 1 2]);
t=0:0.01:30;
step(sys,t)
has the same number of zeros and poles, MATLAB just detects if the number of zeros is bigger than poles not if the system is unstable, the stability is related to the position of the poles.

Sign in to comment.


Alberto
Alberto on 21 Oct 2013
one way to achieve this is to add two poles very far from 0. For example, you can add (s+400) two times, like this:
sys=((400*400)*((s^3)+(s^2)+s+1))/((s+1)*(s+400)*(s*400));
from control theory, these poles are not dominant and don't affect the step response.
bye

Gauri Shankar Prasad
Gauri Shankar Prasad on 5 May 2018
k_dc = 5; Tc = 10; u = 2;
s = tf('s'); sys = k_dc/(Tc*s+1)
step(u*sys)
MATLAB shows
Error using DynamicSystem/step (line 95) Too many input arguments.
Error in Untitled3 (line 8) step(u*sys)
What is the problem??

Categories

Find more on Control System Toolbox in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!