How can I fix this error?

197 次查看（过去 30 天）
sadel
sadel 2011-6-20
回答： Gauri Shankar Prasad 2018-5-5
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 个评论
显示 5更早的评论
Arnaud Miege
Arnaud Miege 2011-6-20
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 2011-6-20
@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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Paulo Silva
Paulo Silva 2011-6-20
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 个评论
显示 7更早的评论
Fangjun Jiang
Fangjun Jiang 2011-6-21
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 2011-6-21
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" :)

请先登录，再进行评论。

更多回答（4 个）

Fangjun Jiang
Fangjun Jiang 2011-6-20
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 2011-6-20
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 个评论
Paulo Silva
Paulo Silva 2011-6-20
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.

请先登录，再进行评论。

Alberto
Alberto 2013-10-21
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
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??

类别

在 Help Center 和 File Exchange 中查找有关 Control System Toolbox 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by