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wda5
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Matching step response from Simulink with Matlab

Asked by wda5
on 17 Mar 2015
Latest activity Commented on by wda5
on 17 Mar 2015
Hi I am new to Simulink and am trying to figure out how to get correct results. I modeled a simple control loop in Matlab and Simulink to see if the step response was the same. It's not. I'm guessing this has to do with the solver settings? I used the default for this.
Can anyone help me understand how to get Simulink to give better results? Thanks!
Here is the Matlab code:
sys = tf(1,[1 1 1]);
syscl = pid(4,2,1,0);
sys2 = series(syscl,sys);
sys3 = feedback(sys2,1);
[y,t] = step(sys3);
figure(2)
plot(t,y,ScopeData.time,ScopeData.signals.values)
Here is the simulink model:
And here is a plot of the results:

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1 Answer

Answer by Arkadiy Turevskiy on 17 Mar 2015
 Accepted Answer

No, it actually has to do with calculating the derivative of a signal. You don't want to use pure derivative block. To match results you see from MATLAB code, use PID Controller block from Simulink library or modify your own PID implementation in Simulink to replace pure derivative block with a feedback loop consisting of a gain in the forward path and an integrator in the feedback path.
This is explained in detail in this thread .

  1 Comment

Thanks! That is exactly what was wrong. Switched to regular PID block. Initially I avoided it because I didn't want to deal with the filter in the system.

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