how to change the rise time of step input in simulink

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Swasthik Baje Shankarakrishna Bhat
Réponse apportée : Paul le 22 Sep 2022
Hello,
I want to change the rise time of the step input in simulink. I tried the transfer function 1/(s+1) but not satifying my requirement. Can someone suggest me some tricks. Thanks in advace.
Regards,
Swasthik
  2 commentaires
Sam Chak
Sam Chak le 19 Sep 2022
Can you specify all the performance requirements?
What did you mean by "tried the transfer function"? What is that transfer function for? For Plant, Actuator, or Compensator, or Prefilter? If possible, please the Plant transfer function.
Swasthik Baje Shankarakrishna Bhat
Hi sam,
i just need a step input with variable rise time. Something like this,
Thanks and regards,
Swasthik

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Réponse acceptée

Sam Chak
Sam Chak le 21 Sep 2022
Modifié(e) : Sam Chak le 22 Sep 2022
Edit: I created a general one so that you can enter the desired ramp up parameters:
% u = min(1, max(0, "Linear Line function"));
ramp_start = 5;
ramp_end = 8;
t = linspace(0, 25, 251);
u = min(1, max(0, 1/(ramp_end - ramp_start)*(t - ramp_start)));
plot(t, u, 'linewidth', 1.5), grid on, ylim([-1 2])
If you have Fuzzy Logic Toolbox license, then you can use this linsmf() function. Here is a demo for a Double Integrator:
% Plant
Gp = tf(1, [1 0 0])
Gp = 1 --- s^2 Continuous-time transfer function.
% PID
kp = 0;
ki = 0;
kd = 0.8165;
Tf = kd;
Gc = pid(kp, ki, kd, Tf)
Gc = s Kp + Kd * -------- Tf*s+1 with Kp = 0, Kd = 0.817, Tf = 0.817 Continuous-time PDF controller in parallel form.
% Closed-loop system
Gcl = feedback(Gc*Gp, 1)
Gcl = s ------------------- s^3 + 1.225 s^2 + s Continuous-time transfer function.
% Saturated Ramp Response
t = linspace(0, 25, 251);
u = linsmf(t, [5 8]); % rise time is from 5 to 8 sec
lsim(Gcl, u, t), ylim([-1 2]), grid on
  3 commentaires
Sam Chak
Sam Chak le 21 Sep 2022
I forgot to mention that although linsmf is user-friendly, it requires the Fuzzy Logic Toolbox.
If you don't have Toolbox, then you can try this alternative. It produces the same time-delayed saturated ramp signal.
% Syntax:
% u = min(1, max(0, "Linear Line function"));
t = linspace(0, 25, 251);
u = min(1, max(0, 1/3*(t - 5)));
plot(t, u, 'linewidth', 1.5), grid on, ylim([-1 2])
Swasthik Baje Shankarakrishna Bhat
Hi Sam,
Sorry to bother you again. Can we have a neative sloped step input with programmable fall time?
Thanks and regards,
Swasthik

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Plus de réponses (3)

Timo Dietz
Timo Dietz le 20 Sep 2022
Modifié(e) : Timo Dietz le 20 Sep 2022
Hello,
what about using a single-sided ramp function: b * (1 - exp(-a*s)) / s^2
The gradient of the rising slope is '1', so after the time 'a' the amplitude 'a' is reached. The factor 'b' should finally allow you to control the steepness of the 'step'.
Does this meet your requirement?
  3 commentaires
Timo Dietz
Timo Dietz le 21 Sep 2022
Modifié(e) : Timo Dietz le 21 Sep 2022
Hello,
I thought you are searching for a solution in the frequency/laplace domain since you mentioned the pT1.
So, with 'H_slope = b/a * (1 - exp(-a*s)) / s^2' you can control the slope (in time domain) via a and b:
syms s;
a = 1.5; % time after which amplitude is reached
b = 2; % amplitude
% frequency domain
H_slope = b/a * (1 - exp(-a*s)) / s^2;
% time domain
f_slope = ilaplace(H_slope);
fplot(f_slope, [0 2]);
grid on;
Swasthik Baje Shankarakrishna Bhat
thanks Timo

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Swasthik Baje Shankarakrishna Bhat
perfect, Thanks sam. Have a nice day
  1 commentaire
Sam Chak
Sam Chak le 21 Sep 2022
It's great to hear that it works. If you find the solution is helpful, please consider accepting ✔ and voting 👍 the Answer. Thanks!

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Paul
Paul le 22 Sep 2022
A 1D Lookup Table seems like a good option.

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