how to change the rise time of step input in simulink

Question

Swasthik Baje Shankarakrishna Bhat 2022-9-19

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回答： Paul 2022-9-22

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

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Sam Chak 2022-9-19

Hi @Swasthik Baje Shankarakrishna Bhat

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 2022-9-21

Hi sam,

i just need a step input with variable rise time. Something like this,

Thanks and regards,

Swasthik

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Answer 1

Sam Chak 2022-9-21

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编辑：Sam Chak 2022-9-22

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Hi @Swasthik Baje Shankarakrishna Bhat

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

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Sam Chak 2022-9-21

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Hi @Swasthik Baje Shankarakrishna Bhat

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 2022-9-21

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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Answer 2

Timo Dietz 2022-9-20

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编辑：Timo Dietz 2022-9-20

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?