how to change the rise time of step input in simulink
26 ビュー (過去 30 日間)
古いコメントを表示
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 件のコメント
Sam Chak
2022 年 9 月 19 日
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.
採用された回答
Sam Chak
2022 年 9 月 21 日
編集済み: Sam Chak
2022 年 9 月 22 日
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])
% PID
kp = 0;
ki = 0;
kd = 0.8165;
Tf = kd;
Gc = pid(kp, ki, kd, Tf)
% Closed-loop system
Gcl = feedback(Gc*Gp, 1)
% 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 件のコメント
Sam Chak
2022 年 9 月 21 日
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])
その他の回答 (3 件)
Timo Dietz
2022 年 9 月 20 日
編集済み: Timo Dietz
2022 年 9 月 20 日
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 件のコメント
Timo Dietz
2022 年 9 月 21 日
編集済み: Timo Dietz
2022 年 9 月 21 日
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
2022 年 9 月 21 日
1 件のコメント
Sam Chak
2022 年 9 月 21 日
It's great to hear that it works. If you find the solution is helpful, please consider accepting ✔ and voting 👍 the Answer. Thanks!
参考
カテゴリ
Help Center および File Exchange で Classical Control Design についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!