problems with using the laplace transform

Question

Eliraz Nahum 2018-12-28

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/437487-problems-with-using-the-laplace-transform

评论： Star Strider 2018-12-30

采纳的回答： Star Strider

hello everyone,

I am trying to plot a system response (y(t)) to an input of u(t)=t*1(t).

The laplace transform of u(t) is U(s)=L{u(t)}= 1/(s^2).

The system is represented in terms of transfer function G(s) = 2/(s^3+5*s^2+4*s+2);

I am trying to create Y(S)=G(s)*U(s) and then convert it to the time domain by ilaplace(Y(s)).

I can't understand why it doesn't work. I get an error:

please help...

clear all
close all
clc
syms t y1(t) s Y1(s)
G_cl_1=2/(s^3+5*s^2+4*s+2);
Y1=G_cl_1*(1/(s^2)); %Finding the Output y(t) while using Laplace Transform
y1=ilaplace(Y1); %Converting Y1(s) to the time space using Opposite Laplace Transform
ezplot (y1)

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Answer 1

Star Strider 2018-12-28

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/437487-problems-with-using-the-laplace-transform#answer_354224

在 MATLAB Online 中打开

First, specify ‘Y1’ and ‘y1’ as functions in your code.

Second, use the vpa function to simplify ‘y1’ so ezplot (or fplot) can plot it.

syms t y1(t) s Y1(s)
G_cl_1=2/(s^3+5*s^2+4*s+2);
Y1(s) = G_cl_1*(1/(s^2)); %Finding the Output y(t) while using Laplace Transform
y1(t) = ilaplace(Y1, s, t); %Converting Y1(s) to the time space using Opposite Laplace Transform
y1 = vpa(y1)
ezplot (y1, [-3  -1])

That works for me. (I specified the limits for ezplot to provide a representative part of the curve. Choose whatever limits you want.)

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Eliraz Nahum 2018-12-30

it still doesn't work for me and the results make no sense...

Thanks, I will try to find out what's wrong.

Star Strider 2018-12-30

在 MATLAB Online 中打开

As always, my pleasure.

Using:

y1 = vpa(y1, 10)

(to make the constants a bit more tractable without losing significant precision), I get:

y1(t) =
t + 0.008162161899*exp(-4.152757602*t) + 1.991837838*exp(-0.423621199*t)*cos(0.5496842464*t) - 0.222527365*exp(-0.423621199*t)*sin(0.5496842464*t) - 2.0

Does that come closer to what you are expecting? To create an anonymous function from it, use the matlabFunction function.

(I am using R2018b. There could be differences with older versions.)

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