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problems with using the laplace transform

Asked by Eliraz Nahum on 28 Dec 2018
Latest activity Commented on by Star Strider
on 30 Dec 2018
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:
error.JPG
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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1 Answer

Answer by Star Strider
on 28 Dec 2018
 Accepted Answer

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.)

  2 Comments

it still doesn't work for me and the results make no sense...
Thanks, I will try to find out what's wrong.
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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