I made a unit square pulse and added it to the Fourier.
The integral formula came out well, but the graph is strange.
What's the problem?(I am sorry for my poor English.)
<Code>
clear;
close all;
clc;
syms t w
y=rectangularPulse(t);
f(t)=y*exp(-i*w*t);
F=int(f(t),t,-100,100)
fplot(w,F)

 Respuesta aceptada

Ameer Hamza
Ameer Hamza el 24 de Mayo de 2020
Editada: Ameer Hamza el 25 de Mayo de 2020

1 voto

This demo shows how to find the fourier and inverse fourier transforms
Fourier transform:
syms t w
y(t) = rectangularPulse(t);
F(w) = int(y(t)*exp(-1i*w*t), t, -0.5, 0.5); % pulse only exist between -0.5 to 0.5
figure;
fplot(F, [-100 100]);
Inverse:
F_inv(t) = int(F(w)*exp(-1i*w*t), w, -100, 100); % integrate frequencies between -100 to 100
t_val = linspace(-2, 2, 1000);
F_val = F_inv(t_val);
figure;
plot(t_val, F_val)

2 comentarios

Jin You
Jin You el 25 de Mayo de 2020
thank you so much!!! It's very helpful!!!!
Ameer Hamza
Ameer Hamza el 25 de Mayo de 2020
I am glad to be of help!!

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