Can you help me? :(
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Jin You
am 24 Mai 2020
Kommentiert: Ameer Hamza
am 25 Mai 2020
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)

Akzeptierte Antwort
Ameer Hamza
am 24 Mai 2020
Bearbeitet: Ameer Hamza
am 25 Mai 2020
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)

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