how to make fft of rectangular function?
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Ts = 0.01; N=2000; t=-20:Ts:(N-1)*Ts; T = 1; fs=1/Ts; f=0:fs/N:(N-1)/N*fs; x1 = rectpuls(t, T); xk=fft(x1); figure(1); plot(t,x1); figure(2); plot(0:N-1, xk); figure(3); plot(f, 1/N*abs(xk));
figure1,2 are not found. what is the problem? how can i fix it?
Answers (2)
Star Strider
on 2 Dec 2017
Small error in figure(3). You do not need figure(2).
This works:
Ts = 0.01; N=2000; t=-20:Ts:(N-1)*Ts;
T = 1;
fs=1/Ts;
f=0:fs/N:(N-1)/N*fs;
x1 = rectpuls(t, T);
xk=fft(x1);
figure(1); plot(t,x1)2
figure(3); plot(f, 1/N*abs(xk(1:length(f))));
4 Comments
YEON HWEE BAE
on 3 Dec 2017
Star Strider
on 3 Dec 2017
My pleasure.
If my Answer helped you solve your problem, please Accept it!
Parth Patel
on 16 Apr 2020
Edited: Parth Patel
on 16 Apr 2020
In this program , i asume, f = frequency.
can the value of frequency be zero?
f=0:fs/N:(N-1)/N*fs;
thanks in adavnce
Star Strider
on 16 Apr 2020
Yes.
A zero frequency signal is d-c, usually present as a constant offset in a signal that is otherwise varying. The d-c component is the mean of the signal.
Tilkesh
on 28 Mar 2022
0 votes
function y = rect(x, D)
% function y = rect(x, D)
if nargin == 1, D = 1;
x = abs(x);
y = double(x<D/2);
y(x == D/2) = 0.5;
end
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