Fourier Transform of tripuls

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Moisa Tedy Gabriel
Moisa Tedy Gabriel 2020-6-24
回答： Priyanshu Mishra 2020-7-29
Below you got what I've tried, but it does not work. The second graph does not match with the actual fft of a triangle function. Can someone help me? The use of tripuls is required! Ty!
clc
t=-2:.01:2;
subplot(2, 1, 1);
x=tripuls(t,2);
plot(t, x);
title('Triangle');
subplot(2, 1, 2);
Ts = mean(diff(t)); % Sampling Interval
Fs = 1/Ts; % Sampling Frequency
Fn = Fs/2; % Nyuquist Frequency
L = length(x);
ftx = fft(x)/L; % Fourier Transform
ftxs = fftshift(ftx); % Shift To Centre
plot(t, ftxs);
title('TdF Triangle');

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回答（1 个）

Priyanshu Mishra
Priyanshu Mishra 2020-7-29
Hi Moisa,
fft is giving you a complex double values . Try taking only absolute values. I tried the following code and getting sinc squared function which is expected for a triangular pulse.
clc
t=-2:.01:2;
subplot(2, 1, 1);
x=tripuls(t,2);
plot(t, x);
title('Triangle');
subplot(2, 1, 2);
Ts = mean(diff(t)); % Sampling Interval
Fs = 1/Ts; % Sampling Frequency
Fn = 2*Fs; % Nyuquist Frequency
L = length(x);
ftx = abs(fft(x)); % Fourier Transform
ftxs = fftshift(ftx); % Shift To Centre
plot(t, ftxs);
title('TdF Triangle')

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