Fourier Transform of tripuls
6 ビュー (過去 30 日間)
古いコメントを表示
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');
0 件のコメント
回答 (1 件)
Priyanshu Mishra
2020 年 7 月 29 日
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')
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Subplots についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!