Hello all, I have a signal defined by the below. I want to take the Matlab FFT of this signal but I got a plot that does is not the Fourier Transform. Can someone direct me on how to be able to compute the FFT of my signal? Thanks Mates!

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Neo 2015-7-5

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https://ww2.mathworks.cn/matlabcentral/answers/228463-hello-all-i-have-a-signal-defined-by-the-below-i-want-to-take-the-matlab-fft-of-this-signal-but-i

评论： Neo 2015-7-6

Signal defined by:

%%Time specifications:
   Fs = 8000;                   % samples per second
   dt = 1/Fs;                   % seconds per sample
   StopTime = 0.25;             % seconds
   t = (0:dt:StopTime-dt)';     % seconds
   %%Sine wave:
   Fc = 60;                     % hertz
   x = 3*sin(x) + sin(.5*x+40) + 2*sin(3*x-60);
   % Plot the signal versus time:
   figure;
   plot(t,x);
   xlabel('time (in seconds)');
   title('Signal versus Time');
   zoom xon;

The signal image is attached.

I did this for the FFT of the signal:

 Y = fft(x);% Plot single-sided amplitude spectrum.
plot(Y) 
title('Fast Fourier Transform of X graph)')
xlabel('Frequency (Hz)')
ylabel('Time (seconds)')

I figured since it was a simple function, I did not need elaborate coding but then I got the "FFT" attached.

Can someone tell me how to get the correct FFT please! Thanks.

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Neo 2015-7-6

I actually typed it in as I wrote it and I got a function (signal anyway) do you know why that is?

Neo 2015-7-6

I used the correction that you noted and I got a straight line instead of a signal.

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Answer 1

Image Analyst 2015-7-5

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Your stop time is so short that your signal is essentially a constant over that period. So the FFT is a just big DC spike as expected, which corresponds to a big "rect" function rather than the sum of 3 sine waves of different periods.

Set a stop time around 12*pi so you can get more of a varying waveform in there, especially more complete cycles instead of just a microscopic sample out of the waveform. That small sample is essentially a constant.

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Neo 2015-7-6

So would this be my new code?

%% Time specifications: Fs = 8000; % samples per second dt = 1/Fs; % seconds per sample StopTime = 12*pi; % seconds t = (0:dt:StopTime-dt)'; % seconds %% Sine wave: Fc = 60; % hertz x = 3*sin(t) + sin(.5*t+40) + 2*sin(3*t-60); % Plot the signal versus time: figure; plot(t,x); xlabel('time (in seconds)'); title('Signal versus Time'); zoom xon;Y = fft(x);% Plot single-sided amplitude spectrum. plot(Y) title('Fast Fourier Transform of X graph)') xlabel('Frequency (Hz)') ylabel('Time (seconds)')

Neo 2015-7-6