problem in Hanning window with FFT

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mouh nyquist 2015-1-1

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评论： mouh nyquist 2015-1-2

采纳的回答： Shoaibur Rahman

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where is the error in my program, why I don't get the amplitude is not 4 .

fs=300;
N=3000;
t = (1:N)/fs;  % Time vector
x = 4*cos(2*pi*100*t); 
% FFT 
fn=hanning(N);  %fenetre de hanning
ft=abs(fft(x'.*fn));%fft
plot(ft);

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Shoaibur Rahman 2015-1-1

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Here are few things: Your plot shows the amplitude of Fourier transform, not the original signal. So, the amplitude may not be 4 (amplitude of the signal). Indeed, the amplitude of fft is determined so that the power of the signal remains same before and after the transform. Lets say, Y = x'.*fn

Y = x'.*fn; 
Power_Y = sum(Y.^2) % power in time domain
fftY = fft(Y); 
Power_fftY = sum(fftY.*conj(fftY))/length(fftY) % power in frequency domain

The amplitude after transformation is set so that Power_Y = Power_fftY

To get the amplitude back, use ifft that is shown in subplot(313) below. Compare these three plots:

subplot(311), plot(Y); % original signal
subplot(312), plot(abs(fftY)); % fft
subplot(313), plot(ifft(fftY)); % ifft

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Shoaibur Rahman 2015-1-2

编辑：Shoaibur Rahman 2015-1-2

在 MATLAB Online 中打开

x = 4*cos(2*pi*100*t)+8*cos(2*pi*10*t);

Its amplitude is 12, range is 12*2 = 24

Now lets include the hanning window, say:

h = hann(N); % as you defined N in your code

Now, x.*h' will have a minimum value of -9.8 and a maximum of 12. You can also plot to see this. So, do you want to show amplitude equal to 21.8? If yes, then instead of multiplying by 2, use:

M=(abs(X)/max(abs(X)))*range(ifft(X));

mouh nyquist 2015-1-2