problem in Hanning window with FFT
18 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
mouh nyquist
le 1 Jan 2015
Commenté : mouh nyquist
le 2 Jan 2015
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);
0 commentaires
Réponse acceptée
Shoaibur Rahman
le 1 Jan 2015
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
13 commentaires
Shoaibur Rahman
le 2 Jan 2015
Modifié(e) : Shoaibur Rahman
le 2 Jan 2015
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));
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Fourier Analysis and Filtering dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!