Wrong FFT for an audio file
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Amirhosein Khanlari
le 21 Jan 2020
Commenté : Star Strider
le 26 Jan 2020
I want to sketch the power spectrum of an audio file but i get wrong answer.human speech should be in range of 50 to 300 hz
[x Fs] = audioread('v0.mp3');
nf=length(x);
Y = fft(x,nf);
Y = Y-mean(Y);
f = Fs/2*linspace(0,1,nf/2+1);
plot(f,abs(Y(1:nf/2+1)));
i should get this:
but instead i get this:
1 commentaire
Walter Roberson
le 21 Jan 2020
Why are you subtracting the mean() of the fft from the fft results? It would make a lot more sense to have subtracted the mean of x from x
Réponse acceptée
Star Strider
le 21 Jan 2020
The plot is correct. You are not considering the exponential multiplication at the right end of the frequency axis.
This makes it a bit more obvious:
[x Fs] = audioread('Amirhosein Khanlari v0.mp3');
nf=size(x,1);
Y = fft(x-mean(x))/nf;
f = Fs/2*linspace(0,1,fix(nf/2)+1);
figure
plot(f,abs(Y(1:nf/2+1))*2)
xlim([0 1.5E+4])
set(gca, 'XTick', (0:2500:15000))
producing:
4 commentaires
Star Strider
le 26 Jan 2020
That is interesting. Subtracting the mean should produce a zero D-C offset. It does in the Fourier transform, however it does not when the time-domain signal is squared first.
According to Parseval’s Theorem, it is correct to square the Fourier transform or the original time-domain signal. Squaring the Fourier transformed signal to create ‘Psd’:
[x Fs] = audioread('Amirhosein Khanlari v0.mp3');
nf=size(x,1);
Y = fft(x-mean(x))/nf;
f = Fs/2*linspace(0,1,fix(nf/2)+1);
Psd = (abs(Y(1:nf/2+1))*2).^2;
[maxAmp,maxAmpidx] = max(Psd); % Maximum Amplitude & Index
figure
plot(f,Psd)
xlim([0 1.5E+4])
set(gca, 'XTick', (0:2500:15000))
text(f(maxAmpidx), maxAmp, sprintf('\\leftarrow %.1f Hz = %.2E Amplitude', f(maxAmpidx), maxAmp), 'HorizontalAlignment','left')
This produces the correct result.
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