FFT in dB scale
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Hi everybody! I'm trying to use dB scale in FFT. I've found the example http://www.dsplib.ru/content/winex/winex.html
I'm trying to get the same results in MATLAB. But I can't. I have different amplitude levels. What's wrong? So, I do:
Fs=1000;
T=1/Fs;
N=1024;
t=(0:N-1)*T;
f=(0:Fs/(N-1):Fs);
A0=1;
A1=0.003;
A2=10e-4;
f0=220;
f1=240;
f2=230;
s=A0*cos(2*pi*f0*t)+A1*cos(2*pi*f1*t)+A2*cos(2*pi*f2*t);
plot(f,mag2db(abs(fft(s))));
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David Goodmanson
2017-1-24
hi Alexander I assume you are looking for a couple of peaks at -6 dB. Otherwise this answer will not be of help. But to get -6 dB you need to divide the fft result by N before taking mag2dB. Before getting into that, I believe your frequency grid with the factor of N-1 is not quite right. It should have a factor of N and should not end at Fs:
f = (0:N-1)*Fs/N;
That change only scales the frequency axis though, and puts the peaks in the right spot. It does not affect the peak value. You will still not get -6 dB (right now it's more like -7) because f0 = 220 does not represent an exact periodic frequency to the fft. To get a sharp peak at -6 dB, the frequency must be a multiple of Fs/N = 1000/1024. You could replace 220 by 225*1000/1024 which is pretty close, but I think the best way is to just use N = 1000. The fft is blazingly fast anyway, so unless you are in some production situation doing a ten million of these, N = 2^n is not really necessary.
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David Goodmanson
2017-2-7
For the fft to produce a sharp spike, there must be an exact integral number of oscillations in the time domain function. Otherwise what is repeating is not (say) a cosine wave but rather a cosine wave whose last oscillation is truncated. So (in your notation) f*t must equal an integer m. But for an N-point fft, t = NT and Fs = 1/T, so f = m Fs/N. and f must be a multiple of Fs/N. That's not the case for Fs = 1000, N = 1024, f = 220 but it is when you change N to 1000.
Also, you have A2 = 10e-4 not 1e-4, so -60 dB.
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