FFT in dB scale

Question

Alexander Voznesensky 2017-1-23

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/321635-fft-in-db-scale

评论： Alexander Voznesensky 2017-2-22

采纳的回答： David Goodmanson

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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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Answer 1

David Goodmanson 2017-1-24

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https://ww2.mathworks.cn/matlabcentral/answers/321635-fft-in-db-scale#answer_251729

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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 2017-1-24

编辑：David Goodmanson 2017-1-24

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Hi Alexandr (got your name right this time), I did run your code yesterday. Thank you for providing runnable code by the way, a lot of people posting questions here don't. I got +53 dB like you mentioned.

Dividing by N is necessary because of the way Matlab scales their fft. I would not be surprised that if they had it to do all over again they might include that factor in the fft (in which case ifft would also change). After dividing by N, the result is 53 - 20*log10(1024) = -7 dB. After going to a frequency that is a multiple of Fs/N, I got a value at the peak equal to mag2db(1/2) good to many decimal places.

As to the -6 dB, let w = (2pi f) and let's say you transform a cosine function with a phase angle g. Then

cos(wt+g) = (1/2) e^(iwt) e^(ig) + (1/2) e^(-iwt) e^(-ig)

If you report out all frequencies you get two complex peaks of magnitude 1/2, i.e. -6 dB. That's what your fft is trying to do. You are never going to get there, though, if you use a frequency that is not a multiple of Fs/N.

For the transform of a real function both peaks are the same size and contain the same information. Sometimes people choose to report out only the positive frequencies, throw away the phase information and double the size of the peak. So you get 0 dB. That's what they did in the link you provided.

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.

Alexander Voznesensky 2017-2-22