MATLAB: FFT in dB scale

fft db scale mag2db

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))));

Best Answer

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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The frequency of the sine wave is defined as 128 Hz (f = 128) in the above code. Hence there will be two impulses (ideally, but here since x is just an approximation of the sine wave you can observe peaks at that frequency) at 128 Hz and -128Hz in the FFT output.

If you take FFT with N = 128 or N = 100, x is truncated to match the length of N and FFT is computed on the resultant x.

If the frequency f is changed to 100, you could observe a peak at 100Hz.

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The essay is correct, Some remarks are :

1. Try to give variables names different than those of already built in functions , like phase.

2.The random variables is scalar, instead use randn(size(n)), to generate a vector.

3.the function abs(U) computes the magnitude if U is complex, no need to square the result.

4.the defined SNR is correct only for real values signal, which is not in this case, for this example use real(sinusoid) instead.

5. You have to scale the fft result in order to obtain the correct power ( std(u)).

6. You can increase the number of points N for FFT computation as follows fft(u,N).

7. With low number of time samples, you can not detect both frequencies f1 and f2, you need larger number of samples, this is related to Heisenberg uncertainty principle in signal processing, as example for n=310, you can detect the frequencies.

   clear all;close all;
   f1=0.115; f2=0.135; n=0:310-1;
   %PHASE=2*pi*rand();
   w=0.1*randn(size(n));
   u=real(exp(j*2*pi*f1*n))+real(exp(j*2*pi*f2*n))+w;
   %Using FFT
   N=600;
   U=2*fft(u,N)/N;
   fvals=(0:N-1)/N;
   mag=abs(U);
   subplot(1,2,1);plot(fvals,mag);
   xlabel('Frenquency'); ylabel('P(f)'); title('Power spectrum by FFT method');
   subplot(1,2,2);plot(fvals,10*log10(mag.^2));
   xlabel('Frequency'); ylabel('P(f)'); title('Power spectrum by FFT method in dB');

Best Answer

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