Find the nonzero maximum frequency corresponding to FFT graph.

I want to find the maximum corresponding frequency for the ftt graph. The graph shows the maximum frequency at zero frequency, this is not what I want. I am expecting to have a value around 664 Hz.

The codes:

so=dat.data(:,2); %original signal

t=dat.data(:,1); %time

dt=t(2)-t(1);

Fs=1/dt; %sampling frequency

f=linspace(0,Fs,length(t));

%Fourier Transform

SO=fft(so);

SOa=abs(SO);

F_fake=f(find(SOa(1:round(length(SOa)/2))==max(SOa))); % I am getting zero instead of 663.9 Hz.

I want a value which is nonzero. How to eliminate the zero Hz frequency ?

FFT graph.

Answer 1

[David Goodmanson]

Hi Zahra,

the peak at zero frequency is because your signal does not have average value of zero, but has a DC offset. To eliminate the zero freq peak you can remove the offset by subracting the mean off of the signal. So you can use not fft(signal) but rather fft(signal-mean(signal)). (This assumes that the DC offset is not of interest in this case).

[Zahra]

Thank you. Would you please explain more about what does DC offset mean?

[David Goodmanson]

DC offset is a bit of a slang term in the sense that it refers to electrical voltage or current signals, not the most general possible signal. Suppose you have a complicated signal and take the fft. The zero frequency term from the fft corresponds to a constant value in the time domain. Now consider a simple example of constant term of value 3 and just one nonzero frequency of amplitude 10. In terms of voltage this is modeled as a 10V cosine signal in series with a 3V battery, 10*cos(2*pi*f) + 3. The cosine term of course has mean value of 0 and the constant term, 3, is called the DC offset.

Answer 2

[Paul]

If the DC offset is of interest (or even if it isn't)

[maxSOa,index] = max(SOa(2:end)); % ignores first point a f=0
F_fake = f(index+1)

This approach will find the the frequency corresponding to the first maximum value in SOa(2:end) should there be multiple occurrences of the maximum value.