# Area of a Spectrum

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### Accepted Answer

Dr. Seis
on 1 Aug 2012

The area under your curve should just be:

N = numel(x);

dt = 1/fs;

df = fs/N;

y = fft(x)*dt;

area_y = sum(abs(y))*df; % which is also equal to: sum(abs(fft(x)))/N

energy_y = sum(abs(y).^2)*df;

If you really want "energy", then energy_y should be equal to energy_x:

energy_x = sum(x.^2)*dt;

##### 4 Comments

Dr. Seis
on 1 Aug 2012

Edited: Dr. Seis
on 1 Aug 2012

Yeah... in your case, you would just need:

area_y = sum(abs(y))/N;

The frequency increment ( df ) comes into play only if you scaled your FFT amplitudes ( y ) by the time increment ( dt ) - since dt*df = 1/ N. If you do not scale your FFT amplitudes inside your srs function, then you should just divide your sum by N to get the area. If you want to find the area between a frequency range, you will have to do something a little different. See below:

Example:

N = 4096;

fs = 2000;

x = randn(1,N);

df = fs/N;

Nyq = fs/2;

y = fft(x);

f = ifftshift(-Nyq : df : Nyq-df);

If your y represents both the negative and positive frequency amplitudes:

area_y_10_800 = sum(abs(y( abs(f) >= 10 & abs(f) <= 800 )))/N;

or if your y represents only the positive frequencies

area_y_10_800 = 2*sum(abs(y( f >= 10 & f <= 800 )))/N;

However, you do not want to double the amount if you are including either your 0 frequency or Nyquist frequency amplitude in the frequency range.

### More Answers (1)

Wayne King
on 1 Aug 2012

Hi Lisa, If you have the Signal Processing Toolbox, you can use the avgpower() method of a spectrum object.

For example:

Fs = 1000;

t = 0:1/Fs:1-(1/Fs);

x = cos(2*pi*50*t)+sin(2*pi*100*t)+randn(size(t));

psdest = psd(spectrum.periodogram,x,'Fs',Fs,'NFFT',length(x));

avgpower(psdest,[25 75])

The final line above integrates under the PSD from 25 to 75 Hz.

Note you can get the fraction of the total power in the specified interval with:

avgpower(psdest,[25 75])/avgpower(psdest)

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