Area of a Spectrum

2012 8 1
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What matlab command can give me the area of a spectrum. I have shock response spectrum but i need to find the area under the curve.

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Dr. Seis
Dr. Seis 2012-8-1

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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;

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Lisa Justin
Lisa Justin 2012-8-1
编辑：Lisa Justin 2012-8-1
Thanks. Is this what you mean?
N = numel(x);
dt = 1/fs;
df = fs/N;
[y,f] = srs(x,fs,fmin,fmax,fno,Q,Nw);
area_y = sum(abs(y))*df;
Lisa Justin
Lisa Justin 2012-8-1
freq range is between 10 to 800
Dr. Seis
Dr. Seis 2012-8-1
编辑：Dr. Seis 2012-8-1
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.
Lisa Justin
Lisa Justin 2012-8-2
Thanks.

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Wayne King
Wayne King 2012-8-1

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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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Lisa Justin
Lisa Justin 2012-8-1
Thanks Wayne. But it is like this: I calculated the shock response spectrum Y and plotted between frequencies 10-500. I need the peaks and the area under the curve max(Y) gives me the peak and i tried area(Y) and i get a shaded plot again but what i really want is the value of the area of Y.

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