Area of a Spectrum

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.

Answer 1

[Dr. Seis]

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;

[Lisa Justin]

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]

freq range is between 10 to 800

[Dr. Seis]

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]

Thanks.

Answer 2

[Wayne King]

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)

[Lisa Justin]

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.