## Area of a Spectrum

on 1 Aug 2012

### Dr. Seis (view profile)

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.

## Products

No products are associated with this question.

### Dr. Seis (view profile)

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;
```

Lisa Justin

### Lisa Justin (view profile)

on 1 Aug 2012

freq range is between 10 to 800

Dr. Seis

### Dr. Seis (view profile)

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.

Lisa Justin

on 2 Aug 2012

Thanks.

### Wayne King (view profile)

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)`

Lisa Justin

### Lisa Justin (view profile)

on 1 Aug 2012

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.

#### Join the 15-year community celebration.

Play games and win prizes!

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi