Calculating the Hann Window & FFT Energy Correction Factor

[Lindemose]

Hi all,

I am currently looking into the energy correction factor of the Hann window.

So far I have found that to correct for the application of Hann-window and Fourier transform the result needs to be multiplied by sqrt(32)/sqrt(3).

Can any of you explain how to get that factor ?

The amplitude correction factor is 2, that much i grasp...

There is a another factor of 2 as the FFT is performed as a complexed valued operation.

These two factors multiplied is 4 (duhh !)

And when removing that factor from sqrt(32)/sqrt(3) the result is ;

4*sqrt(2)/sqrt(3).

the sqrt(2)/sqrt(3) is the mysteries part. Can anyone explain how to reach this factor ?


Best regards

Lindemose

 

[rbj]

dunno if my markup works. lessee...

i guess it does. the definition for the Hann window (centered at zero and normalized in length) is:

w(x) = \begin{cases}<br /> \frac{1}{2}\left(1 + \cos(\pi x) \right) &amp; \mbox{if } |x| \le 1 \\[3pt]<br /> 0 &amp; \mbox{if } |x| &gt; 1<br /> \end{cases}

now, compared to a no window (where w(x) is 1 for the same stretch of data), this window would reduce your mean voltage magnitude (or whatever signal) by a factor of 1/2. but if the issue is power or energy, you have to square it before integrating.

\int_{-1}^{+1} w^2(x) \ dx = \int_{-1}^{+1} \frac{1}{4}\left(1 + \cos(\pi x) \right)^2 \ dx = \frac{3}{4}

if you did it to no window, the integral would be 2. so energy is reduced by a factor of 3/8. now perhaps they are talking about the r.m.s. reduction, then i think the factor is \sqrt{3/8}. that almost looks right, except for a factor of 2.

i can see where the \sqrt{3} comes from, but not the \sqrt{32}. i think it should be \sqrt{8}.

 

[Lindemose]

Now that makes all the sense in the world, thank yoy very much.

\sqrt{3}[/itex] comes from, but not the \sqrt{32}. i think it should be \sqrt{8}.

By the way the sqrt(32)should be sqrt(8) as you point out. I included the the amplitude correction factor of 2 within the number.

Thanks again and best regards !

Lindemose