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Re: inverse fourier transform

From: Quentin H. Spencer
Subject: Re: inverse fourier transform
Date: Wed, 02 Apr 2003 10:35:01 -0700
User-agent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.3) Gecko/20030314

Heber Farnsworth wrote:

Something seems wrong about this. Why does the function need to be symmetric for ifft to give a real answer? On some level this makes sense because I know that even functions have real transforms. But I know this function is not even (for those with experience in statistics I have a distribution which is skewed). But the distribution is real. Why can't I do an inverse transform to recover the distribution from it's fourier transform?

You are correct. My response was based on the assumption that your characteristic function was all real. If it is complex, then the real part must be even and the imaginary part odd (referred to as hermitian symmetric). If you are still getting an imaginary result, then the problem must be in misalignment as I suggested before.

I looked more closely at the fourier transform which I'm trying to invert. I call it phi.


It's real part is symmetric (even)


and it's imaginary part is odd (looks just like sin on [-pi:pi])


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