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Re: Again an FFT question [message #80625] Thu, 28 June 2012 06:07 Go to previous message
Kenneth P. Bowman is currently offline  Kenneth P. Bowman
Messages: 585
Registered: May 2000
Senior Member
In article <27bbd728-a65d-4925-bd54-d003cfad6493@googlegroups.com>,
Helder <helder@marchetto.de> wrote:

> Thank you all for the answers. I guess the difference between changing sign
> of every second element and shift the array is due to a round-off error. This
> becomes clear when analyzing the differences between the two results using
> double precision or not.
> Thanks also for the explanation of how the frequencies in an FFT are
> displayed.
> I guess I'll have to update old code and remove the shift and use instead the
> center option.

I didn't plug it yesterday, but you might want to look at the chapter
on FFTs in my book.

http://www.amazon.com/An-Introduction-Programming-IDL-Intera ctive/dp/012088559X/r
ef=sr_1_1?ie=UTF8&qid=1340888241&sr=8-1&keywords =idl+bowman

I used to center FFTs, but have gotten away from that. For real input
data, the positive and negative frequencies are complex conjugates, so
the negative frequencies are redundant and you can get all of the information
by plotting only the positive frequencies.

For calculations in the spectral domain, such as filtering, I usually
compute the frequencies (or periods) in dimensional units as a function
of the transform array indices and use that. Then you don't have
to remember to uncenter when transforming back.

Ken Bowman
http://idl.tamu.edu/
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