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FFT result double length than expected

To: s-news@lists.biostat.wustl.edu
Subject: FFT result double length than expected
From: Alvaro Aballe Villero <alvaro.aballe@uca.es>
Date: Fri, 29 Nov 2002 12:08:02 +0100 (CET)
Organization: Universidad de Cadiz
Dear all,

Simple question about the "fft" function of S-Plus (for windows):

Let`s consider a vector "ni, i=1,2,..N" (for example a vector with 2048 
points). This vector is the result of sampling a analog signal at rate of "At" 
(for example 1 sec).

If I perfom the a FFT, I expect to obtain another vector "Ck, k=1,2,..N/2" 
(i.e. a half of length of the vector in the time domain). The frequencies 
corresponding with "Ck" will be "k/(N*At)". However, with the function "fft" I 
get a N-point long result.

What does it mean? Can I just ignore the second part in the follwing case?

I have a time record in the time domain, I want transform it in the frequency 
domain and multiply it for a transfer function (which is a equation function of 
frequency); and finally tranform the result in the time domain. Like applying a 
filter!
Should I extent the filter to double or ignore the second part of the FFT 
result?

Thanks in advance.

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