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From: dbd <dbd@ieee.org>
Newsgroups: comp.soft-sys.matlab
Subject: Re: About FFT and Even Function
Date: Sun, 1 Nov 2009 15:11:04 -0800 (PST)
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On Nov 1, 11:19 am, "Nasser M. Abbasi" <n...@12000.org> wrote:
...
> ...
> No, that is not what I mean. In the above, you have N=7, but "my formula"
> said
>
> x(n)=x(N-n)  for n=0..N-1   which means x(7) is not valid part of the

No. Your first post did not say n=0...N-1. In fact, you chose 1 to N
for the fft indices.

> sequence since 7 is more than 7-1 which is 6. So the sequence should have
> been
>
> x(0)=x(7)   (n=0 case)
> x(1)=x(6)   (n=1 case)
> x(2)=x(5)   (n=2 case)
> x(3)=x(4)   (n=3 case)
>
> So, assume, using your numbers, that  x(0)=0, x(1)=1, x(2)=2, x(3)=3 one
> gets
>
> x(0)=0
> x(1)=1
> x(2)=2
> x(3)=3
> x(4)=3
> x(5)=2
> x(6)=1  and you STOP here, since n goes only up to N-1 whichis 6.
>
> So now the x sequence is  [0,1,2,3,3,2,1]
>
> EDU>> fft([0 1 2 3 3 2 1])
>
> ans =
>
>    12.0000   -5.0489   -0.3080   -0.6431   -0.6431   -0.3080   -5.0489
>
> So, I guess Nasser's formula works after all?  ;)

No Nasser, this is a Matlab group. There is no x(0). That is broken
here. :(

You gave 1 to N as the indices of the fft in your original post. That
was the thing you got right at first. :)

Your follow-up posts are different from the content of your original
post. This demonstrates that "There are often mistakes made" as I
originally posted to Oscar. Thank you for the extended
demonstration. :)

>
> --Nasser

Dale B. Dalrymple