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On Sep 20, 8:01 am, "patrickjennings" <patrick.t.jenni...@gmail.com>
wrote:
> I think we can all agree that FFT { x*(n) } = X*(N-k),
actually the folks at The Math Works do not agree. as fatnbafan said,
MATLAB is hard-wired or hard-coded so that the indices of all arrays
begin with 1, not 0 as it should for the DFT or FFT.
so in MATLAB, if N=length(x); y = conj(x); X = fft(x); Y = =
fft(y); then
Y(k+1) = conj( X(mod(N-k+1, N)) ); % for 0 <= k < N
or stated so elegantly that it's amazing we all don't just sing the
praises of MATLAB,
Y(k) = conj( X(mod(N-k+2, N) ); % for 1 <= k <= N
gee, isn't that elegant?
r b-j
> or the FFT of the
> conj of x(n) is the conj of the reversed version of the FFT of x(n).
>
> But in MATLAB if a= [1+2j 3+4j 5+6j 7+8j] then
>
> fft(conj(a)) = [16+20j -8+0j -4-4j 0-8j]
>
> and
>
> conj(fliplr(fft(a))) = [-8+0j -4-4j 0-8j 16+20j]
>
> Any ideas? Another engineer and I spent most of a day looking at the
> model before finding the fundamental problem.
>
> Cheers
>
> /Patrick
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