I think we can all agree that FFT { x*(n) } = X*(N-k), 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

# Strange FFT Behavior in MATLAB

This is due to the fact that MATLAB index starts at 1 while when you say FFT{x*(n)} = X*(N-k), you are assuming that k = 0,...,N-1. Note that when you do X(N-k) when k = 0, the first sample is X(N), which is the same as X(0). However, if you use fliplr, this is no longer true. HTH,>I think we can all agree that FFT { x*(n) } = X*(N-k), 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 > > > >

Patrick wrote:> I think we can all agree that FFT { x*(n) } = X*(N-k), or the FFT of the > conj of x(n) is the conj of the reversed version of the FFT of x(n).It's not exactly the reversed version. Your formula is correct, you just have to look at it closer. Consider X[k] := DFT{ x[n] } X1[k] := DFT{ x*[n] } So for k=0 you get X1[0] = X*[N-0] = X*[N]. However, this index is beyond the limit (indices go from k=0,1,...N-1), so X*[N] is in fact X*[0] (all indices are modulo N). If you write it out you get (X1[0], X1[1], ... , X1[N-1]) = (X*[0], X*[N-1], ...., X*[1]). Presumably Matlab's 1-based indexing doesn't make this matter any easier.> 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]For that a, I get X = fft(a) = [16+20j -8 -4-4j -8j] X1 = fft(conj(a)) = [16-20j 8j -4+4j -8 ] which is just what the formula says.> Any ideas? Another engineer and I spent most of a day looking at the > model before finding the fundamental problem.No wonder. If I were to look all day at a model*, I would never find any problems! Regards, Andor *Favourite model to look at: Claudia.

Ah ha! X(N-k) isn't a "reversal" of X, it's a circular shift... that was a silly mistake, thanks for the quick answers all. If anyone is ever in Dallas, TX beers are owed. Cheers. /ptj>Patrick wrote: >> I think we can all agree that FFT { x*(n) } = X*(N-k), or the FFT ofthe>> conj of x(n) is the conj of the reversed version of the FFT of x(n). > >It's not exactly the reversed version. Your formula is correct, you >just have to look at it closer. Consider > >X[k] := DFT{ x[n] } >X1[k] := DFT{ x*[n] } > >So for k=0 you get > >X1[0] = X*[N-0] = X*[N]. However, this index is beyond the limit >(indices go from k=0,1,...N-1), so X*[N] is in fact X*[0] (all indices >are modulo N). If you write it out you get > >(X1[0], X1[1], ... , X1[N-1]) = (X*[0], X*[N-1], ...., X*[1]). > >Presumably Matlab's 1-based indexing doesn't make this matter any >easier. > >> 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] > >For that a, I get > >X = fft(a) = [16+20j -8 -4-4j -8j] >X1 = fft(conj(a)) = [16-20j 8j -4+4j -8 ] > >which is just what the formula says. > >> Any ideas? Another engineer and I spent most of a day looking at the >> model before finding the fundamental problem. > >No wonder. If I were to look all day at a model*, I would never find >any problems! > >Regards, >Andor > >*Favourite model to look at: Claudia. > >

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

robert bristow-johnson <rbj@audioimagination.com> wrote in news:1190307365.158064.56600@d55g2000hsg.googlegroups.com:> 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.It's an index-- just a conceptual place holder, not a definition. X(1), in a ones indexed language, merely points to the first memory location in array X. Most likely the ones indexing in Matlab has its roots in Fortran. Aside from having to remember what language you're working in and getting bitten every now and again when moving back and forth between languages, it really isn't that problematic.> Y(k) = conj( X(mod(N-k+2, N) ); % for 1 <= k <= N > > gee, isn't that elegant?As compared to what? If you're comparing to finding and downloading the c implemenation of fftw, compiling it, linking to it, then figuring out which of the many arcane fftw routines you need to be calling, and then figuring out what arcane pattern your returned arrays are in, maybe creating complex variable types from two doubles, and then first doing your elegant zero-indexed circular shift, its a veritable tango. If you're talking about doing the same in some zero-indexed matlab clone, probably not as elegant. -- Scott Reverse name to reply