phase of FFT
in [DSP]
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From: HardySpicer on 20 Jan 2010 19:31 On Jan 21, 6:14 am, Greg Berchin <gberc...(a)comicast.net.invalid> wrote: > On Wed, 20 Jan 2010 10:29:37 -0500, Greg Berchin <gberc...(a)comicast.net.invalid> > wrote: > > >I believe that Matlab uses the positive exponent form. > > I just checked, and currently Matlab does use the signal processing standard > (negative exponent on the forward transform). Seems to me that this is a change > from early Matlab versions, but I could be wrong. > > Greg Why do people in modern times leave out the 1/N scaling for direct FFT and put it in the inverse FFT? Makes more sense if you go from time to freq to have the 1/N in eg for dc we get the average and must divide by N. Don't say "it's just scaling" and doesn't matter - it does! Hardy
From: Greg Berchin on 20 Jan 2010 20:04 On Wed, 20 Jan 2010 16:31:25 -0800 (PST), HardySpicer <gyansorova(a)gmail.com> wrote: >Why do people in modern times leave out the 1/N scaling for direct FFT >and put it in the inverse FFT? >Makes more sense if you go from time to freq to have the 1/N in eg for >dc we get the average and must divide by N. >Don't say "it's just scaling" and doesn't matter - it does! I agree, and this has been a topic of discussion here on comp.dsp many times in the past. I think that the only answer is "tradition". The issue becomes even more important when implementing convolution as transform-domain multiplication. It's easy to accidentally end up with an extra N or 1/N factor. Greg
From: robert bristow-johnson on 20 Jan 2010 21:00
On Jan 20, 8:04 pm, Greg Berchin <gberc...(a)comicast.net.invalid> wrote: > On Wed, 20 Jan 2010 16:31:25 -0800 (PST), HardySpicer <gyansor...(a)gmail.com> > wrote: > > >Why do people in modern times leave out the 1/N scaling for direct FFT > >and put it in the inverse FFT? > >Makes more sense if you go from time to freq to have the 1/N in eg for > >dc we get the average and must divide by N. > >Don't say "it's just scaling" and doesn't matter - it does! > > I agree, and this has been a topic of discussion here on comp.dsp many times in > the past. I think that the only answer is "tradition". well, there are a couple of traditions. to me, clearly the most natural fundamental definition would be N-1 x[n] = SUM{ X[k] * exp(+j*2*pi*n*k/N) } k=0 the orthonormal basis functions are exp(j*2*pi*n*k/N) and X[k] are the coefficients. X[0], unscaled, is the DC component. then the inverse of that (the forward DFT) is N-1 X[k] = 1/N * SUM{ x[n] * exp(-j*2*pi*n*k/N) } n=0 the X[k] are now means rather than sums. > The issue becomes even more important when implementing convolution as > transform-domain multiplication. It's easy to accidentally end up with an extra > N or 1/N factor. maybe it should be 1/sqrt(N) in both the forward and inverse DFT. i think, for the most prevalent convention, N-1 X[k] = SUM{ x[n] * exp(-j*2*pi*n*k/N) } n=0 N-1 x[n] = 1/N * SUM{ X[k] * exp(+j*2*pi*n*k/N) } k=0 then, defining X[k] = DFT{ x[n] } Y[k] = DFT{ y[n] } H[k] = DFT{ h[n] } it is true that if Y[k] = H[k] * X[k] (multiplication, not convolution) then N-1 y[n] = SUM{ h[i] * x[n-i] } i=0 with no 1/N scaling. if the DFT were defined the "natural" way, there would be a 1/N factor in that convolution summation. r b-j |