DFT spectra question 2009-02-13 06:32:09 DFT : X(K) = sum(x(n)*exp(-j*2*pi*k*n/N));stft: X(K) = sum(x(n)*w(n)*exp(-j*2*pi*k*n/N) );no window,no overlap.let us decompose stft to be 3 parts:1. x(n) : let x(n) to be a single frequency signal x(n) = cos(2*pi*f*n/fs). f equal to be certain interger multiples of DFT resolution fs/N. Then its spectra must be two dirac functions located at the frequency -f and f respectively. 2. w(n): no window,then w(n) =1 when n = 0--N-1, and 0 otherwise. It's spectra is a sinc function, which is centered at 0hz.3. the filter ,which is exp(-j*2*pi*k*n/N), Here,we only think the (k+1)th filter, that is k = f*N/fs; It's spectra is a dirac functions located at frequency -f .The multiplication at time domian corresponds with convolution at frequency domain.Let us convolve part 1 and 2,then we can move the sinc and get the superpostion of 2 sinc,which are located at -f and f frequency respectively.then,we convolve the result above with part 3. We would achieve 2 sinc ,which located at -2*f and 0 hz. It does not match with truth. The truth is " its spectra must be two dirac functions located at the frequency -f and f respectively. "why?Any comments are appreciated. Cheers HyeeWang

DFT spectra question Reply #1 – 2009-02-23 05:44:33 Quote from: hyeewang on 2009-02-13 06:32:09DFT : X(K) = sum(x(n)*exp(-j*2*pi*k*n/N));stft: X(K) = sum(x(n)*w(n)*exp(-j*2*pi*k*n/N) );no window,no overlap.let us decompose stft to be 3 parts:1. x(n) : let x(n) to be a single frequency signal x(n) = cos(2*pi*f*n/fs). f equal to be certain interger multiples of DFT resolution fs/N. Then its spectra must be two dirac functions located at the frequency -f and f respectively. 2. w(n): no window,then w(n) =1 when n = 0--N-1, and 0 otherwise. It's spectra is a sinc function, which is centered at 0hz.3. the filter ,which is exp(-j*2*pi*k*n/N), Here,we only think the (k+1)th filter, that is k = f*N/fs; It's spectra is a dirac functions located at frequency -f .The multiplication at time domian corresponds with convolution at frequency domain.Let us convolve part 1 and 2,then we can move the sinc and get the superpostion of 2 sinc,which are located at -f and f frequency respectively.then,we convolve the result above with part 3. We would achieve 2 sinc ,which located at -2*f and 0 hz. It does not match with truth. The truth is " its spectra must be two dirac functions located at the frequency -f and f respectively. "why?Any comments are appreciated. Cheers HyeeWang It is not logical. Plz throw it to rubbish.Shame.