Topic: DFT spectra question (Read 1924 times)
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## 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?

Cheers
HyeeWang

## DFT spectra question

##### Reply #1 – 2009-02-23 05:44:33
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?