DFT spectra question

[HyeeWang]

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 attime 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

[Rune Allnor]

On 13 Feb, 07:27, HyeeWang <[email protected]> wrote:
> 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) );

Note that these sums are taken ofer finitely many terms
and that these FTs produce discrete spectra.

> 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.

Nope. Since you deal with discrete spectra, you don't get Dirac
functions, which ara awkward limit cases of continuous functions.
You get Kronecker deltas, which is a different thing.

> 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 .

Nope. It's a set of Kronecker delta located at k/N, k and N integers.
Unless a sinusoidal 'hits' a k/N ration exactly, the energy will
be 'smeared' out across all spectrum coefficients.

> 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. "

Nope. This is only valid when you deal with continuous spectra.
You deal with discrete spectra.

The problem is that you mix in ideas and concpets for continuous
spectra in the discussion of discrete spectra. Continuous spectra
arise when you use the forms of the FT that deal with continuous
or discrete input data of infinite length. You recognize these
forms as the ones that use the -jw term in the exponentials.

Rune