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Given a stationary discrete-time stochastic process x(n) with
discrete-time Fourier transform X(e^jw) and spectrum Sx(w). If x(n) is
downsampled by two to give y(n) it is well known that

Y(e^jw) = 0.5 [X(e^jw/2) + X(e^j(pi+w/2))].  (A)

My question is, what is the resulting spectrum, not just the Fourier
Transform. I know that Sx(w) = E[X(e^jw)X*(e^jw)] and Sy(w) =
E[Y(e^jw)Y*(e^jw)]. From (A), this gives

Sy(w) = 0.25[  Sx(w/2) + Sx(pi+w/2) 
             + E[X(e^jw)X*(e^j(pi+w/2))
             + E[X*(e^jw)X(e^j(pi+w/2))] ]

I understand the first two terms which are the original spectrum in
the first half band and the aliasing from the second half band. Do the
second two terms disappear or are they also contributing to aliasing?

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p...@yahoo.com (porterboy) writes:

> Given a stationary discrete-time stochastic process x(n) with
> discrete-time Fourier transform X(e^jw) 

Wait right here - a discrete-time stochastic process is neither
absolutely summable nor finite-energy, thus the DTFT may not
converge for such a sequence (re: "Signals and Systems", Oppenheim
and Willsky). All the work I've done and seen on these uses
the autocorrelation function, i.e., Wiener and Khinchine say
that the power spectrum of such a signal is Fourier transform
of the autocorrelation.

> and spectrum Sx(w). If x(n) is
> downsampled by two to give y(n) it is well known that
> 
> Y(e^jw) = 0.5 [X(e^jw/2) + X(e^j(pi+w/2))].  (A)

For the following, I'm assuming X(e^jw) exists.

I suppose you mean downsampling without the usual pre-filtering? If
so, then I don't know about the "0.5" factor in front - that would
make Parseval a liar I think. I also think, since this is the DTFT
and not the DFT, that the "pi" should be "pi*Fs". Otherwise I 
think this is right.

> My question is, what is the resulting spectrum, not just the Fourier
> Transform. 

> I know that Sx(w) = E[X(e^jw)X*(e^jw)] 

Really? I've never seen this expression used for a stochastic process, 
probably due to the convergence issue above. Got a reference?

> and Sy(w) =
> E[Y(e^jw)Y*(e^jw)]. From (A), this gives
> 
> Sy(w) = 0.25[  Sx(w/2) + Sx(pi+w/2) 
>              + E[X(e^jw)X*(e^j(pi+w/2))
>              + E[X*(e^jw)X(e^j(pi+w/2))] ]
> 
> I understand the first two terms which are the original spectrum in
> the first half band and the aliasing from the second half band. Do the
> second two terms disappear or are they also contributing to aliasing?

Since I don't the relationship on which you are basing this derivation,
I can't answer your question. I think you're essentially asking this: If
you downsample a signal that has been oversampled without pre-filtering
it, have you reduced the quantization noise (I think you posted this under
another subject yesterday). The answer is no. Intuitively, that noise in 
the upper half of the original spectrum will fold down into the new 
downsampled spectrum.
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
r...@sonyericsson.com, 919-472-1124

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Randy Yates <r...@sonyericsson.com> writes:
> [...]
> Wait right here [...]

Is this thing on?
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
r...@sonyericsson.com, 919-472-1124

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