dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

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?

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

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?

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

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?

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

Eh... The "Fourier transform" is linked to the "spectrum" in the same way
that the "addition" is linked to the "sum". So seeing how the FT reacts
to time scaling explains what happens to the spectrum.

I have a problem with your equation (A) above. I can't see straight away
why you get two terms there. I'd expect the "usual" scaling property
to hold:

If

x(t) <-> X(w)

is a Fourier transform pair, then

x(at) <-> 1/|a|*X(w/a).

Rune

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

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?

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

Eeeep!
Sorry about the multiple post there... the computer froze...

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

XXXX@XXXXX.COM (porterboy) writes:

Eeeep?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
XXXX@XXXXX.COM , 919-472-1124

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

XXXX@XXXXX.COM (porterboy) writes:

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.

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.

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

Since I don't the relationship on which you are basing this derivation,
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
XXXX@XXXXX.COM , 919-472-1124

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

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

Have a look at Gilbert and Strang "wavelets and filterbanks" bottom of
page 91 or Vaidyanathan "multirate systems and filterbanks" top of
page 105. The second term is aliasing because it is a discrete-time
system.

Again, this is continuous time, so there is no spectral wrap-around at
half the sampling frequency as occurs in discrete time.

dsp >> Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).

Randy Yates < XXXX@XXXXX.COM > writes:

Is this thing on?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
XXXX@XXXXX.COM , 919-472-1124

```>hey Bharat, i took a look at your website.  good luck there in
>Bangalore.

thanks. good that your pointed that any stable IIR's impulse response
will be of finite duration and thus will be considered as energy signal.
i just wanted to avoid it for sometime as the reader might get more
confused and hence wanted to illustrate with simple examples.

```

```Le Chaud Lapin wrote:

...

> -Le Chaud Lapin-

Does Le Chaud Lapin go cold turkey?

Jerry
--
Engineering is the art of making what you want from things you can get.
?
```

```Hello friends,
I am little confused that why do we always apply fourier transform fo
amplitude modulation, frequency modulation why not laplace transform.
Second question is that when I studied control system then everywher
Laplace Transform is used.
So please tell me why we use LT in controls and why FT in communication.

```