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?

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?

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?

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?

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?

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

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?

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?

Eeeep!

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

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

XXXX@XXXXX.COM (porterboy) writes:

Eeeep?

--

Randy Yates

Sony Ericsson Mobile Communications

Research Triangle Park, NC, USA

XXXX@XXXXX.COM , 919-472-1124

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,

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

XXXX@XXXXX.COM , 919-472-1124

> 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.

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.

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

Is this thing on?

--

Randy Yates

Sony Ericsson Mobile Communications

Research Triangle Park, NC, USA

XXXX@XXXXX.COM , 919-472-1124

Similar Threads

1. What's the difference between a signal's fourier transform, and its energy spectrum

2. Power spectrum and autocorrelation form a fourier transform pair

3. What's the difference between a signal's fourier transform, and its energy spectrum

>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.

4. Fourier Transform (spectrum) of QPSK and BPSK?

5. OY: why fourier transform in AM and FM why not LT

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. ?

6. The Fourier transform in nature not just time<>frequency domain

7. why fourier transform in AM and FM why not LT

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.

8. The Fourier transform in nature not just timefrequency domain