Sign in

Not a member? | Forgot your Password?

Search compdsp

Search tips

Find us on Facebook!





Free PDF Downloads

A Quadrature Signals Tutorial: Complex, But Not Complicated

Understanding the 'Phasing Method' of Single Sideband Demodulation

Complex Digital Signal Processing in Telecommunications

Introduction to Sound Processing

C++ Tutorial

Introduction of C Programming for DSP Applications

Fixed-Point Arithmetic: An Introduction

Cascaded Integrator-Comb (CIC) Filter Introduction

Discussion Groups

FIR Filter Design Software

Free Online Books

See Also

Embedded SystemsFPGA

Discussion Groups | Comp.DSP | Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform).


There are 5 messages in this thread.

You are currently looking at messages 1 to .


Is this discussion worth a thumbs up?

0

Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform). - porterboy - 2004-08-17 14:24:00

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?


Re: Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform). - Rune Allnor - 2004-08-18 01:45:00

p...@yahoo.com (porterboy) wrote in message news:<c...@posting.google.com>...
> 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.

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 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?

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


Re: Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform). - porterboy - 2004-08-18 03:15:00

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


Re: Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform). - 2004-08-18 08:01:00

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

> Eeeep!

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


Re: Downsampling by 2: What happens the Spectrum? (Not the Fourier Transform). - porterboy - 2004-08-18 10:34:00

> Y(e^jw) = 0.5 [X(e^jw/2) + X(e^j(pi+w/2))].  (A)
> I have a problem with your equation (A) above. I can't see straight away 
> why you get two terms there. 

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.

> If 
> 
>     x(t)    <->    X(w)
> 
> is a Fourier transform pair, then 
> 
>     x(at)   <->   1/|a|*X(w/a).

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