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

Started by porterboy August 17, 2004
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?
porterboy76@yahoo.com (porterboy) wrote in message
news:<c4b57fd0.0408170905.30c13258@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
Eeeep!
Sorry about the multiple post there... the computer froze...
porterboy76@yahoo.com (porterboy) writes:

> Eeeep!
Eeeep? -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
> 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.
> 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.
porterboy76@yahoo.com (porterboy) writes:

> Eeeep!
Eeeep? -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124
Eeeep!
Sorry about the multiple post there... the computer froze...
porterboy76@yahoo.com (porterboy) wrote in message
news:<c4b57fd0.0408170905.30c13258@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
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?