Question about Z transform of decimation

Started by Jeff September 6, 2004
Hi,
I am learning about digital decimation. The problem is like this:
Z-transform of input sequence and filter aree X(z), H(z) respectively.
After the filter H(z), there is a 2 decimation. From one book talking
about decimation, it says the Z-transform of output sequence after
decimation is:

V(z)=(1/2)*[H(z^(1/2)*X(z^(1/2))+H(-z^(1/2)*X(-z^(1/2))]

I have learned Z transform, but no Z transform with exponential of
(1/2). Can you tell me how to get the above result? Or, give me some
link relate to that. I have checked DSPGuru website without success.


Thanks in advance.
freelait2000@yahoo.com (Jeff) wrote in message
news:<6cdef69d.0409061348.159386ca@posting.google.com>...
> Hi, > I am learning about digital decimation. The problem is like this: > Z-transform of input sequence and filter aree X(z), H(z) respectively. > After the filter H(z), there is a 2 decimation. From one book talking > about decimation, it says the Z-transform of output sequence after > decimation is: > > V(z)=(1/2)*[H(z^(1/2)*X(z^(1/2))+H(-z^(1/2)*X(-z^(1/2))] > > I have learned Z transform, but no Z transform with exponential of > (1/2). Can you tell me how to get the above result? Or, give me some > link relate to that. I have checked DSPGuru website without success. > > > Thanks in advance.
We are examining the system [use fixed-width font] +-----+ | | | x(n) ------>| | 2 |------> x'(m) | V | +-----+ A very crude argument would go something like X(k)=ZT{x(n)} <-> x(n) is a Z transform pair. Define the decimated sequence x'(m) such that x'(m)= x(2n), n=...,-2,-1,0,1,2,... which means x(n) <-> x'(m/2), m=...,-4,-2,0,2,4,... Set up the formal exptression for X'(k)=ZT{x'(m)} and then substitute n/2 for m. CAVEAT - The book by Proakis & Manolakis does not mention scaling of the running index under the "Properties of the Z transform" section, so there may be one or two pitfalls in the derivation that I am not aware of. In fact, when I re-read what I just wrote, the line of "arguments" sound so shaky that I am tempted to hit the "cancel" button instead of "send"... Rune