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.

f...@yahoo.com (Jeff) wrote in message
news:<6...@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