Z-transform of a truncated sequence?

Hi all,

Suppose we are given a function F(z) which is the z-transform of a real
sequence with index from 0 to infinity in the time domain, that's to
say,

F(z) is the Z-transform of a0, a1, a2, ... all the way down...

These a's add up to 1. That's to say: a0+a1+a2+a3+ ... =1.

Now suppose I want to truncate the sequence into a finite-length
version, and thus create a new sequence, which is

b0, b1, b2, ... b_n, STOP.

where b0=a0, b1=a1, ... b_n=a_n + a_(n+1) + a_(n+2) + ... + ...

so b0+b1+b2+... + b_n=1, they again add up to 1.

But now b's are a finite sequence with index from 0 to n...

What is the Z-transform of the new truncated sequence b?

In terms of the function F(z), which is the Z-transform of the original
sequence a?

Lucy wrote:

> Hi all,
>
> Suppose we are given a function F(z) which is the z-transform of a real
> sequence with index from 0 to infinity in the time domain, that's to
> say,
>
> F(z) is the Z-transform of a0, a1, a2, ... all the way down...
>
> These a's add up to 1. That's to say: a0+a1+a2+a3+ ... =1.
>
> Now suppose I want to truncate the sequence into a finite-length
> version, and thus create a new sequence, which is
>
> b0, b1, b2, ... b_n, STOP.
>
> where b0=a0, b1=a1, ... b_n=a_n + a_(n+1) + a_(n+2) + ... + ...
>
> so b0+b1+b2+... + b_n=1, they again add up to 1.
>
> But now b's are a finite sequence with index from 0 to n...
>
> What is the Z-transform of the new truncated sequence b?

If you write

B(z) = sum_{k=0}^n b_k z^-k

and

Br(z) =sum_{k=n+1}^infinity a_k z^-k,

then you have

B(z) = F(z) - Br(z) + z^-n (1 - sum_{k=0}^n a_k).

This is obvious, I just wrote down using sums what you wrote in words,
so it's probably not what you are looking for. One hint: if F(z) is a
rational function, then B(z) is also a rational function and has the
same denominator as F(z). This is similar like the tail-cancelling for
TIIR filters, which you can read about here:
http://ccrma.stanford.edu/~jos/tiirts/TIIR_Filters.html.

Regards,
Andor