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?

# Z-transform of a truncated sequence?

Started by ●September 1, 2006

Reply by ●September 3, 20062006-09-03

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