Z-transform of a discrete convolution

1. Jan 12, 2012

samuelandjw

Hi,

Suppose we have these two functions and their z-transforms are
$$P(r,z)=\sum_{t=0}^{\infty}P(r,t)z^t$$
and
$$F(r,z)=\sum_{t=0}^{\infty}F(r,t)z^t$$.
Now we are going to transform the following convolution of P and F:
$$\sum_{t'\le{t}}F(r,t')P(0,t-t')$$.
The result is said to be
$$F(r,z)P(0,z)$$.
But I don't know how to obtain the result. There are two difficulties:(1)the upper limit of t' is t, which is finite. (2)the two summations are coupled.
Anyone can help?

Last edited: Jan 12, 2012
2. Jan 12, 2012

Hawkeye18

It should be $z^t$, not $t^z$; then it is just the standard power series multiplication.

3. Jan 12, 2012

samuelandjw

Sorry about the typo. But my problem remains. Could you show a bit more on the decoupling process?

4. Jan 12, 2012

Hawkeye18

If you multiply 2 power series $\sum_{k=0}^\infty a_k z^k$ and $\sum_{k=0}^\infty b_k z^k$, the coefficients in the power series of of the product are given by the discrete convolution.

To get the term with $z^n$ in the product you need to add up all products $a_k z^k b_{n-k}z^{n-k}$, $0\le k \le n$. Thus the coefficient at $z^n$ is given by $\sum_{k=0}^n a_k b_{n-k}$

In your case $a_k = F(r,k)$, $b_k = P(0, k)$.

5. Jan 12, 2012

samuelandjw

Thanks! Starting from the result is indeed easier!