Z-transform of a discrete convolution

[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?

 

[Hawkeye18]

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

 

[samuelandjw]

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

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

 

[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).

 

[samuelandjw]

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).

Thanks! Starting from the result is indeed easier!