Multiplication Property of Z-Transform

66 Views Asked by At

I have searched a lot online and have yet to find anything that proves the multiplication property of the z-transform ie

$$ x_1[n]x_2[n] \iff \frac{1}{2 \pi j} \oint X_1(u)X_2(\frac{z}{u})u^{-1}du $$

The operation seems very different to the continuous domain counterpart for laplace transforms. Is a knowledge of Complex Analysis necessary to understand this?

1

There are 1 best solutions below

0
On BEST ANSWER

Not an answer, because the paths of integration below are incorrect. However, it illustrates one approach:

\begin{eqnarray} {1 \over 2 \pi i}\int_C (\int_C \hat{x}(w) \hat{y}({ z \over w}) w^{-1} dw) z^{n-1} dz &=& {1 \over 2 \pi i} \int_C \hat{x}(w)(\int_C \hat{y}({ z \over w}) z^{n-1} dz) w^{-1} dw \\ &=& {1 \over 2 \pi i} \int_C \hat{x}(w)(\int_C \hat{y}({ z \over w}) z^{n-1} {1 \over w}dz) dw \\ &=& {1 \over 2 \pi i} \int_C \hat{x}(w)(\int_C \hat{y}(v) v^{n-1}w^{n-1} dv) dw \\ &=& {1 \over 2 \pi i} \int_C \hat{x}(w)(\int_C \hat{y}(v) v^{n-1} dv) w^{n-1}dw \\ &=& x(n)y(n) \end{eqnarray} Using the substitution $v={z \over w}$.