# Complex Convolution Theorem of Z-Transforms

Complex Convolution Theorem :-

Let two signals $x_{1}[n]&space;,&space;\&space;x_{2}[n]$  which are complex signals then the product of these two signals be $x_{3}[n]&space;=&space;x_{1}[n]\&space;x_{2}[n]$ .

$x_{1}[n]\leftrightarrow&space;X_{1}(Z)&space;\&space;\&space;\&space;ROC&space;:a_{1}<&space;\left&space;|&space;Z&space;\right&space;|&space;

$x_{2}[n]\leftrightarrow&space;X_{2}(Z)&space;\&space;\&space;\&space;ROC&space;:a_{2}<&space;\left&space;|&space;Z&space;\right&space;|&space;

$Z\left&space;\{&space;x_{1}[n]\&space;x_{2}[n]&space;\right&space;\}\leftrightarrow&space;\&space;\&space;?$.

we know that  $X(Z)&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}&space;x[n]&space;\&space;Z^{-n}$ .

$Z\left&space;\{&space;x_{3}[n]\right&space;\}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}(x_{1}[n]\&space;x_{2}[n])&space;\&space;Z^{-n}$ .

$Z\left&space;\{&space;x_{3}[n]\right&space;\}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}(\&space;x_{2}[n]\&space;Z^{-n})&space;\&space;\frac{1}{2\pi&space;j}&space;\oint_{c}X_{1}(v)\&space;v^{n-1}\&space;dv$ .

$Z\left&space;\{&space;x_{3}[n]\right&space;\}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}(\&space;x_{2}[n]\&space;(\frac{Z}{v}&space;)^{-n}&space;\&space;\frac{1}{2\pi&space;j}&space;\oint_{c}X_{1}(v)\&space;v^{-1}\&space;dv$ .

$Z\left&space;\{&space;x_{3}[n]\right&space;\}&space;=&space;\&space;\frac{1}{2\pi&space;j}&space;\oint_{c}X_{2}(\frac{Z}{v}&space;)&space;X_{1}(v)\&space;v^{-1}\&space;dv$ .

Parseval’s Relation :-

The two complex valued signals $x_{1}[n]&space;,&space;\&space;x_{2}[n]$   then the Parseval’s relation states that

$\sum_{n=-\infty&space;}^{\infty&space;}&space;x_{1}[n]&space;\&space;x_{2}^{*}[n]&space;=&space;\&space;\frac{1}{2\pi&space;j}&space;\oint_{c}&space;X_{1}(v)\&space;X_{2}^{*}(\frac{1}{v^{*}}&space;)&space;v^{-1}\&space;dv$ .

Proof:-

By using complex convolution theorem

$\sum_{n=-\infty&space;}^{\infty&space;}&space;x_{1}[n]&space;\&space;x_{2}^{*}[n]\&space;Z&space;^{-n}&space;=&space;\&space;\frac{1}{2\pi&space;j}&space;\oint_{c}&space;X_{1}(v)\&space;X_{2}^{*}(\frac{Z^{*}}{v^{*}}&space;)&space;v^{-1}\&space;dv$ .

in the above equation substitute $Z=1$ , then

$\sum_{n=-\infty&space;}^{\infty&space;}&space;x_{1}[n]&space;\&space;x_{2}^{*}[n]&space;=&space;\&space;\frac{1}{2\pi&space;j}&space;\oint_{c}&space;X_{1}(v)\&space;X_{2}^{*}(\frac{1}{v^{*}}&space;)&space;v^{-1}\&space;dv$ .

Hence Parseval’s relation is proved.

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