Complex Convolution Theorem of Z-Transforms

Complex Convolution Theorem :-

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

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

Z\left \{ x_{1}[n]\ x_{2}[n] \right \}\leftrightarrow \ \ ?.

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

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

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

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

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

Parseval’s Relation :-

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

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

 

Proof:-

By using complex convolution theorem

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

in the above equation substitute Z=1 , then

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

Hence Parseval’s relation is proved.

 

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.