Multiplication Property of Z-Transform

Z-Transform

The Z-transform is a mathematical tool which is used to convert the difference equations in discrete time domain into the algebraic equations in z-domain. Mathematically, if $\mathrm{\mathit{x\left ( n \right )}}$ is a discrete time function, then its Z-transform is defined as,

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$

Multiplication Property of Z-Transform

Statement – The multiplication property of Z-transform states that the multiplication of two signals in time domain corresponds to the complex convolution in z-domain. For this reason, the multiplication property is also called the complex convolution property of Z-transform. Therefore, if

$$\mathrm{\mathit{x_{\mathrm{1}}\left ( n \right )\overset{ZT}{\leftrightarrow}X_{\mathrm{1}}\left ( z \right )\:\: \mathrm{and}\:\: x_{\mathrm{2}}\left ( n \right )\overset{ZT}{\leftrightarrow}X_{\mathrm{2}}\left ( z \right ) }}$$

Then, according to the multiplication property

$$\mathrm{\mathit{x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right )\overset{ZT}{\leftrightarrow}\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X_{\mathrm{1}}\left ( p \right )X_{\mathrm{2}}\left ( \frac{z}{p} \right )p^{\mathrm{-1}}dp}}$$

Proof

From the definition of Z-transform, we have,

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$

$$\mathrm{\mathit{\therefore Z\left [ x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right ) \right ]\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }\left [x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right ) \right ]z^{-n}\; \; \; \cdot \cdot \cdot \left ( \mathrm{1} \right )}}$$

But, from the definition of inverse Z-transform, we have,

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X\left ( z \right )z^{\left ( n-\mathrm{1} \right )}dz\; \; \cdot \cdot \cdot \left ( \mathrm{2} \right ) }}$$

Replacing the complex variable z by p in eq. (2), we get,

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X\left ( p \right )p^{\left ( n-\mathrm{1} \right )}dp\; \; \cdot \cdot \cdot \left ( \mathrm{3} \right ) }}$$

Substituting the value of $\mathrm{\mathit{x_{\mathrm{1}}\left ( n \right )}}$ using the eq. (3) into the eq. (1), we get,

$$\mathrm{\mathit{Z\left [ x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right ) \right ]\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }\left [ \frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X_{\mathrm{1}}\left ( p \right )p^{\left ( n-\mathrm{1} \right )}dp \right ]x_{\mathrm{2}}\left ( n \right )z^{-n}}} $$

$$\mathrm{\mathit{\Rightarrow Z\left [ x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right ) \right ]\mathrm{\, =\, }\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X_{\mathrm{1}}\left ( p \right )\left [\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x_{\mathrm{2}}\left ( n \right ) p^{n}p^{\mathrm{-1}}z^{-n} \right ]dp}} $$

$$\mathrm{\mathit{\Rightarrow Z\left [ x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right ) \right ]\mathrm{\, =\, }\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X_{\mathrm{1}}\left ( p \right )\left [\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x_{\mathrm{2}}\left ( n \right ) \left ( p^{\mathrm{-1}}z \right )^{-n} \right ]p^{\mathrm{-1}}\, dp}}$$

$$\mathrm{\mathit{\therefore Z\left [ x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right ) \right ]\mathrm{\, =\, }\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X_{\mathrm{1}}\left ( p \right ) X_{\mathrm{2}}\left ( \frac{z}{p} \right ) p^{\mathrm{-1}}\, dp}}$$

Also, it can be written as,

$$\mathrm{\mathit{x_{\mathrm{1}}\left ( n \right )x_{\mathrm{2}}\left ( n \right )\overset{ZT}{\leftrightarrow}\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint X_{\mathrm{1}}\left ( p \right ) X_{\mathrm{2}}\left ( \frac{z}{p} \right ) p^{\mathrm{-1}}\, dp}}$$