# Multiplication by Exponential Sequence 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 $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is a discrete time function, then its Z-transform is defined as,

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

## Multiplication by Exponential Sequence Property of Z-Transform

Statement - The exponential multiplication property of Z-transform states that the exponential sequence multiplied in time domain corresponds to the scaling in z-domain. The exponential multiplication property is also known as scaling in z-domain property of the Z-transform. Therefore, if

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{z}\right)};\:\mathrm{ROC}\:\mathrm{=}\:\mathit{R}}$$

Then, according to the exponential multiplication property,

$$\mathrm{\mathit{a^{\mathit{n}}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\frac{\mathit{z}}{\mathit{a}}}\right)};\:\mathrm{ROC}\:\mathrm{=}\:\left| \mathit{a}\right|\mathit{R}}$$

Where, a is a complex number.

Proof

From the definition of the Z-transform, we have,

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

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

$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left[\mathit{a^{\mathit{n}}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\frac{\mathit{z^{\mathit{-n}}}}{\mathit{a^{\mathit{-n}}}}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathrm{\left ( \frac{\mathit{z}}{\mathit{a}} \right )}^{-\mathit{n}}}$$

$$\mathrm{\therefore \mathit{Z}\mathrm{\left[\mathit{a^{\mathit{n}}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left ( \frac{\mathit{z}}{\mathit{a}} \right )}}$$

Also, it can be represented as,

$$\mathrm{\mathit{a^{\mathit{n}}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\frac{\mathit{z}}{\mathit{a}}}\right)};\:\mathrm{ROC}\:\mathrm{=}\:\left| \mathit{a}\right|\mathit{R}}$$

Important −

• If the given time domain sequence is multiplied by a growing exponential sequence, i.e., $\mathit{e}^{j\omega n}$,then

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{e}^{\mathit{j\omega n}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]} \:\mathrm{=}\:\mathit{X}\mathrm{\left( \frac{\mathit{z}}{\mathit{e}^{\mathit{j\omega }}} \right )}\:\mathrm{=}\:\mathit{X}\mathrm{\left( \mathit{e}^{-\mathit{j}\omega }\mathit{z}\right )}}$$

• If the given time domain sequence is multiplied by a decaying exponential sequence, i.e., $\mathit{e}^{-j\omega n}$,then

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{e}^{\mathit{-j\omega n}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]} \:\mathrm{=}\:\mathit{X}\mathrm{\left( \frac{\mathit{z}}{\mathit{e}^{\mathit{-j\omega }}} \right )}\:\mathrm{=}\:\mathit{X}\mathrm{\left( \mathit{e}^{\mathit{j}\omega }\mathit{z}\right )}}$$

## Numerical Example (1)

Using the multiplication by exponential property of Z-transform, find the Z-transform of the following signal

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)} \:\mathrm{=}\:\mathrm{\left( 2\right )}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$

Solution

The given sequence is,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)} \:\mathrm{=}\:\mathrm{\left( 2\right )}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$

Since the Z-transform of unit step function is given by,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\frac{\mathit{z}}{\mathit{z}-1};\:\mathrm{ROC}\to\left|\mathit{z} \right|>1}$$

Now, using the scaling in z-domain or exponential property $\mathrm{\left [ i.e,\mathit{a^{\mathit{n}}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\frac{\mathit{z}}{\mathit{a}}}\right)} \right ]}$ of Z-transform, we get,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathrm{\left ( 2 \right )}^{\mathrm{\mathit{n}}}\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{Z}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}_{\mathit{z=\mathrm{\left ( \frac{\mathit{z}}{2} \right )}}}\:\mathrm{=}\:\frac{\mathrm{\left ( \frac{\mathit{z}}{2} \right )}}{\mathrm{\left ( \frac{\mathit{z}}{2} \right )}-1}}$$

$$\mathrm{\therefore \mathit{Z}\mathrm{\left[\mathrm{\left ( 2 \right )}^{\mathrm{\mathit{n}}}\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\frac{\mathit{z}}{\mathit{z}-2};\:\mathrm{ROC}\to\left|\mathit{z} \right|>2}$$

## Numerical Example (2)

Using the scaling in z-domain property of Z-transform, find the Z-transform of the signal

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)} \:\mathrm{=}\:\mathrm{\mathrm{\left ( \frac{1}{2} \right )}}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$

Solution

The given sequence is,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)} \:\mathrm{=}\:\mathrm{\mathrm{\left ( \frac{1}{2} \right )}}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$

Since the Z-transform of the unit step sequence is given by,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\frac{\mathit{z}}{\mathit{z}-1};\:\mathrm{ROC}\to\left|\mathit{z} \right|>1}$$

Now, from the property of multiplication by an exponential $\mathrm{\left [ i.e,\mathit{a^{\mathit{n}}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\frac{\mathit{z}}{\mathit{a}}}\right)} \right ]}$ ,we have,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathrm{\left ( \frac{1}{2} \right )}^{\mathrm{\mathit{n}}}\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{Z}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}_{\mathit{z=\mathrm{\left [ \frac{\mathit{z}}{\frac{1}{2}} \right ]}}=2\mathit{z}}\:\mathrm{=}\:\frac{2\mathit{z}}{\mathrm{2}\mathit{z}-1}}$$

$$\mathrm{\therefore \mathit{Z}\mathrm{\left[\mathrm{\left ( \frac{1}{2} \right )}^{\mathrm{\mathit{n}}}\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathrm{\left ( \frac{\mathit{z}}{\mathit{z}-\frac{1}{2}} \right )};\:\mathrm{ROC}\to\left|\mathit{z} \right|>\frac{1}{2}}$$