Multiplication by Exponential Sequence Property of Z-Transform

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Z-Transform

The Z-transform is a mathematical tool used to convert the difference equations in the discrete-time domain into algebraic equations in the z-domain. Mathematically, if x(n) is a discrete-time function, then its Z-transform is defined as −

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

Multiplication by Exponential Sequence Property of Z-Transform

Statement

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

$$\mathrm{x(n) \:\overset{ZT}\longleftrightarrow\: X(z); \text{ ROC } \:=\: R}$$

Then, according to the exponential multiplication property −

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

Where a is a complex number.

Proof

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

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

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

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

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

Also, it can be represented as −

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

Important

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

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

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

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

Numerical Example 1

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

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

Solution

Given the sequence is,

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

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

$$\mathrm{Z[u(n)] \:=\: \frac{z}{z\:-\:1};\: \text{ ROC } \:\rightarrow\: |z| \:\gt\: 1}$$

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

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

$$\mathrm{\therefore\:Z[(2)^n u(n)] \:=\: \frac{z}{z\:-\:2}\:;\:\: \text{ ROC } \:\rightarrow \:|z| \:\gt\: 2}$$

Numerical Example 2

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

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

Solution

Given the sequence is,

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

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

$$\mathrm{Z[u(n)]\:=\:\frac{z}{z\:-\:1}\:;\: \text{ ROC } \:\rightarrow \:|z| \:\gt\: 1}$$

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

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

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