Convolution Property of Z-Transform

Signals and Systems Electronics & Electrical Digital Electronics

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,

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

Convolution in Time Domain Property of Z-Transform

Statement - The convolution in time domain property of Z-transform states that the Z-transform of the convolution of two discrete time sequences is equal to the multiplication of their Z-transforms. Therefore, if,

$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)};\:\:\mathrm{ROC}\:\mathrm{=}\:\mathit{R}_{1}}$$

$$\mathrm{\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)};\:\:\mathrm{ROC}\:\mathrm{=}\:\mathit{R}_{2}}$$

Then, according to the convolution property,

$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)};\:\:\mathrm{ROC}\:\mathrm{=}\:\mathit{R}_{\mathrm{1}}\cap\mathit{R}_{\mathrm{2}} }$$

Proof

The convolution of two sequences is defined as,

$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n-k}\right)}}$$

Now, from the definition of Z-transform, we have,

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

$$\mathrm{\therefore \mathit{Z}\mathrm{\left [\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathrm{\left [ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\mathit{z}^{-n}}$$

$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathrm{\left[\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n-k}\right)} \right ]}\mathit{z}^{-n}}$$

$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n-k}\right)}\mathit{z}^{-k}\mathit{z}^{-\mathrm{\left ( \mathit{n-k} \right )}}}$$

Rearranging the order of summations, we get,

$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{z}^{-k}\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n-k}\right)}\mathit{z}^{-\mathrm{\left(\mathit{n-k}\right)}}}$$

Substituting $\mathrm{\left(\mathit{n-k}\right)}\:\mathrm{=}\:\mathit{m}$ in the second summation, we have,

$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{z}^{-\mathit{k}}\sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{m}\right)}\mathit{z}^{-\mathrm{ \mathit{m}}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}}}$$

Or it can also be represented as

Numerical Example

Using the convolution property of Z-transform, find the Z-transform of the following signal.

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{\left(\frac{1}{3} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}*\mathrm{\left(\frac{1}{5} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$

Solution

Given signal is,

Let

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}}$$

$$\mathrm{\therefore \mathit{x_{\mathrm{1}}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{\left(\frac{1}{3} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$

Taking Z-transform, we get,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)} \:\mathrm{=}\:\mathit{Z}\mathrm{\left[ \mathrm{\left(\frac{1}{3} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}\right ]}}$$

$$\mathrm{\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)} \:\mathrm{=}\:\frac{\mathit{z}}{\mathrm{\left( \mathit{z-\frac{\mathrm{1}}{\mathrm{3}}}\right )}};\:\mathrm{ROC} \rightarrow \left|\mathit{z} \right|>\frac{1}{3}}$$

Similarly,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)} \:\mathrm{=}\:\mathit{Z}\mathrm{\left[ \mathrm{\left(\frac{1}{5} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}\right ]}}$$

$$\mathrm{\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)} \:\mathrm{=}\:\frac{\mathit{z}}{\mathrm{\left( \mathit{z-\frac{\mathrm{1}}{\mathrm{5}}}\right )}};\:\mathrm{ROC} \rightarrow \left|\mathit{z} \right|>\frac{1}{5}}$$

Now, using the convolution property of Z-transform $\mathrm{\left [ i.e.,\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)} \right ]}$ ,we get,

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)} \mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}}$$

$$\mathrm{\therefore \mathit{Z}\mathrm{\left[\mathrm{\left(\frac{1}{3} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}*\mathrm{\left(\frac{1}{5} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\frac{\mathit{z}}{\mathrm{\left( \mathit{z-\frac{\mathrm{1}}{\mathrm{3}}}\right )}}\frac{\mathit{z}}{\mathrm{\left( \mathit{z-\frac{\mathrm{1}}{\mathrm{5}}}\right )}}}$$

The ROC of the Z-transform of the given sequence is

$$\mathrm{\mathrm{ROC}\rightarrow \mathrm{\left [ \left|\mathit{z} \right|>\frac{1}{3} \right]}\cap \mathrm{\left [ \left|\mathit{z} \right|>\frac{1}{5} \right]}\:\mathrm{=}\:\left|\mathit{z} \right|>\frac{1}{3}}$$

$$\mathrm{\therefore \mathrm{\left(\frac{1}{3} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}*\mathrm{\left(\frac{1}{5} \right)}^{\mathit{n}}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}\:\overset{\mathit{ZT}}{\leftrightarrow}\:\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left ( \mathit{z-\frac{\mathrm{1}}{\mathrm{3}}} \right )}\mathrm{\left ( \mathit{z-\frac{\mathrm{1}}{\mathrm{5}}} \right )}};\:\mathrm{ROC}\to\:\left|\mathit{z} \right|>\frac{1}{3}}$$

Manish Kumar Saini

Updated on: 24-Jan-2022

8K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started