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Correlation 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,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
Correlation Property of Z-Transform
Statement - The correlation property of Z-transform states that if,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\:\mathrm{and}\:\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}}$$
Then
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}^{-\mathrm{1}}\right)}}$$
Where
$$\mathrm{\mathit{R}_{\mathrm{12}}\mathrm{\left ( \mathit{n} \right )}\:\mathrm{=}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}}$$
Proof
From the definition of Z-transform, we have,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
$$\mathrm{\mathit{\therefore \mathit{Z}\mathrm{\left[ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\right ]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathrm{\left[ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\right ]}\mathit{z}^{-n}}\:\:\:\:\:\:...(1)}$$
The correlation of two signals is defined as,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{k-n}\right)}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k-n}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{k}\right)}\:\:\:\:\:\:...(2)}$$
Therefore, from eqns.(1)&(2), we get,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}}\mathrm{\left[\sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{k-n}\right)} \right ]}\mathit{z}^{-n}$$
Rearranging the order of summations, we get,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathrm{\left[\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{k-n}\right)}\mathit{z}^{-n} \right ]}}$$
Now, putting $\mathrm{\left ( \mathit{k-n} \right )}\:\mathrm{=}\:\mathit{m}\:\mathrm{and}\:\mathit{n}\:\mathrm{=}\:\mathrm{\left ( \mathit{k-m} \right )}$ in the second summation, we get,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathrm{\left[\sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{m}\right)}\mathit{z}^\mathrm{-\left(\mathit{k-m}\right)} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\sum_{\mathit{k=-\infty}}^{\infty}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathrm{\left[\sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{m}\right)}\mathit{z}^\mathrm{\mathit{-k}}\mathit{z}^\mathrm{\mathit{m}} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathrm{\left[ \sum_{\mathit{k=-\infty}}^{\infty}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{z}^{-k}\right ]}\mathrm{\left [ \sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{m}\right)}\mathit{z}^\mathrm{\left(\mathrm{-1}\right)-\mathit{m}} \right ]}}$$
$$\mathrm{\therefore \mathit{Z}\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}^{\mathrm{-1}}\right)}}$$
Also, it can be represented as,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}^{-\mathrm{1}}\right)}}$$
Numerical Example
Using the correlation property of Z-transform, find the Z-transform of $\mathrm{\left[\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}$ ,Where,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{sin}\:\mathit{\omega n}\mathit{u}\mathrm{\left(\mathit{n}\right)}\:\mathrm{and}\:\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$
Solution
The given sequences are,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{sin}\:\mathit{\omega n}\mathit{u}\mathrm{\left(\mathit{n}\right)}\:\mathrm{and}\:\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$
The Z-transform of these two sequences are −
$$\mathrm{\mathit{Z}\mathrm{\left [ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{Z}\mathrm{\left [ \mathrm{sin}\:\mathit{\omega n}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\frac{\mathit{z}\mathrm{sin}\:\mathit{\omega}}{\mathit{z}^{\mathrm{2}}-2\mathit{z}\mathrm{cos}\:\mathit{\omega}\mathrm{+1}}}$$
And
$$\mathrm{\mathit{Z}\mathrm{\left [ \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{Z}\mathrm{\left [ \mathit{u}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\frac{\mathit{z}}{\mathit{z-\mathrm{1}}}}$$
Now, using the correlation property of Z-transform $\mathrm{\left [ i.e,\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}^{-\mathrm{1}}\right)} \right ]}$ ,We get,
$$\mathrm{\mathit{Z}\mathrm{\left [ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}^{-\mathrm{1}}\right)}\:\mathrm{=}\:\mathrm{\left[\frac{\mathit{z}\mathrm{sin}\:\mathit{\omega}}{\mathit{z}^{\mathrm{2}}-2\mathit{z}\mathrm{cos}\:\mathit{\omega}\mathrm{+1}}\right ]}\mathrm{\left [ \frac{\mathit{z}}{\mathit{z-\mathrm{1}}} \right ]}_{\mathit{z=z^{-\mathrm{1}}}}}$$
$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left [ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\mathrm{\left[\frac{\mathit{z}\mathrm{sin}\:\mathit{\omega}}{\mathit{z}^{\mathrm{2}}-2\mathit{z}\mathrm{cos}\:\mathit{\omega}\mathrm{+1}}\right ]}\mathrm{\left [ \frac{\mathit{z^{-\mathrm{1}}}}{\mathit{z^{-\mathrm{1}}-\mathrm{1}}} \right ]}}$$
$$\mathrm{\therefore \mathit{Z}\mathrm{\left [ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}\otimes \mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\frac{\mathit{z}\mathrm{sin}\:\mathit{\omega}}{\mathrm{\left ( \mathit{z}^{\mathrm{2}}-2\mathit{z}\mathrm{cos}\:\mathit{\omega}\mathrm{+1} \right )}\mathrm{\left ( 1-\mathit{z} \right )}}}$$
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