Correlation Property of Z-Transform

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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 x(n) is a discrete time function, then its Z-transform is defined as,

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

Correlation Property of Z-Transform

Statement

The correlation property of Z-transform states that if,

$$\mathrm{x_1(n) \:\overset{ZT}\longleftrightarrow\: X_1(z)\: \text{ and }\: x_2(n) \:\overset{ZT}\longleftrightarrow \:X_2(z) }$$

Then

$$\mathrm{x_1(n)\:\otimes\: x_2(n)\: \:\overset{ZT}\longleftrightarrow\: X_1(z) X_2(z^{-1})}$$

Where

$$\mathrm{R_{12}(n) \:=\: x_1(n) \:\otimes\: x_2(n)}$$

Proof

From the definition of Z-transform, we have,

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

$$\mathrm{\therefore\:Z[x_1(n)\:\otimes\:x_2(n)]\:=\:\sum_{n=-\infty}^{\infty}\:[x_1(n)\:\otimes\:x_2(n)]z^{-n}\:\:\dotso\:(1)}$$

The correlation of two signals is defined as,

$$\mathrm{x_1(n)\:\otimes\:x_2(n)\:=\:\sum_{k=-\infty}^{\infty}\:x_1(k)x_2(k\:-\:n)\:=\:\sum_{k=-\infty}^{\infty}\:x_1(k\:-\:n)x_2(k)\:\:\dotso\:(2)}$$

Therefore, from eqns.(1)&(2), we get,

$$\mathrm{Z[x_1(n)\:\otimes\:x_2(n)]\:=\: \sum_{n=-\infty}^{\infty}\: \left[ \sum_{k=-\infty}^{\infty}\:x_1(k) x_2(k\:-\:n) \right] z^{-n}}$$

Rearranging the order of summations, we get,

$$\mathrm{Z[x_1(n)\:\otimes\:x_2(n)]\:=\:\sum_{k=-\infty}^{\infty}\:x_1(k)\left[\sum_{n=-\infty}^{\infty}\:x_2(k\:-\:n)\:z^{-n}\right]}$$

Now, putting (k − n) = m and n = (k − m) in the second summation, we get,

$$\mathrm{Z[x_1(n)\:\otimes\:x_2(n)] \:=\: \sum_{k=-\infty}^{\infty}\: x_1(k) \left[ \sum_{m=-\infty}^{\infty}\: x_2(m) z^{-(k-m)} \right]}$$

$$\mathrm{\Rightarrow\: Z[x_1(n) \:\otimes\: x_2(n)] \:=\: \sum_{k=-\infty}^{\infty}\: x_1(k) \left[ \sum_{m=-\infty}^{\infty}\: x_2(m) z^{-k} z^m \right]}$$

$$\mathrm{\Rightarrow\: Z[x_1(n)\:\otimes\:x_2(n)] \:=\: \left[ \sum_{k=-\infty}^{\infty} \:x_1(k) z^{-k} \right] \left[ \sum_{m=-\infty}^{\infty}\: x_2(m) z^{(-1)-m} \right]}$$

$$\mathrm{\therefore\: Z[x_1(n)\:\otimes\: x_2(n)] \:=\: X_1(z) X_2(z^{-1})}$$

Also, it can be represented as,

$$\mathrm{x_1(n)\:\otimes\:x_2(n)\:\overset{ZT}\longleftrightarrow\: X_1(z) X_2(z^{-1})}$$

Numerical Example

Using the correlation property of Z-transform, find the Z-transform of $\mathrm{[x_1(n)\:\otimes\:x_2(n)]}$ ,Where,

$$\mathrm{x_1(n)\:=\:\sin(\omega)nu(n)\:\text{ and }\:x_2(n)\:=\:u(n)}$$

Solution

The given sequences are,

$$\mathrm{x_1(n)\:=\:\sin(\omega)nu(n)\:\text{ and }\:x_2(n)\:=\:u(n)}$$

The Z-transform of these two sequences are −

$$\mathrm{Z[x_1(n)\:=\:X_1(z)\:=\:Z[sin(\omega)\:nu(n)]\:=\:\frac{z\:sin\omega}{z^2\:-\:2z\:\cos\:\omega\:+\:1}]}$$

And

$$\mathrm{Z[x_2(n)]\:=\:X_2(z)\:=\:Z[u(n)]\:=\:\frac{z}{z\:-\:1}}$$

Now, using the correlation property of Z-transform $\mathrm{\left[\text{i.e., }\: x_1(n)\:\otimes\:x_2(n)\:\overset{ZT} \longleftrightarrow\:X_1(z) X_2(z^{-1})\right]}$ ,We get,

$$\mathrm{Z[x_1(n)\:\otimes\:x_2(n)]\:=\:X_1(z) X_2(z^{-1})\:=\:\left[\frac{z\:\sin\:\omega}{z^2\:-\:2z\:\cos\: \omega \:+\: 1}\right]\:\cdot\:\left[\frac{z}{z^{-1}}\right]_{z = z^{-1}}}$$

$$\mathrm{\Rightarrow\:Z[x_1(n)\:\otimes\:x_2(n)]\:=\:\left[ \frac{z\:\sin \omega}{z^2\:-\:2z\:\cos \omega \:+\: 1} \right] \left[ \frac{z^{-1}}{z^{-1} \:-\: 1} \right]}$$

$$\mathrm{\therefore\:Z[x_1(n)\:\otimes\:x_2(n)]\:=\:\frac{z\: \sin \omega}{(z^2 \:-\: 2z\: \cos \omega \:+\: 1)(1 \:-\: z)}}$$