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- Z Transform
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Convolution 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 x(n) is a discrete time function, then its Z-transform is defined as,
$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:\sum_{n=-\infty}^{\infty}\:x(n)z^{-n}}$$
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,
$$\mathrm{x_1(n)\:\overset{ZT}\longleftrightarrow\:X_1(z);\:\:\text{ROC }\:=\:R_1}$$
$$\mathrm{x_2(n)\:\overset{ZT}\longleftrightarrow\:X_2(z);\:\:\text{ROC }\:=\:R_2}$$
Then, according to the convolution property,
$$\mathrm{x_1(n) \:\cdot\: x_2(n)\:\overset{ZT}\longleftrightarrow\: X_1(z) X_2(z); \quad \text{ROC } \:=\: R_1 \:\cap\: R_2}$$
Proof
The convolution of two sequences is defined as,
$$\mathrm{x_1(n)\:\cdot\:x_2(n) \:=\: \sum_{k=-\infty}^{\infty}\: x_1(k) x_2(n\:-\:k)}$$
Now, from the definition of Z-transform, we have,
$$\mathrm{Z[x(n)] \:=\: \sum_{n=-\infty}^{\infty}\: x(n) z^{-n}}$$
$$\mathrm{\therefore\: Z[x_1(n)\:\cdot\:x_2(n)] \:=\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: \left[x_1(n)\:\cdot\:x_2(n) \right]z^{-n}}$$
$$\mathrm{\Rightarrow\: X(z) \:=\: \sum_{n=-\infty}^{\infty} \left[ \sum_{k=-\infty}^{\infty}\: x_1(k) x_2(n\:-\:k) \right] z^{-n}}$$
$$\mathrm{\Rightarrow\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: \sum_{k=-\infty}^{\infty}\: x_1(k) x_2(n\:-\:k) z^{-k} z^{-(n-k)}}$$
Rearranging the order of summations, we get,
$$\mathrm{\Rightarrow\: X(z) \:=\: \sum_{k=-\infty}^{\infty}\: x_1(k) z^{-k}\: \sum_{n=-\infty}^{\infty}\: x_2(n\:-\:k) z^{-(n-k)}}$$
Substituting (n â k) = m in the second summation, we have,
$$\mathrm{\Rightarrow\: X(z) \:=\: \sum_{k=-\infty}^{\infty}\: x_1(k) z^{-k}\: \sum_{m=-\infty}^{\infty}\: x_2(m) z^{-m} \:=\: X_1(z) X_2(z)}$$
$$\mathrm{\therefore\:Z[x_1(n)\:\cdot\:x_2(n)]\:=\:X_1(z)X_2(z)}$$
Or it can also be represented as
$$\mathrm{x_1(n)\cdot\:x_2(n)\:\overset{ZT}\longleftrightarrow\:X_1(z)X_2(z);\:\:\text{ROC }\:=\:R_1\:\cap\:R_2}$$
Numerical Example
Using the convolution property of Z-transform, find the Z-transform of the following signal.
$$\mathrm{x(n)\:=\:\left(\frac{1}{3}\right)^{n}\:u(n)\:\cdot\:\left(\frac{1}{5}\right)^{n}\:u(n)}$$
Solution
Given signal is,
$$\mathrm{x(n)\:=\:\left(\frac{1}{3}\right)^{n}\:u(n)\:\cdot\:\left(\frac{1}{5}\right)^{n}u(n)}$$
Let
$$\mathrm{x(n)\:=\:x_1(n)\:\cdot\:x_2(n)}$$
$$\mathrm{\therefore\:x_1(n)\:=\:\left(\frac{1}{3}\right)^{n}\:u(n)}$$
Taking Z-transform, we get,
$$\mathrm{Z[x_1(n)]\:=\:X_1(z)\:=\:Z\left[\left(\frac{1}{3}\right)^{n}\:u(n)\right]}$$
$$\mathrm{X_1(z)\:=\:\frac{z}{\left(z\:-\:\frac{1}{3}\right)}\:;\:\:\:\text{ROC }\:\rightarrow\:|z|\:\gt\:\frac{1}{3}}$$
Similarly,
$$\mathrm{Z[x_2(n)]\:=\:X_2(z)\:=\:Z\left[\left(\frac{1}{5}\right)^{n}\:u(n)\right]}$$
$$\mathrm{X_2(z)\:=\:\frac{z}{\left(z\:-\:\frac{1}{5}\right)}\:;\:\:\:\text{ROC }\:\rightarrow\:|z|\:\gt\:\frac{1}{5}}$$
Now, using the convolution property of Z-transform $\mathrm{\left[\text{i.e., }\:x_1(n)\:\cdot\:x_2(n)\:\overset{ZT}\longleftrightarrow\:X_1(z)X_2(z) \right]}$ ,we get,
$$\mathrm{Z[x(n)]\:=\:X_1(z)X_2(z)}$$
$$\mathrm{\therefore\: Z\left[ \left( \frac{1}{3} \right)^n u(n)\:\cdot\:\left( \frac{1}{5} \right)^n u(n) \right] \:=\: \frac{z}{\left(z \:-\: \frac{1}{3}\right)}\frac{z}{\left(z \:-\: \frac{1}{5}\right)}}$$
The ROC of the Z-transform of the given sequence is
$$\mathrm{\text{ROC } \:\rightarrow\: \left[ |z| \:\gt\: \frac{1}{3} \right] \:\cap\: \left[ |z| \:\gt\: \frac{1}{5} \right] \:=\: |z| \:\gt\: \frac{1}{3}}$$
$$\mathrm{\therefore\: \left(\frac{1}{3} \right)^n u(n) \:\cdot\: \left(\frac{1}{5} \right)^n u(n) \overset{ZT}\longleftrightarrow\: \frac{z^2}{\left(z \:-\: \frac{1}{3}\right)\left(z \:-\: \frac{1}{5}\right)}; \quad \text{ROC } \:\rightarrow\: |z| \:\gt\: \frac{1}{3}}$$