Inverse Z-Transform by Convolution Method

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Z-Transform

The Z-transform is a mathematical tool which is 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:

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

Convolution Method to Find Inverse Z-Transform

The inverse Z-transform can be calculated using the convolution theorem. In the convolution integration method, the given Z-transform X(z) is first split into $\mathrm{X_1(z)}$ and $\mathrm{X_2(z)}$ such that:

$$\mathrm{X(z) \:=\: X_1(z) X_2(z)}$$

The signals $\mathrm{x_1(n)}$ and $\mathrm{x_2(n)}$ are then obtained by taking the inverse Z-transform of $\mathrm{X_1(z)}$ and $\mathrm{X_2(z)}$ respectively. Finally, the function x(n) is obtained by performing the convolution of $\mathrm{x_1(n)}$ and $\mathrm{x_2(n)}$ in the time domain.

As per the definition of the Z-transform of the convolution of two signals, we have,

$$\mathrm{Z[x_1(n) \:\cdot\: x_2(n)] \:=\: X_1(z) X_2(z) \:=\: X(z)}$$

Therefore, the inverse Z-transform is obtained as:

$$\mathrm{x(n) \:=\: Z^{-1}[X(z)] \:=\: Z^{-1}[Z\{ x_1(n)\:\cdot\: x_2(n) \}]}$$

$$\mathrm{\therefore\:Z^{-1}[X(z)] \:=\: x(n) \:=\: x_1(n) \:\cdot\: x_2(n) \:=\: \sum_{k=0}^{n} \:x_1(k) x_2(n\:-\:k)}$$

Numerical Example

Using the convolution method, find the inverse Z-transform of,

$$\mathrm{X(z) \:=\: \frac{z^2}{(z\:-\:3)(z\:-\:4)}}$$

Solution

The given Z-transform function is,

$$\mathrm{X(z) \:=\: \frac{z^2}{(z\:-\:3)(z\:-\:4)}}$$

Let,

$$\mathrm{X(z) \:=\: X_1(z) X_2(z) \:=\: \frac{z}{(z\:-\:3)} \cdot \frac{z}{(z\:-\:4)}}$$

Taking the inverse Z-transform of $\mathrm{X_1(z)}$ and $\mathrm{X_2(z)}$ respectively as -

$$\mathrm{Z^{-1}[X_1(z)] \:=\: x_1(n) \:=\: Z^{-1}\left[\frac{z}{(z\:-\:3)}\right] \:=\: 3^n u(n)}$$

Similarly,

$$\mathrm{Z^{-1}[X_2(z)] \:=\: x_2(n) \:=\: Z^{-1}\left[\frac{z}{(z\:-\:4)}\right] \:=\: 4^n u(n)}$$

Now, using the convolution method for finding the inverse Z-transform, we have,

$$\mathrm{Z^{-1}[X(z)] \:=\: x(n) \:=\: x_1(n)\:\cdot\: x_2(n) \:=\: \sum_{k=0}^{n}\: x_1(k)x_2(n\:-\:k)}$$

$$\mathrm{\therefore\:x(n) \:=\: \sum_{k=0}^{n}\:3^{k}u(k)4^{n\:-\:k}u(n\:-\:k)\:=\:\sum_{k=0}^{n}\: 3^{k}u(k)\left(\frac{4^n}{4^k}\right)u(n\:-\:k)}$$

$$\mathrm{\Rightarrow\:x(n)\:=\:4^k\:\sum_{k=0}^{n}\:\left(\frac{3^k}{4^k}\right) \:=\: 4^n\:\sum_{k=0}^{n}\:\left(\frac{3}{4}\right)^k}$$

$$\mathrm{\Rightarrow\:x(n) \:=\: 4^n \left[ \frac{1 \:-\: \left(\frac{3}{4}\right)^{n+1}}{1\:-\:\left(\frac{3}{4}\right)} \right] \:=\: 4^{n+1} \left[ 1 \:-\: \left(\frac{3}{4}\right)^{n+1} \right]}$$

$$\mathrm{\therefore\:x(n) \:=\: 4^{n+1} u(n) \:-\: 3^{n+1} u(n)}$$