Residue Method to Calculate Inverse Z-Transform

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Z-Transform

The Z-transform is a mathematical tool 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:

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

Inverse Z-Transform using Residue Method

The residue method is also known as the complex inversion integral method. As the Z-transform of a discrete-time signal (n) is defined as:

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

Where, z is a complex variable, and if r is the radius of a circle, then it is given by:

$$\mathrm{z \:=\: r e^{j\omega}}$$

$$\mathrm{\therefore\:X(z) \:=\: X(r e^{j\omega}) \:=\:\sum_{n=-\infty}^{\infty}\: x(n) (r e^{j\omega})^{-n}}$$

$$\mathrm{\Rightarrow\:X(r e^{j\omega}) \:=\: \sum_{n=-\infty}^{\infty}\: \left[ x(n) r^{-n} \right]\: e^{-j\omega n}\:\:\dotso\: (1)}$$

Equation (1) is the Fourier transform of the signal $\mathrm{\left[ x(n)\: r^{-n} \right]}$. Therefore, the inverse discrete-time Fourier transform (DTFT) of the function $\mathrm{X(r e^{j\omega})}$ must be $\mathrm{[x(n) r^{-n}]}$.

$$\mathrm{x(n) r^{-n} \:=\: \frac{1}{2\pi}\: \int_{-\pi}^{\pi} \:X(r e^{j\omega})\: e^{j\omega n}\: d\omega}$$

$$\mathrm{\Rightarrow\:x(n) \:=\: \frac{1}{2\pi}\: \int_{-\pi}^{\pi}\: X(r e^{j\omega}) \:(r e^{j\omega})^{n}\: d\omega\:\:\dotso\:(2)}$$

$$\mathrm{\because\:z \:= \:r e^{j\omega}}$$

$$\mathrm{\therefore\:dz \:=\: j r\: e^{j\omega}\: d\omega}$$

$$\mathrm{\Rightarrow \: d\omega \:=\: \frac{dz}{jre^{j\omega}}\:=\: \frac{dz}{j z}}$$

Substituting the values of z and dω in equation (2), we obtain:

$$\mathrm{x(n) \:=\: \frac{1}{2\pi}\: \int_{-\pi}^{\pi}\: X(z) z^{n}\: \frac{dz}{j z}}$$

$$\mathrm{\Rightarrow\:x(n) \:=\: \frac{1}{2\pi j}\: \int_{-\pi}^{\pi}\: X(z) z^{n-1}\: dz}$$

$$\mathrm{\therefore\:x(n) \:=\: \frac{1}{2\pi j}\: \oint_c \:X(z) \:z^{n-1}\: dz\:\:\dotso\:(3)}$$

Where, c is the contour (circle) in the z-plane in the Region of Convergence (ROC) of X(z). The symbol $\mathrm{\left[\oint_c\right]}$ represents the contour integral around the circle of radius $\mathrm{|z| \:=\: r}$, in a counterclockwise direction.

Equation (3) can be evaluated by determining the sum of all residues of the poles that are inside the circle c, i.e.,

$$\mathrm{x(n) \:=\: \sum\left[\text{Residues of }\: X(z) z^{n-1}\: \text{ at the poles inside the circle } c\right]}$$

$$\mathrm{x(n) \:=\: \left[\sum_i (z \:-\: z_i) X(z) z^{n-1} \right]_{z \:=\: z_i}}$$

If $\mathrm{[X(z) z^{n-1}]}$ has no poles inside the contour circle c for one or more values of n, then x(n) = 0 for these values.

Numerical Example

Find the inverse Z-transform of X(z) using the residue method:

$$\mathrm{X(z) \:=\: \frac{1 \:+\: 2z^{-1}}{1 \:+\: 4z^{-1} \:+\:3z^{-2}};\:\text{ ROC }\:\rightarrow\:|z|\:\gt\:3}$$

Solution

The given Z-transform is:

$$\mathrm{X(z) \:=\: \frac{1 \:+\: 2z^{-1}}{1 \:+\: 4z^{-1} \:+\:3z^{-2}}}$$

$$\mathrm{\therefore\:X(z) \:=\: \frac{z(z \:+\: 2)}{z^2 \:+\: 4z \:+\: 3} \:=\: \frac{z(z \:+\: 2)}{(z \:+\: 1)(z \:+\: 3)}}$$

Using the residue method, we have:

$$\mathrm{x(n) \:=\: \sum\: \text{Residues of } X(z)z^{n-1} \:\text{ at the pole inside the circle }c}$$

$$\mathrm{\therefore\:x(n) \:=\: \sum\: \text{Residues of } \left[ \frac{z(z \:+\: 2)z^{n-1}}{(z \:+\: 1)(z \:+\: 3)}\right]\: \text{ at the poles inside the circle } c}$$

$$\mathrm{\Rightarrow\:x(n) \:=\: \sum\:\text{Residue at } \frac{z^n(z\:+\:2)}{(z\:+\:1)(z\:+\:3)}\: \text{ at the pole inside the circle } c}$$

$$\mathrm{\therefore\:x(n)\:=\:\sum\:\text{Residues of } \frac{z^{n}(z + 2) }{(z \:+\: 1)(z \:+\: 2)} \:\text{ at the poles } \:z\:=\:-1\:\text{ and } \:z\:=\:-3}$$

$$\mathrm{\Rightarrow\:x(n) \:=\: \left[(z\:+\:1)\frac{z^{n}(z\:+\:2)}{(z \:+\: 1)(z \:+\: 3)}\right]_{z \:=\: -1} \:+\: \left[(z\:+\:3)\frac{z^{n}(z\:+\:2)}{(z\:+\:1)(z\:+\:3)}\right]_{z\:=\:-3}}$$

$$\mathrm{\therefore\:x(n) \:=\: \frac{1}{2} (-1)^n\: u(n) \:+\: \frac{1}{2} (-3)^n\: u(n)}$$