# Residue Method to Calculate Inverse 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 $\mathrm{\mathit{x\left ( n \right )}}$ is a discrete time function, then its Z-transform is defined as,

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$

## Inverse Z-Transform using Residue Method

The residue method is also known as complex inversion integral method. As the Z-transform of a discrete-time signal $\mathrm{\mathit{x\left ( n \right )}}$ is defined as

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$

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

$$\mathrm{\mathit{z}\:\mathrm{=}\:\mathit{re^{j\:\omega }}}$$

$$\mathrm{\therefore \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\: \mathit{X}\mathrm{\left(\mathit{re^{j\:\omega }}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathrm{\left (\mathit{re^{j\:\omega}} \right )}^{\mathit{-n}}}$$

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

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

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

$$\mathrm{\Rightarrow\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{1}{2\pi }\int_{-\pi }^{\pi}\mathit{X}\mathrm{\left ( \mathit{re^{j\:\omega }}\right )}\mathrm{\left ( \mathit{re^{j\:\omega }} \right )}^{\mathit{n}}\:\mathit{d\omega }\:\:\:\:\:\:...(2)}$$

$$\mathrm{\because \mathit{z}\:\mathrm{=}\:\mathit{re^{j\:\omega }}}$$

$$\mathrm{\therefore \mathit{dz}\:\mathrm{=}\:\mathit{jre^{j\:\omega }}\:\mathit{d\omega }}$$

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

Substituting value of z and $\mathit{d\omega }$ in the equation (2), we get,

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

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

$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\oint_{\mathit{c}}^{}\mathit{X}\mathrm{\left (\mathit{z}\right )}\mathit{z^{\mathit{n-\mathrm{1}}}}\mathit{dz}\:\:\:\:\:\:...(3)}$$

Where, c is the circle in the z-plane in the ROC of X(z). The symbol $\mathrm{\left [ \oint_{\mathit{c}}^{} \right ]}$ represents the integration around the circle of radius $\left| \mathit{z}\right|\:\mathrm{=}\:\mathit{r}$ in a counter clockwise direction.

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

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

$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\: \mathrm{\left [ \sum_{\mathit{i}}^{}\mathrm{\left ( \mathit{z-z_{\mathit{i}}} \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}z^{\mathit{n-\mathrm{1}}} \right ]}_{\mathit{z=z_{\mathit{i}}}}}$$

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

## Numerical Example

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

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{1\:+2\mathit{z^{-\mathrm{1}}}}{1+4\mathit{z^{-\mathrm{1}}\mathrm{\, +\, }\mathrm{3}\mathit{z^{-\mathrm{2}}}}};\:\mathrm{ROC}\to\left|\mathit{z} \right|>3}$$

Solution

The given Z-transform is,

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

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

Using the residue method, we have,

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

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

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

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

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

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