Partial Fraction Expansion Method for Inverse Z-Transform

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Inverse Z-Transform

The inverse Z-transform is defined as the process of finding the time domain signal x(n) from its Z-transform X(z). The inverse Z-transform is denoted as −

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

Partial Fraction Expansion Method to Find Inverse Z-Transform

In order to determine the inverse Z-transform of X(z) using partial fraction expansion method, the denominator of X(z) must be in factored form. In this method, we obtained the partial fraction expansion of $\mathrm{\frac{X(z)}{z}}$ instead of X(z). This is because the Z-transform of time-domain sequences have Z in their numerators.

The partial fraction expansion method is applied only if $\mathrm{\frac{X(z)}{z}}$ is a proper rational function, i.e., the order of its denominator is greater than the order of its numerator.

If $\mathrm{\frac{X(z)}{z}}$ is not a proper function, then it should be written in the form of a polynomial and a proper function before applying the partial fraction method.

The disadvantage of the partial fraction method is that, the denominator of X(z) must be in factored form. Once the $\mathrm{\frac{X(z)}{z}}$ is obtained as a proper function, then using the standard Z-transform pairs and the properties of Z-transform, the inverse Z-transform of each partial fraction can be obtained.

Let a rational function $\mathrm{\frac{X(z)}{z}}$ given as −

$$\mathrm{\frac{X(z)}{z} \:=\: \frac{N(z)}{D(z)} \:=\: \frac{b_0 z^m \:+\: b_1 z^{m-1} \:+\: b_2 z^{m-2} \:+\: \dotso \:+\: b_m z^n}{a_1 z^{n-1} \:+\: a_2 z^{n-2} \:+\: \dotso \:+\: a_n}}$$

When the order of numerator is less than the order of denominator, i.e.,m < n, then $\mathrm{\frac{X(z)}{z}}$ is a proper function. If m &geq; n , then $\mathrm{\frac{X(z)}{z}}$ is not a proper function, then it is to be written as −

$$\mathrm{\frac{X(z)}{z} \:=\: c_0 z^{n-m} \:+\: c_1 z^{n-m-1} \:+\: \dotso \:+\: c_{n-m} \:+\: \frac{N_1(z)}{D(z)}}$$

Where, $\mathrm{\left[c_0 z^{n-m} \:+\: c_1 z^{n-m-1} \:+\: \dotso \:+\: c_{n-m}\right]}$ is a polynomial and $\mathrm{\frac{N_1(z)}{D(z)}}$ is the proper rational function.

Now, there are two cases for the proper rational function $\mathrm{\frac{X(z)}{z}}$ as follows −

Case I - When $\mathrm{\frac{X(z)}{z}}$ has All Distinct Poles

When all the poles of $\mathrm{\frac{X(z)}{z}}$ are distinct, then the function $\mathrm{\frac{X(z)}{z}}$ can be expanded in the form given below −

$$\mathrm{\frac{X(z)}{z} \:=\: \frac{C_1}{z \:-\: K_1} \:+\: \frac{C_2}{z \:-\: K_2} \:+\: \frac{C_3}{z \:-\: K_3} \:+\: \dotso \:+ \:\frac{C_n}{z\:-\:K_n}}$$

Here, the coefficients $\mathrm{C_1,\: C_2,\: C_3,\: \dotso,\: C_n}$ can be determined by using the equation given below −

$$\mathrm{C_i \:=\: \left[(z \:-\: K_i)\frac{X(z)}{z} \right]_{z \:=\: K_i} \quad \text{where,} \quad i \:=\: 1,\: 2,\: 3,\: \dots}$$

Case II - When $\mathrm{\frac{X(z)}{z}}$ has l-repeated Poles and Remaining (n-l) Poles Are Simple

Consider pth pole is repeated l times. Then, the function $\mathrm{\frac{X(z)}{z}}$ can be expressed as,

$$\mathrm{\frac{X(z)}{z} \:=\: \frac{C_1}{z \:-\: K_1} \: +\: \frac{C_2}{z \:-\:K_2} \:+\: \dotso \:+\: \frac{C_{p1}}{z \:-\: K_p} \:+\: \frac{C_{p2}}{(z \:-\: K_p)^2} \:+\: \dotso \:+\: \frac{C_{pl}}{(z \:-\: K_p)^l}}$$

Where,

$$\mathrm{C_{pl} \:=\: \left[(z \:-\: K_p)\frac{l X(z)}{z} \right]_{z \:=\: K_p}}$$

Also, if the Z-transform X(z) has a complex pole, then the partial fraction can be expressed as −

$$\mathrm{\frac{X(z)}{z} \:=\: \frac{C_1}{z \:-\: K_1} \:+\: \frac{C_1^*}{z \:-\:K_1^*} }$$

Where, $\mathrm{C_1^*}$ is the complex conjugate of C1 and $\mathrm{K_1^*}$ is the complex conjugate of K1. Therefore, it is clear that the complex poles result in complex conjugate coefficients in the partial fraction expansion.

Numerical Example

Find the inverse Z-transform of

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

Solution

Given Z-transform is,

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

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

$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{1}{2} \left\{\frac{z}{(z\:-\: 1)\left[z \:-\: \left(\frac{1}{2}\right) \right]} \right\}}$$

By taking partial fraction, we get,

$$\mathrm{\Rightarrow\: \frac{X(z)}{z} \:=\: \frac{A}{(z - 1)} \:+\: \frac{B}{\left[z \:-\: \left(\frac{1}{2}\right)\right]}}$$

Where, A and B are determined as follows −

$$\mathrm{A \:=\: \left[(z \:-\: 1) \frac{X(z)}{z} \right]_{z\:=\:1}}$$

$$\mathrm{=\: (z \:-\: 1) \left[ \frac{1}{2} \frac{z}{(z \:-\: 1) \left[ z \:-\: \left(\frac{1}{2}\right) \right]} \right]_{z\:=\:1}}$$

$$\mathrm{=\: \frac{1}{2} \left[ \frac{1}{1 \:-\: \left(\frac{1}{2}\right) }\right] \:=\: 1}$$

Similarly,

$$\mathrm{B \:=\: \left[ \left(z \:-\: \frac{1}{2}\right) \frac{X(z)}{z} \right]_{z \:=\: \frac{1}{2}}}$$

$$\mathrm{=\: \left(z \:-\: \frac{1}{2}\right) \left[ \frac{1}{2}\frac{z}{(z \:-\: 1) \left[ z \:-\: \left( \frac{1}{2} \right) \right] }\right]_{z \:=\: \frac{1}{2}}}$$

$$\mathrm{=\: \frac{1}{2} \left[ \frac{1}{\left( \frac{1}{2} \right) \:-\: 1} \right] \:=\: -1}$$

$$\mathrm{\therefore\: \frac{X(z)}{z} \:=\: \frac{1}{(z \:-\: 1)} \:-\: \frac{1}{\left[z \:-\: \left(\frac{1}{2}\right) \right]}}$$

$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{z}{(z \:-\: 1)} \:-\: \frac{z}{\left[z \:-\: \left(\frac{1}{2}\right)\right]} \quad \text{ROC}\: \rightarrow\: |z|\: \gt\: 1}$$

Because the region of convergence (ROC) of the given Z-transform is |z| > 1, thus both the sequences must be casual. Hence, by taking the inverse Z-transform, we get,

$$\mathrm{Z^{-1}[X(z)] \:=\: Z^{-1} \left[ \frac{z}{(z \:-\: 1)} \:-\: \frac{z}{\left[z \:-\: (\frac{1}{2})\right]} \right]}$$

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

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