Signals and Systems – Partial Fraction Expansion Method for Inverse Z-Transform

Signals and Systems Electronics & Electrical Digital Electronics

Inverse Z-Transform

The inverse Z-transform is defined as the process of finding the time domain signal $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ from its Z-transform $\mathit{X}\mathrm{\left(\mathit{z}\right)}$. The inverse Z-transform is denoted as −

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

Partial Fraction Expansion Method to Find Inverse Z-Transform

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

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

If $\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}$ 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 $\mathit{X}\mathrm{\left(\mathit{z}\right)}$ must be in factored form. Once the $\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}$ 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 $\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}$ given as −

$$\mathrm{\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}\:\mathrm{=}\:\frac{\mathit{N}\mathrm{\left(\mathit{z} \right)}}{\mathit{D}\mathrm{\left( \mathit{z}\right)}}\:\mathrm{=}\: \frac{\mathit{b}_{\mathrm{0}}\mathit{z}^{\mathit{m}}\mathrm{+}\mathit{b}_{\mathrm{1}}\mathit{z}^{\mathit{m}-\mathrm{1}}\mathrm{+}\mathit{b}_{\mathrm{2}}\mathit{z}^{\mathit{m}-\mathrm{2}}\mathrm{+}...\mathit{b}_{\mathit{m}}}{\mathit{z}^{\mathit{n}}\mathrm{+}\mathit{a_\mathrm{\mathrm{1}}z}^{\mathit{n}-\mathrm{1}}\mathrm{+}\mathit{a_\mathrm{\mathrm{2}}z}^{\mathit{n}-\mathrm{2}}\mathrm{+}...\mathit{a}_{\mathit{n}}}}$$

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

$$\mathrm{\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}\:\mathrm{=}\:\mathit{c}_{\mathrm{0}}\mathit{z}^{\mathit{n-m}}\:\mathrm{+}\:\mathit{c}_{\mathrm{1}}\mathit{z}^{\mathit{n-m}-\mathrm{1}}\:\mathrm{+}\:...\mathrm{+}\:\mathit{c}_{\mathit{n-m}}\:\mathrm{+}\:\frac{\mathit{N}_{\mathrm{1}}\mathrm{\left(\mathit{z} \right)}}{\mathit{D}\mathrm{\left( \mathit{z}\right)}}}$$

Where, $\mathrm{\left[\mathit{c}_{\mathrm{0}}\mathit{z}^{\mathit{n-m}}\:\mathrm{+}\:\mathit{c}_{\mathrm{1}}\mathit{z}^{\mathit{n-m}-\mathrm{1}}\:\mathrm{+}\:...\mathrm{+}\:\mathit{c}_{\mathit{n-m}}\right]}$ is a polynomial and $\frac{N_{\mathrm{1}}\mathrm{\left(\mathit{z} \right)}}{\mathit{D}\mathrm{\left( \mathit{z}\right)}}$ is the proper rational function.

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

Case I - When $\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}$ has all distinct poles −

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

Here, the coefficients $\mathit{C}_{\mathrm{1}},\mathit{C}_{\mathrm{2}},\mathit{C}_{\mathrm{3}},...,\mathit{C}_{\mathit{n}}$ can be determined by using the equation given below −

$$\mathrm{\mathit{C}_{\mathit{i}}\:\mathrm{=}\:\mathrm{\left[ \mathrm{\left ( \mathit{z-K_{\mathit{i}}}\right)}\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}\right]}_{\mathit{z=K_{\mathit{i}}}}\:;\mathrm{Where},\mathit{i}\:\mathrm{=}\:\mathrm{1,2,3..}}$$

Case II - When $\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}$ has l-repeated poles and the remaining (n-l) poles are simple −

Consider $\mathit{p}^{\mathit{th}}$ pole is repeated l times. Then, the function $\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}$ can be expressed as,

$$\mathrm{\frac{\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}\:\mathrm{=}\:\frac{\mathit{C}_{\mathrm{1}}}{\mathit{z-K}_{\mathrm{1}}}\:\mathrm{+}\:\frac{\mathit{C}_{\mathrm{2}}}{\mathit{z-K}_{\mathrm{2}}}\:\mathrm{+}\:...\:\mathrm{+}\:\frac{\mathit{C}_{\mathit{p\mathrm{1}}}}{\mathit{z-K_{\mathit{p}}}}\:\mathrm{+}\:\frac{\mathit{C_{\mathit{p\mathrm{2}}}}}{\mathrm{\left( \mathit{z-\mathit{k_{p}}}\right )^{\mathrm{2}}}}\:\mathrm{+}\:...\:\mathrm{+}\:\frac{\mathit{C_{\mathit{p\mathit{l}}}}}{\mathrm{\left( \mathit{z-\mathit{k_{p}}}\right )^{\mathit{l}}}}}$$

Where,

$$\mathrm{\mathit{C}_{\mathit{pl}}\:\mathrm{=}\:\mathrm{\left[ \mathrm{\left ( \mathit{z-K_{\mathit{p}}}\right)}\frac{\mathit{l} \:\mathit{X}\mathrm{\left(\mathit{z}\right)}}{\mathit{z}}\right]}_{\mathit{z=K_{\mathit{p}}}}}$$

Also, if the Z-transform $\mathit{X}\mathrm{\left(\mathit{z}\right)}$ has a complex pole, then the partial fraction can be expressed as −

Where, $\mathit{C}_{\mathrm{1}}^{*}$ is the complex conjugate of $\mathit{C}_{\mathrm{1}}$ and $\mathit{K}_{\mathrm{1}}^{*}$ is the complex conjugate of $\mathit{K}_{\mathrm{1}}$. 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

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

Solution

Given Z-transform is,

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

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

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

By taking partial fraction, we get,

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

Where, A and B are determined as follows −

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

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

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

Similarly,

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

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

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

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

$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\:\frac{\mathit{z}}{\mathrm{\left ( \mathit{z-\mathrm{1}}\right)}}-\frac{\mathit{z}}{\mathrm{\left [ \mathit{z}-\mathrm{\left(\frac{1}{2}\right )}\right]}};\:\mathrm{ROC}\to \left|\mathit{z}\right|>\:\mathrm{1}}$$

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

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

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

Manish Kumar Saini

Updated on: 11-Jan-2022

8K+ Views

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