---
title: "Methods of inverse z-transform"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:methods-inverse-z-transform"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/methods-inverse-z-transform
site: BCT Engineering
---

# Methods of inverse z-transform

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2071 Bhadra, 6 marks
- Video: [Direct Computation Method to Calculate Inverse Z-Transform](https://www.youtube.com/watch?v=XRWaKZoJVZY) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/methods-inverse-z-transform
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

What are the methods of inverse z-transform? Explain each of them using suitable example.

## Short answer

Methods: direct division (coefficients of z^{−k}), computational (recursive difference equation), partial-fraction expansion of X(z)/z, and inversion integral (sum of residues). E.g., X(z)=z/((z−1)(z−0.5)) gives x(k)=2−2(0.5)^k.

## Answer

The **inverse z-transform** recovers the sequence $x(k)$ (the sample values $x(kT)$) from $X(z)$. It gives only the values at the sampling instants, not the continuous signal between them. There are four common methods.

#### 1. Direct division method

Write $X(z)$ as a ratio of polynomials in $z^{-1}$ and divide the numerator by the denominator. The coefficient of $z^{-k}$ is $x(k)$. This is simple and good for the first few values, but it gives no closed form.

**Example:** $X(z) = \dfrac{10z + 5}{(z-1)(z-0.2)} = \dfrac{10z^{-1} + 5z^{-2}}{1 - 1.2z^{-1} + 0.2z^{-2}}$

$$
X(z) = 10z^{-1} + 17z^{-2} + 18.4z^{-3} + 18.68z^{-4} + \dots
$$

So $x(0) = 0$, $x(1) = 10$, $x(2) = 17$, $x(3) = 18.4$.

#### 2. Computational method

Treat $X(z)$ as a transfer function excited by a unit pulse. Either write its **difference equation** and solve it recursively (or with MATLAB `filter`), or simulate it.

**Example (same $X(z)$):** $x(k) = 1.2x(k-1) - 0.2x(k-2) + 10\delta(k-1) + 5\delta(k-2)$. This gives $x(1) = 10$, $x(2) = 1.2(10) + 5 = 17$, $x(3) = 1.2(17) - 0.2(10) = 18.4$.

#### 3. Partial-fraction expansion method

Expand $X(z)/z$ into partial fractions, multiply back by $z$, and use the standard table ($\frac{z}{z-a} \leftrightarrow a^k$). This gives a closed form.

**Example:**

$$
\frac{X(z)}{z} = \frac{10z + 5}{z(z-1)(z-0.2)} = \frac{25}{z} + \frac{18.75}{z-1} - \frac{43.75}{z-0.2}
$$

$$
x(k) = 25\delta(k) + 18.75 - 43.75(0.2)^k
$$

Check: $x(0) = 25 + 18.75 - 43.75 = 0$, and $x(1) = 18.75 - 8.75 = 10$.

#### 4. Inversion integral method

$$
x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum\text{residues of } X(z)z^{k-1}
$$

The contour $C$ encloses all the poles. This works for repeated poles too, and needs no table.

**Example:** $X(z) = \dfrac{z}{(z-1)(z-0.5)}$

$$
x(k) = \left[\frac{z^k}{z-0.5}\right]_{z=1} + \left[\frac{z^k}{z-1}\right]_{z=0.5} = 2 - 2(0.5)^k
$$

#### Comparison

| Method | Gives | Best for |
|---|---|---|
| Direct division | A few numerical values | Quick checks |
| Computational | Values by recursion or computer | Long sequences, simulation |
| Partial fractions | Closed form | Simple, distinct poles |
| Inversion integral | Closed form | Repeated or awkward poles |

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The inverse z-transform gives:
   A. The sample values x(kT) only, not the signal between samples
   B. The continuous signal x(t)
   C. The Laplace transform
   D. Only the final value
   Answer: A. It recovers the sequence at the sampling instants. (Key point: What the inverse gives)

2. In the direct division method, x(k) is:
   A. The coefficient of z^{-k} in the quotient
   B. The coefficient of z^{k}
   C. The residue at z = 0
   D. The remainder
   Answer: A. Dividing polynomials in z^{-1} gives a series whose z^{-k} coefficient is x(k); it gives no closed form. (Key point: Direct division)

3. In the partial-fraction method, which function is expanded?
   A. X(z)/z
   B. X(z)
   C. X(z) z^{k-1}
   D. zX(z)
   Answer: A. Expanding X(z)/z and multiplying back by z yields terms z/(z-a) found in the table. (Key point: Partial fractions)

4. Which method is best for repeated or awkward poles and needs no table?
   A. Inversion integral
   B. Direct division
   C. Computational method
   D. Partial fractions with distinct poles
   Answer: A. Residues handle repeated poles directly. (Key point: Best use)

## Related in Digital Control System

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- [Inverse Z-transform of z(z+2)/(z−1)²](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-z-plus-2) — Digital Control System, chapter 2, asked 10×
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- [Inverse Z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-pfe-2z3-z) — Digital Control System, chapter 2, asked 5×

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