---
title: "X(z) via convolution integral, repeated pole"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:convolution-integral-three-poles"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/convolution-integral-three-poles
site: BCT Engineering
---

# X(z) via convolution integral, repeated pole

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2067 Mangsir, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/convolution-integral-three-poles
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Given X(s)=1/((s+1)²(s+3)(s+2)) and T=1. Obtain X(z) by using convolution integral in left half plane.

## Short answer

Sum residues of X(s)z/(z-e^{sT}) at poles s=-1 (double), s=-2, s=-3. For T=1: X(z)=0.1839z/(z-0.3679)^2 - 0.75z/(z-0.3679) + z/(z-0.1353) - 0.25z/(z-0.0498).

## Answer

Use the **convolution integral closed in the left half-plane**. The z-transform is the sum of residues of $X(s)\dfrac{z}{z-e^{sT}}$ at the poles of $X(s)$. This is valid because $X(s)$ has 4 poles and no zeros.

$$
X(z) = \sum_{\text{poles of } X(s)}\text{Res}\left[\frac{1}{(s+1)^2(s+2)(s+3)}\cdot\frac{z}{z - e^{sT}}\right]
$$

Poles: a double pole at $s = -1$, and simple poles at $s = -2$ and $s = -3$.

#### Residue at $s = -2$

$$
R_2 = \left[\frac{1}{(s+1)^2(s+3)}\cdot\frac{z}{z-e^{sT}}\right]_{s=-2} = \frac{1}{(1)(1)}\cdot\frac{z}{z - e^{-2T}} = \frac{z}{z - e^{-2T}}
$$

#### Residue at $s = -3$

$$
R_3 = \left[\frac{1}{(s+1)^2(s+2)}\cdot\frac{z}{z-e^{sT}}\right]_{s=-3} = \frac{1}{(4)(-1)}\cdot\frac{z}{z - e^{-3T}} = -\frac{1}{4}\cdot\frac{z}{z - e^{-3T}}
$$

#### Residue at the double pole $s = -1$

Let $F(s) = \dfrac{1}{(s+2)(s+3)}$, so $F(-1) = \dfrac12$ and

$$
F'(s) = -\frac{2s+5}{(s+2)^2(s+3)^2} \;\Rightarrow\; F'(-1) = -\frac{3}{4}
$$

$$
\begin{aligned}
R_1 &= \frac{d}{ds}\left[F(s)\frac{z}{z-e^{sT}}\right]_{s=-1} = F'(-1)\frac{z}{z-e^{-T}} + F(-1)\frac{Tze^{-T}}{(z-e^{-T})^2} \\
&= -\frac{3}{4}\cdot\frac{z}{z-e^{-T}} + \frac{T e^{-T}}{2}\cdot\frac{z}{(z-e^{-T})^2}
\end{aligned}
$$

#### General result

$$
X(z) = \frac{Te^{-T}}{2}\cdot\frac{z}{(z-e^{-T})^2} - \frac{3}{4}\cdot\frac{z}{z-e^{-T}} + \frac{z}{z-e^{-2T}} - \frac{1}{4}\cdot\frac{z}{z-e^{-3T}}
$$

#### Substitute $T = 1$

$e^{-1} = 0.36788$, $e^{-2} = 0.13534$, $e^{-3} = 0.04979$, and $\frac{Te^{-T}}{2} = 0.18394$:

$$
X(z) = \frac{0.18394z}{(z-0.36788)^2} - \frac{0.75z}{z-0.36788} + \frac{z}{z-0.13534} - \frac{0.25z}{z-0.04979}
$$

Combined into a single ratio:

$$
X(z) = \frac{0.030919z^3 + 0.023057z^2 + 0.000939z}{z^4 - 0.92088z^3 + 0.27828z^2 - 0.030011z + 0.000912}
$$

The denominator is $(z-0.36788)^2(z-0.13534)(z-0.04979)$.

#### Check

The inverse Laplace transform is $x(t) = \tfrac14\left[(2t-3)e^{-t} + 4e^{-2t} - e^{-3t}\right]$. Its samples are:

| $k$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $x(k)$ | 0 | 0.03092 | 0.05153 | 0.03979 |

Long division of $X(z)$ gives $0.03092z^{-1} + 0.05153z^{-2} + 0.03979z^{-3} + \dots$, which matches.

**Answer:** $X(z) = \dfrac{0.18394z}{(z-0.3679)^2} - \dfrac{0.75z}{z-0.3679} + \dfrac{z}{z-0.1353} - \dfrac{0.25z}{z-0.0498}$.

## Check yourself

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

1. What are the poles of 1/((s+1)^2(s+2)(s+3))?
   A. Double pole at -1, simple poles at -2 and -3
   B. Simple poles at -1, -2, -3
   C. Double pole at -2
   D. Double poles at -1 and -2
   Answer: A. (s+1) is squared, so -1 is a double pole. (Key point: Pole structure)

2. The residue at s = -2 is:
   A. z/(z - e^{-2T})
   B. -z/(z - e^{-2T})
   C. -(1/4) z/(z - e^{-2T})
   D. 2z/(z - e^{-2T})
   Answer: A. 1/((s+1)^2(s+3)) at s=-2 is 1/(1 x 1) = 1. (Key point: Residue at s = -2)

3. The coefficient multiplying z/(z - e^{-3T}) is:
   A. -1/4
   B. 1/4
   C. -1
   D. 1/2
   Answer: A. 1/((s+1)^2(s+2)) at s=-3 is 1/((4)(-1)) = -1/4. (Key point: Residue at s = -3)

4. The residue at the double pole s = -1 requires:
   A. Differentiating F(s) z/(z-e^{sT}), with F=1/((s+2)(s+3)), at s = -1
   B. Evaluating without differentiation
   C. Only F(-1)
   D. Integrating F(s)
   Answer: A. A double pole needs one derivative, giving a term with z/(z-e^{-T})^2 as well. (Key point: Double pole handling)

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