---
title: "Convolution and Z-transform via convolution integral"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:convolution-meaning-z-transform"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/convolution-meaning-z-transform
site: BCT Engineering
---

# Convolution and Z-transform via convolution integral

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2068 Magh, 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-meaning-z-transform
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

What do you mean by convolution? Obtain Z-transform of X(s)=1/s²(s+1) by use of the convolution integral in the left half s-plane.

## Short answer

Convolution combines an input with a system's impulse response to yield the output; in the s or z domain it becomes multiplication. For X(s)=1/s^2(s+1), the convolution integral method gives X(z)=Tz/(z-1)^2 - z/(z-1) + z/(z-e^{-T}).

## Answer

#### Convolution

**Convolution** combines two signals to give the response of a linear time-invariant system. If $g$ is the impulse response and $x$ is the input, the output is the convolution $y = g * x$.

- Continuous time: $y(t) = \displaystyle\int_0^t g(t-\tau)\,x(\tau)\,d\tau$
- Discrete time (convolution sum): $y(k) = \displaystyle\sum_{h=0}^{k} g(k-h)\,x(h)$

The key property is that **convolution in time becomes multiplication in the transform domain**: $Y(s) = G(s)X(s)$ and $Y(z) = G(z)X(z)$. This is why pulse transfer functions are useful.

#### Convolution integral for the z-transform

The impulse-sampled signal is $x^*(t) = x(t)\sum_k\delta(t-kT)$, which is a product in time. Its Laplace transform is therefore a **complex convolution** of $X(s)$ with the transform of the impulse train, $\dfrac{1}{1 - e^{-Ts}}$:

$$
X^*(s) = \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}X(p)\,\frac{1}{1 - e^{-T(s-p)}}\,dp
$$

The contour is closed in the **left half of the p-plane**, which encloses the poles of $X(p)$. This is valid when $X(s)$ has at least two more poles than zeros. Setting $e^{Ts} = z$ gives:

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

#### Apply to $X(s) = \dfrac{1}{s^2(s+1)}$

Poles: a double pole at $s = 0$ and a simple pole at $s = -1$.

**Simple pole $s = -1$:**

$$
R_1 = \left[\frac{1}{s^2}\cdot\frac{z}{z - e^{sT}}\right]_{s=-1} = \frac{z}{z - e^{-T}}
$$

**Double pole $s = 0$:**

$$
\begin{aligned}
R_2 &= \frac{d}{ds}\left[\frac{1}{s+1}\cdot\frac{z}{z - e^{sT}}\right]_{s=0} \\
&= \left[-\frac{1}{(s+1)^2}\cdot\frac{z}{z-e^{sT}} + \frac{1}{s+1}\cdot\frac{zTe^{sT}}{(z-e^{sT})^2}\right]_{s=0} \\
&= -\frac{z}{z-1} + \frac{Tz}{(z-1)^2}
\end{aligned}
$$

#### Sum of residues

$$
X(z) = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - e^{-T}}
$$

Over the common denominator:

$$
X(z) = \frac{z\left[(T - 1 + e^{-T})z + (1 - e^{-T} - Te^{-T})\right]}{(z-1)^2(z - e^{-T})}
$$

#### Check

$x(t) = t - 1 + e^{-t}$ (from partial fractions $\frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$). Its z-transform is $\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$, which is the same.

For $T = 1$ s: $X(z) = \dfrac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}$.

**Answer:** $X(z) = \dfrac{z[(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})]}{(z-1)^2(z-e^{-T})}$.

## Check yourself

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

1. What does convolution of impulse response g and input x give?
   A. The output of an LTI system, y = g * x
   B. The product g(t)x(t)
   C. The derivative of x
   D. The inverse of g
   Answer: A. Convolution combines the two to give the output of a linear time-invariant system. (Key point: Meaning of convolution)

2. Convolution in time corresponds to what in the transform domain?
   A. Multiplication, Y = GX
   B. Addition
   C. Differentiation
   D. Convolution again
   Answer: A. Y(s)=G(s)X(s) and Y(z)=G(z)X(z), which makes pulse transfer functions useful. (Key point: Convolution-multiplication property)

3. Why is X*(s) a complex convolution?
   A. The sampled signal is the product of x(t) and an impulse train, and products in time become convolutions in s
   B. Sampling adds noise
   C. The impulse train is a ramp
   D. Because x(t) is periodic
   Answer: A. x*(t) = x(t) times the impulse train, so its Laplace transform is a convolution in the p-plane. (Key point: Why a complex convolution appears)

4. For X(s)=1/(s^2(s+1)), the residue at the double pole s = 0 is:
   A. Tz/(z-1)^2 - z/(z-1)
   B. z/(z-e^{-T})
   C. Tz/(z-1)^2 + z/(z-1)
   D. z/(z-1)^2
   Answer: A. Differentiating [1/(s+1)] z/(z-e^{sT}) at s=0 gives -z/(z-1) + Tz/(z-1)^2. (Key point: Residue at the double pole)

## Related in Digital Control System

- [X(z) via convolution integral, repeated pole](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aconvolution-integral-three-poles) — Digital Control System, chapter 2
- [X(z) from convolution integral: 1/(s²(s+1))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aconvolution-integral-1-s2-s1) — Digital Control System, chapter 2
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