---
title: "Z-transform of a running sum"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-of-sum"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-of-sum
site: BCT Engineering
---

# Z-transform of a running sum

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2082 Kartik, 5 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/z-transform-of-sum
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Consider the function y(k), which is a sum of functions x(h), where h=0,1,2,…,k, such that y(k) = Σ x(h) (sum from h = 0 to k), k=0,1,2,…, where y(k)=0 for k<0. Obtain Y(z).

## Short answer

The running sum satisfies y(k)-y(k-1)=x(k) with y(k)=0 for k<0. Taking the z-transform gives Y(z)=X(z)/(1-z^{-1}) = zX(z)/(z-1).

## Answer

$y(k)$ is the **running sum (accumulation)** of $x(k)$. Its z-transform is $Y(z) = \dfrac{X(z)}{1-z^{-1}} = \dfrac{z}{z-1}X(z)$.

#### Step 1: Write a difference equation

$$
\begin{aligned}
y(k) &= x(0) + x(1) + \dots + x(k-1) + x(k) \\
y(k-1) &= x(0) + x(1) + \dots + x(k-1)
\end{aligned}
$$

Subtracting:

$$
y(k) - y(k-1) = x(k), \qquad k = 0, 1, 2, \dots
$$

This also holds at $k = 0$ because $y(-1) = 0$ (given $y(k) = 0$ for $k < 0$), so $y(0) = x(0)$.

#### Step 2: Take the z-transform

Use the right-shift (delay) theorem $\mathcal{Z}[y(k-1)] = z^{-1}Y(z)$, which needs $y(k) = 0$ for $k < 0$ (given):

$$
Y(z) - z^{-1}Y(z) = X(z)
$$

$$
Y(z) = \frac{1}{1-z^{-1}}X(z) = \frac{z}{z-1}X(z)
$$

#### Alternative view (convolution)

$y(k) = \sum_{h=0}^{k} x(h)\cdot 1(k-h)$ is the convolution of $x(k)$ with the unit step $1(k)$. Convolution in time is multiplication in $z$:

$$
Y(z) = X(z)\cdot \mathcal{Z}[1(k)] = X(z)\cdot\frac{1}{1-z^{-1}}
$$

#### Example

Let $x(k) = 1$ for all $k \ge 0$ (unit step), so $X(z) = \dfrac{z}{z-1}$. Then $y(k) = k+1$ and

$$
Y(z) = \frac{z}{z-1}\cdot\frac{z}{z-1} = \frac{z^2}{(z-1)^2}
$$

Check: $\mathcal{Z}[k+1] = \dfrac{z}{(z-1)^2} + \dfrac{z}{z-1} = \dfrac{z + z(z-1)}{(z-1)^2} = \dfrac{z^2}{(z-1)^2}$, which agrees.

#### Remarks

- $\dfrac{1}{1-z^{-1}}$ is the pulse transfer function of a **digital accumulator** (summer). It has a pole at $z = 1$.
- This result is the discrete version of integration ($1/s$) and is used in the integral term of a digital PID controller.

**Answer:** $Y(z) = \dfrac{X(z)}{1 - z^{-1}} = \dfrac{z\,X(z)}{z-1}$.

## Check yourself

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

1. What difference equation does y(k) = sum x(h) satisfy?
   A. y(k) - y(k-1) = x(k)
   B. y(k) = x(k) - x(k-1)
   C. y(k+1) = x(k)
   D. y(k) + y(k-1) = x(k)
   Answer: A. Subtracting y(k-1) from y(k) leaves the last term x(k). (Key point: Difference equation)

2. Why can Z[y(k-1)] be written as z^-1 Y(z)?
   A. Because y(k) = 0 for k < 0
   B. Because x(k) is a step
   C. Because z = 1
   D. Because the sum is infinite
   Answer: A. The right-shift theorem has no extra terms when y vanishes for negative k. (Key point: Why zero initial)

3. What is Y(z)?
   A. X(z)/(1 + z^-1)
   B. (1 - z^-1) X(z)
   C. zX(z)
   D. X(z)/(1 - z^-1) = zX(z)/(z-1)
   Answer: D. Y - z^-1 Y = X gives Y = X/(1 - z^-1). (Key point: Result)

4. The running sum is the convolution of x(k) with what?
   A. Unit ramp
   B. Exponential 0.5^k
   C. Unit impulse
   D. Unit step 1(k)
   Answer: D. Convolution in time multiplies by Z[1(k)] = 1/(1 - z^-1). (Key point: Convolution view)

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