---
title: "Z-transform of a step delayed by 4"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-delayed-step"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-delayed-step
site: BCT Engineering
---

# Z-transform of a step delayed by 4

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2076 Bhadra, 4 marks
- Video: [Z-Transform of Unit Step Function](https://www.youtube.com/watch?v=b7kM3ZcBiUs) — 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/z-transform-delayed-step
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find the z-transform of the following: x(t)={0, t<0; 0, t≤ 4; 1, t>4}

## Short answer

With T=1 s, x(k)=0 for k≤4, x(k)=1 for k≥5. X(z)=Σ_{k=5}∞ z^{-k}=z^{-5}/(1–z^{-1})=1/[z^4(z–1)].

## Answer

$x(t)$ is a unit step that switches on just after $t = 4$. No sampling period is given, so take $T = 1$ s (the usual choice for such problems).

#### Step 1: Sampled sequence

$x(kT) = 1$ only when $kT > 4$. At $t = 4$ exactly the value is 0 (since $x = 0$ for $t \le 4$).

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | ... |
|---|---|---|---|---|---|---|---|---|
| $x(k)$ | 0 | 0 | 0 | 0 | 0 | 1 | 1 | ... |

So $x(k) = 1(k-5)$, a unit step delayed by 5 samples.

#### Step 2: Z-transform from the definition

$$
\begin{aligned}
X(z) &= \sum_{k=0}^{\infty}x(k)z^{-k} = z^{-5} + z^{-6} + z^{-7} + \dots \\
&= z^{-5}\left(1 + z^{-1} + z^{-2} + \dots\right) = \frac{z^{-5}}{1 - z^{-1}}, \quad |z| > 1
\end{aligned}
$$

The same result follows from the shift theorem: $\mathcal{Z}[1(k-5)] = z^{-5}\mathcal{Z}[1(k)]$.

$$
X(z) = \frac{z^{-5}}{1-z^{-1}} = \frac{1}{z^4(z-1)}
$$

#### Remarks

- If the step were defined as 1 for $t \ge 4$, the sample at $k = 4$ would be 1, and $X(z) = \dfrac{z^{-4}}{1-z^{-1}}$. The strict inequality $t > 4$ is what moves the first nonzero sample to $k = 5$.
- For a general $T$ with $4 = nT$ ($n$ an integer), the first nonzero sample is $k = n+1$, so $X(z) = \dfrac{z^{-(n+1)}}{1-z^{-1}}$.

**Answer ($T = 1$ s):** $X(z) = \dfrac{z^{-5}}{1 - z^{-1}} = \dfrac{1}{z^4(z-1)}$.

## Check yourself

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

1. x = 1 only for t > 4 with T = 1. First nonzero sample?
   A. k = 3
   B. k = 0
   C. k = 5
   D. k = 4
   Answer: C. At t = 4 the value is 0, so k = 5 is first. (Key point: First nonzero sample)

2. x(k) equals?
   A. 1(k-4)
   B. delta(k-5)
   C. k 1(k-5)
   D. 1(k-5)
   Answer: D. A unit step delayed by 5 samples. (Key point: Representation)

3. X(z)?
   A. z^-5/(1-z^-1) = 1/(z^4(z-1))
   B. z^-4/(1-z^-1)
   C. z^-5
   D. 1/(z^5(z-1))
   Answer: A. Geometric series from z^-5: z^-5/(1 - z^-1) = 1/(z^4(z-1)). (Key point: Result)

4. If the step were 1 for t >= 4, X(z) would be?
   A. unchanged
   B. z^-4/(1-z^-1)
   C. z^-5/(1-z^-1)
   D. z^-3/(1-z^-1)
   Answer: B. Then the sample at k = 4 is 1. (Key point: Effect of t >= 4)

## Related in Digital Control System

- [Z-transform of ramp minus delayed ramp](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Az-transform-ramp-minus-delayed-ramp) — Digital Control System, chapter 2
- [Z-transform of k² e^(−ak) from unit step](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Az-transform-k2-e-ak) — Digital Control System, chapter 2
- [Difference equation with step input, x(1)=0.3679](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adifference-equation-step-0-3679) — Digital Control System, chapter 2, asked 2×

## Related in other subjects

- [Fourier transform of unit step](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-unit-step) — Signals and Systems, chapter 3, asked 3×
- [Hilbert transform properties; energy and power of unit step](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ahilbert-properties-unit-step-energy) — Communication Systems, chapter 2