---
title: "Prove shifting theorem of z-transform"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:shifting-theorem-prove"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/shifting-theorem-prove
site: BCT Engineering
---

# Prove shifting theorem of z-transform

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2074 Bhadra, 1+4 marks; 2070 Magh, 5 marks; 2068 Bhadra (asked 3 times)
- Video: [Time Shifting Property of Z-Transform](https://www.youtube.com/watch?v=MOvn9DNZeAg) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/shifting-theorem-prove
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

If x(t) = 0 for t < 0 and Z[x(t)] = X(z) for t ≥ 0, then prove that Z[x(t + nT)] = zⁿ [X(z) - Σ_{k=0}^{n-1} x(kT) z^{-k}].

## Short answer

By definition, Z[x(t+nT)] = sum_{k=0}^∞ x((k+n)T) z^{-k}. Substituting m=k+n gives z^n sum_{m=n}^∞ x(mT) z^{-m} = z^n [X(z) - sum_{k=0}^{n-1} x(kT) z^{-k}], assuming x(t)=0 for t<0.

## Answer

#### Proof (shift to the left / advance)
By the definition of the z-transform, with x(t) = 0 for t < 0,

Z[x(t + nT)] = Σ (k = 0 to ∞) x((k + n)T) z⁻ᵏ

Put m = k + n, so k = m − n and m runs from n to ∞:

Z[x(t + nT)] = Σ (m = n to ∞) x(mT) z^−(m − n) = zⁿ Σ (m = n to ∞) x(mT) z⁻ᵐ

The sum from n to ∞ equals the full sum minus the first n terms:

Σ (m = n to ∞) x(mT) z⁻ᵐ = Σ (m = 0 to ∞) x(mT) z⁻ᵐ − Σ (m = 0 to n − 1) x(mT) z⁻ᵐ = X(z) − Σ (m = 0 to n − 1) x(mT) z⁻ᵐ

Therefore

**Z[x(t + nT)] = zⁿ [ X(z) − Σ (k = 0 to n − 1) x(kT) z⁻ᵏ ]**

#### Special cases
- n = 1: Z[x(t + T)] = z X(z) − z x(0)
- n = 2: Z[x(t + 2T)] = z² X(z) − z² x(0) − z x(T)

These are the forms used to solve difference equations.

## Check yourself

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

1. How does the proof start?
   A. Σ x(kT) z^{k+n}
   B. Σ x(kT+n) z⁻ᵏ
   C. Σ x(kT) z^{-(k+n)}
   D. Z[x(t+nT)] = Σ_{k≥0} x((k+n)T) z⁻ᵏ
   Answer: D. Definition applied to the shifted signal. (Key point: Definition)

2. What substitution is used?
   A. m = k+n, giving zⁿ Σ_{m=n}^{∞} x(mT) z⁻ᵐ
   B. m = k-n, giving z^{-n}
   C. m = kn
   D. m = k, no change
   Answer: A. z^-(m-n) = zⁿ z⁻ᵐ. (Key point: Re-indexing)

3. Why is the sum from 0 to n-1 subtracted?
   A. Because x(t)=0 for t<0
   B. Σ from n to ∞ equals full sum minus the first n terms
   C. Because it tends to 0
   D. To make the series converge
   Answer: B. Gives X(z) - Σ_{k<n} x(kT) z⁻ᵏ. (Key point: Subtraction)

4. What does n=2 give?
   A. z²X(z) - z x(0) - x(T)
   B. z²X(z) - z²x(T)
   C. z²X(z) - z²x(0) - z x(T)
   D. z²X(z)
   Answer: C. z²[X - x(0) - x(T)/z]. (Key point: n = 2)

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