---
title: "Convolution theorem for z-transform"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:convolution-theorem-z-transform"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/convolution-theorem-z-transform
site: BCT Engineering
---

# Convolution theorem for z-transform

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2069 Bhadra
- Video: [To Prove Convolution Property of Z Transform](https://www.youtube.com/watch?v=nOx0djpDoS4) — Canara Engineering College Mangalore 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/convolution-theorem-z-transform
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

State and prove convolution theorem for z-transform.

## Short answer

For causal x(k), y(k) with transforms X(z), Y(z), the convolution (x*y)(k)=sum_{m=0}^{k} x(m)y(k-m) satisfies Z[x*y] = X(z)Y(z). Proof interchanges sums using causal support.

## Answer

#### Statement

If x(k) and y(k) are causal sequences with Z[x(k)] = X(z) and Z[y(k)] = Y(z), then the z-transform of their convolution is the product of the transforms:

Z[x(k) * y(k)] = X(z) · Y(z)

where the convolution is (x * y)(k) = Σ_{m=0}^{k} x(m) y(k − m).

Equivalently, Z⁻¹[X(z)Y(z)] = x(k) * y(k).

#### Proof

By definition,

Z[x * y] = Σ_{k=0}^{∞} (x * y)(k) z⁻ᵏ = Σ_{k=0}^{∞} Σ_{m=0}^{k} x(m) y(k − m) z⁻ᵏ

Since x and y are causal (y(n) = 0 for n < 0), the inner sum can be taken over all m from 0 to ∞. Interchange the order of summation (valid inside the common region of convergence):

```text
Z[x * y] = Σ_{m=0}^{∞} x(m) Σ_{k=m}^{∞} y(k − m) z⁻ᵏ
```

Put n = k − m, so k = n + m and z⁻ᵏ = z⁻ᵐ z⁻ⁿ:

```text
Z[x * y] = Σ_{m=0}^{∞} x(m) z⁻ᵐ  ·  Σ_{n=0}^{∞} y(n) z⁻ⁿ
         = X(z) · Y(z)
```

The region of convergence contains the intersection of the ROCs of X(z) and Y(z). ∎

#### Use

It turns the convolution of two sequences into a simple product, so inverse transforms of products can be found. Example: Z⁻¹[z/(z−1) · z/(z−2)] = 1ᵏ * 2ᵏ = Σ_{m=0}^{k} 2^{k−m} = 2^{k+1} − 1.

## Check yourself

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

1. The convolution theorem states
   A. Z[x * y] = X(z) Y(z)
   B. Z[x * y] = X(z) + Y(z)
   C. Z[x * y] = X(z)/Y(z)
   D. Z[x * y] = X(z) * Y(z) (convolution in z)
   Answer: A. Convolution becomes multiplication. (Key point: Statement)

2. (x * y)(k) is
   A. sum x(m) y(m)
   B. sum for m = 0 to k of x(m) y(k - m)
   C. x(k) y(k)
   D. sum for m = 0 to k of x(m) y(k + m)
   Answer: B. Causal convolution sum. (Key point: Definition)

3. What is done in the proof after the double sum is written?
   A. Differentiate with respect to z
   B. Take limits z -> infinity
   C. Interchange order of summation, then put n = k - m
   D. Use the initial value theorem
   Answer: C. Then z^-k = z^-m z^-n splits. (Key point: Proof step)

4. Inverse of z/(z-1) * z/(z-2) is
   A. 1 + 2^k
   B. 2^k - 1
   C. 2^(k+1) + 1
   D. 1^k * 2^k = 2^(k+1) - 1
   Answer: D. Sum of 2^(k-m) for m = 0..k. (Key point: Use)

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