---
title: "Derive the convolution sum"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-sum-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-sum-derivation
site: BCT Engineering
---

# Derive the convolution sum

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2082 Kartik, 5 marks; 2081 Asoj, 6 marks; 2080 Asoj, 3 marks; 2078 Baisakh, 4 marks; 2080 Chaitra, 6 marks; 2076 Baisakh, 4 marks; 2075 Baisakh, 5 marks; 2083 Baisakh (new course), 4 marks (asked 8 times)
- Video: [Tutorial: Convolution sum](https://www.youtube.com/watch?v=U3BwStzKvs0) — Rose-Hulman Online
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-sum-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Derive the convolution sum for a discrete-time LTI system.

## Short answer

Express x[n] as sum x[k] delta[n-k]. By linearity and time invariance, each shifted impulse gives h[n-k], so y[n]=sum x[k]h[n-k]=x[n]*h[n]. The convolution sum output has N1+N2-1 samples when x has N1 and h N2 samples.

## Answer

The **convolution sum** expresses the output of a discrete-time LTI system in terms of the input $x[n]$ and the impulse response $h[n]$ (the output when the input is $\delta[n]$).

#### Step 1: Any sequence as a sum of shifted impulses

Since $\delta[n-k] = 1$ only at $n = k$, each sample can be picked out:

$$
x[n] = \sum_{k=-\infty}^{\infty} x[k]\,\delta[n-k]
$$

For example, $x[n] = \{2, \underset{\uparrow}{1}, 3\}$ gives $x[n] = 2\delta[n+1] + \delta[n] + 3\delta[n-1]$.

#### Step 2: Apply the system properties

Let $T\{\cdot\}$ be the system and $h[n] = T\{\delta[n]\}$.

$$
y[n] = T\{x[n]\} = T\left\{\sum_{k} x[k]\,\delta[n-k]\right\}
$$

- **Linearity (additivity and homogeneity):** the $x[k]$ are just constants (weights), so
$$
y[n] = \sum_{k=-\infty}^{\infty} x[k]\,T\{\delta[n-k]\}
$$
- **Time invariance:** the response to $\delta[n-k]$ is $h[n-k]$.

#### Step 3: Result

$$
y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k] = x[n] * h[n]
$$

This is the **convolution sum**. Putting $m = n-k$ gives the equivalent form

$$
y[n] = \sum_{k=-\infty}^{\infty} h[k]\,x[n-k]
$$

showing convolution is commutative.

**Special cases:**
- Causal system ($h[n] = 0$, $n < 0$) and causal input: $y[n] = \sum_{k=0}^{n} x[k]\,h[n-k]$.
- Lengths: if $x$ has $N_1$ samples and $h$ has $N_2$, $y$ has $N_1 + N_2 - 1$ samples.

**Procedure:** (1) fold $h[k]$ to $h[-k]$; (2) shift by $n$ to get $h[n-k]$; (3) multiply by $x[k]$ sample by sample; (4) sum the products to get $y[n]$; (5) repeat for every $n$.

**Example:** $x[n] = \{\underset{\uparrow}{1}, 2\}$, $h[n] = \{\underset{\uparrow}{1}, 1, 1\}$:

| $n$ | Calculation | $y[n]$ |
|---|---|---|
| 0 | $1\cdot1$ | 1 |
| 1 | $1\cdot1 + 2\cdot1$ | 3 |
| 2 | $1\cdot1 + 2\cdot1$ | 3 |
| 3 | $2\cdot1$ | 2 |

So $y[n] = \{\underset{\uparrow}{1}, 3, 3, 2\}$, with $2 + 3 - 1 = 4$ samples.

## Check yourself

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

1. How is a DT sequence x[n] written as sum of impulses?
   A. x[n] = sum over k of x[k] delta[n - k]
   B. x[n] = sum over k of x[n] delta[k]
   C. x[n] = sum over k of delta[n - k]
   D. x[n] = sum over k of x[k] u[n - k]
   Answer: A. delta[n-k] picks out the sample at n = k, so the sum reproduces x[n]. (Key point: Impulse decomposition)

2. Which property gives the response to delta[n - k] as h[n - k]?
   A. Linearity
   B. Causality
   C. Stability
   D. Time invariance
   Answer: D. Delaying the input by k delays the response by k. (Key point: Use of time invariance)

3. Why can x[k] be taken outside T{.}?
   A. Because x[k] is a constant weight for each k, and the system is linear
   B. Because the system is memoryless
   C. Because delta[n-k] is zero
   D. Because the system is causal
   Answer: A. Homogeneity and additivity allow scaling and summation to pass through T. (Key point: Use of linearity)

4. If x has N1 samples and h has N2 samples, how many samples does y have?
   A. N1 x N2
   B. N1 + N2 - 1
   C. N1 + N2
   D. max(N1, N2)
   Answer: B. For x = {1,2}, h = {1,1,1} the result has 2 + 3 - 1 = 4 samples. (Key point: Output length)

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