FOUNDATION · CHAPTER 4 OF 8

Convolution, slowly

This chapter works convolution out by hand: flip and slide with every interval written down, discrete convolution as polynomial multiplication, and the three laws read as cascaded and parallel systems. An engineer needs it because an LTI system's output is its input convolved with its impulse response.

Updated 12 Oct 2026

What you will be able to do

You will compute the convolution of two short sequences by hand and as a polynomial product. You will predict the length and the support of the result before you start. You will carry out flip-and-slide for two pulses with every interval written out. You will read the three laws of convolution as statements about cascaded and parallel systems, and get the step response as a running sum.

Before you start

Chapter 3, What a system is, and why linear and time-invariant matter
Sums with a variable upper limit and the geometric series
Definite integrals of a constant over an interval, and a change of index in a sum
Multiplying two polynomials term by term

Convolution of sequences, by hand

Chapter 3 showed that an LTI system with impulse response $h [n]$ turns an input $x [n]$ into a sum of shifted copies of $h [n]$ , one per input sample, each scaled by that sample. Written out for one output instant, that is a sum over the input index. Here $*$ means convolution, and discrete time uses square brackets.

DISCRETE CONVOLUTION

y [n] = (x * h) [n] = k = - \infty \sum \infty x [k] h [n - k]

Holds for each integer $n$ at which the sum converges; with finitely many nonzero samples it always does. $x [k]$ is the input sample at index $k$ . $h [n - k]$ is $h$ flipped (the index runs backwards) and slid so that its origin sits at $k = n$ . $y$ carries the unit of $x$ times the unit of $h$ .

For sequences with a handful of nonzero samples there is a faster route. Give each sequence a placeholder $w$ and write it as a polynomial whose coefficient of $w^{k}$ is the sample at index $k$ (a negative index gives a negative power). Multiplying two polynomials collects, in the coefficient of $w^{n}$ , every product $x [k] h [m]$ with $k + m = n$ . That is exactly the sum above.

CONVOLUTION AS POLYNOMIAL MULTIPLICATION

(k \sum x [k] w^{k}) (m \sum h [m] w^{m}) = n \sum y [n] w^{n}

Holds for sequences with finitely many nonzero samples, so every sum is finite; negative indices give negative powers of $w$ . The coefficient of $w^{n}$ on the right is $\sum_{k} x [k] h [n - k]$ because the powers add: $k + m = n$ . $w$ is only a bookkeeping symbol.

LENGTH AND SUPPORT OF THE RESULT

if x [n] \neq = 0 only on N_{1} \leq n \leq N_{2}, h [n] \neq = 0 only on M_{1} \leq n \leq M_{2} : y [n] \neq = 0 only on N_{1} + M_{1} \leq n \leq N_{2} + M_{2}

Holds for sequences of finite length. By the polynomial rule, the lowest power of $w$ in the product is $N_{1} + M_{1}$ and the highest is $N_{2} + M_{2}$ : $L_{x} + L_{h} - 1$ samples, with $L_{x} = N_{2} - N_{1} + 1$ , $L_{h} = M_{2} - M_{1} + 1$ . The end samples are $x [N_{1}] h [M_{1}]$ and $x [N_{2}] h [M_{2}]$ ; interior samples can still be zero.

WORKED EXAMPLE

Let $x [n] = 1, 2, 3$ for $n = 0, 1, 2$ and zero elsewhere, and let $h [n] = 2, 1$ for $n = 0, 1$ and zero elsewhere. Find $y = x * h$ .

Solution

Support first. $x$ lives on $0 \leq n \leq 2$ (3 samples) and $h$ on $0 \leq n \leq 1$ (2 samples), so $y$ lives on $0 \leq n \leq 3$ , which is $3 + 2 - 1 = 4$ samples.
Write the polynomials: $x \to 1 + 2 w + 3 w^{2}$ and $h \to 2 + w$ . Each sample becomes the coefficient of the power of $w$ that matches its index.
Multiply $x$ 's polynomial by each term of $h$ in turn. By $2$ : $2 + 4 w + 6 w^{2}$ . By $w$ : $w + 2 w^{2} + 3 w^{3}$ . There is one row per term of $h$ .
Add like powers: $2 + (4 + 1) w + (6 + 2) w^{2} + 3 w^{3} = 2 + 5 w + 8 w^{2} + 3 w^{3}$ . So $y [0..3] = 2, 5, 8, 3$ .
Check one value with the sum: $y [2] = x [0] h [2] + x [1] h [1] + x [2] h [0] = 0 + 2 \cdot 1 + 3 \cdot 2 = 8$ . It matches.
Check the totals: setting $w = 1$ in the product shows $\sum y = (\sum x) (\sum h)$ . Here $6 \cdot 3 = 18$ and $2 + 5 + 8 + 3 = 18$ .

Answer: $y [n] = 2, 5, 8, 3$ for $n = 0, 1, 2, 3$ and zero elsewhere.

Flip and slide for continuous time

In continuous time the sum becomes an integral and the dummy index becomes $τ$ . Continuous time uses round brackets, $x (t)$ . The picture is the same. Flip $h$ about $τ = 0$ , shift it right by $t$ , multiply it with $x$ point by point, and take the area of the product. The answer at time $t$ is that area. The work is in finding, for each $t$ , where the two shapes overlap.

CONTINUOUS CONVOLUTION

y (t) = (x * h) (t) = \int_{- \infty}^{\infty} x (τ) h (t - τ) d τ

Holds for each $t$ at which the integral converges. $τ$ is the integration variable, in seconds like $t$ . If $h$ is nonzero only for $a < t < b$ , then $h (t - τ)$ , its flipped copy shifted right by $t$ , is nonzero only for $t - b < τ < t - a$ . $y$ carries the unit of $x$ times that of $h$ times seconds.

Write the support of $x (τ)$ and, in terms of $t$ , the support of $h (t - τ)$ .
Slide from $t = - \infty$ . Find the value of $t$ at which the two supports first touch. Before it, $y = 0$ .
Each time an edge of the sliding shape passes an edge of the fixed shape, a new interval begins. For shapes made of pieces, a point where the formula changes counts as an edge. Write down every such value of $t$ .
On each interval write the limits of the integral as the overlap, and integrate.
After the last overlap ends, $y = 0$ again. So $y$ is zero outside the range from the sum of the left edges to the sum of the right edges.

WATCH OUT FOR

Confusion: sliding $h (τ - t)$ , which is not flipped, instead of $h (t - τ)$ . Test: at $t = 0$ the sliding shape must be $h (- τ)$ , the mirror image of $h$ .

WORKED EXAMPLE

Let $x (t) = 1$ for $0 < t < 1$ and zero elsewhere, and let $h (t) = 1$ for $0 < t < 1$ and zero elsewhere, with $t$ in seconds and $x$ , $h$ dimensionless. Find $y (t) = (x * h) (t)$ on every interval.

Solution

$x (τ)$ is $1$ on $0 < τ < 1$ . $h (t - τ)$ is $1$ when $0 < t - τ < 1$ , that is on $t - 1 < τ < t$ . So the product is $1$ only where these two ranges overlap, and $0$ elsewhere.
The supports first touch at $t = 0$ and last touch at $t = 2$ . These are the sums of the left edges ( $0 + 0$ ) and of the right edges ( $1 + 1$ ). So $y = 0$ for $t < 0$ and for $t > 2$ .
For $0 < t < 1$ the sliding window $(t - 1, t)$ has its left edge below $0$ , so the overlap is $0 < τ < t$ . Then $y = \int_{0}^{t} 1 d τ = t$ , in seconds.
For $1 < t < 2$ the window's right edge is beyond $1$ , so the overlap is $t - 1 < τ < 1$ . Then $y = \int_{t - 1}^{1} 1 d τ = 2 - t$ , in seconds.
Check at $t = 1$ : both formulas give $1$ s, so $y$ has no jump there. This is the peak, where the window covers the whole pulse: area $1 \times 1 \times 1$ s.

Answer: $y (t) = 0$ for $t < 0$ ; $y (t) = t$ for $0 < t < 1$ ; $y (t) = 2 - t$ for $1 < t < 2$ ; $y (t) = 0$ for $t > 2$ ( $y$ in seconds). It is a triangle of height 1 s and base 2 s.

The three laws as systems, and the step response

The algebra of convolution is the algebra of connecting systems. Each law below is a fact about LTI boxes with impulse responses $h_{1}$ and $h_{2}$ . Each holds for sequences with finitely many nonzero samples. For longer sequences it holds whenever every sum involved stays finite when each term is replaced by its absolute value. The continuous versions follow the same steps with integrals.

COMMUTATIVE LAW

x * h = h * x

Substitute $m = n - k$ in $\sum_{k} x [k] h [n - k]$ . As $k$ runs over all integers so does $m$ , giving $\sum_{m} h [m] x [n - m]$ . Reading: input $x$ to a system $h$ gives the same output as input $h$ to a system $x$ . So you may flip and slide either signal: flip the simpler one.

ASSOCIATIVE LAW AND THE CASCADE

(x * h_{1}) * h_{2} = x * (h_{1} * h_{2})

For finite sequences it follows from the polynomial rule: a product of three polynomials does not depend on the grouping. Reading: $x$ through $h_{1}$ and then $h_{2}$ equals $x$ through one LTI system with $h = h_{1} * h_{2}$ . Since $h_{1} * h_{2} = h_{2} * h_{1}$ , swapping the two stages leaves the output unchanged.

DISTRIBUTIVE LAW AND THE PARALLEL PAIR

x * (h_{1} + h_{2}) = x * h_{1} + x * h_{2}

Distribute $x [k] (h_{1} + h_{2}) [n - k]$ and split the sum in two. Reading: two LTI systems fed the same input, with their outputs added, act as one system with $h = h_{1} + h_{2}$ . In series the impulse responses convolve; in parallel they add.

Now feed the unit step $u [n]$ , which is $1$ for $n \geq 0$ and $0$ for $n < 0$ . By the commutative law the output is $\sum_{k} h [k] u [n - k]$ . The factor $u [n - k]$ is $1$ exactly when $k \leq n$ and $0$ otherwise. So the sum keeps every $h [k]$ up to index $n$ and drops the rest.

STEP RESPONSE AS A RUNNING SUM

s [n] = k = - \infty \sum n h [k], h [n] = s [n] - s [n - 1]

Holds for each $n$ at which the sum converges; when $h [k] = 0$ for $k < 0$ it runs from $k = 0$ to $n$ , a finite sum. $s [n]$ is the output for input $u [n]$ . Subtracting $s [n - 1]$ from $s [n]$ leaves the single term $h [n]$ . In continuous time the same steps give $s (t) = \int_{- \infty}^{t} h (τ) d τ$ where it converges.

WATCH OUT FOR

Confusion: swapping stages when one is not LTI. The swap rests on $h_{1} * h_{2} = h_{2} * h_{1}$ , so it is guaranteed only when both stages are LTI. Test: one short input that gives different outputs in the two orders shows the swap fails.

WORKED EXAMPLE

A moving average has $h [n] = \frac{1}{2}$ for $n = 0, 1$ and zero elsewhere. Find its step response $s [n]$ for all $n$ .

Solution

Use $s [n] = \sum_{k \leq n} h [k]$ . This is valid because $u [n - k] = 1$ exactly when $k \leq n$ .
For $n < 0$ no nonzero $h [k]$ has $k \leq n$ , so $s [n] = 0$ .
For $n = 0$ the sum contains $h [0]$ only: $s [0] = \frac{1}{2}$ .
For $n \geq 1$ the sum contains both $h [0]$ and $h [1]$ : $s [n] = \frac{1}{2} + \frac{1}{2} = 1$ . No further samples of $h$ exist to add.
Recover $h$ from $s$ as a check: $s [0] - s [- 1] = \frac{1}{2}$ , $s [1] - s [0] = \frac{1}{2}$ , $s [2] - s [1] = 0$ . This is the original $h$ .

Answer: $s [n] = 0$ for $n < 0$ , $s [0] = \frac{1}{2}$ and $s [n] = 1$ for $n \geq 1$ .

Check yourself

CHECK YOURSELF 1

A sequence $x [n]$ is zero outside $- 2 \leq n \leq 1$ , with $x [- 2]$ and $x [1]$ nonzero. A sequence $h [n]$ is zero outside $3 \leq n \leq 6$ , with $h [3]$ and $h [6]$ nonzero. What are the first and last indices at which $y = x * h$ is nonzero?

(A)
$1$ and $7$
(B)
$1$ and $6$
(C)
$3$ and $7$
(D)
$0$ and $8$

Answer and solution

Answer: (A)

$y$ is zero outside the sums of the end indices: $- 2 + 3 = 1$ and $1 + 6 = 7$ .
$y [1] = x [- 2] h [3]$ and $y [7] = x [1] h [6]$ are products of nonzero samples, so both are nonzero.
Check the count: $4 + 4 - 1 = 7$ samples, and $1$ to $7$ has $7$ samples.

CHECK YOURSELF 2

Let $x [n] = 1, 2, 1$ and $h [n] = 1, - 1, 2$ , each for $n = 0, 1, 2$ and zero elsewhere. Find $y [2]$ where $y = x * h$ .

Answer and solution

Answer: $1$ (accepted range $\pm 0.01$ )

Use $y [2] = \sum_{k} x [k] h [2 - k]$ with $k = 0, 1, 2$ .
$x [0] h [2] = 1 \cdot 2 = 2$ .
$x [1] h [1] = 2 \cdot (- 1) = - 2$ .
$x [2] h [0] = 1 \cdot 1 = 1$ . The total is $2 - 2 + 1 = 1$ .

CHECK YOURSELF 3One or more options are correct

Two LTI systems have impulse responses $h_{1} [n]$ and $h_{2} [n]$ , each with finitely many nonzero samples. Which statements are true?

(A)
Connected in cascade, they act as one LTI system with impulse response $h_{1} * h_{2}$ .
(B)
In a cascade, swapping the order of the two systems changes the output for some inputs.
(C)
Fed the same input with their outputs added, they act as one LTI system with impulse response $h_{1} + h_{2}$ .
(D)
Fed the same input with their outputs added, they act as one LTI system with impulse response $h_{1} * h_{2}$ .

Answer and solution

Answer: (A), (C)

A is true by the associative law: $(x * h_{1}) * h_{2} = x * (h_{1} * h_{2})$ .
B is false: $h_{1} * h_{2} = h_{2} * h_{1}$ , so for two LTI stages the order does not matter.
C is true by the distributive law: $x * h_{1} + x * h_{2} = x * (h_{1} + h_{2})$ , which also makes D false.

CHECK YOURSELF 4

Let $x (t) = 1$ for $0 < t < 2$ and zero elsewhere, and $h (t) = 3$ for $0 < t < 1$ and zero elsewhere, with $t$ in seconds and $x$ , $h$ dimensionless. Find $y$ at $t = 2.5$ s, in seconds, where $y = x * h$ .

Answer and solution

Answer: $1.5$ (accepted range $\pm 0.01$ )

$y (t) = \int x (τ) h (t - τ) d τ$ , and $h (t - τ) = 3$ when $t - 1 < τ < t$ .
At $t = 2.5$ s that window is $1.5 < τ < 2.5$ . The signal $x (τ)$ is $1$ only for $0 < τ < 2$ .
The overlap is $1.5 < τ < 2$ , a width of $0.5$ s. So $y (2.5) = 1 \cdot 3 \cdot 0.5 s = 1.5$ s.

Convolution, slowly

What you will be able to do

Before you start

Convolution of sequences, by hand

Flip and slide for continuous time

The three laws as systems, and the step response

Check yourself

Used in these GATE topics