FOUNDATION · CHAPTER 3 OF 8

What a system is, and why linear and time-invariant matter

This chapter treats a system as a rule from input signal to output signal, tests three models for linearity and time invariance, and derives convolution from those two properties. An engineer needs it because, for such a system, one measured response is enough to predict the output.

Updated 12 Oct 2026

What you will be able to do

You will describe a system as a rule that turns an input signal into an output signal. You will test any such rule for linearity and time invariance. You will see why those two properties fix the output once a single response is known, and you will derive the convolution sum and integral yourself.

Before you start

Chapter 1, Complex numbers, Euler's formula and the complex exponential
Sums with a changing index (sigma notation), and an integral as the limit of a sum
Derivatives, the capacitor law (current equals C times the rate of change of voltage) and Kirchhoff's voltage law for a series loop
A first-order linear differential equation with a given starting value has exactly one solution

A system is a rule from input to output

A signal is a quantity that changes with time: a function $x (t)$ of continuous time $t$ , with round brackets, or a sequence $x [n]$ over integer steps $n$ , with square brackets. A system takes a whole signal in and gives a whole signal out. Write $y = T {x}$ , where $T$ is the rule. The output at one instant may depend on the input at other instants, so $T$ acts on the entire signal. Even when written $T {x [n]}$ , the bracket holds the whole sequence, not one number.

Take a series loop of a source $x (t)$ , a resistor $R$ in ohms and a capacitor $C$ in farads, with $y (t)$ the capacitor voltage. The current is $i = C y^{'} (t)$ and the loop law gives $x = R i + y$ . Substituting gives the first model below, with $R C$ in seconds. The other two are written directly as rules. All three start at rest: for an input that is zero before some time, the output stays zero until the input first becomes nonzero.

Model	Rule	Input	Output
RC circuit	$R C y^{'} (t) + y (t) = x (t)$	source voltage $x (t)$ in V	capacitor voltage $y (t)$ in V; capacitor uncharged until the source first turns on
Moving average	$y [n] = \frac{1}{3} (x [n] + x [n - 1] + x [n - 2])$	a reading on day $n$	the average of day $n$ 's reading and the two before it
Bank balance	$y [n] = 1.01 y [n - 1] + x [n]$	deposit in month $n$ , in rupees	balance after month $n$ at 1% monthly interest; zero before the first deposit

Linearity and time invariance, tested

Linearity means scaling and adding inputs scales and adds outputs. Time invariance means delaying the input only delays the output. Each is settled by feeding the rule a combination or a delayed copy and comparing.

LINEARITY

T {a x_{1} + b x_{2}} = a T {x_{1}} + b T {x_{2}}

Must hold for every pair of inputs $x_{1}, x_{2}$ and all constants $a, b$ . In particular a zero input must give a zero output. $T {x_{1}}$ and $T {x_{2}}$ are the two separate responses.

TIME INVARIANCE

T {x [n]} = y [n] ⟹ T {x [n - n_{0}]} = y [n - n_{0}]

Must hold for every input and every integer delay $n_{0}$ . Inside $T {}$ is the whole signal: if input $x [n]$ gives output $y [n]$ , the delayed input $x [n - n_{0}]$ gives $y [n - n_{0}]$ at every $n$ . In continuous time replace $n$ by $t$ and $n_{0}$ by any delay $t_{0}$ in seconds.

Linearity test: feed $a x_{1} + b x_{2}$ into the rule. Each rule adds terms that are a constant times an input value, an output value or a derivative, so $a y_{1} + b y_{2}$ satisfies the rule, and it is zero before the input starts. A rule started at rest has only one output: the recurrences compute each value from earlier ones, and the RC equation with a fixed starting value has one solution. So $a y_{1} + b y_{2}$ is the output.
Time-invariance test: feed $x [n - n_{0}]$ . The constants ( $\frac{1}{3}$ , $1.01$ , $R C$ ) do not depend on $n$ or $t$ , so $y [n - n_{0}]$ satisfies the rule with the delayed input and is zero until that input starts. By the same uniqueness it is the output.

System	Linear?	Time-invariant?	Reason
RC circuit	Yes	Yes	linear equation, constant $R$ and $C$ , at rest
Moving average	Yes	Yes	fixed weights $\frac{1}{3}$
Bank balance	Yes	Yes	fixed rate, at rest
$y [n] = x [n]^{2}$	No	Yes	doubling $x$ quadruples $y$ ; nothing depends on $n$ , so a delay passes through
$y [n] = n x [n]$	Yes	No	scaling and adding pass through; the weight $n$ changes with time

WATCH OUT FOR

A bank account opened with a balance of 1000 rupees, or a charged capacitor, is not linear. Test: give it zero input. If the output is not zero, linearity fails.

WORKED EXAMPLE

The bank system $y [n] = 1.01 y [n - 1] + x [n]$ starts at rest. Input A is a deposit of 1000 rupees in month 0 only. Input B is a deposit of 500 rupees in month 1 only. Check linearity and time invariance for months 0 to 2.

Solution

Input A: month 0 has no earlier balance, so it gives $1000$ . With no more deposits, each month adds 1% interest: $1.01 \times 1000 = 1010$ , then $1.01 \times 1010 = 1020.1$ . So $y_{A} = 1000, 1010, 1020.1$ .
Input B: nothing arrives in month 0, so it gives $0$ . Month 1 adds the deposit, $500$ . Month 2 adds interest only, $1.01 \times 500 = 505$ . So $y_{B} = 0, 500, 505$ .
Input A plus B, through the same rule: month 0 gives $1000$ , month 1 gives $1.01 \times 1000 + 500 = 1510$ , month 2 gives $1.01 \times 1510 = 1525.1$ .
Superposition asks whether this equals the sum of the separate outputs: $1000 + 0, 1010 + 500, 1020.1 + 505 = 1000, 1510, 1525.1$ . It does.
B is A delayed by one month and halved, so its response should be $y_{A}$ delayed and halved. Delaying gives $0, 1000, 1010$ ; halving gives $0, 500, 505$ , which is $y_{B}$ .

Answer: Both properties hold for these inputs: the response to A plus B is $1000, 1510, 1525.1$ rupees, and B's response is A's delayed one month and halved. A check on particular inputs supports the properties; the general test above proves them.

Why linear and time-invariant means convolution

The unit pulse $δ [n]$ is 1 at $n = 0$ and 0 at every other $n$ . Any sequence is a pile of such pulses, one at each time, scaled by the signal's value there. If the system scales, adds and shifts faithfully, its output is the same pile of responses, all shifted copies of one response. So one measured response describes the system.

SIFTING

x [n] = k = - \infty \sum \infty x [k] δ [n - k]

Holds for any sequence. Since $δ [n - k] = 0$ for $k \neq = n$ , only the term $k = n$ survives, and it equals $x [n]$ . Here $x [k]$ are plain numbers and $δ [n - k]$ is a pulse at time $k$ .

IMPULSE RESPONSE

h [n] = T {δ [n]}, T {δ [n - k]} = h [n - k]

The second equality uses time invariance with delay $k$ , any integer. $h [n]$ is the output when the input is the unit pulse $δ [n]$ and the system starts at rest; it has the unit of $y$ per unit of $x$ .

CONVOLUTION SUM

y [n] = T {k \sum x [k] δ [n - k]} = k \sum x [k] T {δ [n - k]} = k = - \infty \sum \infty x [k] h [n - k] = (x * h) [n]

Needs linearity and time invariance. The middle step passes $T$ through the sum: linearity gives this for finitely many terms, and for infinitely many it is assumed. $*$ denotes convolution.

WATCH OUT FOR

The weight is $h [n - k]$ , not $h [k - n]$ . Test: let $x [k] = δ [k]$ . The sum must return $h [n]$ . The wrong form returns $h [- n]$ .

Continuous time is the same argument with the pulses made thin. Let $p_{Δ} (t)$ be a pulse of height $1/Δ$ from $t = 0$ to $t = Δ$ , so its area is 1. As $Δ \to 0$ it shrinks to the unit impulse $δ (t)$ : zero width, unit area. A staircase of such pulses approximates the signal. Write $h_{Δ} (t)$ for the response to $p_{Δ} (t)$ .

STAIRCASE APPROXIMATION

x (t) \approx k \sum x (k Δ) p_{Δ} (t - k Δ) Δ

Valid for a signal that changes little over a width $Δ$ (in s). The pulse $p_{Δ} (t - k Δ)$ is $p_{Δ}$ delayed by $k Δ$ , so time invariance gives the response $h_{Δ} (t - k Δ)$ . Linearity scales it by the number $x (k Δ) Δ$ and adds: $y (t) \approx \sum_{k} x (k Δ) h_{Δ} (t - k Δ) Δ$ .

CONVOLUTION INTEGRAL

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

Needs linearity, time invariance, and that the limit may be passed through the system. As $Δ \to 0$ , $k Δ$ becomes $τ$ (in s), the sum becomes an integral, $p_{Δ}$ becomes $δ (t)$ and $h_{Δ}$ becomes the impulse response $h (t)$ . $h$ has the unit of $y$ per unit of $x$ , per second.

WORKED EXAMPLE

The moving average $y [n] = \frac{1}{3} (x [n] + x [n - 1] + x [n - 2])$ receives $x [0] = 3$ , $x [1] = 6$ , $x [2] = 9$ and zero at every other $n$ . Find $y [n]$ by convolution.

Solution

Impulse response: put $x = δ [n]$ into the rule. Then $h [0] = h [1] = h [2] = \frac{1}{3}$ and $h [n] = 0$ elsewhere. The system is LTI, so convolution applies.
Write $y [n] = \sum_{k} x [k] h [n - k]$ . Only $k = 0, 1, 2$ have $x [k] \neq = 0$ , and $h [n - k] = \frac{1}{3}$ only when $0 \leq n - k \leq 2$ .
$n = 0$ : only $k = 0$ contributes, so $y [0] = 3 \times \frac{1}{3} = 1$ .
$n = 1$ : $k = 0, 1$ contribute, so $y [1] = (3 + 6) \times \frac{1}{3} = 3$ . $n = 2$ : $y [2] = (3 + 6 + 9) \times \frac{1}{3} = 6$ .
$n = 3$ : $k = 1, 2$ contribute, so $y [3] = (6 + 9) \times \frac{1}{3} = 5$ . $n = 4$ : only $k = 2$ , so $y [4] = 9 \times \frac{1}{3} = 3$ . For $n \geq 5$ or $n < 0$ no term survives, so $y = 0$ .
Check $n = 3$ straight from the rule: $\frac{1}{3} (x [3] + x [2] + x [1]) = \frac{1}{3} (0 + 9 + 6) = 5$ . It agrees.

Answer: $y [n] = 1, 3, 6, 5, 3$ for $n = 0, \dots, 4$ , and zero elsewhere.

Check yourself

CHECK YOURSELF 1One or more options are correct

Which of these systems are linear? (Take all of them to start at rest.)

(A)
$y (t) = 3 x (t - 2)$
(B)
$y [n] = x [n]^{2}$
(C)
$y [n] = n x [n]$
(D)
$y (t) = \frac{d x ( t )}{d t}$

Answer and solution

Answer: (A), (C), (D)

A: a delay and a constant gain are both linear.
B: squaring fails scaling, since $(2 x)^{2} = 4 x^{2} \neq = 2 x^{2}$ .
C: the weight $n$ is fixed for each $n$ , so scaling and adding pass through. It is linear but not time-invariant.
D: the derivative of $a x_{1} + b x_{2}$ is $a x_{1}^{'} + b x_{2}^{'}$ , so differentiation is linear.

CHECK YOURSELF 2

An LTI system has $h [n] = δ [n] + 2 δ [n - 1]$ and receives $x [n] = δ [n] + 3 δ [n - 1]$ . Find $y [1]$ .

Answer and solution

Answer: $5$ (accepted range $\pm 0.01$ )

$y [1] = \sum_{k} x [k] h [1 - k]$ and only $k = 0, 1$ give nonzero $x [k]$ .
$k = 0$ : $x [0] h [1] = 1 \times 2 = 2$ . $k = 1$ : $x [1] h [0] = 3 \times 1 = 3$ .
Total $2 + 3 = 5$ .

CHECK YOURSELF 3

The bank system $y [n] = 1.01 y [n - 1] + x [n]$ starts at rest. A single deposit of 2000 rupees is made in month 0 and nothing after. Find the balance in rupees after month 3.

Answer and solution

Answer: $2060.6$ (accepted range $\pm 0.05$ )

For a unit deposit at month 0: $y [0] = 1$ , and each later month multiplies by $1.01$ , so $y [n] = 1.0 1^{n}$ for $n \geq 0$ .
By linearity a 2000-rupee deposit gives $2000 \times 1.0 1^{n}$ . At $n = 3$ : $1.0 1^{3} = 1.030301$ .
$2000 \times 1.030301 = 2060.602$ rupees.

CHECK YOURSELF 4

An LTI system has impulse response $h [n]$ . What is its response to $2 δ [n - 3]$ ?

(A)
$2 h [n]$
(B)
$2 h [n - 3]$
(C)
$h [2 n - 3]$
(D)
$2 h [n + 3]$

Answer and solution

Answer: (B)

Time invariance delays the response by 3: $δ [n - 3]$ gives $h [n - 3]$ .
Linearity scales it by 2, giving $2 h [n - 3]$ .

Used in these GATE topics

LTI systems: properties, impulse response, convolution and correlation