FOUNDATION · CHAPTER 5 OF 8

Why sinusoids: eigenfunctions and the frequency idea

This chapter shows, from convolution, why a complex exponential leaves a linear time-invariant system changed only in gain and phase. It then finds Fourier coefficients as projections onto orthogonal harmonics. This is why engineers describe filters and channels by their response at each frequency.

Updated 12 Oct 2026

What you will be able to do

You will show, from convolution alone, why a complex exponential leaves an LTI system with its shape intact and only a complex number attached. You will read that number as a gain and a phase shift at one frequency. You will treat harmonics as perpendicular directions and find each Fourier coefficient as a projection. You will build the square wave's series one step at a time.

Before you start

Chapter 1, Complex numbers, Euler's formula and the complex exponential (including conjugates and the real part)
Chapter 4, Convolution, slowly
The unit step $u (t)$ and the unit impulse $δ (t)$
Linearity: a sum of inputs gives the sum of their outputs (superposition)
Integrating an exponential, with $a$ real or complex and $a \neq = 0$ : $\int e^{a t} d t = e^{a t} / a$
The dot product, and perpendicular unit vectors

The complex exponential passes through unchanged

Convolution, $y = h * x$ , gives the output of an LTI system. Here $h$ is the impulse response and $*$ is convolution, as in Chapter 4; continuous time is written $x (t)$ , discrete time $x [n]$ , and $j$ is the imaginary unit. Try an input that behaves simply under a shift: $x (t) = e^{s t}$ with $s = σ + j ω$ (Chapter 1). Delaying it by $τ$ only multiplies it by a constant, $e^{s (t - τ)} = e^{s t} e^{- s τ}$ . That constant slides out of the convolution integral.

COMPLEX EXPONENTIAL IN, SAME EXPONENTIAL OUT

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

Holds for an LTI system, for those complex $s$ where the integral for $H (s)$ converges (its region of convergence, ROC), with the input present for all time. $H (s)$ is the system function, a complex number for each $s$ ; $t$ and $τ$ are in seconds, $s$ in 1/s.

An input that comes out as itself times a constant is an eigenfunction of the system, and $H (s)$ is its eigenvalue. In discrete time, with $z$ a complex number and $h [n]$ the impulse response, the same slide-out step gives $z^{n}$ in, $H (z) z^{n}$ out, where $H (z) = \sum_{k} h [k] z^{- k}$ wherever that sum converges. Now use superposition: if $x (t) = \sum_{i} c_{i} e^{s_{i} t}$ , a finite sum with each $s_{i}$ in the ROC, the output is $\sum_{i} c_{i} H (s_{i}) e^{s_{i} t}$ . Convolution has become one multiplication per term. That is the reason to write signals as sums of exponentials.

WATCH OUT FOR

$H (s)$ is not defined for every $s$ , only where its integral converges: for $h (t) = e^{t} u (t)$ , only for $Re s > 1$ . Test: for an exponential $h$ , check that $h (τ) e^{- s τ}$ decays as $τ \to + \infty$ and as $τ \to - \infty$ .

What frequency response means

Set $σ = 0$ , so $s = j ω$ and the input is a pure oscillation. The number $H (j ω)$ , also written $H (ω)$ , is the frequency response; it exists when the $j ω$ axis lies in the ROC. At each frequency its size is the gain and its angle is the phase shift (a negative angle is a lag). Physically, it is what you measure when you drive a stable system with a long-running sine and compare the output's amplitude and phase with the input's. A real cosine is half the sum of $e^{j ω_{0} t}$ and $e^{- j ω_{0} t}$ (Chapter 1), so each half goes through separately.

TWO HALVES OF A COSINE THROUGH THE SYSTEM

y (t) = \frac{A}{2} [H (j ω_{0}) e^{j (ω_{0} t + ϕ)} + H (- j ω_{0}) e^{- j (ω_{0} t + ϕ)}] = A Re {H (j ω_{0}) e^{j (ω_{0} t + ϕ)}}

Input $x (t) = A cos (ω_{0} t + ϕ)$ , $h$ real, $j ω$ axis in the ROC. Then $H (- j ω_{0}) = \int h (τ) e^{j ω_{0} τ} d τ$ is the conjugate of $H (j ω_{0})$ , so the two halves are conjugates and add to twice a real part. $A$ is in input units, $ω_{0}$ in rad/s, $ϕ$ in rad.

RESPONSE TO A REAL SINUSOID

y (t) = A ∣ H (j ω_{0}) ∣ cos (ω_{0} t + ϕ + ∠ H (j ω_{0}))

Same conditions: real $h$ , $j ω$ axis in the ROC, input present for all time. Put $H (j ω_{0}) = ∣ H ∣ e^{j ∠ H}$ into the second line of the previous formula: the real part of $∣ H ∣ e^{j (ω_{0} t + ϕ + ∠ H)}$ is this cosine. $∣ H ∣$ is the gain (output units per input unit); $∠ H$ is the phase shift in rad. The frequency does not change.

Take $h (t) = 2 e^{- 3 t} u (t)$ , with $t$ in seconds. Its response is $H (j ω) = 2/ (3 + j ω)$ , worked out in the example below. The gain falls smoothly from $2/3$ at $ω = 0$ to zero, and the phase lag deepens towards $- 9 0^{\circ}$ . So this system passes slow waves and weakens fast ones.

$ω$ (rad/s)	$∣ H (j ω) ∣$	$∠ H (j ω)$
0	2/3 ≈ 0.667	0
3	√2/3 ≈ 0.471	−45°
4	2/5 = 0.4	−53.1°
∞	0	−90°

WORKED EXAMPLE

An LTI system has $h (t) = 2 e^{- 3 t} u (t)$ with $t$ in seconds. The input is $x (t) = cos (3 t)$ , present for all time. Find $y (t)$ .

Solution

Find $H (j ω)$ from its integral. Since $h (τ) = 0$ for $τ < 0$ , $H (j ω) = \int_{0}^{\infty} 2 e^{- (3 + j ω) τ} d τ = 2/ (3 + j ω)$ , using $\int_{0}^{\infty} e^{- a τ} d τ = 1/ a$ , valid because $Re a = 3 > 0$ . This also shows the $j ω$ axis is inside the ROC.
Evaluate at $ω_{0} = 3$ rad/s, the input's frequency: $H (j 3) = 2/ (3 + j 3)$ . Multiply top and bottom by $3 - j 3$ to clear the complex denominator: $2 (3 - j 3) /18 = (1 - j) /3$ .
Read the size: $∣ H ∣ = 1 + 1 /3 = 2 /3 \approx 0.471$ . Read the angle: the real part is positive and the imaginary part is negative and equal in size, so $∠ H = - 4 5^{\circ} = - π /4$ rad.
The input has $A = 1$ and $ϕ = 0$ , so the response formula gives the output directly.

Answer: $y (t) \approx 0.471 cos (3 t - π /4)$ : the same frequency, amplitude scaled by 0.471, phase lagging by $4 5^{\circ}$ . At 3 rad/s that lag is a time delay of $(π /4) /3 \approx 0.262$ s.

WATCH OUT FOR

The result describes a wave present for all time. A cosine switched on at $t = 0$ also produces a start-up transient; in a stable system it dies away and leaves this response. Test: ask whether the input exists for every $t$ .

Harmonics as perpendicular directions

For perpendicular unit vectors $e_{1}, e_{2}, e_{3}$ , any vector is $v = \sum_{i} (v \cdot e_{i}) e_{i}$ : each coefficient comes from one dot product, untouched by the other directions, because they are perpendicular. Signals with period $T_{0}$ work the same way. The directions are the harmonics $e^{j k ω_{0} t}$ , with $ω_{0} = 2 π / T_{0}$ rad/s and integer $k$ . The dot product becomes the inner product $⟨ x, g ⟩ = \int_{0}^{T_{0}} x (t) g^{*} (t) d t$ ; the conjugate makes $⟨ x, x ⟩ = \int_{0}^{T_{0}} ∣ x (t) ∣^{2} d t$ a squared length, never negative. Orthogonal (perpendicular) means $⟨ x, g ⟩ = 0$ .

HARMONICS ARE ORTHOGONAL

\int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t = {T_{0}, 0, k = m k \neq = m

$k, m$ are integers; $T_{0}$ is the period in seconds and $ω_{0} T_{0} = 2 π$ . This is $⟨ e^{j k ω_{0} t}, e^{j m ω_{0} t} ⟩$ . For $k = m$ the integrand is 1. For $k \neq = m$ the antiderivative $e^{j (k - m) ω_{0} t} / (j (k - m) ω_{0})$ takes the same value at $0$ and $T_{0}$ , because $e^{j 2 π (k - m)} = 1$ , so the integral is zero.

FOURIER COEFFICIENT AS A PROJECTION

x (t) = k = - \infty \sum \infty a_{k} e^{j k ω_{0} t} ⟹ a_{m} = \frac{1}{T _{0}} \int_{0}^{T_{0}} x (t) e^{- j m ω_{0} t} d t

Holds for $x (t)$ periodic with period $T_{0}$ , assuming the series converges well enough to integrate term by term. Multiply the series by $e^{- j m ω_{0} t}$ and integrate over one period: each term with $k \neq = m$ gives zero and $k = m$ gives $a_{m} T_{0}$ ; divide by $T_{0}$ . $a_{0}$ is the average value of $x$ , in the units of $x$ .

So $a_{m}$ is the inner product of $x$ with the $m$ -th harmonic, divided by that harmonic's squared length $T_{0}$ : a projection. Now do the square wave slowly. Let $x (t) = 1$ for $0 < t < T_{0} /2$ and $x (t) = - 1$ for $T_{0} /2 < t < T_{0}$ , repeating every $T_{0}$ .

SQUARE WAVE COEFFICIENTS

a_{k} = \frac{1}{T _{0}} [\int_{0}^{T_{0} /2} e^{- j k ω_{0} t} d t - \int_{T_{0} /2}^{T_{0}} e^{- j k ω_{0} t} d t] = \frac{1}{T _{0}} [\frac{1 - e ^{- j k π}}{j k ω _{0}} - \frac{e ^{- j k π} - e ^{- j 2 k π}}{j k ω _{0}}] = \frac{( 1 - ( - 1 ) ^{k} ) - ( ( - 1 ) ^{k} - 1 )}{j k ω _{0} T _{0}} = \frac{2 ( 1 - ( - 1 ) ^{k} )}{j k ω _{0} T _{0}} = \frac{1 - ( - 1 ) ^{k}}{j π k}

For $k \neq = 0$ . Line 2: $\int_{a}^{b} e^{c t} d t = (e^{c b} - e^{c a}) / c$ with $c = - j k ω_{0}$ , and $ω_{0} T_{0} /2 = π$ . Line 3: $e^{- j k π} = (- 1)^{k}$ , $e^{- j 2 k π} = 1$ . Line 4: collect terms, then $ω_{0} T_{0} = 2 π$ . For $k = 0$ the halves cancel: $a_{0} = 0$ . Odd $k$ : $a_{k} = 2/ (j π k)$ . Even $k \neq = 0$ : $a_{k} = 0$ .

SQUARE WAVE AS A SINE SERIES

x (t) = \frac{4}{π} (sin ω_{0} t + \frac{sin 3 ω _{0} t}{3} + \frac{sin 5 ω _{0} t}{5} + \dots)

Because $x$ is real, $a_{- k}$ is the conjugate of $a_{k}$ , so the pair gives $2 Re {\frac{2}{j π k} e^{j k ω_{0} t}} = \frac{4}{π k} sin k ω_{0} t$ for odd $k$ . Holds at every $t$ away from the jumps; at a jump the sum is 0, halfway between $- 1$ and $1$ .

Terms kept	Highest harmonic	Sum at $t = T_{0} /4$ (true value 1)
1	1	1.273
2	3	0.849
3	5	1.103
4	7	0.922

The sums swing above and below 1 and settle towards it. Near a jump they behave differently: this is the Gibbs phenomenon, described here without proof. Once many terms are kept, the partial sums overshoot by about 9% of the jump, so by about 0.18 here, where the jump is 2. More terms squeeze the overshoot closer to the jump, but its height does not shrink to zero. Smooth sines cannot switch instantly, so a finite sum rings.

WORKED EXAMPLE

A signal has period $T_{0} = 2$ s: $x (t) = 1$ for $0 < t < 1$ and $x (t) = 0$ for $1 < t < 2$ . Find $a_{0}$ and $a_{1}$ .

Solution

Here $ω_{0} = 2 π / T_{0} = π$ rad/s. $a_{0}$ is the average value; $x = 1$ for half the period, so $a_{0} = \frac{1}{2} \int_{0}^{1} 1 d t = 0.5$ .
For $a_{1}$ , use the projection formula with $m = 1$ . Since $x = 0$ on $1 < t < 2$ , only $0 < t < 1$ contributes: $a_{1} = \frac{1}{2} \int_{0}^{1} e^{- j π t} d t$ .
Integrate the exponential with $c = - j π$ : $\int_{0}^{1} e^{- j π t} d t = (e^{- j π} - 1) / (- j π) = (1 - e^{- j π}) / (j π)$ .
Use $e^{- j π} = - 1$ (Chapter 1): the numerator is 2, so $a_{1} = \frac{1}{2} \cdot \frac{2}{j π} = \frac{1}{j π} = - \frac{j}{π}$ , since $1/ j = - j$ .

Answer: $a_{0} = 0.5$ and $a_{1} = - j / π$ , of size $1/ π \approx 0.318$ .

Check yourself

CHECK YOURSELF 1

An LTI system is stable, meaning $\int_{- \infty}^{\infty} ∣ h (t) ∣ d t$ is finite. Which input comes out of every such system as the same signal multiplied only by a (complex) constant?

(A)
$u (t)$ , the unit step
(B)
$e^{j 3 t}$
(C)
$t^{2}$
(D)
$δ (t)$

Answer and solution

Answer: (B)

Stability makes $∣ H (j 3) ∣ \leq \int ∣ h (τ) ∣ d τ$ finite, so the slide-out step gives $H (j 3) e^{j 3 t}$ for every stable system.
One system rules out the others: $h (t) = e^{- t} u (t)$ turns $u (t)$ into $(1 - e^{- t}) u (t)$ , $δ (t)$ into $e^{- t} u (t)$ , and $t^{2}$ into $t^{2} - 2 t + 2$ .
None of those is a constant times its input, so the answer is B.

CHECK YOURSELF 2

A system with a real impulse response has frequency response $H (j ω) = 4/ (6 + j ω)$ , with $ω$ in rad/s. The input is $5 cos (8 t)$ with $t$ in seconds, present for all time. Find the amplitude of the output.

Answer and solution

Answer: $2$ (accepted range $\pm 0.01$ )

The gain at $ω_{0} = 8$ rad/s is $∣ H (j 8) ∣ = 4/∣6 + j 8∣$ .
$∣6 + j 8∣ = 36 + 64 = 10$ , so the gain is $0.4$ .
The output amplitude is $5 \times 0.4 = 2$ .

CHECK YOURSELF 3One or more options are correct

Let $ω_{0} = 2 π / T_{0}$ . Over one period $0 \leq t \leq T_{0}$ , which pairs of signals are orthogonal, meaning the integral of their product is zero?

(A)
$cos ω_{0} t$ and $sin ω_{0} t$
(B)
$cos ω_{0} t$ and $cos 2 ω_{0} t$
(C)
$cos ω_{0} t$ and $cos ω_{0} t$
(D)
$sin ω_{0} t$ and $sin 3 ω_{0} t$

Answer and solution

Answer: (A), (B), (D)

Expand each product into terms $e^{j n ω_{0} t}$ . By the orthogonality result with $m = 0$ , a term with $n \neq = 0$ integrates to zero over a period; only a constant term survives.
A gives $n = \pm 2$ , B gives $\pm 1, \pm 3$ , D gives $\pm 2, \pm 4$ : no constant term, so each integral is zero.
C: $cos^{2} ω_{0} t$ contains the constant $\frac{1}{2}$ , so its integral is $T_{0} /2 \neq = 0$ .

CHECK YOURSELF 4

The square wave of this chapter (values $\pm 1$ , so each jump is 2) is rebuilt from its first $N$ nonzero sine terms. As $N$ grows large, what happens to the largest overshoot of the partial sum beside a jump?

(A)
It shrinks towards zero
(B)
It settles near 0.18 and does not shrink
(C)
It grows without limit
(D)
It vanishes once $N$ is large enough

Answer and solution

Answer: (B)

Near a jump the Gibbs phenomenon applies: for large $N$ the overshoot is about 9% of the jump.
The jump is 2, so the overshoot is about $0.09 \times 2 = 0.18$ .
More terms move the peak closer to the jump but do not lower it, so the answer is B.

Why sinusoids: eigenfunctions and the frequency idea

What you will be able to do

Before you start

The complex exponential passes through unchanged

What frequency response means

Harmonics as perpendicular directions

Check yourself

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