# Fourier Series & Transforms
> Any reasonable signal is a sum of pure rotations $e^{i\omega t}$ — and the complex exponentials are
> *orthogonal*, so each rotation's weight can be read off independently by an inner product. That
> single idea, "project onto a pure frequency," is the whole of Fourier analysis. We build it here in
> continuous time so the discrete and *quantum* versions later feel inevitable.
## Learning Objectives
By the end of this lesson you will be able to:
1. **Prove** the orthogonality of complex exponentials on an interval and use it to extract Fourier coefficients.
2. **Compute** the complex Fourier series of a periodic function and state the convergence guarantees you can rely on.
3. **Define** the continuous Fourier transform and its inverse, and **apply** linearity, time/frequency shift, and the convolution theorem.
4. **Derive** Parseval/Plancherel and interpret it as conservation of energy between time and frequency domains.
5. **Implement** a numerical Fourier-series approximation and **demonstrate** the convolution theorem with `np.fft`.
---
## Intuition
Picture a signal — a sound, a voltage, an image row — as a recipe. The Fourier idea is that the
recipe is always a *blend of pure tones*: each tone is a complex rotation $e^{i\omega t}$ spinning at
its own frequency $\omega$, weighted by a complex amplitude that carries both *how much* (modulus)
and *what phase* (argument). The Fourier transform is the machine that hands you the recipe; the
inverse transform stirs the tones back into the signal.
Why does reading off one tone not disturb the others? Because the pure tones are **orthogonal**: over
a full period, $e^{i m t}$ and $e^{i n t}$ have zero overlap unless $m=n$. This is the continuous
sibling of the roots-of-unity vanishing sum from [lesson 1](01-complex-numbers-functions.md) —
distinct frequencies average to nothing. Orthogonality is what makes "project onto frequency $\omega$"
a clean, independent measurement, exactly as orthonormal basis vectors let you read off coordinates
in linear algebra.
Two domains, one object. The **time domain** answers "what is the value now?"; the **frequency
domain** answers "what is it made of?" Operations that are tangled in one domain become trivial in
the other — most spectacularly, *convolution* (smearing/filtering, expensive in time) becomes plain
*multiplication* in frequency. Keep this dual picture in mind; the discrete version in
[lesson 3](03-discrete-fourier-transform.md) turns it into an algorithm, and the quantum version
turns that algorithm into a circuit.
---
## Theory
### Orthogonality of complex exponentials
Work on the interval $[0,T]$ and define, for integer $n$, the functions $e_n(t) = e^{i 2\pi n t/T}$.
These are exactly the *tones whose period divides $T$*. Equip the space of (square-integrable)
functions on $[0,T]$ with the Hermitian inner product (conjugate the left factor, matching
[Appendix C §1](../../appendix/C-notation-glossary.md)):
$$
\langle f, g\rangle := \int_0^T f(t)^{*}\,g(t)\,dt.
$$
**Lemma (orthogonality).**
$$
\langle e_m, e_n\rangle = \int_0^T e^{-i2\pi m t/T}\,e^{i2\pi n t/T}\,dt = T\,\delta_{mn}.
$$
*Proof.* The integrand is $e^{i2\pi(n-m)t/T}$. If $m=n$ it equals $1$ and integrates to $T$.
Otherwise let $k = n-m\neq 0$ and integrate the exponential:
$$
\int_0^T e^{i2\pi k t/T}\,dt
= \left[\frac{T}{i2\pi k}\,e^{i2\pi k t/T}\right]_0^T
= \frac{T}{i2\pi k}\big(e^{i2\pi k} - 1\big) = \frac{T}{i2\pi k}\big(1 - 1\big) = 0,
$$
since $e^{i2\pi k} = 1$ for integer $k$. $\blacksquare$
This is the continuous analog of the roots-of-unity orthogonality $\tfrac1N\sum_x \omega^{x(j-k)} =
\delta_{jk}$ ([Appendix E §6](../../appendix/E-math-identities.md)): replace the sum over $N$ sample
points by an integral over the period. The $\{e_n/\sqrt T\}$ form an *orthonormal* family.
### Fourier series of a periodic function
Let $f$ be $T$-periodic and square-integrable on a period. Because $\{e_n\}$ is a complete
orthogonal system (a theorem of analysis we use, not reprove), $f$ has the **complex Fourier series**
$$
f(t) = \sum_{n=-\infty}^{\infty} c_n\, e^{i2\pi n t/T}.
$$
The coefficients fall out of orthogonality: take the inner product of both sides with $e_m$ and use
the lemma. Only the $n=m$ term survives,
$$
\langle e_m, f\rangle = \sum_n c_n\langle e_m, e_n\rangle = \sum_n c_n\,T\delta_{mn} = c_m\,T,
$$
so the **analysis formula** is
$$
\boxed{\,c_n = \frac{1}{T}\int_0^T f(t)\,e^{-i2\pi n t/T}\,dt\,}.
$$
Each $c_n$ is the (normalized) overlap of $f$ with the pure tone of frequency $n/T$ — the amount of
that tone in the blend. Note $c_0 = \tfrac1T\int_0^T f\,dt$ is the average (DC) value.
**Reality and symmetry.** If $f$ is real then $c_{-n} = c_n^{*}$ (conjugate-symmetric spectrum),
because conjugating the integrand flips the sign of the exponent. Pairing $\pm n$ then recovers the
familiar real sine/cosine series:
$$
f(t) = c_0 + \sum_{n=1}^{\infty}\Big[a_n\cos\tfrac{2\pi n t}{T} + b_n\sin\tfrac{2\pi n t}{T}\Big],
\quad a_n = 2\operatorname{Re}c_n,\ \ b_n = -2\operatorname{Im}c_n.
$$
**Convergence (what you can rely on).** For $f\in L^2[0,T]$ the partial sums converge to $f$ *in the
$L^2$ (mean-square) sense*: $\|f - \sum_{|n|\le M} c_n e_n\|_2 \to 0$. If $f$ is piecewise-$C^1$, the
series converges *pointwise* to $f(t)$ where $f$ is continuous and to the midpoint
$\tfrac12\big(f(t^+)+f(t^-)\big)$ at jumps; near a jump the partial sums overshoot by a fixed
$\approx 9\%$ (the **Gibbs phenomenon**), which never disappears but moves toward the jump as $M$
grows. We will *see* Gibbs in the code.
### The continuous Fourier transform
Let the period $T\to\infty$: the discrete frequencies $n/T$ crowd into a continuum $\xi$, the sum
becomes an integral, and the Fourier *series* becomes the Fourier *transform*. For an integrable (or
$L^2$) function $f:\mathbb R\to\mathbb C$ we use the convention
$$
\boxed{\;\hat f(\xi) = \mathcal F[f](\xi) = \int_{-\infty}^{\infty} f(t)\,e^{-i2\pi\xi t}\,dt\;}
$$
with frequency $\xi$ in cycles per unit time, and the **inversion formula**
$$
\boxed{\;f(t) = \mathcal F^{-1}[\hat f](t) = \int_{-\infty}^{\infty} \hat f(\xi)\,e^{+i2\pi\xi t}\,d\xi\;}.
$$
The $e^{\mp 2\pi i\xi t}$ kernel and the $\pm$ sign convention match the DFT forward/inverse pair in
[lesson 3](03-discrete-fourier-transform.md) and the program-wide choice in
[Appendix C §4](../../appendix/C-notation-glossary.md) (no stray $1/2\pi$ factors with this "ordinary
frequency" convention — a convenience we exploit below).
### Key properties
Let $\hat f = \mathcal F[f]$, $\hat g = \mathcal F[g]$, and $a,b\in\mathbb C$. The following are
direct consequences of the integral definition.
**Linearity.** $\mathcal F[a f + b g] = a\hat f + b\hat g$ (the integral is linear).
**Time shift.** $\mathcal F[f(t - t_0)](\xi) = e^{-i2\pi\xi t_0}\,\hat f(\xi)$.
*Proof.* Substitute $u = t - t_0$, $du = dt$:
$\int f(t-t_0)e^{-i2\pi\xi t}\,dt = \int f(u)e^{-i2\pi\xi(u+t_0)}\,du = e^{-i2\pi\xi t_0}\hat f(\xi)$.
A delay in time is a *linear phase* in frequency — modulus unchanged, phase tilted. (The
relative-phase intuition from lesson 1 reappears: shifting a signal does not change its energy
spectrum, only its phases.)
**Frequency shift (modulation).** $\mathcal F[e^{i2\pi\xi_0 t}f(t)](\xi) = \hat f(\xi - \xi_0)$ —
multiplying by a tone in time slides the spectrum (dual of the time-shift rule).
**Convolution theorem.** Define the **convolution** $(f * g)(t) = \int_{-\infty}^{\infty}
f(\tau)\,g(t-\tau)\,d\tau$. Then
$$
\boxed{\;\mathcal F[f * g] = \hat f \cdot \hat g\;}\qquad\text{and dually}\qquad
\mathcal F[f\cdot g] = \hat f * \hat g.
$$
*Proof.* Write out and swap the order of integration (justified for $L^1$ functions by Fubini):
$$
\mathcal F[f*g](\xi) = \int_t\!\!\int_\tau f(\tau)g(t-\tau)\,d\tau\;e^{-i2\pi\xi t}\,dt
= \int_\tau f(\tau)\Big[\int_t g(t-\tau)e^{-i2\pi\xi t}\,dt\Big]d\tau.
$$
The inner integral is the time-shift rule: $\int_t g(t-\tau)e^{-i2\pi\xi t}dt = e^{-i2\pi\xi\tau}
\hat g(\xi)$. Hence
$$
\mathcal F[f*g](\xi) = \hat g(\xi)\int_\tau f(\tau)e^{-i2\pi\xi\tau}\,d\tau = \hat f(\xi)\hat g(\xi).
\qquad\blacksquare
$$
This is the property that makes the FFT a workhorse of computation: convolution costs $O(N^2)$
directly but $O(N\log N)$ via "transform, multiply, inverse-transform." We use it in the code.
**Parseval / Plancherel — derivation.** With the ordinary-frequency convention above, the transform
*preserves energy*:
$$
\boxed{\;\int_{-\infty}^{\infty}|f(t)|^2\,dt = \int_{-\infty}^{\infty}|\hat f(\xi)|^2\,d\xi\;}.
$$
*Proof.* Start from the left and write $|f|^2 = f\,f^{*}$, expressing $f$ by inversion and $f^{*}$ by
the conjugate of inversion:
$$
\int_t |f(t)|^2\,dt
= \int_t f(t)\Big(\int_\xi \hat f(\xi) e^{i2\pi\xi t}\,d\xi\Big)^{\!*}\,dt
= \int_t f(t)\int_\xi \hat f(\xi)^{*} e^{-i2\pi\xi t}\,d\xi\,dt.
$$
Swap the order of integration (Fubini) and group the $t$-integral:
$$
= \int_\xi \hat f(\xi)^{*}\Big[\int_t f(t)e^{-i2\pi\xi t}\,dt\Big]d\xi
= \int_\xi \hat f(\xi)^{*}\,\hat f(\xi)\,d\xi
= \int_\xi |\hat f(\xi)|^2\,d\xi. \qquad\blacksquare
$$
The bracket is just the forward transform $\hat f(\xi)$. The polarized version
$\int f^{*}g = \int \hat f^{*}\hat g$ (Plancherel) follows identically with $g$ in place of the
second $f$ — i.e. **the Fourier transform is unitary on $L^2$**, an inner-product-preserving map.
This is the continuous shadow of the fact, proved in [lesson 3](03-discrete-fourier-transform.md),
that the normalized DFT matrix is unitary — and ultimately of why the *Quantum* Fourier Transform is
a legal (unitary) quantum operation.
### Reading the frequency domain
The spectrum $\hat f$ is itself complex: $|\hat f(\xi)|$ is the **magnitude spectrum** (how much
energy at frequency $\xi$) and $\arg\hat f(\xi)$ is the **phase spectrum** (the alignment of that
component). Sharp features in time spread out in frequency and vice versa — a delta in time is flat
in frequency; a single tone in time is a spike in frequency. This reciprocity (formalized by the
uncertainty principle and by the sampling theorem in lesson 3) is *the* recurring theme: localizing
in one domain costs spread in the other.
---
## Worked Examples
### Example 1 — Fourier series of a square wave
Let $f$ be the odd square wave of period $T = 2\pi$ with $f(t) = +1$ on $(0,\pi)$ and $-1$ on
$(-\pi,0)$ (so $T$-periodic). Compute its complex coefficients $c_n = \tfrac1T\int_{-\pi}^{\pi}
f(t)e^{-int}\,dt$ (here $2\pi n/T = n$).
For $n=0$: $c_0 = \tfrac1{2\pi}\int_{-\pi}^{\pi} f = 0$ (equal $+$ and $-$ areas; zero average).
For $n\neq 0$, split the integral and use $\int e^{-int}dt = e^{-int}/(-in)$:
$$
c_n = \frac{1}{2\pi}\Big[\int_0^{\pi}(+1)e^{-int}\,dt + \int_{-\pi}^{0}(-1)e^{-int}\,dt\Big].
$$
The first integral is $\frac{1}{-in}(e^{-in\pi}-1)$; the second is $-\frac{1}{-in}(1 - e^{in\pi})$.
Using $e^{\pm in\pi} = (-1)^n$ and simplifying,
$$
c_n = \frac{1}{2\pi}\cdot\frac{1}{-in}\Big[\big((-1)^n - 1\big) - \big(1 - (-1)^n\big)\Big]
= \frac{1}{2\pi}\cdot\frac{2\big((-1)^n - 1\big)}{-in}
= \frac{1 - (-1)^n}{i\pi n}.
$$
So $c_n = 0$ for even $n$ and $c_n = \dfrac{2}{i\pi n}$ for odd $n$. Reassembling the real series
(pairing $\pm n$, only odd $n$ survive) gives the classic
$$
f(t) = \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin\big((2k+1)t\big)}{2k+1}
= \frac{4}{\pi}\Big(\sin t + \tfrac13\sin 3t + \tfrac15\sin 5t + \cdots\Big).
$$
Only odd harmonics appear (by the symmetry $f(t+\pi) = -f(t)$), and their amplitudes decay like
$1/n$ — slow decay characteristic of a jump discontinuity, which is why truncating the series
produces Gibbs overshoot near $t=0,\pi$.
Truncate it yourself below. Beside the jump the partial sum peaks near
$\tfrac{2}{\pi}\operatorname{Si}(\pi) \approx 1.179$, about 9% of the jump of 2, however many
terms you keep; the overshoot only narrows. The triangle and the plucked string, which have no
jumps, show the contrast: their coefficients fall like $1/n^2$ and the error goes to zero.
```qm-demo fourier-synthesis
```
### Example 2 — Transform of a Gaussian (a fixed point)
Let $f(t) = e^{-\pi t^2}$. Its transform is itself:
$$
\hat f(\xi) = \int_{-\infty}^{\infty} e^{-\pi t^2}e^{-i2\pi\xi t}\,dt = e^{-\pi\xi^2}.
$$
*Sketch.* Complete the square in the exponent: $-\pi t^2 - i2\pi\xi t = -\pi(t + i\xi)^2 - \pi\xi^2$.
Pulling out $e^{-\pi\xi^2}$ leaves $\int e^{-\pi(t+i\xi)^2}dt$, which equals $\int e^{-\pi u^2}du = 1$
(the Gaussian integral, with a contour-shift justification). Hence $\hat f(\xi) = e^{-\pi\xi^2}$. The
Gaussian is the eigenfunction with eigenvalue $1$: maximally "spread-balanced" between domains, the
extremal case of the time–frequency reciprocity. (Wide Gaussians in time become narrow in frequency
and vice versa, as the scaling property $\mathcal F[f(at)] = \tfrac1{|a|}\hat f(\xi/a)$ shows.)
### Example 3 — Convolution as filtering
Let $g$ be a normalized box of width $w$: $g(t) = \tfrac1w$ for $|t|\le w/2$, else $0$. Then $(f *
g)(t) = \tfrac1w\int_{t-w/2}^{t+w/2} f(\tau)\,d\tau$ is the *moving average* of $f$ — a smoother. Its
transform is, by the convolution theorem, $\widehat{f*g} = \hat f\cdot\hat g$ where $\hat g(\xi) =
\operatorname{sinc}(w\xi) = \sin(\pi w\xi)/(\pi w\xi)$. Because $\hat g$ decays away from $\xi=0$,
multiplication *suppresses high frequencies*: smoothing in time = low-pass filtering in frequency.
This is the entire logic of digital filtering, and the next code cell verifies it numerically.
---
## Hands-on (Python)
```python
import numpy as np
import matplotlib.pyplot as plt
# =====================================================================
# 1. Numerically approximate a Fourier series (square wave) and see Gibbs
# =====================================================================
def square_wave(t):
"""Odd square wave, period 2π: +1 on (0, π), -1 on (-π, 0)."""
return np.sign(np.sin(t))
def fourier_partial_sum(t, M):
"""Sum the odd harmonics through (2M-1): (4/π) Σ sin((2k+1)t)/(2k+1)."""
out = np.zeros_like(t)
for k in range(M):
n = 2 * k + 1
out += np.sin(n * t) / n
return (4 / np.pi) * out
t = np.linspace(-np.pi, np.pi, 2000, endpoint=False)
plt.figure(figsize=(8, 4))
plt.plot(t, square_wave(t), "k--", lw=1, label="target square wave")
for M in (1, 3, 10, 50):
plt.plot(t, fourier_partial_sum(t, M), lw=1, label=f"M={M} harmonics")
plt.title("Fourier partial sums → square wave (note Gibbs overshoot at jumps)")
plt.xlabel("t"); plt.ylabel("f(t)"); plt.legend(fontsize=8); plt.tight_layout(); plt.show()
# Cross-check the *coefficients* numerically against the closed form 2/(iπn) for odd n.
# square_wave(t) = sign(sin t) is exactly the Example-1 signal: +1 on (0,π), -1 on (π,2π).
T = 2 * np.pi
M = 200_000
ts = np.linspace(0, T, M, endpoint=False) # one period, uniform grid
dt = T / M
f = square_wave(ts)
for n in (1, 2, 3, 5):
# cn = (1/T)∫_0^T f e^{-int} dt, as a Riemann sum (version-robust; np.trapz was
# removed in NumPy 2.0). For odd n this matches 2/(iπn); for even n it is 0.
cn = np.sum(f * np.exp(-1j * n * ts)) * dt / T
closed = (1 - (-1) ** n) / (1j * np.pi * n) # = 2/(iπn) for odd n, 0 for even n
print(f"n={n}: numeric c_n = {cn: .4f} closed form = {closed: .4f}")
# =====================================================================
# 2. Demonstrate the convolution theorem with np.fft
# Claim: conv(f, g) == IFFT( FFT(f) * FFT(g) ) (circular convolution)
# =====================================================================
rng = np.random.default_rng(0)
N = 256
f = rng.standard_normal(N) # a noisy signal
# A small normalized box kernel (moving-average / low-pass filter), zero-padded to length N.
w = 9
g = np.zeros(N); g[:w] = 1.0 / w
# (a) Direct CIRCULAR convolution (matches what the DFT computes):
direct = np.array([np.sum(f * np.roll(g[::-1], k + 1)) for k in range(N)])
# (b) Via the convolution theorem: multiply spectra, transform back.
via_fft = np.fft.ifft(np.fft.fft(f) * np.fft.fft(g)).real
print("\nconvolution theorem holds:",
np.allclose(direct, via_fft, atol=1e-10)) # True
# Show that convolving with the box suppresses high-frequency content (low-pass).
F = np.fft.rfft(f)
FG = np.fft.rfft(via_fft)
freq = np.fft.rfftfreq(N)
plt.figure(figsize=(8, 4))
plt.plot(freq, np.abs(F), label="|FFT(signal)|")
plt.plot(freq, np.abs(FG), label="|FFT(signal ∗ box)|")
plt.title("Convolution with a box = multiplication by a sinc = low-pass filter")
plt.xlabel("frequency"); plt.ylabel("magnitude"); plt.legend(); plt.tight_layout(); plt.show()
```
What you should observe: the partial sums visibly converge to the square wave, with the persistent
~9% Gibbs overshoot hugging the jumps as $M$ grows; the numerically computed $c_n$ match $2/(i\pi n)$
for odd $n$ and vanish for even $n$; `np.allclose(direct, via_fft)` prints `True`, confirming
$\mathcal F[f*g] = \hat f\hat g$; and the filtered spectrum is visibly attenuated at high
frequencies — convolution in time is multiplication in frequency.
> **Note on circular vs. linear convolution.** `np.fft` computes *circular* convolution (the DFT
> treats data as one period of a periodic signal — see [lesson 3](03-discrete-fourier-transform.md)).
> For *linear* convolution, zero-pad both inputs to length $\ge \text{len}(f)+\text{len}(g)-1$ before
> transforming. This periodicity is the bridge from the continuous theory here to the discrete
> machinery next.
---
## Exercises
**1. (Easy) Linearity and shift.** Given $\hat f(\xi)$, write the transform of $h(t) = 3f(t) -
f(t-5)$ in terms of $\hat f$.
Solution
By linearity and the time-shift rule, $\hat h(\xi) = 3\hat f(\xi) - e^{-i2\pi\xi\cdot 5}\hat f(\xi)
= \big(3 - e^{-i10\pi\xi}\big)\hat f(\xi)$. The delay contributes a frequency-dependent phase factor
$e^{-i10\pi\xi}$; the magnitude spectrum is reshaped by $|3 - e^{-i10\pi\xi}|$.
Solution
$c_0 = \tfrac1T\int_0^T f(t)e^{0}\,dt = \tfrac1T\int_0^T f$, which is the mean by definition. For
$f = 2 + 3\cos(2\pi t/T)$ the cosine integrates to $0$ over a full period, leaving $c_0 = 2$. (The
$\cos$ contributes $c_{\pm 1} = 3/2$ each, since $\cos\theta = \tfrac12(e^{i\theta}+e^{-i\theta})$.)
Solution
Write $f = \sum_n c_n e_n$ and $f^{*} = \sum_m c_m^{*} e_m^{*}$ with $e_n(t)=e^{i2\pi nt/T}$. Then
$$
\tfrac1T\int_0^T |f|^2\,dt = \tfrac1T\int_0^T \sum_{n,m} c_n c_m^{*}\, e_n e_m^{*}\,dt
= \sum_{n,m} c_n c_m^{*}\,\tfrac1T\langle e_m, e_n\rangle = \sum_{n,m} c_n c_m^{*}\,\delta_{mn}
= \sum_n |c_n|^2,
$$
using the orthogonality lemma $\tfrac1T\langle e_m, e_n\rangle = \delta_{mn}$. Energy in time =
energy in the coefficients. $\blacksquare$
Solution
$(f*g)(t) = \int \mathbf 1_{[0,1]}(\tau)\mathbf 1_{[0,1]}(t-\tau)\,d\tau$ = length of the overlap of
$[0,1]$ and $[t-1,t]$. This is the **triangle** function: $0$ for $t\le 0$, rising as $t$ for
$0\le t\le 1$, falling as $2-t$ for $1\le t\le 2$, and $0$ for $t\ge 2$. For the transform,
$\hat f(\xi) = \int_0^1 e^{-i2\pi\xi t}dt = \frac{1-e^{-i2\pi\xi}}{i2\pi\xi} = e^{-i\pi\xi}
\operatorname{sinc}(\xi)$ (with $\operatorname{sinc}(\xi)=\sin\pi\xi/\pi\xi$). By the convolution
theorem $\widehat{f*g} = \hat f^2 = e^{-i2\pi\xi}\operatorname{sinc}^2(\xi)$, the well-known
transform of a triangle (a squared sinc) — consistent with the direct computation.
Solution
Repeat the Parseval derivation with $g$ in the second slot: expand $\langle f,g\rangle = \int_t
f(t)^{*}g(t)\,dt$, substitute $g(t) = \int_\xi \hat g(\xi)e^{i2\pi\xi t}d\xi$, swap integrals
(Fubini), and recognize $\int_t f(t)^{*}e^{i2\pi\xi t}dt = \big(\int_t f(t)e^{-i2\pi\xi t}dt\big)^{*}
= \hat f(\xi)^{*}$. This leaves $\int_\xi \hat f(\xi)^{*}\hat g(\xi)\,d\xi = \langle\hat f,\hat
g\rangle$. Preserving the inner product is exactly the defining property of a unitary operator
($U^\dagger U = I$); in finite dimensions the same statement is "the normalized DFT matrix $U$
satisfies $U^\dagger U = I$," proved in [lesson 3](03-discrete-fourier-transform.md).
Solution
Integrate the coefficient formula by parts. Each integration by parts moves a derivative off the
exponential and onto $f$: $\int_0^T f(t)e^{-i2\pi n t/T}dt = \frac{T}{i2\pi n}\int_0^T
f'(t)e^{-i2\pi n t/T}dt$ (boundary terms vanish by periodicity). Hence $c_n = \frac{T}{i2\pi n}\cdot
c_n^{(f')}$, i.e. differentiating multiplies the coefficient by $i2\pi n/T$; equivalently a smooth
$f'$ has bounded coefficients, so $c_n = O(1/|n|)$ for $C^1$. Repeating $k$ times gives $c_n =
O(|n|^{-k})$ provided $f^{(k)}\in L^2$. The square wave is *not* continuous ($k=0$ effectively, with
a jump), so its coefficients decay only like $1/n$ — the slowest case — which is precisely why its
truncated series converges slowly and exhibits Gibbs overshoot. Smoothness in time ⇔ rapid decay in
frequency, the reciprocity theme again.
1. State the orthogonality relation for $e_n(t)=e^{i2\pi nt/T}$ on $[0,T]$ and why it matters.
$\int_0^T e_m^{*}e_n\,dt = T\delta_{mn}$. It lets you read off each Fourier coefficient
independently by an inner product: $c_n = \tfrac1T\langle e_n, f\rangle$ — projecting $f$ onto one
pure frequency without interference from the others.
2. Write the analysis and synthesis formulas for the continuous Fourier transform (this lesson's convention).
Analysis: $\hat f(\xi) = \int f(t)e^{-i2\pi\xi t}dt$. Synthesis (inverse): $f(t) = \int \hat
f(\xi)e^{+i2\pi\xi t}d\xi$. The $\mp$ signs and the absence of $1/2\pi$ factors match the program's
DFT/QFT conventions.
3. What does the convolution theorem say, and why is it computationally important?
$\mathcal F[f*g] = \hat f\cdot\hat g$: convolution in time is pointwise multiplication in frequency.
It turns an $O(N^2)$ convolution into "FFT, multiply, inverse-FFT," which is $O(N\log N)$.
4. State Parseval/Plancherel and interpret it physically.
$\int|f|^2\,dt = \int|\hat f|^2\,d\xi$ (and $\langle f,g\rangle = \langle\hat f,\hat g\rangle$).
Total energy is conserved between time and frequency domains; equivalently, $\mathcal F$ is unitary
on $L^2$.
5. Why does a time delay leave the magnitude spectrum unchanged?
Because $\mathcal F[f(t-t_0)] = e^{-i2\pi\xi t_0}\hat f(\xi)$: the delay multiplies by a
unit-modulus phase factor, so $|\hat f|$ is unchanged and only the phase spectrum tilts.
6. How does this continuous theory foreshadow the quantum Fourier transform?
The Fourier transform is unitary (Plancherel), built from orthogonal complex exponentials, and
diagonalizes convolution. Its finite-dimensional version (the DFT, lesson 3) is a unitary matrix,
and the *quantum* version implements that same unitary on amplitudes — see
[Term 2.4](../../term-2-qc-core/c04-qft-qpe/01-quantum-fourier-transform.md).