Fourier Series and Transforms

7.1Even and Odd Functions

7.2Periodic Functions and Fourier Series

7.2.1Fundamental Periodic Basis

For the exponential function \( e^{ikx} \) to have period \(L\),

Since \( e^{i2\pi n}=1 \) for integers \(n\), we have

Thus, the functions

form a complete orthogonal set over one period.

7.2.2Complex Fourier Series

7.2.3Real Form of the Fourier Series

7.2.4Dirichlet Conditions and Convergence of Fourier Series

7.2.5Meaning and Interpretation of the Fourier Series

The Fourier series allows us to express any sufficiently smooth periodic function \( f(x) \) (with period \(L\)) as a sum of simple oscillatory components — sine and cosine waves. Mathematically,

Each term in the sum represents a harmonic component (or mode) of the signal.

Thus,

Each harmonic contributes a specific frequency, amplitude, and phase to the total waveform. The original function \( f(x) \) is reconstructed by summing all harmonic components:

As more harmonics are included, the approximation of \( f(x) \) becomes more accurate. Lower harmonics (\(n=0,1,2\)) describe the overall shape, while higher harmonics capture finer details and sharp transitions.

Applications in Signal Processing

Example - Square step on \(\lbrack{}-\pi,\pi\rbrack{}\)

Find the Fourier series of

\begin{equation} f(x)=\begin{cases} 0,& -\pi<x<0,\\[2pt] 1,& 0<x<\pi, \end{cases} \qquad\text{(period }2\pi\text{)}. \end{equation}

Fourier form on \([-\pi,\pi]\). With period \(L=2\pi\), the series is

\begin{equation} f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\big[a_n\cos(nx)+b_n\sin(nx)\big], \end{equation}

where, from the coefficient formulas with \(L=2\pi\) so that \(\tfrac{2}{L}=\tfrac1\pi\),

\begin{equation} a_0=\frac{1}{\pi}\!\int_{-\pi}^{\pi}\!f(x)\,dx,\quad a_n=\frac{1}{\pi}\!\int_{-\pi}^{\pi}\!f(x)\cos(nx)\,dx,\quad b_n=\frac{1}{\pi}\!\int_{-\pi}^{\pi}\!f(x)\sin(nx)\,dx. \end{equation}

Solution.

Step 1: \(a_0\). The constant term of the series is \(\tfrac{a_0}{2}\), so we expect it to equal the average of \(f\), namely \(\tfrac12\).

\begin{equation} a_0=\frac{1}{\pi}\!\int_{-\pi}^{\pi}\!f(x)\,dx = \frac{1}{\pi}\!\int_{0}^{\pi}\!1\,dx =1 \quad\Rightarrow\quad \frac{a_0}{2}=\frac{1}{2}. \end{equation}

Step 2: \(a_n\) (all \(n\ge1\)).

\begin{equation} a_n=\frac{1}{\pi}\!\int_{0}^{\pi}\!\cos(nx)\,dx =\frac{1}{\pi}\cdot\frac{\sin(nx)}{n}\Big|_{0}^{\pi}=0. \end{equation}

Step 3: \(b_n\) (all \(n\ge1\)).

\begin{equation} b_n=\frac{1}{\pi}\!\int_{0}^{\pi}\!\sin(nx)\,dx =\frac{1}{\pi}\cdot\frac{-\cos(nx)}{n}\Big|_{0}^{\pi} =\frac{1}{n\pi}\big(1-\cos(n\pi)\big) =\frac{1-(-1)^n}{n\pi}. \end{equation}

Hence

\begin{equation} b_n= \begin{cases} 0,& n\ \text{even},\\[4pt] \displaystyle \frac{2}{n\pi},& n\ \text{odd}. \end{cases} \end{equation}

Result.

\begin{equation} \boxed{ f(x)=\frac{1}{2}+\frac{2}{\pi}\!\left(\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\cdots\right) } \end{equation}

(i.e., the sum runs over odd \(n\)). At the jump points \(x\equiv 0,\pi\ (\mathrm{mod}\ 2\pi)\), the series converges to the average \(\tfrac{1}{2}\) (Dirichlet’s theorem).

Example - Periodic ramp on \((0,\pi)\) (period \(2\pi\))

Fourier form on \([-\pi,\pi]\) (period \(2\pi\)).

\begin{equation} f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\big[a_n\cos(nx)+b_n\sin(nx)\big], \quad \begin{aligned} a_0&=\frac{1}{\pi}\!\int_{-\pi}^{\pi} f(x)\,dx,\\ a_n&=\frac{1}{\pi}\!\int_{-\pi}^{\pi} f(x)\cos(nx)\,dx,\\ b_n&=\frac{1}{\pi}\!\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx. \end{aligned} \end{equation}

Solution.

\begin{equation} a_0=\frac{1}{\pi}\int_{0}^{\pi}\!x\,dx =\frac{1}{\pi}\Big[\frac{x^{2}}{2}\Big]_0^{\pi} =\frac{\pi}{2} \quad\Rightarrow\quad \frac{a_0}{2}=\boxed{\frac{\pi}{4}}. \end{equation}

Step 2: Cosine coefficients (\(n\ge1\)).

\begin{equation} a_n=\frac{1}{\pi}\int_{0}^{\pi}\!x\cos(nx)\,dx \quad(\text{IBP: }u=x,\ dv=\cos nx\,dx \Rightarrow v=\tfrac{1}{n}\sin nx). \end{equation}

\begin{equation} a_n=\frac{1}{\pi}\Big[x\frac{\sin(nx)}{n}\Big]_{0}^{\pi} -\frac{1}{\pi}\int_{0}^{\pi}\frac{\sin(nx)}{n}\,dx =\frac{1}{\pi n^2}\big(\cos n\pi-1\big) =\boxed{\frac{(-1)^n-1}{\pi n^2}} = \begin{cases} 0,& n\ \text{even},\\[4pt] -\dfrac{2}{\pi n^2},& n\ \text{odd}. \end{cases} \end{equation}

Step 3: Sine coefficients (\(n\ge1\)).

\begin{equation} b_n=\frac{1}{\pi}\int_{0}^{\pi}\!x\sin(nx)\,dx \quad(\text{IBP: }u=x,\ dv=\sin nx\,dx \Rightarrow v=-\tfrac{1}{n}\cos nx). \end{equation}

\begin{equation} b_n=\frac{1}{\pi}\Big[-x\frac{\cos(nx)}{n}\Big]_{0}^{\pi} +\frac{1}{\pi}\int_{0}^{\pi}\frac{\cos(nx)}{n}\,dx =-\frac{(-1)^n}{n} =\boxed{\frac{(-1)^{n+1}}{n}}. \end{equation}

Final series.

\begin{equation} \boxed{ f(x)=\frac{\pi}{4} +\sum_{n=1}^{\infty}\left[ \frac{(-1)^n-1}{\pi n^2}\cos(nx) +\frac{(-1)^{n+1}}{n}\sin(nx) \right] } \end{equation}

or, separating even/odd indices,

\begin{equation} \boxed{ f(x)=\frac{\pi}{4} -\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\cos\big((2k+1)x\big)}{(2k+1)^2} +\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx). } \end{equation}

At the jump points \(x\equiv 0,\pi\ (\mathrm{mod}\ 2\pi)\), the series converges to the midpoint value \(\tfrac{1}{2}(0+\pi)=\tfrac{\pi}{2}\).

Example - Periodic half-ramp

Let

\begin{equation} f(t)= \begin{cases} 0, & -\dfrac{T}{2}< t < 0,\\[4pt] t, & 0< t < \dfrac{T}{2}, \end{cases} \qquad \text{extended periodically with period }T. \end{equation}

We find its real Fourier series

\begin{equation} f(t)=\dfrac{a_0}{2}+\sum_{n=1}^{\infty}\Big[ a_n\cos(\omega_n t)+ b_n\sin(\omega_n t)\Big], \qquad \omega_n=\dfrac{2\pi n}{T}. \end{equation}

Solution.

\begin{equation} a_0=\frac{2}{T}\int_{-T/2}^{T/2}f(t)\,dt =\frac{2}{T}\int_{0}^{T/2} t\,dt =\frac{2}{T}\cdot\frac{(T/2)^2}{2} =\boxed{\frac{T}{4}}. \end{equation}

Hence the DC term is \(a_0/2=\boxed{T/8}\).

Step 2: Cosine coefficients.

\begin{equation} a_n=\frac{2}{T}\int_{-T/2}^{T/2} f(t)\cos(\omega_n t)\,dt =\frac{2}{T}\int_{0}^{T/2} t\cos(\omega_n t)\,dt. \end{equation}

Integration by parts with \(u=t\), \(dv=\cos(\omega_n t)\,dt\) gives

\begin{equation} \int t\cos(\omega_n t)\,dt=\frac{t\sin(\omega_n t)}{\omega_n} +\frac{\cos(\omega_n t)}{\omega_n^2}. \end{equation}

Therefore

\begin{equation} \begin{aligned} a_n&=\frac{2}{T}\left[ \frac{t\sin(\omega_n t)}{\omega_n} +\frac{\cos(\omega_n t)}{\omega_n^2} \right]_{0}^{T/2} =\frac{2}{T}\left[ \frac{(T/2)\sin(\omega_n T/2)}{\omega_n} +\frac{\cos(\omega_n T/2)-1}{\omega_n^2} \right]. \end{aligned} \end{equation}

Using \(\sin(\omega_n T/2)=\sin(n\pi)=0\) and \(\cos(\omega_n T/2)=\cos(n\pi)=(-1)^n\),

\begin{equation} \boxed{\,a_n=\frac{2}{T}\cdot\frac{(-1)^n-1}{\omega_n^2}\,} =\boxed{\,\frac{T}{2\pi^2 n^2}\Big[(-1)^n-1\Big]\,}. \end{equation}

Hence

\begin{equation} \boxed{\,a_n=0\ \text{for even }n,\qquad a_n=-\frac{T}{\pi^2 n^2}\ \text{for odd }n.\,} \end{equation}

Step 3: Sine coefficients.

\begin{equation} b_n=\frac{2}{T}\int_{-T/2}^{T/2} f(t)\sin(\omega_n t)\,dt =\frac{2}{T}\int_{0}^{T/2} t\sin(\omega_n t)\,dt. \end{equation}

With \(u=t\), \(dv=\sin(\omega_n t)\,dt\),

\begin{equation} \int t\sin(\omega_n t)\,dt =-\frac{t\cos(\omega_n t)}{\omega_n} +\frac{\sin(\omega_n t)}{\omega_n^2}. \end{equation}

Thus

\begin{equation} b_n=\frac{2}{T}\left[ -\frac{t\cos(\omega_n t)}{\omega_n} +\frac{\sin(\omega_n t)}{\omega_n^2} \right]_{0}^{T/2} =-\frac{2}{T}\cdot\frac{(T/2)\cos(\omega_n T/2)}{\omega_n} =\boxed{\,\frac{(-1)^{n+1}}{\omega_n}\,} =\boxed{\,\frac{(-1)^{n+1}T}{2\pi n}\,}. \end{equation}

Result (real Fourier series).

\begin{equation} \boxed{ f(t)=\frac{T}{8} +\sum_{\substack{n=1\\ \text{odd }n}}^{\infty}\!\left(-\frac{T}{\pi^2 n^2}\right)\cos\!\left(\frac{2\pi n t}{T}\right) +\sum_{n=1}^{\infty}\frac{(-1)^{n+1}T}{2\pi n}\,\sin\!\left(\frac{2\pi n t}{T}\right) }. \end{equation}

• \(a_n\) vanish for even \(n\); cosine terms appear only for odd harmonics and decay as \(1/n^2\).

• \(b_n\) decay as \(1/n\) with alternating sign, reflecting the sharp corner at \(t=0\) (Gibbs effect upon truncation).

• The average (DC) level over a period is \(T/8\).

Example - Complex Fourier Series of a Periodic Ramp Function \(f(x)=x\) (period \(L\))

Consider a spatially periodic function

\begin{equation} f(x) = x, \qquad -\frac{L}{2} < x < \frac{L}{2}, \end{equation}

which repeats every interval of length \(L\). This represents a periodic ramp waveform, analogous to a sawtooth pattern in signal processing.

Solution.

For a periodic function \(f(x)\) of period \(L\), the complex Fourier coefficients are

\begin{equation} c_n = \frac{1}{L}\int_{-L/2}^{L/2} f(x)\, e^{-i\,k_n x}\,dx, \qquad \text{where } k_n = \frac{2\pi n}{L}. \end{equation}

Substitute \(f(x)=x\):

\begin{equation} \boxed{ c_n = \frac{1}{L}\int_{-L/2}^{L/2} x\, e^{-i k_n x}\,dx. } \end{equation}

Step 2. Integration by parts.

Let

\begin{equation} u = x, \quad dv = e^{-i k_n x}\,dx \quad\Rightarrow\quad du = dx, \quad v = \frac{e^{-i k_n x}}{-i k_n} = \frac{i}{k_n} e^{-i k_n x}. \end{equation}

Then

\begin{equation} \begin{aligned} c_n &= \frac{1}{L}\Big[u v\Big]_{-L/2}^{L/2} - \frac{1}{L}\int_{-L/2}^{L/2} v\,du \\[4pt] &= \frac{i}{L k_n}\Big[x\,e^{-i k_n x}\Big]_{-L/2}^{L/2} - \frac{i}{L k_n}\int_{-L/2}^{L/2} e^{-i k_n x}\,dx. \end{aligned} \end{equation}

Step 3. Simplify the second term.

The exponential integral gives

\begin{equation} \int_{-L/2}^{L/2} e^{-i k_n x}\,dx = \frac{2\sin(k_n L/2)}{k_n} = \frac{2\sin(\pi n)}{k_n} = 0, \end{equation}

since \(\sin(\pi n)=0\) for all integer \(n\). Hence, only the boundary term contributes.

Step 4. Evaluate the boundary term.

\begin{equation} \begin{aligned} c_n &= \frac{i}{L k_n} \Big[\frac{L}{2} e^{-i k_n L/2} - \left(-\frac{L}{2}\right)e^{i k_n L/2}\Big] \\[4pt] &= \frac{i}{2 k_n}\Big(e^{-i\pi n} + e^{i\pi n}\Big) = \frac{i}{k_n}\cos(\pi n). \end{aligned} \end{equation}

Using \(\cos(\pi n)=(-1)^n\), we have

\begin{equation} \boxed{ c_n = \frac{i}{k_n}(-1)^n = \frac{iL}{2\pi n}(-1)^n, \qquad n\neq 0. } \end{equation}

Step 5. DC (zero-frequency) coefficient.

\begin{equation} c_0 = \frac{1}{L}\int_{-L/2}^{L/2} x\,dx = 0, \end{equation}

since \(f(x)=x\) is an odd function and its average over one period is zero.

Step 6. Complete complex Fourier series.

\begin{equation} \boxed{ f(x) = \sum_{\substack{n=-\infty \\ n\neq 0}}^{\infty} \frac{iL}{2\pi n}(-1)^n\, e^{i\,k_n x}, \qquad k_n = \frac{2\pi n}{L}. } \end{equation}

Interpretation.

• The amplitude of each harmonic decreases as \(1/n\), meaning higher-frequency components contribute less to the total waveform.

• Because \(f(x)\) is an odd function, the coefficients \(c_n\) are purely imaginary (\(c_{-n}=c_n^*\)), corresponding to a sine-only spectrum.

• Truncating the series to a few terms yields a smooth approximation of the ramp—analogous to applying a low-pass filter that removes sharp discontinuities.

7.3Fourier Transform

The Fourier transform can be viewed as the limiting case of the complex Fourier series when the period \(L \to \infty\). In this limit, the function \( f(x) \) becomes nonperiodic, and the discrete spectrum of wave numbers becomes continuous.

Transition from Discrete to Continuous Spectrum

Consider a periodic function \(f(x)\) with period \(L\):

When \(L \to \infty\), the spacing between successive wave numbers becomes infinitesimal:

The discrete index \(n\) is replaced by a continuous variable \(k\):

Substituting into the series and taking the limit \(L \to \infty\):

Example - Rectangular pulse of width \(2a\)

Let

\begin{equation} f(x)= \begin{cases} 1, & -a<x<a,\\[2pt] 0, & \text{otherwise}, \end{cases} \qquad a>0. \end{equation}

This is a nonperiodic rectangular pulse centered at the origin with total width \(2a\) and unit height.

We use the transform pair

\begin{equation} g(k)=\frac{1}{2\pi}\!\int_{-\infty}^{\infty} f(x)\,e^{-ikx}\,dx, \qquad f(x)=\int_{-\infty}^{\infty} g(k)\,e^{ikx}\,dk. \end{equation}

Solution.

\begin{equation} g(k)=\frac{1}{2\pi}\int_{-a}^{a} e^{-ikx}\,dx. \end{equation}

Step 2. Evaluate the integral.

\begin{equation} \int e^{-ikx}\,dx = \frac{e^{-ikx}}{-ik} \quad\Rightarrow\quad \int_{-a}^{a} e^{-ikx}\,dx = \frac{e^{-ika}-e^{ika}}{-ik} = \frac{2\sin(ka)}{k}. \end{equation}

Step 3. Simplify and express the result.

\begin{equation} \boxed{ g(k) = \frac{1}{2\pi}\cdot\frac{2\sin(ka)}{k} = \frac{1}{\pi}\,\frac{\sin(ka)}{k} = \frac{a}{\pi}\,\mathrm{sinc}(ak), } \qquad \mathrm{sinc}(u)=\frac{\sin u}{u}. \end{equation}

At \(k=0\), we take the limit \(\displaystyle g(0)=\lim_{k\to0}\frac{a}{\pi}\,\mathrm{sinc}(ak)=\frac{a}{\pi}.\)

Key features.

• \(g(k)\) is real, even, and has zeros at \(k=\pm n\pi/a\) for \(n=1,2,3,\ldots\).

• The main lobe width \(\Delta k \sim 2\pi/a\) — increasing \(a\) (wider pulse) narrows the spectrum.

• The transform pair summarizes the width–bandwidth tradeoff: a narrow function in \(k\)-space corresponds to a wide function in \(x\)-space and vice versa.

Example - Fourier Transform of \( f(t) = e^{-a|t|} \)

Let

\begin{equation} f(t) = e^{-a|t|}, \qquad a > 0. \end{equation}

This function decays exponentially on both sides of \(t=0\):

We use the Fourier transform convention:

\begin{equation} f(t) = \int_{-\infty}^{\infty} g(\omega)e^{i\omega t}\,d\omega, \qquad g(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(t)e^{-i\omega t}\,dt. \end{equation}

Solution.

\begin{equation} g(\omega) = \frac{1}{2\pi}\!\left[ \int_{-\infty}^{0} e^{at} e^{-i\omega t}\,dt + \int_{0}^{\infty} e^{-at} e^{-i\omega t}\,dt \right]. \end{equation}

Step 2. Evaluate each integral.

\begin{equation} \int_{0}^{\infty} e^{-at} e^{-i\omega t}\,dt = \frac{1}{a+i\omega}, \qquad \int_{-\infty}^{0} e^{at} e^{-i\omega t}\,dt = \frac{1}{a-i\omega}. \end{equation}

Step 3. Combine the results:

\begin{equation} g(\omega) = \frac{1}{2\pi}\left(\frac{1}{a+i\omega} + \frac{1}{a-i\omega}\right) = \frac{1}{\pi}\,\frac{a}{a^2+\omega^2}. \end{equation}

Thus, the Fourier transform of \( e^{-a|t|} \) is

\begin{equation} \boxed{ g(\omega) = \frac{1}{\pi}\,\frac{a}{a^2+\omega^2} } \end{equation}

which is known as the Lorentzian function.

Shape of \( g(\omega) \):

As \(a\) increases, \(g(\omega)\) becomes broader and shorter; as \(a\) decreases, it becomes narrower and taller.

7.4Where Fourier Analysis Is Used

The single idea of this chapter — that a function can be rebuilt from sines and cosines, or from complex exponentials — is one of the most widely used in all of physics and engineering. The reason is that the basis functions \(e^{ikx}\) are the natural language of anything that oscillates, and of the linear equations that govern waves, heat and quantum systems.

• Solving partial differential equations. Fourier's original motive was heat flow. Because each mode \(e^{ikx}\) evolves independently under the heat, wave and Schrödinger equations, expanding the initial state in modes turns a partial differential equation into a set of ordinary ones. This is the method of Chapter 10.

• Signal processing. A signal's Fourier coefficients are its frequency content. Filtering, noise removal and compression are all operations on that spectrum, as the low-pass filter of this chapter showed.

• Spectroscopy and optics. The rectangular-pulse and Lorentzian transforms computed above are exactly the line shapes measured in spectroscopy: the width of a spectral line and the lifetime of the state that produced it are a Fourier transform pair.

• Quantum mechanics. Position and momentum wavefunctions are Fourier transforms of one another. The width–bandwidth tradeoff seen in the pulse example — narrow in \(x\) means wide in \(k\) — is, with \(p=\hbar k\), the Heisenberg uncertainty principle.

• Normal modes. The harmonics of a Fourier series are the standing waves of a vibrating string or air column; their frequencies \(\nu_n = n/L\) are why instruments have a well-defined pitch.

Summary Table

For the Interested Reader

Fourier analysis is a subject where a single good picture replaces pages of algebra: once you see a Fourier series as a sum of rotating vectors, the coefficient formulas stop looking arbitrary. The videos below are the best route to that picture. Everything listed is free.

The idea, seen

3Blue1Brown builds the Fourier series and then the transform from the geometry of rotating vectors, motivated — as Fourier himself was — by heat flow:

Video

But what is a Fourier series? From heat flow to drawing with circles — 3Blue1Brown

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Video

But what is the Fourier transform? A visual introduction — 3Blue1Brown

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A complementary take, arriving at the transform from the physics rather than the geometry:

Video

To Understand the Fourier Transform, Start From Quantum Mechanics — Physics with Elliot

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Worked technique and practice

• Paul's Online Math Notes. Fourier Series and Fourier Sine Series — careful worked examples of exactly the coefficient integrals in this chapter, including the even/odd shortcuts.