CHAPTER 5
Fourier Transform
Fourier series handle periodic signals with a discrete list of frequencies. Let the period go to infinity and the list fills in to a continuous spectrum \(F(\omega)\). That is the Fourier transform, and it works on the whole real line.
5.1 From Fourier series to Fourier transform
Stretch the period of a pulse train and watch the spectral lines merge into a curve
Period → ∞
one pulse repeated every Tline spectrum T·cₙ at ωₙ = 2πn/Tlimit F(ω) = 2 sin ω / ω
Theorem 5.1 · Fourier integral theoremIf \(f\) is absolutely integrable and \(f,f'\) are piecewise continuous, then
\[\frac{f(t^-)+f(t^+)}2=\frac1{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}d\omega,\qquad F(\omega)=\int_{-\infty}^{\infty}f(\tau)e^{-i\omega\tau}d\tau\]
At a jump it gives the midpoint, just as Fourier series do.
5.2 Definition and the transform pair
Definitions 5.3 and 5.4, the convention used in every MA2014 paper
Forward\[F(\omega)=\mathcal F\{f(t)\}=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\]
Inverse\[f(t)=\mathcal F^{-1}\{F\}=\frac1{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}d\omega\]
\(f=u(t)e^{-t}\): \(F=\int_0^\infty e^{-t}e^{-i\omega t}dt=\int_0^\infty e^{-(1+i\omega)t}dt\)
\(=\left[\frac{-e^{-(1+i\omega)t}}{1+i\omega}\right]_0^\infty=\boxed{\frac{1}{1+i\omega}}=\frac{1-i\omega}{1+\omega^2}\)
\(F\) is complex. Its magnitude \(|F|=\frac1{\sqrt{1+\omega^2}}\) says how much of each frequency is present. Its angle is the phase. Pick it in the viewer below.
5.3 Transform pairs (Table 5.1 and the appendix)
Tap to flip. The generalized pairs (δ, constants, sin, cos, sgn) can be quoted directly in exams.
\(Au(t)e^{-\alpha t}\)tap
\(\dfrac{A}{\alpha+i\omega}\)
\(1\) for \(|t|\le\alpha\), else 0tap
\(\dfrac{2\sin\alpha\omega}{\omega}\)
\(e^{-\alpha|t|}\)tap
\(\dfrac{2\alpha}{\alpha^2+\omega^2}\)
\(e^{-\alpha t^2}\)tap
\(\sqrt{\dfrac\pi\alpha}\,e^{-\omega^2/4\alpha}\)
\(\delta(t-a)\)tap
\(e^{-i\omega a}\)
constant \(A\)tap
\(2\pi A\,\delta(\omega)\)
\(\cos at\)tap
\(\pi\big(\delta(\omega+a)+\delta(\omega-a)\big)\)
\(\sin at\)tap
\(\frac\pi i\big(\delta(\omega-a)-\delta(\omega+a)\big)\)
\(\operatorname{sgn}t\)tap
\(\dfrac2i\,\text{p.v.}\dfrac1\omega\)
\(Au(t)\)tap
\(A\left(\pi\delta(\omega)-i\,\text{p.v.}\frac1\omega\right)\)
\(\text{p.v.}\frac1t\)tap
\(-i\pi\operatorname{sgn}\omega\)
\(e^{iat}\)tap
\(2\pi\delta(\omega-a)\)
Exercise 5.8\(\mathcal F\{2025u(t)\}=2025\left(\pi\delta(\omega)-\frac{i}{\omega}\right)\) · \(\mathcal F\{\cos100t\}=\pi(\delta(\omega+100)+\delta(\omega-100))\) · \(\mathcal F\{e^{-9t^2}\}=\frac{\sqrt\pi}{3}e^{-\omega^2/36}\)
Transform viewer: time on the left, frequency on the right
f(t)Re FIm F|F|
Remark 5.20, the uncertainty principleMake the pulse narrower (bigger α for the Gaussian, smaller α for the rectangle) and the spectrum gets wider. You cannot be short in time and narrow in frequency at once.
5.4 Properties of the Fourier transform
Learn these and most exam transforms take two lines
| property | time domain | frequency domain |
|---|---|---|
| Linearity (5.9) | \(af+bg\) | \(aF+bG\) |
| Frequency shift (5.11) | \(e^{iat}f(t)\) | \(F(\omega-a)\) |
| Time shift (5.13) | \(f(t-t_0)\) | \(e^{-i\omega t_0}F(\omega)\) |
| Scaling (5.18) | \(f(\alpha t)\) | \(\frac1\alpha F(\omega/\alpha)\) |
| Time reversal (5.21) | \(f(-t)\) | \(F(-\omega)\) |
| Conjugation (5.23) | \(\overline{f(t)}\) | \(\overline{F(-\omega)}\) |
| Differentiation (5.25) | \(f'(t)\) | \(i\omega F(\omega)\) |
| Multiply by t (5.29) | \(tf(t)\) | \(i\,F'(\omega)\) |
| Convolution (5.33) | \(f*g=\int f(\lambda)g(t-\lambda)d\lambda\) | \(F\,G\) |
| Correlation (5.39) | \(f\star g=\int f(\lambda)g(\lambda-t)d\lambda\) | \(F(\omega)\overline{G(\omega)}\) |
| Duality (5.41) | \(F(t)\) | \(2\pi f(-\omega)\) |
Property explorer: apply an operation to a Gaussian pulse
originalafter operationRe FIm F|F|
2025 exam Q2(b) (10 marks)
\(\mathcal F\{f'\}=\int_{-\infty}^\infty f'(t)e^{-i\omega t}dt\). Integrate by parts: \(=\Big[f(t)e^{-i\omega t}\Big]_{-\infty}^{\infty}+i\omega\int f(t)e^{-i\omega t}dt\).
Since \(f\to0\) at \(\pm\infty\) and \(|e^{-i\omega t}|=1\), the bracket is 0. So \(\boxed{\mathcal F\{f'\}=i\omega F(\omega)}\).
Spot the square: \(-t^2+2t-1=-(t-1)^2\). So the function is \((t-1)e^{-(t-1)^2}=g(t-1)\) with \(g(t)=te^{-t^2}\).
\(\mathcal F\{e^{-t^2}\}=\sqrt\pi e^{-\omega^2/4}\). Note that \(te^{-t^2}=-\frac12\frac{d}{dt}e^{-t^2}\), so by part (i): \(G=-\frac12\,i\omega\sqrt\pi e^{-\omega^2/4}\).
Time shift by 1: \(\boxed{\mathcal F=-\frac{i\sqrt\pi}{2}\,\omega\,e^{-\omega^2/4}e^{-i\omega}}\)
More worked properties
Ex 5.10: \(\mathcal F\{2u(t)e^{-t}+u(t)e^{-2t}\}=\frac{2}{1+i\omega}+\frac1{2+i\omega}\).
Ex 5.15: \(3e^{-it}\) on \([-5,5]\) is \(3e^{-it}\) times a box with \(\alpha=5\), so \(F=\frac{6\sin5(\omega+1)}{\omega+1}\).
Ex 5.17 / Tut 2(i): \(\mathcal F\{4e^{-|t|}-5e^{-3|t+2|}\}=\frac{8}{1+\omega^2}-e^{2i\omega}\frac{30}{9+\omega^2}\).
Ex 5.34: \(\frac1{(9+i\omega)^2}=\frac1{9+i\omega}\cdot\frac1{9+i\omega}\), so \(f=(u e^{-9t})*(ue^{-9t})=te^{-9t}u(t)\).
Ex 5.43 (duality): \(\mathcal F\{\frac{2\sin t}{t}\}=2\pi\) for \(|\omega|<1\), \(\pi\) at \(|\omega|=1\), and 0 otherwise.
Ex 5.15: \(3e^{-it}\) on \([-5,5]\) is \(3e^{-it}\) times a box with \(\alpha=5\), so \(F=\frac{6\sin5(\omega+1)}{\omega+1}\).
Ex 5.17 / Tut 2(i): \(\mathcal F\{4e^{-|t|}-5e^{-3|t+2|}\}=\frac{8}{1+\omega^2}-e^{2i\omega}\frac{30}{9+\omega^2}\).
Ex 5.34: \(\frac1{(9+i\omega)^2}=\frac1{9+i\omega}\cdot\frac1{9+i\omega}\), so \(f=(u e^{-9t})*(ue^{-9t})=te^{-9t}u(t)\).
Ex 5.43 (duality): \(\mathcal F\{\frac{2\sin t}{t}\}=2\pi\) for \(|\omega|<1\), \(\pi\) at \(|\omega|=1\), and 0 otherwise.
5.5 Inverse transforms (Exercise 5.45)
Reshape \(F\) until it matches a table entry, then undo the properties
\(\frac{2025}{5-i\omega}\). Since \(\mathcal F\{e^{at}u(-t)\}=\frac{1}{a-i\omega}\), the answer is \(2025e^{5t}u(-t)\).
\(\frac{\sin\omega\cos\omega}{\omega}=\frac14\cdot\frac{2\sin2\omega}{\omega}\), so the answer is \(\frac14\) on \(|t|\le2\) and 0 elsewhere.
\(e^{-\omega^2/12}\): Gaussian with \(4\alpha=12\), so \(\alpha=3\). Answer: \(\sqrt{\frac3\pi}e^{-3t^2}\).
\(e^{20i\omega}=e^{-i\omega(-20)}\), so the answer is \(\delta(t+20)\).
\(\frac{1}{\omega^2+3}=\frac1{2\sqrt3}\cdot\frac{2\sqrt3}{3+\omega^2}\), so the answer is \(\frac{1}{2\sqrt3}e^{-\sqrt3|t|}\).
\(\delta(\omega)\): \(\frac1{2\pi}\int\delta(\omega)e^{i\omega t}d\omega=\frac1{2\pi}\).
\(\frac1{(\omega+2)^2+9}\) is \(\frac16\cdot\frac{6}{9+\omega^2}\) shifted to \(\omega+2\), so the answer is \(\frac16e^{-2it}e^{-3|t|}\).
\(\frac{e^{-5i\omega}}{i\omega}\): since \(\mathcal F\{\frac12\operatorname{sgn}t\}=\frac1{i\omega}\), delaying by 5 gives \(\frac12\operatorname{sgn}(t-5)\).
\(\frac{2i\omega}{3+i\omega}=2-\frac{6}{3+i\omega}\), so the answer is \(2\delta(t)-6e^{-3t}u(t)\).
\(\frac{1}{15+8i\omega-\omega^2}=\frac1{(3+i\omega)(5+i\omega)}=\frac12\left(\frac1{3+i\omega}-\frac1{5+i\omega}\right)\), so the answer is \(\frac12(e^{-3t}-e^{-5t})u(t)\).
5.6 Even and odd functions, sine and cosine transforms
Half-line problems \((x\ge0)\) use \(F_s\) or \(F_c\)
Even f\(\mathcal F\{f\}=2\int_0^\infty f\cos\omega t\,dt\), which is real and even.
\(F_c(\omega)=\int_0^\infty f\cos\omega t\,dt\)
\(f=\frac2\pi\int_0^\infty F_c\cos\omega t\,d\omega\)
\(F_c(\omega)=\int_0^\infty f\cos\omega t\,dt\)
\(f=\frac2\pi\int_0^\infty F_c\cos\omega t\,d\omega\)
Odd f\(\mathcal F\{f\}=-2i\int_0^\infty f\sin\omega t\,dt\), which is imaginary and odd.
\(F_s(\omega)=\int_0^\infty f\sin\omega t\,dt\)
\(f=\frac2\pi\int_0^\infty F_s\sin\omega t\,d\omega\)
\(F_s(\omega)=\int_0^\infty f\sin\omega t\,dt\)
\(f=\frac2\pi\int_0^\infty F_s\sin\omega t\,d\omega\)
Ex 5.53 · f = e^{−at}\(F_c=\frac{a}{a^2+\omega^2}\), \(F_s=\frac{\omega}{a^2+\omega^2}\).
Tutorial 6: box \(|t|<1\) gives \(F=\frac{2\sin\omega}\omega\). Inverting at \(t=0\): \(1=\frac1{2\pi}\int\frac{2\sin\omega}{\omega}d\omega\), so \(\int_0^\infty\frac{\sin x}xdx=\frac\pi2\).
Tutorial 6: box \(|t|<1\) gives \(F=\frac{2\sin\omega}\omega\). Inverting at \(t=0\): \(1=\frac1{2\pi}\int\frac{2\sin\omega}{\omega}d\omega\), so \(\int_0^\infty\frac{\sin x}xdx=\frac\pi2\).
5.7 Solving PDEs with Fourier transforms
Transform in x. The PDE becomes an ODE in t (or y). Solve it, then invert.
PDE in (x, t)
→ 𝓕ₓ
ODE in t
\(\partial_x^2\to-\omega^2\)
\(\partial_x^2\to-\omega^2\)
→
solve ODE
→ 𝓕⁻¹
u(x, t)
Heat on an infinite rod (Exercise 5.54)
\(\theta_t=k\theta_{xx}\) gives \(\Theta_t=-k\omega^2\Theta\), so \(\Theta=\Theta(\omega,0)e^{-k\omega^2t}\).
\(\Theta(\omega,0)=\mathcal F\{e^{-ax^2}\}=\sqrt{\frac\pi a}e^{-\omega^2/4a}\), so \(\Theta=\sqrt{\frac\pi a}e^{-\omega^2\left(\frac1{4a}+kt\right)}\).
This is again a Gaussian, with \(\frac1{4b}=\frac1{4a}+kt\). Inverting gives \(\boxed{\theta=\frac{1}{\sqrt{1+4akt}}\exp\left(-\frac{ax^2}{1+4akt}\right)}\). It gets wider and lower, and the total heat is conserved.
Diffusion and wave on the whole linet =
2025 exam Q3(b): infinite strip, Fourier transform in x (12 marks)
\(U_{xx}+U_{yy}=0\) on \(0<y<1\), \(U(x,0)=0\), \(U(x,1)=e^{-x^2}\).
Transform in \(x\): \(-\omega^2\hat U+\hat U_{yy}=0\), so \(\hat U=A\cosh\omega y+B\sinh\omega y\).
\(\hat U(\omega,0)=0\Rightarrow A=0\). \(\hat U(\omega,1)=\sqrt\pi e^{-\omega^2/4}\Rightarrow B=\frac{\sqrt\pi e^{-\omega^2/4}}{\sinh\omega}\).
\(\hat U=\sqrt\pi e^{-\omega^2/4}\dfrac{\sinh\omega y}{\sinh\omega}\)
Invert (the integrand is even in \(\omega\)):\[U=\frac{1}{\sqrt\pi}\int_0^\infty e^{-\omega^2/4}\frac{\sinh\omega y}{\sinh\omega}\cos\omega x\,d\omega\]
Numerical evaluation of the boxed integral. The top edge carries the Gaussian, and the heat fades toward the cold bottom edge.