The Fourier Transform
Ch 24 — Part V — Fourier Transform
Taking the \(L\to\infty\) limit of Fourier series on \([-L,L]\) gives the Fourier transform, which decomposes non-periodic functions into a continuum of frequencies.
Definition
$$\hat f(\xi) = \mathcal{F}\{f\}(\xi) = \int_{-\infty}^\infty f(x) e^{-i \xi x}\,dx, \qquad f(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat f(\xi) e^{i \xi x}\,d\xi.$$ (Conventions differ by factors of \(2\pi\) — Stanford uses the convention above.)
Key properties
- Linearity: \(\mathcal{F}\{\alpha f + \beta g\} = \alpha \hat f + \beta \hat g\).
- Derivative: \(\mathcal{F}\{f'\}(\xi) = i\xi \hat f(\xi)\). (Derivatives become multiplication by \(i\xi\).)
- Translation: \(\mathcal{F}\{f(x-a)\}(\xi) = e^{-i a \xi} \hat f(\xi)\).
- Modulation: \(\mathcal{F}\{e^{i\alpha x} f\}(\xi) = \hat f(\xi - \alpha)\).
- Convolution: \(\mathcal{F}\{f * g\} = \hat f \cdot \hat g\).
Why it helps PDEs
Taking FT in \(x\) converts \(\partial_x^2\) into \((i\xi)^2 = -\xi^2\), turning a PDE into an ODE in \(t\) that can be solved explicitly.
Practice Problems
Hint
Solution
\(\hat f(\xi) = \int_0^\infty e^{-ax} e^{-i\xi x} dx + \int_{-\infty}^0 e^{ax} e^{-i\xi x} dx = \frac{1}{a + i\xi} + \frac{1}{a - i\xi} = \frac{2a}{a^2 + \xi^2}\).
Answer: \(\dfrac{2a}{a^2 + \xi^2}.\)
Hint
Solution
\(\mathcal{F}\{f'\} = \frac{2 a i \xi}{a^2 + \xi^2}\).
Answer: See solution.
Hint
Solution
\(\int f(x-a) e^{-i\xi x} dx = \int f(u) e^{-i\xi(u+a)} du = e^{-i a \xi} \hat f(\xi).\)
Answer: See solution.