10.4: Basic Properties of the Fourier Transform
- Page ID
- 34575
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
The Fourier transform has several important properties. These can all be derived from the definition of the Fourier transform; the proofs are left as exercises.
- The Fourier transform is linear: if we have two functions \(f(x)\) and \(g(x)\), whose Fourier transforms are \(F(k)\) and \(G(k)\) respectively, then for any constants \(a, b \in \mathbb{C}\), \[a f(x) + b g(x) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; a F(k) + b G(k).\]
- Performing a coordinate translation on a function causes its Fourier transform to be multiplied by a phase factor: \[f(x+b) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; e^{ikb} \, F(k).\] As a consequence, translations leave the Fourier spectrum \(|F(k)|^2\) unchanged.
- If the Fourier transform of \(f(x)\) is \(F(k)\), then \[f^*(x) \quad \overset{\mathrm{FT}}{\longrightarrow} \;\; F^*(-k).\] As a consequence, the Fourier transform of a real function must satisfy the symmetry relation \(F(k) = F^*(-k)\), meaning that the Fourier spectrum is symmetric about the origin in k-space: \(\big|\,F(k)\,\big|^2 = \big|\,F(-k)\,\big|^2.\)
- When you take the derivative of a function, that is equivalent to multiplying its Fourier transform by a factor of \(ik\): \[\frac{d}{dx} f(x) \,\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; ik F(k).\]
For functions of time, because of the difference in sign convention discussed in Section 10.3, there is an extra minus sign: \[\frac{d}{dt} f(t) \;\;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; -i\omega F(\omega).\]