The Fourier Transform on $\mathbb{R}$
The Fourier Transform on $\mathbb{R}$
The Fourier transform $\hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\,dx$ extends Fourier series to non-periodic functions on $\mathbb{R}$. The inversion formula reconstructs $f$ from $\hat{f}$, and Plancherel's theorem $\|f\|_2=\|\hat{f}\|_2$ establishes the Fourier transform as a unitary operator on $L^2(\mathbb{R})$. The convolution theorem $\widehat{f*g}=\hat{f}\cdot\hat{g}$ makes the Fourier transform the natural tool for LTI systems and differential equations.
Fourier Transform & Inversion
For $f\in L^1(\mathbb{R})$, the Fourier transform is $\hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\,dx$ (in physics notation: $\hat{f}(\omega)=\int f(t)e^{-i\omega t}\,dt$). Inversion theorem: if $f,\hat{f}\in L^1$, then $f(x)=\int\hat{f}(\xi)e^{2\pi i\xi x}\,d\xi$ a.e. Plancherel: the Fourier transform extends to a unitary map $\mathcal{F}:L^2(\mathbb{R})\to L^2(\mathbb{R})$ with $\|\hat{f}\|_2=\|f\|_2$ and $\mathcal{F}^{-1}=\mathcal{F}^*$ (conjugate is the inverse). Key pairs: Gaussian $e^{-\pi x^2}\leftrightarrow e^{-\pi\xi^2}$; rect $\leftrightarrow$ sinc.
Convolution Theorem
$(f*g)(x)=\int f(y)g(x-y)\,dy$. The convolution theorem: $\widehat{f*g}=\hat{f}\cdot\hat{g}$ and $\widehat{fg}=\hat{f}*\hat{g}$. Consequences: LTI system response $y=h*x$ has transform $\hat{y}=\hat{h}\hat{x}$ (multiplication in frequency domain). Differentiation: $\widehat{f'}(\xi)=2\pi i\xi\hat{f}(\xi)$, converting differential operators to polynomial multiplication. Correlation: $\widehat{f\star g}=\overline{\hat{f}}\cdot\hat{g}$ (matched filtering).
Example 1
Compute the Fourier transform of $f(x)=e^{-a|x|}$ for $a>0$.
Solution: $\hat{f}(\xi)=\int_{-\infty}^\infty e^{-a|x|}e^{-2\pi i\xi x}\,dx=\int_0^\infty e^{-ax}e^{-2\pi i\xi x}\,dx+\int_{-\infty}^0 e^{ax}e^{-2\pi i\xi x}\,dx=\frac{1}{a+2\pi i\xi}+\frac{1}{a-2\pi i\xi}=\frac{2a}{a^2+(2\pi\xi)^2}$. This is a Lorentzian/Cauchy distribution in frequency. Note: $f$ is in $L^1\cap L^2$; $\hat{f}$ is real, even, and $\in L^2$ but not $L^1$.
Example 2
Use the convolution theorem to solve $\frac{d}{dt}y(t)+ay(t)=f(t)$, $a>0$.
Solution: Take Fourier transform: $(2\pi i\omega+a)\hat{y}=\hat{f}$, so $\hat{y}(\omega)=\hat{h}(\omega)\hat{f}(\omega)$ where $\hat{h}(\omega)=1/(a+2\pi i\omega)$. The inverse transform of $\hat{h}$ is $h(t)=e^{-at}u(t)$ (causal exponential). Thus $y(t)=(h*f)(t)=\int_0^t e^{-a(t-s)}f(s)\,ds$, recovering Green's function solution.
Practice
- Prove Plancherel's theorem $\|\hat{f}\|_2=\|f\|_2$ using the Fourier inversion formula.
- Compute $\hat{f}$ for $f(x)=\text{rect}(x)$ ($=1$ for $|x|<1/2$, $0$ otherwise) and show the result is a sinc function.
- Use Parseval to evaluate $\int_{-\infty}^\infty\frac{\sin^2(\pi\xi)}{\pi^2\xi^2}\,d\xi$.
- Derive the uncertainty principle $\sigma_x\sigma_\xi\geq\frac{1}{4\pi}$ using $\|f\|^2=\int|f|^2$ and Cauchy-Schwarz.
Show Answer Key
1. Start from Fourier inversion $f(x)=\int\hat{f}(\xi)e^{2\pi i\xi x}d\xi$. Then $\|f\|_2^2=\int f(x)\overline{f(x)}dx=\int\int\hat{f}(\xi)\overline{\hat{f}(\eta)}\delta(\xi-\eta)d\xi\,d\eta=\int|\hat{f}(\xi)|^2d\xi=\|\hat{f}\|_2^2$.
2. $\hat{f}(\xi)=\int_{-1/2}^{1/2}e^{-2\pi i\xi x}dx=\frac{\sin(\pi\xi)}{\pi\xi}=\text{sinc}(\xi)$.
3. By Parseval: $\int|\hat{f}|^2d\xi=\int|f|^2dx$. With $f=\text{rect}$: LHS $=\int\frac{\sin^2(\pi\xi)}{\pi^2\xi^2}d\xi$, RHS $=\int_{-1/2}^{1/2}1\,dx=1$. So the integral equals $1$.
4. Define $\sigma_x^2=\int x^2|f|^2dx$ and $\sigma_\xi^2=\int\xi^2|\hat{f}|^2d\xi$. By Cauchy-Schwarz on $xf$ and $f'$: $\|f\|_2^2\le2(\int x^2|f|^2dx)^{1/2}(\int|f'|^2dx)^{1/2}$. Using $\|f'\|_2=2\pi\|\xi\hat{f}\|_2$: $\sigma_x\sigma_\xi\ge1/(4\pi)$.