James A. Brofos quantitative strategies researcher

The Fourier Transform (I)

We say that a function $f:\mathbb{R}\to\mathbb{C}$ is integrable if \begin{align} \int_{-\infty}^\infty \vert f(x)\vert ~\mathrm{d}x < +\infty. \end{align}
Let $f: \mathbb{R}\to\mathbb{C}$ be an integrable function. We define the Fourier transform of $f$ to be the function $\hat{f} : \mathbb{R}\to\mathbb{R}$ given by \begin{align} \hat{f}(\xi) = \int_{-\infty}^\infty f(x) \exp(-2\pi i \xi x) ~\mathrm{d}x. \end{align}
Let's compute the Fourier transform of the function $f(x) = \exp(-x^2)$. Applying the definitions yields, \begin{align} \hat{f}(\xi) &= \int_{-\infty}^\infty f(x) \exp(-2\pi i x \xi) ~\mathrm{d}x \\ &= \int_{-\infty}^\infty \exp(-x^2) \exp(-2\pi i x \xi) ~\mathrm{d}x \\ &= \int_{-\infty}^\infty \exp(-x^2) \left[\cos(-2\pi x\xi) + i \sin(-2\pi x\xi)\right] ~\mathrm{d}x \\ &= \int_{-\infty}^\infty \exp(-x^2) \cos(2\pi x\xi) ~\mathrm{d}x. \end{align} To solve this equation, we differentiate the Fourier transform with respect to $\xi$ yielding and apply integration by parts noting that $(fg)' - f g' = f' g$, \begin{align} \frac{\mathrm{d}}{\mathrm{d}\xi}\hat{f}(\xi) &= -2\pi \int_{-\infty}^\infty x\exp(-x^2) \sin(2\pi x\xi)~\mathrm{d}x \\ &= \pi \int_{-\infty}^\infty \left(\frac{\mathrm{d}}{\mathrm{d}x} \exp(-x^2)\right) \sin(2\pi x\xi) ~\mathrm{d}x \\ &= \pi \left(\exp(-x^2)\sin(2\pi x\xi)\bigg|_{-\infty}^\infty - 2\pi\xi \int_{-\infty}^\infty \exp(-x^2) \cos(2\pi x\xi) ~\mathrm{d}x \right) \\ &= -2\pi^2 \xi \hat{f}(\xi) \end{align} One verifies then that the solution of this differential equation is of the form $\hat{f}(\xi) = C \exp(-\pi^2 \xi^2)$. To determine the unknown constant $C$ one computes the Fourier transform at the zero frequency to obtain, \begin{align} C &= \int_{-\infty}^\infty \exp(-x^2)~\mathrm{d}x \\ &= \sqrt{\left(\int_{-\infty}^\infty \exp(-x^2)~\mathrm{d}x\right)\left(\int_{-\infty}^\infty \exp(-y^2)~\mathrm{d}y\right)} \\ &= \sqrt{\int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-(x^2 + y^2)) ~\mathrm{d}x~\mathrm{d}y} \\ &= \sqrt{\int_0^\infty \int_0^{2\pi} r \exp(-r^2) ~\mathrm{d}\theta ~\mathrm{d}r} \\ &= \sqrt{2\pi \left[-\frac{\exp(-r^2)}{2}\right]_0^\infty} \\ &= \sqrt{\pi}. \end{align} Therefore, $\hat{f}(\xi) = \sqrt{\pi} \exp(-\pi^2\xi^2)$.
Consider the rectangular pulse function defined by \begin{align} f(x) = \begin{cases} 0 & ~\mathrm{if}~ |x| > 1/2 \\ 1 &~\mathrm{if}~ |x| \leq 1/2 \end{cases}. \end{align} We compute the Fourier transform of the rectangular impulse by applying the definition: \begin{align} \hat{f}(\xi) &= \int_{-\infty}^\infty f(x) \exp(-2\pi i x\xi) ~\mathrm{d}x \\ &= \int_{-1/2}^{1/2} \exp(-2\pi i x \xi) ~\mathrm{d}x \\ &= \int_{-1/2}^{1/2} \cos(-2 \pi x \xi) + i \sin(-2\pi x \xi) ~\mathrm{d}x \\ &= \int_{-1/2}^{1/2} \cos(2\pi x \xi) ~\mathrm{d}x \\ &= \left[\frac{\sin(2\pi x\xi)}{2\pi \xi}\right]_{-1/2}^{1/2} \\ &= \frac{\sin(\pi \xi)}{2\pi\xi} - \frac{\sin(-\pi\xi)}{2\pi\xi} \\ &= \frac{\sin(\pi\xi)}{\pi\xi} \\ &= \mathrm{sinc}(\xi). \end{align} Thus we have shown that the Fourier transform of the rectangular pulse is the normalized sinc function.
The functions $e_n(x) = \exp(2\pi i nx / T)$ are orthogonal on the Hilbert space $L^2([-T/2, T/2])$.
Let $m, n\in \mathbb{N}$. Then the Hilbert space inner product is, \begin{align} \langle e_m, e_n\rangle &= \int_{-T/2}^{T/2} \overline{e_m(x)} e_n(x) ~\mathrm{d}x \\ &= \int_{-T/2}^{T/2} \exp(-2\pi i mx / T) \exp(2\pi i nx/T) ~\mathrm{d}x \\ &= \int_{-T/2}^{T/2} \exp(2\pi i nx/T - 2\pi i m x/T) ~\mathrm{d}x \\ &= \int_{-T/2}^{T/2} \exp(2\pi i x /T (n - m)) ~\mathrm{d}x. \end{align} In the case $m=n$, we have, \begin{align} \langle e_n, e_n\rangle &= \int_{-T/2}^{T/2} ~\mathrm{d}x \\ &= \frac{T}{2} + \frac{T}{2} \\ &= T. \end{align} Now suppose that $m\neq n$. Then we have \begin{align} \langle e_m, e_n \rangle &= \int_{-T/2}^{T/2} \cos(2\pi x(n-m) /T) + i\sin(2\pi x (n-m) / T) ~\mathrm{d}x \\ &= \int_{-T/2}^{T/2} \cos(2\pi x (n-m) / T) ~\mathrm{d}x \\ &= \left[\frac{T \sin(2\pi x (n-m) / T)}{2\pi (n-m)}\right]_{-T/2}^{T/2} \\ &= \frac{T}{2\pi (n-m)} \left[\sin(\pi (n-m)) - \sin(-\pi (n-m)) \right] \\ &= \frac{T \sin(\pi (n-m))}{\pi(n-m)} \\ &= 0. \end{align}

We may therefore easily create an orthonormal set of functions for the Hilbert space by properly normalizing the functions $e_n$; indeed:

The functions $\tilde{e}_n(x) = \exp(2\pi i nx / T) / \sqrt{T}$ are orthonormal on the Hilbert space $L^2([-T/2, T/2])$.
Let $f \in L^2([-T/2, T/2])$. The orthogonal projection of $f$ onto the function $\tilde{e}_n$ is \begin{align} \tilde{f}_n(x) = \langle \tilde{e}_n, f\rangle ~ \tilde{e}_n(x) \end{align} where \begin{align} \langle \tilde{e}_n, f\rangle &= \frac{1}{\sqrt{T}} \int_{-T/2}^{T/2} f(x) \exp(-2\pi i x n / T) ~\mathrm{d}x \\ &= \frac{1}{\sqrt{T}} \hat{f}(n / T). \end{align}

This shows the connection between the Fourier transform of a function and its Fourier series coefficients when orthogonally projected onto the function $\tilde{e}_n$.