MATH 414 Lecture 15

« previous | Monday, February 17, 2014 | next »

Fourier Transforms

Given $f(t)$, we define $\hat{f}(\lambda )$ as

$\begin{array}{rl}{ℱ}[f](\lambda )=\hat{f}(\lambda )& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t\\ {{ℱ}}^{-1}[\hat{f}](t)=\frac{f(t+)+f(t-)}{2}& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}(\lambda ){\mathrm{e}}^{i\lambda t}\mathrm{d}t\end{array}$

Example

For example, $f(t)=\theta (t){\mathrm{e}}^{-t}$, where $\theta (t)$ is the Heaviside function.

$\begin{array}{rl}{ℱ}[f(t)](\lambda )& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\theta (t){\mathrm{e}}^{-t}{\mathrm{e}}^{-i\lambda t}\mathrm{d}t\\ & =\frac{1}{\sqrt{2\pi }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-t-i\lambda t}\mathrm{d}t\\ & =\frac{1}{\sqrt{2\pi }}\left(\frac{1}{1+i\lambda }\right)-\frac{1}{1+i\lambda }\underset{t\to \mathrm{\infty }}{\lim }{\mathrm{e}}^{-(1+i\lambda )t}\\ & =\frac{1}{\sqrt{2\pi }}\frac{1}{1+i\lambda }\end{array}$

Properties

Linearity

${ℱ}$ and ${{ℱ}}^{-1}$ are linear transformations. That is, if $\alpha$ and $\beta$ are scalars, then

Theorem. ${ℱ}[\alpha f+\beta g]=\alpha {ℱ}\left[f\right]+\beta {ℱ}\left[g\right]$

Proof.

$\begin{array}{rl}{ℱ}[\alpha f+\beta g]& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\alpha f+\beta g){\mathrm{e}}^{-i\lambda t}\mathrm{d}t\\ & =\alpha (\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t)+\beta (\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t)\\ & =\alpha {ℱ}[f]+\beta {ℱ}[g]\end{array}$

quod erat demonstrandum

Product of Powers

(Proven by Leibniz's Rule)

${ℱ}[t^{n}f(t)]=i^n\frac{{\mathrm{d}}^n\hat{f}}{\mathrm{d}{\lambda }^n}$

(and its inverse)

${{ℱ}}^{-1}[{\lambda }^n\hat{f}(\lambda )]=(-i{)}^n\frac{{\mathrm{d}}^{n}f}{\mathrm{d}{t}^n}$

Derivatives

(Proven with Integration by parts)

${ℱ}[f^{(n)}(t)]={\left(i\lambda \right)}^n\hat{f}(\lambda )$

(and its inverse)

${{ℱ}}^{-1}\left[\frac{{\mathrm{d}}^n\hat{f}}{\mathrm{d}{\lambda }^n}\right](t)={(-it)}^{n}f(t)$

Translation / Shift

Given $f(t)$, suppose we want to find $f(t-a)$ (shift to right by $a$ units)

Theorem. ${ℱ}[f(t-a)]={\mathrm{e}}^{-i\lambda a}{ℱ}\left[f\right]$

Proof.

${ℱ}[f(t-a)]=\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t-a){\mathrm{e}}^{-i\lambda t}\mathrm{d}t$

Let $\tau =t-a$, then

${ℱ}[f(\tau )]=\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(\tau ){\mathrm{e}}^{-i\lambda (\tau +a)}\mathrm{d}t={\mathrm{e}}^{-i\lambda \tau }\hat{f}(\tau )$

quod erat demonstrandum

Translation in spatial domain is change of phase in time domain

Scaling

$\begin{array}{rl}{ℱ}[f(bt)]& =\frac{1}{b}\hat{f}\left(\frac{\lambda }{b}\right)\\ {{ℱ}}^{-1}[\hat{f}(c\lambda )]& =\frac{1}{c}f\left(\frac{t}{c}\right)\end{array}$

Laplace Transform

Let $f(t)=\theta (t)g(t)$, where $\theta (t)$ is the Heaviside function. In other words, if $f(t)=0$ for $t<0$, then

Theorem. ${ℱ}[f(t)]=\frac{1}{\sqrt{2\pi }}{ℒ}[f(t)]\left(i\lambda \right)$

Proof.

${ℱ}[f(t)]=\frac{1}{\sqrt{2\pi }}\underset{0}{\overset{\mathrm{\infty }}{\int }}g(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t$

if we let $s=i\lambda$, we see that ${ℱ}[f]=\frac{1}{\sqrt{2\pi }}{ℒ}\left\{f\right\}(s)=\frac{1}{\sqrt{2\pi }}{ℒ}\left\{f\right\}\left(i\lambda \right)$

quod erat demonstrandum

Example

The fourier transform of the tent function $f(t)=\pi -\left|x\right|$ for $x\in [-\pi ,\pi ]$ is ${ℱ}\left[f\right]=\sqrt{\frac{2}{\pi }}\left(\frac{1-\cos (\pi \lambda )}{{\lambda }^2}\right)$.

find the Fourier transform of $g(t)=\{\begin{array}{ll}1& x\in [-\pi ,0]\\ -1& x\in [0,\pi ]\\ 0& \text{otherwise}\end{array}$

Observe that $g(t)=f^{\prime }(t)$, so ${ℱ}\left[g\right]=i\lambda {ℱ}\left[f\right]=i\sqrt{\frac{2}{\pi }}\left(\frac{1-\cos (\pi \lambda )}{\lambda }\right)$