MATH 414 Lecture 14

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Fourier Transforms

In fourier series, we construct $f(x)$ as a $2a$-periodic function.

1. Analysis: ${\alpha }_n=\frac{1}{2a}\underset{-a}{\overset{a}{\int }}f(x){\mathrm{e}}^{-\frac{\pi ni}{a}}\mathrm{d}x$
2. Synthesis: $f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{\mathrm{e}}^{i\frac{\pi n}{a}x}$

Fundamental Frequency

• ${\nu }_{fund}=\frac{1}{2a}$ (natural)
• ${\omega }_{circular}={\omega }_1=\frac{2\pi }{\text{period}}=\frac{\pi }{a}$ (circular frequency; this is what we will be using)
• ${\omega }_n=n{\omega }_1$

All allowed frequencies are integer multiples of the fundamental frequency.

Gap

${\omega }_{n+1}-{\omega }_n=(n+1){\omega }_1-n{\omega }_1={\omega }_1$

Spacing gets smaller as $n$ gets larger

What happens as spacing goes to $0$? We approach a continuum of frequencies.

Derivation

${\lambda }_n=\frac{\pi n}{a}$

$\begin{array}{rl}F(\lambda )& =\underset{-a}{\overset{a}{\int }}f(x){\mathrm{e}}^{-i\lambda x}\mathrm{d}x\\ F\left(\frac{n\pi }{a}\right)& =\underset{-a}{\overset{a}{\int }}f(x){\mathrm{e}}^{-\frac{i\pi n}{a}x}\mathrm{d}x\\ {\alpha }_n& =\frac{1}{2a}F\left(\frac{n\pi }{a}\right)\end{array}$

Now

$f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}F({\lambda }_n)\frac{1}{2a}{\mathrm{e}}^{i{\lambda }_{n}x}$

Let $\mathrm{\Delta }\lambda ={\lambda }_{n+1}-{\lambda }_n=\frac{\pi (n+1)}{a}-\frac{\pi n}{a}=\frac{\pi }{a}$. Then

$f(x)=\frac{1}{2}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}F({\lambda }_n){\mathrm{e}}^{i{\lambda }_{n}x}\mathrm{\Delta }\lambda$

This is a riemann sum for

$\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}F(\lambda ){\mathrm{e}}^{i\lambda x}\mathrm{d}x$

As $a\to \mathrm{\infty }$, we get

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

This is our synthesis step, and

$\hat{f}(\lambda )=\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x){\mathrm{e}}^{i\lambda x}\mathrm{d}x$

is our corresponding analysis step.

Definition

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

Example

Let $f(x)=\{\begin{array}{ll}1& x\in [-\pi ,\pi ]\\ 0& \text{otherwise}\end{array}$

$\begin{array}{rl}\hat{f}(\lambda )& =\frac{1}{\sqrt{2\pi }}\underset{-\pi }{\overset{\pi }{\int }}{\mathrm{e}}^{i\lambda x}\\ & =\frac{2}{\sqrt{2\pi }}\left(\frac{{\mathrm{e}}^{i\lambda \pi }-{\mathrm{e}}^{-i\lambda \pi }}{2i\lambda }\right)\\ & =\sqrt{\frac{2}{\pi }}\frac{\sin (\lambda \pi )}{\lambda }\\ \hat{f}(0)& =\sqrt{2\pi }\end{array}$

Let $\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(z):=\frac{\sin (\pi z)}{\pi z}$. In this example, we have

$\hat{f}=\sqrt{2\pi }\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(\lambda )$

Inverse Fourier Transform

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

A Quartet of Functions

Function	Fourier Transform
$f(x)$	$\hat{f}(\lambda )$
$\hat{f}(x)$	$f(-\lambda )$

When you know one transform, you know another.