MATH 414 Lecture 14

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Fourier Transforms

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

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

Fundamental Frequency

$\nu _{fund}={\frac {1}{2a}}$ (natural)
$\omega _{circular}=\omega _{1}={\frac {2\pi }{\mbox{period}}}={\frac {\pi }{a}}$ (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

$\omega _{n+1}-\omega _{n}=\left(n+1\right)\,\omega _{1}-n\,\omega _{1}=\omega _{1}$

Spacing gets smaller as $n$ gets larger

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

Derivation

$\lambda _{n}={\frac {\pi \,n}{a}}$

${\begin{aligned}F(\lambda )&=\int _{-a}^{a}f(x)\,\mathrm {e} ^{-i\,\lambda \,x}\,\mathrm {d} x\\F\left({\frac {n\,\pi }{a}}\right)&=\int _{-a}^{a}f(x)\,\mathrm {e} ^{-{\frac {i\,\pi \,n}{a}}\,x}\,\mathrm {d} x\\\alpha _{n}&={\frac {1}{2a}}\,F\left({\frac {n\,\pi }{a}}\right)\end{aligned}}$

Now

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

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

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

This is a riemann sum for

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

As $a\to \infty$ , we get

${\begin{aligned}f(x)&={\frac {1}{2\pi }}\,\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }f(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\right)\mathrm {e} ^{i\,\lambda \,x}\,\mathrm {d} x\\&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }{\hat {f}}(\lambda )\,\mathrm {e} ^{i\,\lambda \,x}\,\mathrm {d} x\end{aligned}}$

This is our synthesis step, and

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

is our corresponding analysis step.

Definition

{\begin{aligned}{\mathcal {F}}\left[f\right]\left(\lambda \right)={\hat {f}}(\lambda )&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(x)\,\mathrm {e} ^{-i\,\lambda \,x}\,\mathrm {d} x\end{aligned}}

Example

Let $f(x)={\begin{cases}1&x\in \left[-\pi ,\pi \right]\\0&{\mbox{otherwise}}\end{cases}}$

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