10/09 - Fourier Transform

$⟨e^{j2\pi nt/T}\mid e^{j2\pi mt/T}⟩$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi nt/T}{e}^{-j2\pi mt/T}dt$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi (n-m)t/T}dt$
	$=\underset{-T/2}{\overset{T/2}{\int }}{e}^{j2\pi (n-m)t/T}dt$	Assuming the function is perodic with the period T
	$=T{\delta }_{m,n}$

Fourier Transform

Remember from 10/02 - Fourier Series

• ${\alpha }_m=\frac{1}{T}\underset{-T/2}{\overset{T/2}{\int }}x(t)e^{-j2\pi mt/T}dt$
• $x(t)=x(t+T)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_m{e}^{j2\pi m/T}$

If we let $T\to \mathrm{\infty }$

$\frac{1}{T}$	$\to df$
$\frac{n}{T}$	$\to f$	Remember $f=\frac{2\pi n}{T}$
$T$	$\to \mathrm{\infty }$
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T}$	$\to \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}()df$

Definitions

$F[x(t)]$	$=X(f)$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt$	$={⟨x(t)\mid e^{j2\pi ft}⟩}_t$
$F^{-1}[x(t)]$	$=x(t)$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi ft}df$	$={⟨X(f)\mid e^{-j2\pi ft}⟩}_f$

Examples

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi ft}{e}^{-j2\pi f\lambda }df$	$={⟨e^{j2\pi ft}\mid e^{j2\pi ft}⟩}_f$	$=\delta (t-\lambda )$
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi tf}{e}^{-j2\pi t{f}_0}dt$	$={⟨e^{j2\pi tf}\mid e^{j2\pi t{f}_0}⟩}_t$	$=\delta (f-f_0)$

$F^{-1}[F[x(t)]]$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )e^{-j2\pi f\lambda }d\lambda ]e^{j2\pi ft}df$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi ft}df$	$=x(t)$
		$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-\lambda )}dfd\lambda$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\delta (t-\lambda )d\lambda$	$=x(t)$
	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )e^{-j\omega \lambda }d\lambda ]e^{j\omega t}\frac{1}{2\pi }d\omega$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\left[\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j(t-\omega )\lambda }d\omega \right]d\lambda$	$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\delta (t-\omega )d\lambda$	$=x(t)$

Sifting property of the delta function

The dirac delta function is defined as any function, denoted as $\delta (t-u)$, that works for all variables that makes the following equation true: $x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)\delta (t-u)du$

• When dealing with $\omega$, it behaves slightly different than dealing with $f$. When dealing with $=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )\left[\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j(t-\omega )\lambda }d\omega \right]d\lambda$, note that the delta function is $\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j(t-\omega )\lambda }d\omega$. The $\frac{1}{2\pi }$ is tacked onto the front. Thus, when dealing with $\omega$, you will often need to multiply it by $2\pi$ to cancel out the $\frac{1}{2\pi }$.