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

${\displaystyle }$