## The Signal

$X(j \omega) = \cos(4 \omega + \frac{\pi}{3})$

Taken from 4.22.b from the course book, it looks interesting and I want to try it.

## The Inverse Fourier Transform

$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j \omega)e^{j\omega t}d\omega$

For this problem I will not be using the above equation but in stead be using duality.

$x(t) = \cos(4 t + \frac{\pi}{3})$

note

$x(t) = \sum^{\infty}_{k = -\infty} a_k e^{j k \omega_o t}$

and

$\cos(4 t + \frac{\pi}{3}) = \frac{e^{j(4 t + \frac{\pi}{3})}}{2} + \frac{e^{-j(4 t + \frac{\pi}{3})}}{2}$

$\omega_o = 4$

$a_1 = e^{j \frac{\pi}{3}}$

$a_{-1} = e^{-j \frac{\pi}{3}}$

$= 2 \pi e^{j \frac{\pi}{3}} \delta(\omega - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(\omega + 4)$

duality applied

$\frac{1}{2 \pi}( 2 \pi e^{j \frac{\pi}{3}} \delta(-t - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(-t + 4))$

$e^{j \frac{\pi}{3}} \delta(-t - 4) + e^{-j \frac{\pi}{3}} \delta(-t + 4)$

## Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.