| Line 38: | Line 38: | ||
| − | <math>\frac{1}{2 \pi}()</math> | + | duality applied |
| + | |||
| + | <math>\frac{1}{2 \pi}( 2 \pi e^{j \frac{\pi}{3}} \delta(-t - 4) + 2 \pi e^{-j \frac{\pi}{3}} \delta(-t + 4))</math> | ||
<math>Insert formula here</math> | <math>Insert formula here</math> | ||
Revision as of 12:18, 8 October 2008
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)) $
$ Insert formula here $
