(The Inverse Fourier Transform)
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<math>(e^{j \frac{\pi}{3}} \delta(-t - 4) + e^{-j \frac{\pi}{3}} \delta(-t + 4)</math>
+
<math>e^{j \frac{\pi}{3}} \delta(-t - 4) + e^{-j \frac{\pi}{3}} \delta(-t + 4)</math>

Revision as of 13:21, 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)) $


$ 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.

Buyue Zhang