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<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{j k \omega_o t}</math> | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{j k \omega_o t}</math> | ||
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| + | <math>\cos(4 t + \frac{\pi}{3}) = \frac{e^{j(4 t + \frac{\pi}{3})}}{2} + \frac{e^{-j(4 t + \frac{\pi}{3})}}{2}</math> | ||
Revision as of 11:19, 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} $
