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| − | <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{ e^{j(4 \omega + \frac{\pi}{3})} + e^{j(4 \omega + \frac{\pi}{3})}}{2}e^{j\omega t}d\omega</math> | + | <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{ e^{j(4 \omega + \frac{\pi}{3})} + e^{-j(4 \omega + \frac{\pi}{3})}}{2}e^{j\omega t}d\omega</math> |
Revision as of 16:06, 7 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 $
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\cos(4 \omega + \frac{\pi}{3})e^{j\omega t}d\omega $
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{ e^{j(4 \omega + \frac{\pi}{3})} + e^{-j(4 \omega + \frac{\pi}{3})}}{2}e^{j\omega t}d\omega $
