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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> | ||
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| + | I do not know how to go from here because I believe it would go to Infinity. I will visit Mimi to find out what to do. | ||
Revision as of 17:16, 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 $
I do not know how to go from here because I believe it would go to Infinity. I will visit Mimi to find out what to do.
