(New page: <math>X(\omega)=cos{(6\omega)+ (\pi/6)}</math> Start by guessing the solution: <math>\,\mathcal{X}(t)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math> <math>\,\mathcal{X}(\omeg...)
(No difference)

Revision as of 18:04, 8 October 2008

$ X(\omega)=cos{(6\omega)+ (\pi/6)} $

Start by guessing the solution:


$ \,\mathcal{X}(t)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $

$ \,\mathcal{X}(\omega)=\int_{0}^{+\infty}e^{-5t}cos{(2t)}e^{-j\omega t}\,dt, $

$ \,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-5t}e^{2jt}e^{-j\omega t}\,dt + 1/2\int_{0}^{+\infty}e^{-5t}e^{-2jt}e^{-j\omega t}\,dt\, $

$ \,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-t(5-2j+j\omega)}\,dt + 1/2\int_{0}^{+\infty}e^{-t(5+2j+j\omega)}\,dt\, $

$ \,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-(j\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $

$ \,\mathcal{X}(\omega)=\left. 1/2\frac{e^{-t(5-2j+j\omega)}}{-(5-2j+j\omega)}\right]_{0}^{+\infty} + \left. 1/2\frac{e^{-t(5+2j+j\omega)}}{-(5+2j+j\omega)}\right]_{0}^{+\infty} $


$ \,\mathcal{X}(\omega)=\frac{1}{2(5-2j+j\omega)} + \frac{1}{2(5+2j+j\omega)} $

Alumni Liaison

Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics