(New page: ==CT Periodic Signal== :<math>x(t)=2cos(\dfrac{\pi}{2}t) \,</math> :<math>T=\dfrac{2\pi}{\pi/2} \,</math> :<math>T=4 \,</math> :<math>x(t)=2\dfrac{e^{.5 j t \pi}-e^{-.5 j t \pi}}{2}</math>...)
(No difference)

Revision as of 12:11, 23 September 2008

CT Periodic Signal

$ x(t)=2cos(\dfrac{\pi}{2}t) \, $
$ T=\dfrac{2\pi}{\pi/2} \, $
$ T=4 \, $
$ x(t)=2\dfrac{e^{.5 j t \pi}-e^{-.5 j t \pi}}{2} $
$ x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi} \, $
$ ak=\dfrac{1}{T}\int_0^T x(t) e^{-jk\dfrac{2\pi}{T}t} dt $
$ a1=\dfrac{1}{4}\int_0^4 x(t) e^{-j\dfrac{2\pi}{4}t} dt $
$ a1=\dfrac{1}{4}\int_0^4 (e^{\dfrac{\pi}{2} j t }-e^{-\dfrac{\pi}{2} j t }) e^{-jt\dfrac{\pi}{2}} dt $
$ a1=\dfrac{1}{4}\int_0^4 -e^{-\pi j t } dt $
$ a1=\dfrac{1}{4}[\dfrac{e^{-\pi j t }}{\pi}]|_0^4 $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010