Revision as of 17:55, 26 September 2008 by Cdleon (Talk)

$ \ h(t) = 5e^{-t} $

$ \ H(jw) = 5\int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau} $

$ \ H(jw) = 5[-\frac{1}{1 + jw}e^{-\tau}e^{-jwr} ]^{\infty}_0 $

$ \ H(jw) = \frac{5}{1+ jw} $

So,

$ \ b_{0} = 0 $

$ \ b_{1} = (\frac{1 + 2j}{2}) (\frac{5}{1+jw}) $

$ \ b_{-1}= (\frac{1 + 2j}{2}) (\frac{5}{1 - j}) $

$ \ b_{2} = \frac{5}{2j} \frac{5}{1 +5j} $

$ \ b_{-2}= \frac{5}{2j} \frac{5}{1 - 5j} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang