Revision as of 18:43, 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,

$ \ a_{0} = b_{0} $

$ \ b_{1} = a_{1} * (frac{1}{1+jw} $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010