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CT LTI system Part a

$ h(t) = e^{-t}u(t) $

$ H(jw) = \int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau} $
$ = [-{1 \over 1 + jw}e^{-\tau}e^{-jwr} ]^{\infty}_0 $

$ = {1 \over 1+ jw} $



CT LTI system Part b

$ y(t) = \sum_{i=m}^n x_i = x_m + x_{m+1} + x_{m+2} +\dots+ x_{n-1} + x_n. $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010