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<math>\ h(t) = 5e^{-t} </math>
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<math> H(jw) = 5\int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau}  </math>

Revision as of 17:37, 26 September 2008


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

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

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010