(New page: == part A == <math>e^{st}\rightarrow h(t)\rightarrow H(S)e^{st}</math> <math>H(s) = \int_{-\infty}^\infty h(t)e^{-st}dt</math>)
 
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<math>H(s) = \int_{-\infty}^\infty h(t)e^{-st}dt</math>
 
<math>H(s) = \int_{-\infty}^\infty h(t)e^{-st}dt</math>
 +
 +
assume
 +
 +
<math>h(t) = 5u(t-3)</math>

Revision as of 19:28, 25 September 2008

part A

$ e^{st}\rightarrow h(t)\rightarrow H(S)e^{st} $

$ H(s) = \int_{-\infty}^\infty h(t)e^{-st}dt $

assume

$ h(t) = 5u(t-3) $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010