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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

Rewriting the periodic signal in Question 1,

$ x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4}) $

$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) [e^{j 2w_0 t}]+ {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) [e^{-j 2w_0 t}] $

$ x(t) = \sum_{i=-2}^2 a_ie^{jkw_0 t} $

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BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman