Revision as of 15:18, 26 September 2008 by Choi88 (Talk)

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CT Signal

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

This is a signal with period $ T = {2\pi \over w_0} $

$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j w_0 t + {\pi \over 4})}+e^{-(j w_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} [e^{j w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j w_0 t} e^{-j{\pi \over 4}}] $

$ e^{-j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} $
$ e^{j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} $



$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j w_0 t} e^{-j{\pi \over 4}}] $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin