(CT Signal)
(CT Signal)
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== CT Signal ==
 
== CT Signal ==
:<math> x(t) = 1 + sin(w_0 t) + 3cos(w_0 t + {\pi \over 4})  </math><br><br>
+
:<math> x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4})  </math><br><br>
 
This is a signal with period <math>T = {2\pi \over w_0}</math><br><br>
 
This is a signal with period <math>T = {2\pi \over w_0}</math><br><br>
:<math> 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})}]</math><br><br>
+
:<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j 2w_0 t + {\pi \over 4})}+e^{-(j 2w_0 t + {\pi \over 4})}]</math><br><br>
:<math> 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}}]</math><br><br>
+
:<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}]</math><br><br>
 
:<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br>
 
:<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br>
 
:<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </math>
 
:<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </math>
 
<br><br>
 
<br><br>
:<math> 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}}]</math><br><br>
+
:<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}]</math><br><br>
:<math> 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 w_0 t}]+ {3 \over 2}{1 \over \sqrt{2}} - j{1 \over \sqrt{2}} [e^{-j w_0 t}]</math><br><br>
+
:<math> 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}]</math><br><br>

Revision as of 15:23, 26 September 2008

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CT Signal

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

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

Kuei-Nuan Lin