(New page: Given:<br> <math>e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math><br> <math>e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt}</math><br> Solve: <br> <math>cos(2t) \rightarrow SYSTEM...)
 
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
Given:<br>
 
Given:<br>
<math>e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math><br>
+
<math>e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math><br><br>
 
<math>e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt}</math><br>
 
<math>e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt}</math><br>
  
 
Solve: <br>
 
Solve: <br>
<math>cos(2t) \rightarrow SYSTEM \rightarrow ?</math><br>
+
<math>cos(2t) \rightarrow SYSTEM \rightarrow ?</math><br><br>
 +
At this point, we must use Euler's relation to expand cos(2t) into exponentials. Then, we will be able to use the given inputs and corresponding outputs to come to a conclusion.<br><br>
 +
<math>cos(2t) = \frac{e^{2jt}+e^{-2jt}}{2}</math><br><br>
 +
<math>cos(2t) \rightarrow SYSTEM \rightarrow \frac{1}{2}te^{-2jt}+\frac{1}{2}te^{2jt}</math><br><br>
 +
The above answer can then be simplified.

Latest revision as of 10:09, 17 September 2008

Given:
$ e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt} $

$ e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt} $

Solve:
$ cos(2t) \rightarrow SYSTEM \rightarrow ? $

At this point, we must use Euler's relation to expand cos(2t) into exponentials. Then, we will be able to use the given inputs and corresponding outputs to come to a conclusion.

$ cos(2t) = \frac{e^{2jt}+e^{-2jt}}{2} $

$ cos(2t) \rightarrow SYSTEM \rightarrow \frac{1}{2}te^{-2jt}+\frac{1}{2}te^{2jt} $

The above answer can then be simplified.

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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