Difference between revisions of "HW3.B Jeremiah Wise ECE301Fall2008mboutin" - Rhea

Revision as of 10:09, 19 September 2008

Given:

For a linear system we have:

$e^{j2t} \rightarrow [system] \rightarrow te^{-j2t}\!$
$e^{-j2t} \rightarrow [system] \rightarrow te^{j2t}\!$

To find the response of the system above we first note that $e^{j2t} = cos(2t) + jsin(2t)$

@@ Line 1: / Line 1: @@
-<math>e^{2jt} \rightarrow </math><box>system</box> \rightarrow te^{-2jt}\!</math><br>
+----
-<math>e^{-2jt} \rightarrow system \rightarrow te^{2jt}\!</math><br>
+'''Given:'''
+For a linear system we have:
+<math>e^{j2t} \rightarrow [system] \rightarrow te^{-j2t}\!</math><br>
+<math>e^{-j2t} \rightarrow [system] \rightarrow te^{j2t}\!</math><br>
+----
+To find the response of the system above we first note that <math>e^{j2t} = cos(2t) + jsin(2t)