Difference between revisions of "HW3.B Caleb Tan ECE301Fall2008mboutin" - Rhea

Revision as of 08:03, 18 September 2008

$e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt}$

$e^{-2jt} \rightarrow linear-system \rightarrow te^{2jt}$

The input, cos(2t) is equal to $\frac{1}{2}(e^{j2t} + e^{-j2t})$

@@ Line 5: / Line 5: @@
 <math>e^{-2jt} \rightarrow linear-system \rightarrow te^{2jt} </math>
-From these two transformations we can tell that the system is as below:
+The input, cos(2t) is equal to <math>\frac{1}{2}(e^{j2t} + e^{-j2t})</math>
-<math>x(t) \rightarrow linear-system \rightarrow y(t) = tx(-t)</math>
-Therefore, when the input, x(t) = cos(2t), the output is as follows:
-<math>cost(2t) \rightarrow linear-system \rightarrow y(t) = tcos(-2t)</math>