(New page: First let's take a quick look at the example systems' results to determine what the system does. What we know:<br> <math>input \rightarrow system \rightarrow output\!</math> <br> <math>e^{...) |
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It looks like the system is performing the following operation:<br> | It looks like the system is performing the following operation:<br> | ||
<math>x(t) \rightarrow system \rightarrow tx(-t)\!</math><br> | <math>x(t) \rightarrow system \rightarrow tx(-t)\!</math><br> | ||
| − | When the preceding system is applied to <math>cos(2t)</math>, we the get result: | + | When the preceding system is applied to <math>cos(2t)</math>, we the get result:<br> |
| − | <math>cos(2t) \rightarrow system \rightarrow tcos(-2t)\!</math><br> | + | <math>cos(2t) \rightarrow system \rightarrow tcos(-2t)\!</math><br><br> |
Therefore, the system's response will be <math>t*cos(-2t)</math>. | Therefore, the system's response will be <math>t*cos(-2t)</math>. | ||
Revision as of 13:49, 14 September 2008
First let's take a quick look at the example systems' results to determine what the system does.
What we know:
$ input \rightarrow system \rightarrow output\! $
$ e^{2jt} \rightarrow system \rightarrow te^{-2jt}\! $
$ e^{-2jt} \rightarrow system \rightarrow te^{2jt}\! $
It looks like the system is performing the following operation:
$ x(t) \rightarrow system \rightarrow tx(-t)\! $
When the preceding system is applied to $ cos(2t) $, we the get result:
$ cos(2t) \rightarrow system \rightarrow tcos(-2t)\! $
Therefore, the system's response will be $ t*cos(-2t) $.
