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<math>x(t) \rightarrow linear-system \rightarrow y(t) = tx(-t)</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> | ||
Revision as of 08:57, 18 September 2008
Part B: The basics of linearity
$ e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt} $
$ e^{-2jt} \rightarrow linear-system \rightarrow te^{2jt} $
From these two transformations we can tell that the system is as below:
$ x(t) \rightarrow linear-system \rightarrow y(t) = tx(-t) $
Therefore, when the input, x(t) = cos(2t), the output is as follows:
$ cost(2t) \rightarrow linear-system \rightarrow y(t) = tcos(-2t) $
