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<math>x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} </math> | <math>x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} </math> | ||
| − | The input, cos(2t) | + | The input, <math>cos(2t) = \frac{1}{2}(e^{j2t} + e^{-j2t})</math> |
From the properties of a linear system <math>ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t)</math>, | From the properties of a linear system <math>ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t)</math>, | ||
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<math>y(t) = \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math> | <math>y(t) = \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math> | ||
| + | |||
<math>y(t) = t[\frac{1}{2}(e^{-2jt}+e^{2jt})]</math> | <math>y(t) = t[\frac{1}{2}(e^{-2jt}+e^{2jt})]</math> | ||
| + | |||
<math>y(t) = tcos(2t)</math> | <math>y(t) = tcos(2t)</math> | ||
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| + | Therefore, <math>x(t) = cos(2t) \rightarrow linear-system \rightarrow y(t) = tcos(2t) </math> | ||
Latest revision as of 08:15, 18 September 2008
Part B: The basics of linearity
$ x_1(t) = e^{2jt} \rightarrow linear-system \rightarrow y_1(t) = te^{-2jt} $
$ x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} $
The input, $ cos(2t) = \frac{1}{2}(e^{j2t} + e^{-j2t}) $
From the properties of a linear system $ ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t) $, where $ a = b = \frac{1}{2} $
The response to cos(2t) is:
$ y(t) = \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $
$ y(t) = t[\frac{1}{2}(e^{-2jt}+e^{2jt})] $
$ y(t) = tcos(2t) $
Therefore, $ x(t) = cos(2t) \rightarrow linear-system \rightarrow y(t) = tcos(2t) $
