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<math>x(t) \to system \to t x(-t)</math> | <math>x(t) \to system \to t x(-t)</math> | ||
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| + | We know that | ||
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| + | <math>x(t) = cos(2t) = \frac{e^{2jt} + e^{-2jt}}{2}</math>. | ||
| + | |||
| + | So if we take the known system and apply it to the equation right above, the response will be: | ||
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| + | <math>\frac{te^{-2jt} + te^{2jt}}{2} = t \frac{e^{-2jt} + e^{2jt}}{2}</math> | ||
| + | |||
| + | Which is then equivalent to <font size ="3">'''<math>tcos(2t)</math>'''</font> | ||
| + | |||
| + | The response to <font size ="3">'''<math>cos(2t)</math>'''</font> is <font size ="3">'''<math>tcos(2t)</math>'''</font>. | ||
Latest revision as of 14:55, 16 September 2008
Basics of Linearity
Systems given:
$ e^{2jt} \to system \to t e^{-2jt} $
$ e^{-2jt} \to system \to t e^{2jt} $
The basic concept known from the systems given is:
$ x(t) \to system \to t x(-t) $
We know that
$ x(t) = cos(2t) = \frac{e^{2jt} + e^{-2jt}}{2} $.
So if we take the known system and apply it to the equation right above, the response will be:
$ \frac{te^{-2jt} + te^{2jt}}{2} = t \frac{e^{-2jt} + e^{2jt}}{2} $
Which is then equivalent to $ tcos(2t) $
The response to $ cos(2t) $ is $ tcos(2t) $.
