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<math>e^{(2jt)} = cos{(2t)} + jsin{(2t)} -> system -> t*{(cos{(2t)} - jsin{(2t)})}\,</math><br> | <math>e^{(2jt)} = cos{(2t)} + jsin{(2t)} -> system -> t*{(cos{(2t)} - jsin{(2t)})}\,</math><br> | ||
<math>e^{(-2jt)} = cos{(2t)} - jsin{(2t)} -> system -> t*{(cos{(2t)} + jsin{(2t)})}\,</math><br><br> | <math>e^{(-2jt)} = cos{(2t)} - jsin{(2t)} -> system -> t*{(cos{(2t)} + jsin{(2t)})}\,</math><br><br> | ||
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| + | <math>\frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} =</math><br><math> \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) = cos{(2t)}</math> | ||
Revision as of 16:49, 19 September 2008
Part B: The basics of linearity
System’s response to cos(2t)
Using Euler's formula, we get
$ e^{(2jt)} = cos{(2t)} + jsin{(2t)} -> system -> t*{(cos{(2t)} - jsin{(2t)})}\, $
$ e^{(-2jt)} = cos{(2t)} - jsin{(2t)} -> system -> t*{(cos{(2t)} + jsin{(2t)})}\, $
$ \frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} = $
$ \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) = cos{(2t)} $
