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<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> | ||
| − | <math>\frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} =</math><math> \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) | + | When the following equation, |
| + | <math>cos{(2t)} = \frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} =</math><math> \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)})</math> | ||
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| + | goes through the system, we get | ||
| + | <math>\frac{1}{2}(t*{(cos{(2t)} - jsin{(2t)})}) + \frac{1}{2}t*{(cos{(2t)} + jsin{(2t)})} = \frac{1}{2}tcos{(2t)} + \frac{1}{2}tcos{(2t)} = tcos(2t)</math> | ||
Latest revision as of 16:51, 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)})}\, $
When the following equation, $ cos{(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)}) $
goes through the system, we get $ \frac{1}{2}(t*{(cos{(2t)} - jsin{(2t)})}) + \frac{1}{2}t*{(cos{(2t)} + jsin{(2t)})} = \frac{1}{2}tcos{(2t)} + \frac{1}{2}tcos{(2t)} = tcos(2t) $
