| Line 3: | Line 3: | ||
Using Euler's formula, we get | Using Euler's formula, we get | ||
| − | <math>e^{(2jt)} = cos{(2t)} + jsin{(2t)} | + | <math>e^{(2jt)} = cos{(2t)} + jsin{(2t)} -> system -> t*{(cos{(2t)} - jsin{(2t)})}\,</math><br> |
| − | <math>e^{(-2jt)} = cos{(2t)} - jsin{(2t)} | + | <math>e^{(-2jt)} = cos{(2t)} - jsin{(2t)} -> system -> t*{(cos{(2t)} + jsin{(2t)})}\,</math><br><br> |
Revision as of 16:47, 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)})}\, $
