Difference between revisions of "HW3.B Ronny Wijaya ECE301Fall2008mboutin" - Rhea

Revision as of 10:37, 16 September 2008

Since $ e^{2jt} \rightarrow system \rightarrow te^{-2jt}\! $
$ e^{-2jt} \rightarrow system \rightarrow te^{2jt}\! $

and using euler formula, we can replace exponent expressions with

Euler's formula: $ e^{iy}=cos(y)+isin(y)\, $


They will change into:
$ e^{(2jt)} = cos{(2t)} + jsin{(2t)} --> system --> t*{(cos{(2t)} - jsin{(2t)})}\, $
$ e^{(-2jt)} = cos{(2t)} - jsin{(2t)} --> system --> t*{(cos{(2t)} + jsin{(2t)})}\, $

It indicates that the system changes the expression on the middle of cos and sin. While cos(2t) function can be found with this equation:
$ \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)} $
|
|
|
V
$ System \, $
|
|
|
V

$ \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)} $