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Finally,<br> | Finally,<br> | ||
| − | <math>cos{(2t)} \</math> --> == system == <math>tcos({(2t)}\</math> | + | <math>cos{(2t)} \,</math> --> == system == <math>tcos({(2t)}\,</math> |
--> | --> | ||
Revision as of 10:04, 13 September 2008
- I am going to use the definition of Linearity that I learned in class.
- The definition
if x1(t) --> system --> y1(t)
x2(t) --> system --> y2(t)
Then ax1(t) + bx2(t) --> system --> ay1(t) + by2(t) , for any complex constants a,b
$ e^{(2jt)} = cos{(2t)} + jsin{(2t)} --> system --> t*{(cos{(2t)} - jsin{(2t)})}\, $
$ e^{(-2jt)} = cos{(2t)} - jsin{(2t)} --> system --> t*{(cos{(2t)} + jsin{(2t)})}\, $
Now, suppose a and b are 1/2.
$ \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)} $
Then output is
$ \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)} $
Finally,
$ cos{(2t)} \, $ --> == system == $ tcos({(2t)}\, $
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