| Line 10: | Line 10: | ||
<math>\cos(2t) = \frac {e^{i2t} + e^{-i2t}} {2} </math> | <math>\cos(2t) = \frac {e^{i2t} + e^{-i2t}} {2} </math> | ||
| + | |||
| + | from the given input we know: | ||
| + | |||
| + | y(t) = t*x(-t) | ||
| + | |||
| + | If we multiply both t's in the equivalent cos(2t) by -1, then they simply switch places leaving the function unchanged. Next, multiply by t and you are given your result. | ||
| + | |||
| + | y(t) = t*cos(2t) | ||
Latest revision as of 14:27, 19 September 2008
We are told that a system is linear and given inputs
$ \,x_1(t)=e^{2jt}\, $ yields $ \,y_1(t)=te^{-2jt}\, $
$ \,x_2(t)=e^{-2jt}\, $ yields $ \,y_2(t)=te^{2jt}\, $
- initial given statements copied from Jeff Kubascik
Since,
$ \cos(2t) = \frac {e^{i2t} + e^{-i2t}} {2} $
from the given input we know:
y(t) = t*x(-t)
If we multiply both t's in the equivalent cos(2t) by -1, then they simply switch places leaving the function unchanged. Next, multiply by t and you are given your result.
y(t) = t*cos(2t)
