(did it) |
m (realized i made a minor mistake (input instead of output)) |
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Since <math>\cos(2t) = {e^{j2t} + e^{-j2t} \over 2}</math> | Since <math>\cos(2t) = {e^{j2t} + e^{-j2t} \over 2}</math> | ||
| − | Then we can suggest that <math>{1 \over | + | Then we can suggest that <math>{1 \over 2}x1 + {1 \over 2}x2 -> |system| -> {1 \over 2}y1 + {1 \over 2}y2 </math> |
| − | This leads to the conclusion that the | + | This leads to the conclusion that the response to cos(2t) would be <math>{1 \over 2}y1 + {1 \over 2}y2 </math> |
| − | Which, simplified, is: <math>\cos(2t) | + | Which, simplified, is: <math>t\cos(2t)</math>. |
Latest revision as of 10:59, 19 September 2008
Homework 3 Ben Horst: A :: B :: C
Answer
If the system is linear (it is) then the following should be true:
Thus, if we say
x1 = exp(2jt) and x2 = exp(-2jt)
then
y1 = t exp(-2jt) and y2 = t exp(2jt)
That means that:
x1 + x2 ->|system|-> y1 + y2
Since $ \cos(2t) = {e^{j2t} + e^{-j2t} \over 2} $
Then we can suggest that $ {1 \over 2}x1 + {1 \over 2}x2 -> |system| -> {1 \over 2}y1 + {1 \over 2}y2 $
This leads to the conclusion that the response to cos(2t) would be $ {1 \over 2}y1 + {1 \over 2}y2 $
Which, simplified, is: $ t\cos(2t) $.
