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 2t}x1 + {1 \over 2t}x2 -> |system| -> {1 \over 2t}y1 + {1 \over 2t}y2 $
This leads to the conclusion that the input of cos(2t) would be $ {1 \over 2t}x1 + {1 \over 2t}x2 $
Which, simplified, is: $ \cos(2t) \over t $.
