Part B: The basics of linearity

System’s response to cos(2t)

Using Euler's formula, we get

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

$ cos{(2t)} = \frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} = $$ \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) $