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)})}\, $

When the following equation, $ 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)}) $

goes through the system, we get $ \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) $