Basics of Linearity
We are given the following information:
- For input $ x(t) = e $ (2jt) the output $ y(t) = te $(-2jt).
- For input $ x(t) = e $ (-2jt) the output $ y(t) = te $(2jt).
- The system is Linear.
We can break down the input $ \,\ x(t) = cos(2t) $ into $ \,\ x(t) = \frac{1}{2} * (e $(j2t) $ \,\ + e $(-j2t)$ \,\ ) $.
Now we can use the property of linearity to determine the output.
We already know the outputs of the two individual parts of $ cos(2t) $. All we have to do is add them together to find the output $ z(t) $.
$ \,\ \frac{1}{2} * (e $(j2t) $ \,\ + e $(-j2t)$ \,\ ) \rArr $ SYSTEM $ \,\ \rArr \frac{1}{2} * te $(-2jt) + $ \,\ \frac{1}{2} * te $(2jt)
$ \,\ \frac{1}{2} * te $(-2jt) + $ \,\ \frac{1}{2} * te $(2jt) $ = t * \frac{1}{2} * (e $(j2t) $ \,\ + e $(-j2t)$ \,\ ) $
$ t * \frac{1}{2} * (e $(j2t) $ \,\ + e $(-j2t)$ \,\ ) = t*cos(2t) $
Therefore, the input signal $ \,\ x(t) = cos(2t) $ yields an output signal of $ \,\ z(t) = tcos(2t) $
