Problem: A linear system’s response to exp(2jt) is t exp(-2jt), and its response to exp(-2jt) is t exp(2jt). What is the system’s response to cos(2t)?
Solution: Call the system f, i.e., $ f(e^{2jt}) = te^{-2jt} $, and $ f(e^{-2jt}) = te^{2jt} $. (In other words, input x generates output y = f(x).)
Since we know the response of f for the two above complex exponential inputs and the fact that f is linear, we can obtain f(cos(2t)) as long as we can find a linear combination of the first two inputs:
$ cos(2t) = \frac{1}{2}(e^{2jt}+e^{-2jt}) $ (a result of Euler's Formula)
$ \ ( = \frac{1}{2}( (\cos(2t)+i\sin(2t)) + (\cos(-2t)+i\sin(-2t)) ) = \frac{1}{2}( (\cos(2t)+i\sin(2t)) + (\cos(2t)-i\sin(2t)) ) $
$ = \frac{1}{2}( 2\cos(2t) ) = \cos(2t) \ ) $ (as a sanity check)
$ \Rightarrow f(\cos(2t)) = f(\frac{1}{2}(e^{2jt}+e^{-2jt})) $ (by substitution)
$ = \frac{1}{2}f(e^{2jt})+\frac{1}{2}f(e^{-2jt}) $ (because f is linear)
$ = \frac{1}{2}(te^{-2jt})+\frac{1}{2}(te^{2jt}) $ (from the given information)
$ = t \frac{1}{2}(e^{-2jt} + e^{2jt}) $ (simplification)
So we have our answer! But wait -- this looks strangely familiar! In fact, we calculated above that $ cos(2t) = \frac{1}{2}(e^{2jt}+e^{-2jt}) $. So, after one final substitution, our answer is:
$ f(\cos(2t)) = t\cos(2t) $
