Help figuring out output signal of LTI system. Would greatly appreciate any help in figuring out the output signal of my discrete time LTI system.
My input signal is cos(ωn) and my frequency response is H(e^jω)=(1+e^−jω)/2.
 A: The steady state response to an input $n \mapsto e^{i \alpha n}$ is
$n \mapsto \hat{H}(e^{i \alpha}) e^{i \alpha n}$.
Note that $\cos  ( n \omega) = {1 \over 2} (e^{in \omega} + e^{-in \omega})$.
Hence the response to the $\cos$ input will be
$n \mapsto {1 \over 2} (\hat{H}(e^{i \omega}) e^{i n \omega} + \hat{H}(e^{-i \omega}) e^{-i n \omega})$.
We see that $\hat{H}( e^{-i \omega}) = \overline{\hat{H}( e^{i \omega})}$, so the
response is $n \mapsto \operatorname{re} (\hat{H}( e^{i \omega}) e^{i n \omega})$.
Note that $\hat{H}( e^{-i \omega}) = {1 \over 2} (1+e^{-i \omega}) = e^{-i { \omega \over 2}} \cos \omega$, so the response is
$n \mapsto \operatorname{re} (e^{-i { \omega \over 2}} (\cos \omega )  e^{i n \omega}) = \cos \omega \operatorname{re} (  e^{i (n-{1\over 2}) \omega}) = \cos \omega \cos((n-{1\over 2}) \omega)$.
Note: A simpler solution would be to note that the given $\hat{H}$ corresponds to the system $y_n = {1 \over 2} (u_n + u_{n-1})$. Then a direct computation yields the above answer.
