Define a DT LTI system
$ \,y[n]=2x[n-1]+3x[n] $
The unit impluse response $ \,h[n] $ is just
$ h[n]=2\delta [n-1] +3\delta[n]\, $
So take laplace transform of this function to get $ H(s)\, $, the system function.
$ H(z) = \sum_{n=-\infty}^{\infty}h[n] e^{-sn} =\sum_{n=-\infty}^{\infty}2\delta [n-1] +\sum_{n=-\infty}^{\infty}3\delta[n] e^{-sn} =2e^{-s}+3 $
$ \,H[z]=3+2e^{-j\omega} $
Compute the response
My original function:
$ x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t} $
I have to multiply each of these fourier coefficients by my $ H[z]\, $, also noting that my $ \,\omega = \pi $
amazingly enough with my $ \,\omega=\pi $, my function turns to
$ \,H(\pi \omega) = 3+2e^{-j\pi}=3-2=1 $
So if my transfer function $ \,H(\pi \omega)=1 $, then my $ \,y(t)=x(t) $
