Definition of a CT LTI system
if $ \,y(t)=Cx(t) $ where C is any constant, then
$ \,h(t)=k\delta (t) $
Example of Laplace transform
Use the definition of laplace transforms to convert an h(t) function to an H(s) function:
$ H(s)=\mathcal{L}\left \{ x(t) \right \} \ = \int_{-\infty}^{\infty} h(t) e^{-st}\, dt $
if our system is $ h(t)=k\delta (t)\, $, then
$ H(s)=\int_{-\infty}^{\infty} k \delta (t) e^{-st}\, dt $
$ H(s)=\int_{-\infty}^{\infty} k \delta (t) e^{-st}\, dt $
$ H(s)=k\int_{-\infty}^{\infty}\delta (t) e^{-st}\, dt $
$ H(s)=k*2e^0 = 2\, $
Compute the new response
Using the formula :$ y(t)=H(j\omega)x(t)\, $ given that
$ 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} $
so simply multiply everything by my $ H(jw)\, $
$ y(t)=(1+j)e^{j\pi t}+(1+j)e^{-j\pi t}+\frac{1+j}{j}e^{j2\pi t}+\frac{-1-j}{j}e^{-j2\pi t} $
