Define a CT LTI System
$ y(t)=2x(t)-3x(t-2)\! $
Unit Impulse Response
The unit impulse response is simply the systems response to an input $ \delta(t)\! $. Thus, in our case, the unit impulse response is simply $ h(t)=2\delta(t)-3\delta(t-2)\! $
System Function
To find the system function $ H(s)\! $ we use the formula:
$ H(s)=\int_{-\infty}^{+\infty} h(t)e^{st}dt $ where $ s=j\omega\! $.
$ H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-2)]e^{-st}dt $
$ =\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-2)e^{-st}dt $ which by the sifting property,
$ =2-3e^{-2s}\! $
System Response
The signal found in Question 1 was:
$ x(t)=\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t} $
To find the system's response to the signal, simply multiply it by the system function:
$ y(t)=H(jw)*x(t)\! $
$ y(t)=(2-3e^{-2jw})*[\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t}] $
$ =\frac{4}{j}e^{j3t}-\frac{4}{j}e^{-j3t}+(1+2)e^{j2t}+(1+2j)e^{-j2t}-\frac{6}{j}e^{3j(t-2)}+\frac{6}{j}e^{-3j(t-2)}-\frac{3+6j}{2}(e^{2j(t-2)})-\frac{3+6j}{2}(e^{-2j(t-2)} $
