CT LTI System:
$ y(t)=x(t)+x(t-3) $.
Unit Impulse Response
Let $ x(t)=\delta(t) $. Then $ h(t)=delta(t)+\delta(t-3) $.
System Function:
$ H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
On applying the sifting property we get
$ H(j\omega)=1+e^{-3j\omega} $
This is also the Laplace transform of the impulse response evaulated .
Response to a Signal :
$ x(t)=15\sin(4t)+(2+j)\cos(6t) =\frac{2+j}{2}e^{-6jt}-\frac{15}{2j}e^{-4jt}+\frac{15}{2j}e^{4jt}+\frac{2+j}{2}e^{6jt} $
multiplying each term of input signal by $ H(j\omega) $
$ y(t)=H(jw)x(t) $
$ y(t)=(1+e^{18j})(\frac{2+j}{2}e^{-6jt})-(1+e^{12j})(\frac{15}{2j}e^{-4jt})+(1+e^{-12j})(\frac{15}{2j}e^{4jt})+(1-e^{-12j})(\frac{2+j}{2}e^{6jt}) $
final $ y(t) $ is the sum of the doubled input $ y'(t)=2x(t) $ and the shifted input $ y''(t)=x(t-2) $.
Thus the system is both linear and time invariant.
Source-: * HW4.3 Ben Laskowski_ECE301Fall2008mboutin
