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

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