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The impulse response of a continuous time system is given by \$h(t) = \delta(t – 1) + \delta(t – 3)\$. What is the value of the step response at t = 2 ?
\$H(S)=\frac{C(s)}{R(s)}=e^{-s}+e^{-3s}\$
Here, \$R(s)= \mathcal{L} u(t-2) \$
Here, \$R(s)= \frac{e^{-2s}}{s}\$
\$C(s)=R(S)(e^{-s}+e^{-3s})\$
\$C(s)=\frac{e^{-2s}}{s}(e^{-s}+e^{-3s})\$
\$C(s)=\frac{e^{-3s}}{s}+\frac{e^{-5s}}{s}\$
Response, \$c(t)=\mathcal{L^{-1}}(\frac{e^{-3s}}{s}+\frac{e^{-5s}}{s})\$
Response, \$c(t)=u(t-3)+u(t-5)\$
\$H(S)=\frac{C(s)}{R(s)}=e^{-s}+e^{-3s}\$
Here, \$R(s)= \mathcal{L} u(t) \$
Here, \$R(s)= \frac{1}{s}\$
\$C(s)=R(S)(e^{-s}+e^{-3s})\$
\$C(s)=\frac{1}{s}(e^{-s}+e^{-3s})\$
\$C(s)=\frac{e^{-s}}{s}+\frac{e^{-3s}}{s}\$
Response, \$c(t)=\mathcal{L^{-1}}(\frac{e^{-s}}{s}+\frac{e^{-3s}}{s})\$
Response, \$c(t)=u(t-1)+u(t-3)\$
