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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)\$

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  • \$\begingroup\$ Have you considered substituting the value "2" for the symbol "t" in the last line? \$\endgroup\$ – Scott Seidman Apr 16 '13 at 13:29
  • \$\begingroup\$ I think there's some confusion about the wording of the question. The OP has interpreted it to mean "what is the response of the system to a step that occurs at t=2?", and the answer he came up with is correct. However, the more common interpretation of the question would be "what is the response of the system at t=2 for a step that occurred at t=0?". Usually in this sort of question, it is implied that the stimulus occurs at t=0. \$\endgroup\$ – Dave Tweed Apr 16 '13 at 14:15
  • \$\begingroup\$ @ Dave Tweed : Y(y)ou A(a)re R(r)ight. \$\endgroup\$ – HOLYBIBLETHE Apr 16 '13 at 14:24
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\$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)\$

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  • \$\begingroup\$ Good. Now, put the value 2 in for t and you'll have the numeric answer that is expected. \$\endgroup\$ – Dave Tweed Apr 16 '13 at 15:47

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