# What is the value of the step response at t = 2?

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, $c(t)=u(t-3)+u(t-5)$

• Have you considered substituting the value "2" for the symbol "t" in the last line? Apr 16, 2013 at 13:29
• 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. Apr 16, 2013 at 14:15
• @ Dave Tweed : Y(y)ou A(a)re R(r)ight. Apr 16, 2013 at 14:24

$H(S)=\frac{C(s)}{R(s)}=e^{-s}+e^{-3s}$

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, $c(t)=u(t-1)+u(t-3)$

• Good. Now, put the value 2 in for t and you'll have the numeric answer that is expected. Apr 16, 2013 at 15:47