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Step response of system is

\$c(t) = 5 - 0.5e^{-2t} + 8e^{-t}\$

then find steady state gain. Answer is -7.5

Now in book solution is given as :

Since this is step response we will find impulse response by differentiating with time, then "new c(t)" = y(t)

\$y(t) = e^{-2t} + -8^{-t}\$

convert into s-domain

\$Y(s) = \frac{1}{s + 2} - \frac{8}{s + 1}\$,

by final value theorem taking s->0 DC gain is -7.5

I understood this much. But my doubt is that why can't I just use \$\frac{C(s)}{R(s)}\$ where \$R(s) = \frac{1}{s}\$ as that is step response. Then by taking s->0 I get DC gain as 5 which is not the correct answer. Please illuminate!

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  • \$\begingroup\$ \$c(0)\neq 0\$, hence this is not a step response in a conventional sense. \$\endgroup\$ – davyjones May 5 '16 at 20:34
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Because when taking C(s)/R(s) you are not calculating the gain, but the final DC operating point, which is trivial to find. Just take \$t\rightarrow\infty\$ and see that \$c(\infty) = 5\$. The DC gain is not the same as the response at DC conditions!

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  • \$\begingroup\$ what is "dc operating point"? And then what is expression for the gain? \$\endgroup\$ – Master Chief Sep 17 '13 at 14:43
  • \$\begingroup\$ The DC operating point is the final state of the system, at \$t = \infty\$. The DC gain is \$c(t=\infty)/i(t=\infty)\$, where i(t) is the input signal. However, in this equation you are not given the input signal, so you don't know its amplitude. The trick is that even without knowing the input signal, you can still deduce the gain of this transfer function by studying the response to a step input. That is why you are getting this exercise. \$\endgroup\$ – user36129 Sep 18 '13 at 8:31
  • \$\begingroup\$ I can't seem to understand what you are saying. How can I deduce gain by studying the step response? This was my initial question if I was not clear before. Setting t=infinity or taking laplace with s=0 gives me 5, but diffrentiating and taking laplace with s=0 gives me -7.5 which is the answer.This is not a homework but just my curosity. \$\endgroup\$ – Master Chief Sep 18 '13 at 17:43
  • \$\begingroup\$ Do you want to understand it conceptually or mathematically? \$\endgroup\$ – user36129 Sep 19 '13 at 7:17
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Imagine a circuit that has at zero volts input an output (offset) of 12.5 volts. When its input is raised to 1V, its output goes to 5V. The gain of this circuit is -7.5, the output change divided by the input change.

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I think impulse response is not the derivative of step response. Why don't you try to get step response from \$Y(s)\$. You will not achieve \$c(t)\$ again.

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  • \$\begingroup\$ Could you explain why you state that the impulse response is not the derivative of the step response? \$\endgroup\$ – Joe Hass Dec 6 '13 at 12:21
  • \$\begingroup\$ Let h(t) be the impulse response and g(t) be the step response. Suppose that h(t) = dg/dt. Then Laplace transform of both sides must be equal. Note that g(t) = inverse Laplace transform of H(s)/s. However, Laplace transform of dg/dt is sG(s) - g(0) =/ sG(s)=H(s) unless g(0)=0. So we have a contradiction. \$\endgroup\$ – BARK Dec 7 '13 at 10:40
  • \$\begingroup\$ @BARK, i think you mean that the given impulse response is not the derivative of the given "step response". (In general, the impulse response is indeed the derivative of the step response, as both of them have zero initial conditions by convention) The question from OP is problematic itself. \$\endgroup\$ – davyjones May 5 '16 at 20:43

protected by Dave Tweed Oct 14 '14 at 15:52

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