I was trying to solve a question in which the transfer function of a system was asked, its unit step response was given as:

$$c(t) = 1-10e^{-t}$$

The method that the book followed was to first find out $C(s)$ i.e.

$$\mathcal{L}(c(t)) = \frac{1-9s}{s(s+1)}$$

Then they find out the Laplace Transform of the input i.e. $R(s) = \frac{1}{s}$ (since the input was step input) and finally the transfer function $G(s)$ was,

$$G(s) = \frac{C(s)}{R(s)} = \frac{1-9s}{s+1} $$ That was the answer.

But I tried to find out the transfer function by first calculating the impulse response ($h(t)$) of the system, which is equal to the time domain differentiation of unit step response ($u(t)$). So,

$$h(t) = \frac{d(1-10e^{-t})}{dt} = (\delta(t))+10e^{-t}) $$

now the transfer function will be Laplace Transform of Impulse response, So Transfer function = $1+\frac{10}{s+1}$

I can't figure out where is the mistake, why the answers differ when we apply a different approach.


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  • $\begingroup$ I don't doubt someone can you help you with your question here but maybe in Mathematica you can find more help. $\endgroup$ – Darwing Sep 23 '13 at 21:40
  • $\begingroup$ I've changed (untagged) to (laplace-transform). It would be good, if someone who understands this post better, would check that tag and add other tags, if needed. $\endgroup$ – Martin Sleziak Oct 17 '13 at 8:50

In the given question, system is assumed to be causal, input is unit step ,therefore step response will be $[1-10e^{-t}]u(t)$. Now , let us apply both the methods :

1: $c(t) =[ 1-10e^{-t}]u(t)$

$H(s)= \frac{C(s)}{X(s)} = s\left[\frac1s-\frac{10}{s+1}\right] = \frac{1-9s}{s+1}$

2: $c(t) = [1-10e^{-t}]u(t)$ \begin{align*} h(t) &= \delta[c(t)]/dt \\ &= \delta[u(t)]/dt -10 \delta[e^{-t}u(t)]/dt) &\text{ UV form differentiation} \\ &= \delta(t) + 10e^{-t}u(t) - 10e^{-t}\delta(t) \\ &= \delta(t)-10\delta(t)+ 10e^{-t}u(t) & x(t)d(t) = x(0)d(t) \\ &= 10e^{-t}u(t) - 9\delta(t) \\ \end{align*}

$H(s) = \frac{10}{s+1} - 9 = \frac{1-9s}{s+1}$

  • $\begingroup$ Yeah, it's correct, but the notation in this proof is pretty confusing (in particular between derivative and Dirac delta, as well inconsistent on the former), so no upvote from me. $\endgroup$ – Fizz Nov 27 '15 at 17:27
  • $\begingroup$ Well, since I upvoted the other somewhat incorrect answer I might as well upvote this too. The derivation of the derivative is correct. $\endgroup$ – Fizz Nov 27 '15 at 17:51

But I tried to find out the transfer function by first calculating the impulse response of the system, which is equal to the time domain differentiation of unit step response.

Oh, no. Just because unit impulse function is the time differentiation of unit step function, it does not follow that impulse response is the derivative of the step response.

Instead, the step response is the convolution of unit impulse response with the step function. See here for the precise reference. So to go in the other direction, you have to find out which function gives the convolution. This is easier to do in the domain of Laplace transforms, since convolutions become multiplications, and so to go in the other direction we just have to perform division.

  • $\begingroup$ For LTI systems however the impulse response [in the time domain] is the derivative of the step response [also in the time domain], according to every reliable source I checked 1 2. This is indeed not the case for LTV systems for instance. The statement on Wikipedia is in the wrong section probably, or not particular enough. And doesn't actually have a reference. $\endgroup$ – Fizz Nov 27 '15 at 16:36
  • $\begingroup$ Note that I've improved/fixed Wikipedia in the mean time, so it contradicts you on the level of effort required for LTI now. $\endgroup$ – Fizz Nov 27 '15 at 17:55

If $Y(s)$ is the impulse response, and $C(s) = \frac{1-9s}{s(s+1)}$ was the unit step input response. Its given that $c(t) = 1 - 10e^{-t}$.

$$Y(s) = sC(s)$$

Converting to time domain from s domain,

$$y(t) = \frac{d c(t)}{dt} + c(0^-)\delta(t)$$

As $$\frac{d(f(t))}{dt} <=> sF(s) - f(0^-)$$. proof

Usually, we assume zero initial conditions, so the second term is 0. i.e. $c(t = 0) = 0$.

So performing the differentiation we get,

$$y(t) = (0 +10e^{-t}) + (1-10e^0) \delta(t) = 10e^{-t} - 9\delta(t)$$

transforming this to the s domain

$$Y(s) = G(s) = \frac{10}{s+1} - 9 = \frac{1-9s}{s+1}$$

So the idea is that when you apply the step input $u(t)$ you get the output as $c(t) = u(t)(1-10e^{-t})$. Because the output can only start after t = 0.

$$y(t) = \frac{d c(t)}{dt} = u(t)\frac{d (1-10e^{-t})}{dt} + (1-10e^{-t})\frac{d u(t)}{dt}$$

$$y(t) = u(t)(0 +10e^{-t}) + (1-10e^{-t}) \delta(t) $$

$$y(t) = 10u(t)e^{-t} + \delta(t) - 10\delta(t)e^{-t} $$

Now since the dirac delta function zeros everything out (it's an impulse centered around zero) $f(t)\delta(t) = f(0)\delta(t)$

$$y(t) = 10u(t)e^{-t} + \delta(t) - 10\delta(t) $$

$$y(t) = 10u(t)e^{-t} - 9\delta(t)$$

$$Y(s) = \frac{1-9s}{s+1}$$

The following is my rough work to solve this problem.

rough work


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