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.
Best Answer
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*}
$H(s) = \frac{10}{s+1} - 9 = \frac{1-9s}{s+1}$