Is impulse response always differentiation of unit step response of a system?

Question

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

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

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

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

(Answer).

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 differetiation of unit step response. so, impulse response = \$ \frac{d}{dt}(1-10^{-t}) = (\delta(t))+10^{-t} \$ now 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.

Answer 1

Impulse response is not always a derivative of unit step response - it is the case in linear systems only. I believe your book is dealing with linear systems, therefore the method you suggest must work.

Assuming that your system is causal in addition to being linear, the unit step response is (probably) given as:

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

Differentiating the above equation leads to:

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

Since Dirac's delta is zero for \$t \neq 0\$, the equation can be rewritten as:

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

Perform Laplace Transform and you'll get to the same result as your book.

Summary:

Dealing with systems don't ever forget to explicitly write the complete form of responses. Keep track of your \$\delta\$'s, \$u\$'s and etc. Differentiate the complete forms of functions.