# why transfer function of derivative controller is $s$ while Laplace of derivative is not only $s$?

For derivative controller transfer function, it's $K_ds$ but the Laplace Transform of $\frac{df(x)}{dx}$ is $sF(s)-f(0)$. For example if

$$f(x) = \frac{dg(x)}{dx}$$ hence its Laplace will be

$$F(s) = sG(s)-g(0)$$

If the transfer function of derivative is $s$ then

$$\frac{F(s)}{G(s)}=s \Leftrightarrow F(s) = sG(s)$$

how can inverse Laplace transform of $F(s) = sG(s)$ give $f(x) = \frac{dg(x)}{dx}$

• I don't understand what is the question here. – Marko Buršič Jul 5 '16 at 18:41
• The assumption is that the initial condition $g(0) = 0$. That is all. Zero initial conditions are often assumed in control theory. – Captainj2001 Jul 5 '16 at 19:29
• A transfer function requires zero initial conditions. – Chu Jul 5 '16 at 21:41

The bilateral Laplace Transform of $f(t)$ is $F(s) = sG(s)$