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}\$

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

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

See: https://en.wikipedia.org/wiki/Two-sided_Laplace_transform#Properties


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