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\$ Commented Jul 5, 2016 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\$ Commented Jul 5, 2016 at 19:29
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    \$\begingroup\$ A transfer function requires zero initial conditions. \$\endgroup\$
    – Chu
    Commented Jul 5, 2016 at 21:41

2 Answers 2


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

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


It is due to the fact that there is an assumption that all initial conditions are zero. This is necessary, as the transfer function is based on linear time invariant systems. If the initial conditions are not zero, there will be a non-linear effect present, due to adding constants.


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