6. Laplace Transform of Derivatives

\[\boxed{\mathcal{L}\left[\frac{df}{dt}\right] = sF(s) - f(0)}\] Similarly, for higher derivatives: \[\begin{gathered} \mathcal{L}\left[\frac{d^nf}{dt^n}\right] = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - s^{n-3}f''(0) - \\ \cdots - sf^{(n-2)}(0) - f^{(n-1)}(0) \end{gathered}\] If, \(f(0) = f'(0) = f''(0) = \cdots = f^{(n-1)}(0)=0\), then \[\mathcal{L}\left[\frac{d^nf}{dt^n}\right] = s^nF(s)\] In process control problems, we usually assume zero initial conditions. This corresponds to the nominal steady state when “deviation variables” are used.

Laplace transform of the first two derivatives: \begin{align*} \mathcal{L}\left[\frac{dy}{dt}\right] &= sY(s) - y(0) \\ \mathcal{L}\left[\frac{d^2y}{dt^2}\right] &= s^2Y(s) - sy(0)-y'(0) \end{align*}