You can take the Laplace transform of the lead-lag compensator to convert it into the state-space form. The frequency domain equations is
$$\frac{Y(s)}{U(s)}=K\frac{s-z}{s-p}$$
We can represent this as the multiplication of two transfer functions
$$\frac{Y(s)}{U(s)}=K(s-z)\cdot\frac{1}{s-p}=\frac{Y(s)}{X(s)}\cdot\frac{X(s)}{U(s)}$$
If $$\frac{X}{U}=\frac{1}{s-p}$$
then we can rearrange this equation to solve for U
$$Xs-Xp=U$$
This gives us the laplace transform of
$$\mathscr{L}^{-1}{Xs+xXp}=\mathscr{L}^{-1}{U}$$
which is just
$$\dot{x}(t)=px(t)+u(t)$$
We do this again with
$$\frac{Y}{X}=K(s-z)$$
Rearrange to solve for y
$$Y=KXs-Xz$$
We can then take the Laplace transform
$$\mathscr{L}^{-1}{Y}=\mathscr{L}^{-1}{KXs-Xz}$$
which gives us
$$y(t)=K\dot{x}(t)-zx(t)=K(p-z)x(t)+Ku(t)$$
Therefore, our final state-space equation is
$$\begin{matrix}
\dot{x(t)}=px(t)+u(t) \\
y(t)=K(p-z)x(t)+Ku(t)
\end{matrix}$$
[[Machine Precision]] – used to determine the equivalence of systems
[[HL-10 SAS]] – used lead-compensators as part of a structural mode filter.
- bendickinsonHowTransformLead2023