how they got such a transfer function?

Question

Could anyone explain to me the paragraph? How they have defined the $PI$ controller in discrete-time and how they got such a transfer function? and why the matrix will not be Schur? Thanks.

Answer 1

1. Part: Discrete formulation of PI-Controller: The continuous Laplace transform of a PI-Controller is given by

$$C(s) = \dfrac{\pi(s)}{e(s)}=k_\text{P}+k_\text{I}\dfrac{1}{s}$$ $$s\pi(s)=k_\text{P}se(s)+k_\text{I}e(s)$$

this implies the differential equation

$$\dot{\pi}(t)=k_\text{P}\dot{e}(t)+k_\text{I}e(t)$$

Now, apply the Euler Approximation for the derivative

$$\dot{x}(t)\approx \dfrac{x(k)-x(k-1)}{T_\text{S}},$$ in which $T_\text{S}$ is the sampling time.

$$\implies \pi(k)=\pi(k-1)+k_\text{P}e(k)-k_\text{P}e(k-1)+k_\text{I}T_\text{S}e(k)$$ $$\implies \pi(k)=\pi(k-1)+\left(k_\text{P}+k_\text{I}T_\text{S}\right)e(k)-k_\text{P}e(k-1)$$ $$\implies \pi(k)=\pi(k-1)+\left(k_\text{P}+k_\text{I}T_\text{S}\right)\left[e(k)-\dfrac{k_\text{P}}{k_\text{P}+k_\text{I}T_\text{S}}e(k-1)\right]$$

Substituting $\kappa = k_\text{P}+k_\text{I}T_\text{S}$ and $\alpha=\dfrac{k_\text{P}}{k_\text{P}+k_\text{I}T_\text{S}}$ results in the form that was given.

2. Part: Determining the $\mathcal{Z}$-Transform: There are rules for getting the $\mathcal{Z}$-Transform of expressions like $f(k) \leftarrow f(z)$ and $f(k-1) \implies f(z)/z$ (Shifting rule). If you apply these rules to equation $(9)$ you will obtain equation $(10)$.

$$\pi(z)=\dfrac{\pi(z)}{z}+\kappa\left[e(z)-\alpha\dfrac{e(z)}{z} \right]$$

$$\implies \pi(z)\left[1-z^{-1} \right]=\kappa e(z)\left[1-\alpha z^{-1} \right]$$

$$C(z)=\dfrac{\pi(z)}{e(z)}=\kappa\dfrac{1-\alpha z^{-1}}{1-z^{-1}}.$$

3. Part: For the last part of the question ... I must admit that I don't know what the Schur property implies. But I guess that there is some theorem which implies the statement in the question. A matrix is Schur iff absolute value of eigenvalues are strictly less than one.