E.3 Exercise 3

Ex. 1: Given the discrete-time system:

\[ G(z) = \begin{bmatrix} \frac{z-1}{(z-0.5)^2} \\ \frac{z}{(z-0.5)^2} \end{bmatrix} \]

and the state space representation

\[ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.25 & 1 \end{bmatrix} \begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u(k) \]

\[ \begin{bmatrix} y_1(k) \\ y_2(k) \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} \]

Find

1. Find the poles and zeros of \(G(s)\),

   1. Poles

      1.1 Find Minors

      \[ M_{11} = \frac{z-1}{(z-0.5)^2} \]
      \[ M_{12} = \frac{z}{(z-0.5)^2} \]

      1.2 Least Common Denominator \(\rightarrow\) \(\varphi(z)\)

      \[ \varphi(z) = (z-0.5)^2 \]

      1.3 Roots

      \[ z_{1,2} = 0.5 \]

   2. Zeros

      2.1 Normal order

      1

      2.2 All minors of order 1

      \(M_{11}\), \(M_{12}\)

      2.3 Max common devisor when \(\varphi(z)\) is denominators

      \(z(z) = 1\), no zeros

2. Determine how many outputs can be regulated to constant references,

   Control

   \[ \begin{aligned} p\leq m \\ \text{No derivative action} \begin{aligned} s=0 \\ z=1 \end{aligned} \end{aligned} \]

   considor only \(y_1\)

   check conditions integration

   \[ G_1(z) = \frac{z-1}{(z-0.5)^2} \left\{\begin{aligned} &1 \leq 1 \\ &\text{Derivative action} \end{aligned}\right. \]

   consider \(y_2\)

   \[ G_2(z) = \frac{z}{(z-0.5)^2} \left\{\begin{aligned} &1 \leq 1 \\ &\text{No derivative action} \end{aligned}\right. \]

   Static gains, find \(u_\infty\)

   \[ G(z=1) = \begin{bmatrix} 0\\4 \end{bmatrix} \]
   \[ y_\infty = G(1)u_\infty \]
   \[ y_\infty = [0]u_\infty \text{(No)} \quad y_\infty = 4u_\infty \text{(Yes)} \]

3. Verify there are no invariant zeros in \(z=1\), using the state space model.

   \[ P(z) = \begin{bmatrix} zI - A & -B \\ C & D \end{bmatrix} = \begin{bmatrix} z & -1 & 0 \\ 0.25 & z-1 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix} \]
   \[ P(z=1) = \begin{bmatrix} 1 & -1 & 0 \\ 0.25 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix} \]

   No zeros in 1

4. Assume the states are measurable, show how to design a pole placement scheme guaranteeing zero error robust regulation (refer to Figure 1)

   Enlarged system

   \[ \dot{\tilde x} = \tilde A \tilde x + \tilde B u + \tilde M d \leftarrow \text{goal} \]
   \[ \tilde x = \begin{bmatrix} x_1 \\ x_2 \\ \eta \end{bmatrix} \quad \dot{\tilde x} = \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot\eta \end{bmatrix} \]
   \[ \eta = \frac{1}{z-1}(y_2^\circ - y_2) = \frac{e_2}{z-1} \]
   \[ \begin{aligned} \eta(k+1) &- \eta(k) = e_2(k) \\ \eta(k+1) &= y_2^\circ - y_2(k) + \eta(k) \\ &= y_2^\circ + \eta(k) - C_2x(k) \end{aligned} \]
   \[ \begin{aligned} x(k+1) &= Ax(k) + Bu(k) \\ \eta(k+1) &= y_2^\circ + \eta(k) - C_2x(k) \end{aligned} \]
   \[ \tilde x(k+1) = \begin{bmatrix} A & \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ C_2 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \eta \end{bmatrix} + \begin{bmatrix} B \\ 0 \end{bmatrix}u + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} y_2^\circ \]

   To do the pole-palcement, we need to check the reachablility of \((\tilde A, \tilde B)\), which is equilvant to \((A, B)\),

   \[ \begin{aligned} M_R &= \begin{bmatrix} B & AB \end{bmatrix} \\ &= \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} \text{(ok)} \end{aligned} \]

   We can use the matlab function K = place(A_tilde, B_tilde, [P_1, P_2, P_3]) toplace the poles,

   \[ K = \begin{bmatrix} K_1 & K_2 & K_\eta \end{bmatrix} \]

5. In case the state are not measurable, can we use a static observer?

Staic Observer

\[ y = Cx + Du = Cx \]

\[ \hat x = C^{-1} y \]

Conditions of static observer: \(C\) is suqare, \(C\) is non-singular

\[ C = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \quad C^{-1} = C \]

\[ \hat x = Cy \]

Ex. 3: Consider the continous system

\[ \left\{\begin{aligned} &\dot x = x^3 -xu^2 \\ &y = x \end{aligned}\right. \]

and the equilivrium point \(\bar x = 1\), \(\bar u = 1\). Design a pole-placement

Point \(\bar x = 1\), \(\bar u = 1\),

\[ \delta \dot x = \frac{\partial f}{\partial x}|_{\bar x, \bar u} \delta x + \frac{\partial u}{\partial x}|_{\bar u, \bar x} \delta u = (3\bar x^2 - \bar u^2) \delta x - 2\bar x \bar u \delta u = 2 \delta x - 2\delta u \]

The system is unstable since we have the pole \(=2\),

\(A = 2\), \(B = -2\), \(C = 1\), \(D = 0\),

Pole-placement requires the reachability that \(M_R = [B] = [-2]\), (ok)

\[ \begin{aligned} &\delta u = v - K_x \delta x \\ &\delta \dot x = 2 \delta x - 2 \delta u = (2\delta x - 2 v + 2 K_x \delta x) = \underbrace{(2 + 2K_x)}_{A_{pp}} \delta x - 2 v \end{aligned} \]

\(p = -1 = 2+2K_x\), \(K_x = -\frac32\)

Design PI for performances

We treat the feedback controlled system as \(\tilde G = \frac{-2}{s+1}\),

The PI regulator have the TF: \(K_p \frac{sT_i + 1}{sT_i}\), and we want \(R(s) = \frac{s+1}{s}\frac{\omega_c}{\mu}\)

And \(L(s) = RG = \frac{\omega_c}{s}\)

Point 2

Conditions for the enlarged systems

\(p = 1 \leq m = 1\), (ok),

derivative action (no) \(G(s) = \frac{-2}{s-2}\)

\[ \dot{\tilde x} = \tilde A \underbrace{\tilde x}_{\begin{bmatrix} x \\ \eta \end{bmatrix}} + \tilde B u + \tilde M d \]

\[ \dot \eta = e = y^\circ - C\delta x \]

\[ \begin{bmatrix} \delta \dot x \\ \dot \eta \end{bmatrix} = \begin{bmatrix} 2&0 \\ -1&0 \end{bmatrix} \begin{bmatrix} \delta x \\ \eta \end{bmatrix} + \begin{bmatrix} -2 \\ 0 \end{bmatrix} u + \begin{bmatrix} 0 \\ 1 \end{bmatrix}y^\circ \]

Point 2

...

Poles = \([-1, -1]\)

\[ \begin{aligned} \varphi^*(s) &= (s+1)^2 = s^2 + 2s + 1 \\ \varphi(s) &= \det(sI+A^*) = \det\begin{bmatrix} s-(2+2K_x)&-2K_\eta \\ 1&s \end{bmatrix} \end{aligned} \]

Compute \(K\)

\[ \begin{aligned} a^* &= a &\rightarrow &1=1\\ b^* &= b &\rightarrow &2=2-2K_x\\ c^* &= c &\rightarrow &1=K_\eta \end{aligned} \]

1. Poles All minous of \(G(s)\)

   \[ M_{21} = \det \begin{bmatrix} \frac{1}{s+1}&\frac{1}{s-0.5} \\ 0&\frac{1}{s+2} \end{bmatrix} \]

   Least common denominator

   \[ \varphi(s) = \frac{1}{(s+1)(s+2)(s-0.5)} \]

   Poles: \(s=-1\), \(s=-0.5\), \(s=-2\)

   Zeros:

   normal orders \(=2\),

   Minors of order 2 with \(\varphi(s)\) at denominator.

   \[ M_{21} = \frac{1}{(s+1)(s+2)}\frac{s-0.5}{s-0.5} \]

   \(s = 0.5\)

   Does the cancellation happen?

   \[ L(s) = G(s)R(s) = \begin{bmatrix} \frac{1}{s+1}&\frac{1}{s-0.5} \\ 0&\frac{1}{s+2} \end{bmatrix} \begin{bmatrix} 1&1 \\ 0&\frac{s-0.5}{s-0.5} \end{bmatrix} = \begin{bmatrix} \frac{1}{s+1}&0 \\ 0&\frac{s-0.5}{(s+1)(s+2)} \end{bmatrix} \]

   Poles

   \[ M_2 = \frac{(s-0.5)}{(s+1)^2(s+2)} \]

   Poles \(= -1, -1, -2\)

   No it is not enough due to control condition

   check all 5 TF: \(S(s)\), \(T(s)\), \(\frac{U(s)}{\partial u}\), \(\frac{U(s)}{y}\), \(\frac{y}{\partial u}\)

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