Lecture 12-Design Via State Space

Design Via State Space

formula often used

$$\zeta = \frac{-ln(\%)}{\sqrt{ln(\%)^2+\pi^2}}\\ T_s = \frac{4}{\zeta \omega_n}\\ T_p = \frac{\pi}{\sqrt{1-\zeta^2} \omega_n}$$

where T_s represent settling time; T_p represent peak time\
so for the polynomial:

$$s^2 + 2\zeta \omega_n s + \omega_n^2\\ 2\zeta \omega_n = \frac{8}{T_s} = \frac{-2 ln(\%)}{T_p}\\ \omega_n^2 = \frac{16 (ln(\%)^2+\pi^2)}{ln(\%)^2 T_s^2} = \frac{ln(\%)^2+\pi^2}{T_p^2}$$

Controllability

The basic equation set: (D alaways = 0)

$$\dot{X} = A X + B U\\ Y = C X + D$$

Introduce the Controller K always 1xN, \
where U always 1x1

$$U = r - KX$$

So, we obtain

$$\dot{X} = AX + B(r - K U) = (A-BK)X + Br$$

if we could manipulate th poles of $|sI- (A-BK)|$
Thus means

$$Y = C \mathbb{L^{-1}}[(sI - (A-BK))^{-1}] * \mathbb{L^{-1}}[BR(s)]$$

Transformation

Here

$$Z = PX$$

We have

$$\dot{Z} = AZ + BU \\ Y = C Z\\ U = r - KZ$$

That means we have

$$\dot{X} = P^{-1}APX + P^{-1}BU\\ Y = CPX\\ U = r - KPX$$

Compare with

$$\dot{X} = A_x X + B_x U\\ Y = C_x X\\ U = r - K_x X$$

So we have:

$$A_x = P^{-1}AP\\ B_x = P^{-1}B\\ C_x = CP\\ K_x = KP$$

Then we have

$$C_{Mx} = [B_x \space A_xB_x \cdots A_x^{N-1}B_x] = P^{-1}C_{Mz}$$

where X is observer canonical form\
Z is other form (like phase variable form, cascade form)

$$P = C_{Mz}C_{Mx}^{-1}\ K_z = K_xP^{-1}$$

Observability

$$\dot{\hat{X}} = A \hat{X} + BU + L(Y - \hat{Y})\\ \hat{Y} = C \hat{X}$$

so with

$$\dot{X} = A X + B U\\ Y = C X$$

then obtain

$$\dot{X} - \dot{\hat{X}} = A(X - \hat{X}) - LC(X- \hat{X})\\ =(A-LC) (X - \hat{X})$$

define $e_X \equiv (X - \hat{X})$, we have

$$\dot{e}_X = (A-LC) e_X$$

If all poles of (A-LC) in the left plane

$$\lim_{t\to \infty} e_X = (X - \hat{X}) = 0$$

Then we could use $\hat{X}$ to estimate $X$\
regardless the influence of initial value $\hat{X}(0)$ and $X(0)$

Transformation

$$Z = PX$$

where X is observer canonical form\
Z is other form (like phase variable form, cascade form)

$$(\dot{Z} - \dot{\hat{Z}}) = (A-LC)(Z - \hat{Z})$$

Then we have

$$(\dot{X} - \dot{\hat{X}}) = P^{-1}(A-LC)P(X - \hat{X})$$

So we have:

$$A_x = P^{-1}AP\\ B_x = P^{-1}B\\ C_x = CP\\ L_x = P^{-1}L$$

Now calculate $O_{Mx}$

$$O_{Mx} = \begin{bmatrix} C_x\\ C_xA_x\\ \vdots\\ C_xA_x^{N-1} \end{bmatrix} = O_{Mz} P$$

So, in conclusion:

$$P = O_{Mz}^{-1}O_{Mx}\ L_z = PL_x$$

Integral Control with 0 Steady-State Error

$$U = V - KX\\ \frac{(R-Y)}{s} K_e =X_N K_e = V \equiv \frac{Y}{T(s)}$$

So

$$\frac{Y}{R} = \frac{K_e \frac{T(s)}{s}}{1+K_e \frac{T(s)}{s}}$$

Then

$$e_{ss} = \lim_{s\to 0+} sR(s) (1 - \frac{Y(s)}{R(s)})\\ = \lim_{s\to 0+} \frac{1}{1+K_e \frac{T(s)}{s}}\\ = \lim_{s\to 0+} \frac{s}{s+K_e T(s)} = 0$$

Because

$$\dot{x}_{N}=R- Y = R- CX = [-C \space 0] \begin{bmatrix} X \\ x_N \end{bmatrix} + R$$

so

$$\begin{bmatrix} \dot{X} \\ \dot{x}_N \end{bmatrix}= \begin{bmatrix} A & 0\\ -C & 0 \end{bmatrix} \begin{bmatrix} X \\ x_N \end{bmatrix}+ \begin{bmatrix} B \\ 0 \end{bmatrix}U+ \begin{bmatrix} 0 \\ 1 \end{bmatrix}R$$

because

$$U = K_e x_N - K X=[-K\space K_e] \begin{bmatrix} X \\ x_N \end{bmatrix}$$

we have

$$\begin{bmatrix} \dot{X} \\ \dot{x}_N \end{bmatrix}=( \begin{bmatrix} A & 0\\ -C & 0 \end{bmatrix}+ \begin{bmatrix} B \\ 0 \end{bmatrix} [-K\space K_e] ) \begin{bmatrix} X \\ x_N \end{bmatrix}+ \begin{bmatrix} 0 \\ 1 \end{bmatrix}R\\ =\begin{bmatrix} A-BK & BK_e\\ -C & 0 \end{bmatrix} \begin{bmatrix} X \\ x_N \end{bmatrix}+ \begin{bmatrix} 0 \\ 1 \end{bmatrix}R$$

Why zero of T(s) Not change with Controller

we know

$$G(s) = \frac{C \text{adj}(sI - A)B}{|sI - A|}\\ T(s) = \frac{C \text{adj}(sI - A + BK)B}{|sI - A + BK|}$$

why the numurator of G(S), T(s) is the same, because $\forall C$, so must prove

$$\text{adj}(sI - A)B = \text{adj}(sI - A + BK)B$$

that is mean $\forall A$ (replace sI-A with A)

$$\text{adj}(A)B = \text{adj}(A + BK)B$$

lemma: Cramer’s Rule

for $AX = B$, where

$$X = \begin{bmatrix} x_1\\ \vdots\\ x_i\\ \vdots\\ x_N \end{bmatrix}$$

We have

$$X = A^{-1}B = \frac{\text{adj}(A)B}{|A|} = \frac{|A \stackrel{i}{\leftarrow} B|\text{for x_i}}{|A|}\\ x_i |A| = \text{adj}(A)B \quad \text{ith element}= |A \stackrel{i}{\leftarrow} B|$$

Here

$$A =\left(\mathbf{a}_{1} \cdots \mathbf{a}_{n}\right)\\ \left(A^{i} \leftarrow B\right) \stackrel{\text { def }}{=}\left(\begin{array}{llllll}{\mathbf{a}_{1}} & {\cdots} & {\mathbf{a}_{i-1}} & B & {\mathbf{a}_{i+1}} & {\cdots} & {\left.\mathbf{a}_{n}\right)}\end{array}\right.$$

proof

$$\text{adj}(A+BK)B \quad \text{ith element} = |(A+BK) \stackrel{i}{\leftarrow} B|\\ = \left|\begin{array}{llllll}{\mathbf{a}_{1}}+k_1B & {\cdots} & {\mathbf{a}_{i-1}}+k_{i-1}B & B & {\cdots} & {\mathbf{a}_{N}+k_NB}|\end{array}\right.\\ = \left|\begin{array}{llllll}{\mathbf{a}_{1}} & {\cdots} & {\mathbf{a}_{i-1}} & B & {\cdots} & {\mathbf{a}_{N}}|\end{array}\right.\\ = |A \stackrel{i}{\leftarrow} B| = \text{adj}(A)B \quad \text{ith element}\\$$

Another rule $|A+BK| = |A| + Kadj(A)B$

$$|A+BK| = \left|\begin{array}{llllll}{\mathbf{a}_{1}}+k_1B & {\cdots} & {\mathbf{a}_{i}}+k_{i}B & {\cdots} & {\mathbf{a}_{N}+k_NB}|\end{array}\right.\\ = \left|\begin{array}{llllll}{\mathbf{a}_{1}} & {\cdots} & {\mathbf{a}_{i-1}} & {\cdots} & {\mathbf{a}_{N}}|\end{array}\right. \\ + \sum_i k_i \left|\begin{array}{llllll}{\mathbf{a}_{1}} & {\cdots} & {\mathbf{a}_{i-1}} & B & {\cdots} & {\mathbf{a}_{N}}|\end{array}\right.\\ = |A| + \sum_i k_i \space [\text{adj}(A)B \quad \text{ith element}]\\ = |A| + K \text{adj}(A)B$$

Conclusion

$$G(s) = \frac{C \text{adj}(sI - A)B}{|sI - A|} = \frac{N(s)}{D_1(s)}\\ T(s) = \frac{C \text{adj}(sI - A + BK)B}{|sI - A + BK|}\\ = \frac{C \text{adj}(sI - A)B}{|sI - A| + K \text{adj}(sI-A)B} = \frac{N(s)}{D_2(s)}$$

if we introduce K_e

$$\frac{Y(s)}{R(s)} \equiv T'(s) = \frac{K_e \frac{T(s)}{s}}{1+K_e \frac{T(s)}{s}}\\ = \frac{K_e N(s)}{sD_2(s) + K_e N(s)}$$

So no matter introduce K and K_e, zeros of T(s) Not change