Lecture 3-Signal Flow Graph

5.6 Signal-Flow Graphs of State Equations

P254 of edition 7 book
Forms:

1. Phase variable
2. Cascade
3. Parallel
4. Controller canonical
5. Observer canonical

$$G(s) = \frac{b_m s^m + ... + b_0}{s^n + a_{n-1} s^{n-1} + ... + a_0 }$$

Phase variable

define $X_1(s) \equiv R(s)/(s^n + a_{n-1} s^{n-1} + … + a_0)$ \
Then define $X_2(s) = sX_1(s),\cdots, X_{n}(s)=sX_{n-1}(s)$\
So $X_{k}(s) = s^{k-1} X_1(s)$

then

$$R(s) = (s^n + a_{n-1} s^{n-1} + ... + a_0)X_1(s)\\ =sX_n(s) + (a_{n-1}X_n(s) + ... + a_0 X_1(s))$$

rearrange it

$$sX_{k}(s) = X_{k+1}(s)\\ sX_n(s) = -(a_0X_1(s) + ... + a_{n-1} X_n(s)) + R(s)$$

Moreover

$$C(s) = (b_m s^m + ... + b_0)X_1(s)\\ = b_m X_{m+1}(s) + ... + b_0 X_1(s)$$

So

$$A= \begin{bmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1\\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{bmatrix} B = \begin{bmatrix} 0\\ 0\\ \vdots\\ 0\\ 1 \end{bmatrix} \\ C = \begin{bmatrix} b_0 & \cdots & b_{m+1} & \cdots & 0 \end{bmatrix}$$

and

$$\dot{X} = AX + Br\\ Y = CX$$

Cascade

$$G(s) = \prod \frac{1}{s-s_k} \prod \frac{s-p_k}{s-s_k} * K$$

define $X_n(s) = R(s) \frac{1}{s-s_n}$, here $sX_n(s) = s_n X_n(s) + R(s)$

define $A_k(s) = A_{k+1}(s) \frac{s-p_k}{s-s_k}$

moreover

$$A_{k+1}(s) + s_k X_k(s) = sX_k(s)\\ sX_k(s) - p_kX_k(s) = A_k(s)$$

so

$$sX_k(s) = p_kX_k(s) + A_k(s)\\ = p_kX_k(s) + (s-s_{k-1})X_{k-1}(s)\\ s[X_k(s)-X_{k-1}(s)] = p_kX_k(s) -s_{k-1}X_{k-1}(s)$$

Parallel

$$G(s) = \sum \frac{t_k}{s- s_k}$$

then define $X_k(s) \equiv R(s)\frac{t_k}{s- s_k}$

so

$$sX_k(s) = s_kX_k(s) + t_kR(s)\\ Y(s) = \sum X_k(s)$$

Controller canonical

reverse $X_1, …, X_n$ to $X_n,…, X_1$\
So 中心对称反射

$$A= \begin{bmatrix} -a_{n-1} & \cdots & -a_2 & -a_1 & -a_0\\ 1 & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & \cdots & 1 & 0\\ \end{bmatrix} B = \begin{bmatrix} 1\\ 0\\ \vdots\\ 0\\ 0 \end{bmatrix} \\ C = \begin{bmatrix} 0 & 0 & \cdots & b_{m+1} & \cdots& b_0 \end{bmatrix}$$

and

$$\dot{X} = AX + Br\\ Y = CX$$

Observer canonical

reverse $C(s)<=>R(s)$\
reverse $X<=>\dot{X}$\
change direction of arrows

$$G(s) = C(sI-A)^{-1}B\\ =G(s)^T = B^T(sI-A^T)^{-1}C^T$$

So here $A \Leftarrow A^T$\
$B \Leftarrow C^T$\
$C \Leftarrow B^T$\
So

$$A= \begin{bmatrix} -a_{n-1} & 1 & 0 & 0 & 0\\ -a_{n-2} & 0 & 1 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ -a_1 & 0 & \cdots & 0 & 1\\ -a_0 & 0 & \cdots & 0 & 0\\ \end{bmatrix} B = \begin{bmatrix} 0\\ 0\\ \vdots\\ b_{m+1}\\ \vdots\\b_0 \end{bmatrix} \\ C = \begin{bmatrix} 1& 0& \cdots& 0& 0 \end{bmatrix}$$

and

$$\dot{X} = AX + Br\\ Y = CX$$