Various standard types of automatic control systems (with examples)

1. Paving the way for state space-related knowledge

Before reading this article, the default reader already knows the relevant knowledge of the state space, including:

• There are more than one form of expression of the state space, which can be expressed in $\tilde{x}=$Transform on the basis of$T$$x\; ;$
• The state space is a means to study the internal variables of the system dynamic process, and the selected state variables $x_i$in more than one way;
• All state variables xi x_i in the system$x_i$It does not have to have a physical meaning, sometimes it is only established for the convenience of mathematical representation; also not every state variable will be observed or measured.

All state space expressions in this paper are established based on the following differential equation:
$$\begin{gathered} y^{(n)}+a_{n-1}{y}^{(n-1)}+a_{n-2}{y}^{(n-2)}+\cdots +a_2\ddot{y}+a_1\dot{y}+a_{0}y= \\ {b}_n{u}^{\left(n\right)}+b_{n-1}{u}^{(n-1)}+\cdots +b_1\dot{u}+b_{0}u\text{\hspace{1em}( 1 )} \end{gathered}$$ Note thatin the above formula,$y^{}$Coefficient an = 1before${}^{(}$${}^n$${}^{)\; a\_n\; =\; 1}$$a_n=1$。

Instead, the goal is to write it in the following state-space form:
$$\begin{gathered} \dot{\mathbf{X}}=\mathbf{A}\mathbf{X}+BU \\ \mathbf{Y}=\mathbf{C}\mathbf{X}+DU\_ \end{gathered}$$

2. [Does not contain] derivative item in the system input

That is, the right side of formula (1) only contains $u$，式(1)变为：
$$y^{(n)}+a_{n-1}{y}^{(n-1)}+a_{n-2}{y}^{(n-2)}+\cdots +a_2\ddot{y}+a_1\dot{y}+a_{0}y=b_{0}u\text{\hspace{1em}( 2 )}$$ At this time, the state quantity is selected as:
$$\{\begin{array}{l}{\dot{x}}_1=x_2\\ {\dot{x}}_2=x_3\\ ⋮\\ {\dot{x}}_{n-1}=x_n\\ {\dot{x}}_n=-a_0{x}_1-a_1{x}_2-\cdots -a_{n-1}{x}_n+b_{0}u\\ y=x_1\end{array}$$其状态空间矩阵为：
$$\mathbf{A}=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ -a_0& -a_1& -a_2& \cdots & -a_{n-1}\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\\ b_0\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}1& 0& \cdots & 0\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(3)}$$

3. [Contains] the derivative item in the system input

In this case, the differential equation of the system has the form of formula (1):
$$\begin{gathered} y^{(n)}+a_{n-1}{y}^{(n-1)}+a_{n-2}{y}^{(n-2)}+\cdots +a_2\ddot{y}+a_1\dot{y}+a_{0}y= \\ {b}_n{u}^{\left(n\right)}+b_{n-1}{u}^{(n-1)}+\cdots +b_1\dot{u}+b_{0}u\text{\hspace{1em}(1)} \end{gathered}$$此时状态量选择为：
$$\{\begin{array}{l}{\dot{x}}_1=x_2+h_{1}u\\ {\dot{x}}_2=x_3+h_{2}u\\ ⋮\\ {\dot{x}}_{n-1}=x_n+h_{n-1}u\\ {\dot{x}}_n=-a_0{x}_1-a_1{x}_2-\cdots -a_{n-1}{x}_n+h_{n}u\\ y=x_1+h_{0}u\end{array}$$where the new parameter $h_i$为：
$$\begin{gathered} h_0=b_n \\ {h}_i=b_{n-i}-\sum _{j=1}^i{a}_{n-j}{h}_{i-j}\text{\hspace{1em}(4)} \end{gathered}$$其状态空间矩阵为：
$$\mathbf{A}=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ -a_0& -a_1& -a_2& \cdots & -a_{n-1}\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}{h}_1\\ h_2\\ ⋮\\ h_{n-1}\\ h_n\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}1& 0& \cdots & 0\end{array}\right],\mathbf{D}=h_0\text{\hspace{1em}( 5 )}$$ Visible matrix$A$ is the same as in formula (3).

4. The system input [contains] derivative term, but ${\mathbf{b}}_n=0$

At this time, the right side of the differential equation (1) of the system starts from $b_{n-1}{u}^{(n-1)}$开始：
$$\begin{gathered} y^{(n)}+a_{n-1}{y}^{(n-1)}+a_{n-2}{y}^{(n-2)}+\cdots +a_2\ddot{y}+a_1\dot{y}+a_{0}y= \\ {b}_{n-1}{u}^{(n-1)}+\cdots +b_1\dot{u}+b_{0}u\text{\hspace{1em}( 1 )} \end{gathered}$$ Substituting formula (4) is easy to know,$h_0=0$ , then a new set of state variables can be selected according to the following rules:
$$\{\begin{array}{l}{\dot{x}}_1=-a_0{x}_n+b_{0}u\\ {\dot{x}}_2=x_1-a_1{x}_n+b_{1}u\\ ⋮\\ {\dot{x}}_{n-1}=x_{n-2}-a_{n-2}{x}_n+b_{n-2}u\\ {\dot{x}}_n=x_{n-1}-a_{n-1}{x}_n+b_{n-1}u\\ y=x_n\end{array}$$其状态空间矩阵为：
$$\mathbf{A}=\left[\begin{array}{ccccc}0& 0& \cdots & 0& -a_0\\ 1& 0& \cdots & 0& -a_1\\ 0& 1& \cdots & 0& -a_2\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 1& -a_{n-1}\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}{b}_0\\ b_1\\ ⋮\\ b_{n-1}\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}0& 0& \cdots & 1\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(6)}$$

5. Controllable standard type

In this case, the differential equation of the system has the form of formula (1):
$$\begin{gathered} y^{(n)}+a_{n-1}{y}^{(n-1)}+a_{n-2}{y}^{(n-2)}+\cdots +a_2\ddot{y}+a_1\dot{y}+a_{0}y= \\ {b}_n{u}^{\left(n\right)}+b_{n-1}{u}^{(n-1)}+\cdots +b_1\dot{u}+b_{0}u\text{\hspace{1em}(1)} \end{gathered}$$此时状态量选择为：
$$\{\begin{array}{l}{\dot{x}}_1=x_2\\ {\dot{x}}_2=x_3\\ ⋮\\ {\dot{x}}_n=-a_0{x}_1-a_1{x}_2-\cdots -a_{n-1}{x}_n+u\\ y=-b_0{x}_1-b_1{x}_2-\cdots -b_{n-1}{x}_n\end{array}$$where the new parameter $b_i$为：
$$\{\begin{array}{l}{b}_0=b_0-a_0{b}_n\\ b_1=b_1-a_1{b}_n\\ ⋮\\ b_i=b_i-a_i{b}_n\\ ⋮\\ b_{n-2}=b_{n-2}-a_{n-2}{b}_n\\ b_{n-1}=b_{n-1}-a_{n-1}{b}_n\end{array}\text{\hspace{1em}(7)}$$其状态空间矩阵为：
$$\mathbf{A}=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ -a_0& -a_1& -a_2& \cdots & -a_{n-1}\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}{b}_0& b_1& \cdots & b_{n-1}\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(8)}$$

6. Considerable Standard

The observable canonical form is the transpose of the controllable canonical form:
$${\mathbf{A}}_c={A_o}^T,B_c={C_o}^T,C_c={B_o}^T$$ where the subscripts$c,o$ denote controllable and observable matrices, respectively. This gives:
$$\mathbf{A}=\left[\begin{array}{ccccc}0& 0& \cdots & 0& -a_0\\ 1& 0& \cdots & 0& -a_1\\ 0& 1& \cdots & 0& -a_2\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 1& -a_{n-1}\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}{b}_0\\ b_1\\ b_2\\ ⋮\\ b_{n-1}\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}0& 0& \cdots & 1\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(9)}$$

7. The equivalent type of the transfer function characteristic equation [when there is no repeated root]

Note: This type can only be used when the characteristic equation of the transfer function has no repeated roots!

According to the different selection methods of the state variables, there are generally two expressions of the equivalent type.

From formula (1), the transfer function relation in the complex plane can be written:
$$W(s)=\frac{b_n{s}^n+b_{n-1}{s}^{n-1}+\cdots +b_{1}s+b_0}{s^n+a_{n-1}{s}^{n-1}+\cdots +a_{1}s+a_0}=\frac{N(s)}{D(s)}$$where $D\left(s\right)=0$ is the characteristic equation of the transfer function, and its characteristic root is$l_i$. Then the denominator can be written as:
$$D\left(s\right)=\left(s-l_1\right)\left(s-l_2\right)\cdots \left(s-l_n\right)$$此时传递函数即：
$$W(s)=\frac{N(s)}{D(s)}=\frac{b_n{s}^n+b_{n-1}{s}^{n-1}+\cdots +b_{1}s+b_0}{\left(s-l_1\right)\left(s-l_2\right)\cdots \left(s-l_n\right)}$$展开即得：
$$W(s)=\frac{N(s)}{D(s)}=\frac{and(s}{U(s)}=\sum _{i=1}^n\frac{c_i}{s-l_i}\text{\hspace{1em}( 10 )}$$ ci$c_i$is the transfer function $W\left(s\right)$ at the pole$l_i$处的留数：
$$c_i=\left(s-l_i\right)W(s){|}_{s={\lambda }_i}\text{\hspace{1em}(11)}$$从而
$$and\left(s\right)=\sum _{i=1}^n\frac{c_i}{s-l_i}U(s)$$

7.1 Type I

Design change is
$$X_i(s)=\frac{1}{s-l_i}U(s)$$则有
$$\{\begin{array}{l}{\dot{x}}_1={\lambda }_1{x}_1+u\\ {\dot{x}}_2={\lambda }_2{x}_2+u\\ ⋮\\ {\dot{x}}_n={\lambda }_n{x}_n+u\\ y=c_1{x}_1+c_2{x}_2+\cdots +c_n{x}_n\end{array}$$其状态空间矩阵为：
$$\mathbf{A}=\left[\begin{array}{ccccc}{\lambda }_1& & & & \\ & {\lambda }_2& & & \\ & & \ddots & & \\ & & & {\lambda }_{n-1}& \\ & & & & {\lambda }_n\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}1\\ 1\\ ⋮\\ 1\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}{c}_1& c_2& \cdots & c_n\end{array}\right],\mathbf{D}=0\text{\hspace{1em}( 12 )}$$ where$A$ isa diagonal matrix, and the elements on the main diagonal are all characteristic roots$l_i$, while the rest of the elements are all 0 .

7.2 Type II

Design change is
$$X_i(s)=\frac{c_i}{s-l_i}U\left(s\right)$$ compared with the first type,$X_{i}The\; numerator\; of\; (s)$ is no longer 1 but$c_i$。则有
$$\{\begin{array}{l}{\dot{x}}_1=l_1{x}_1+c_{1}u\\ {\dot{x}}_2=l_2{x}_2+c_{2}u\\ ⋮\\ {\dot{x}}_n=l_n{x}_n+c_{n}u\\ y=x_1+x_2+\cdots +x_n\end{array}$$Specify the boundary value:
$$\mathbf{A}=\left[\begin{array}{ccccc}{l}_1& & & & \\ & l_2& & & \\ & & \ddots & & \\ & & & l_{n-1}& \\ & & & & l_n\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_{n-1}\\ c_n\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}1& 1& \cdots & 1\end{array}\right],\mathbf{D}=0\text{\hspace{1em}( 13 )}$$ where$A$ is a diagonal matrix, the same as the first type.

8. The equivalent type of the transfer function characteristic equation [when there are multiple roots]

Here we take a characteristic equation with a triple root as an example. Let the characteristic equation at this time be
$$D(s)={\left(s-l_1\right)}^3\left(s-l_4\right)\cdots \left(s-l_n\right)$$ where$l_1$is a triple real pole, and the remaining $l_i(i\ne 1)$ is a simple real pole. Then the transfer function can be decomposed into:
$$W(s)=\frac{N(s)}{D(s)}=\frac{and(s}{U(s)}=\frac{c_{11}}{{\left(s-l_1\right)}^3}+\frac{c_{12}}{{\left(s-l_1\right)}^2}+\frac{c_{13}}{\left(s-l_1\right)}+\sum _{i=4}^n\frac{c_i}{s-l_i}\text{\hspace{1em}( 14 )}$$ The selection method of the state variable is the same as that of the first type and the second type. The state space matrix at this time is:
$$\mathbf{A}=\left[\begin{array}{cccccc}{l}_1& 1& & & & \\ & l_1& 1& & 0& \\ & & l_1& & & \\ & & & l_4& & \\ & 0& & & \ddots & \\ & & & & & l_n\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ 1\\ 1\\ ⋮\\ 1\end{array}\right]$$$$\mathbf{C}=\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& c_4& \cdots & c_n\end{array}\right],\mathbf{D}=0\text{\hspace{1em}( 15 )}$$ Specify the equations:
$$\mathbf{A}=\left[\begin{array}{cccccc}{l}_1& & & & & \\ 1& l_1& & & 0& \\ & 1& l_1& & & \\ & & & l_4& & \\ & 0& & & \ddots & \\ & & & & & l_n\end{array}\right],\mathbf{B}=\left[\begin{array}{c}{c}_{11}\\ c_{12}\\ c_{13}\\ c_4\\ ⋮\\ c_n\end{array}\right]$$$$\mathbf{C}=\left[\begin{array}{cccccc}0& 0& 1& 1& \cdots & 1\end{array}\right],\mathbf{D}=0\text{\hspace{1em}( 16 )}$$ It should be explained that, in formulas (15) and (16),$A$ is no longer a complete diagonal matrix, due to the repeated root$l_1$The existence of , $l_1$在A{\bmA}There is a correspondingsub-block$in\; A$ , and its interior is a matrix with 1 in the upper triangle or lower triangle (such as$subblock\; in\; the\; upper\; left\; corner\; of\; A$ ). In addition, in B or C, corresponding sub-blocks also have their residues. $\mathbf{A}$、$\mathbf{B}$、 C {\bm C} The sub-block order in$C$$l_1$have the same multiplicity, as in this example $l_1$is a triple root, then A {\bm A}The sub-block in the upper left corner$of\; A$$\mathbf{B}$、 C {\bm C} The sub-blocks in$C\; are\; also\; of\; third\; order.$In this paper, the sub-blocks are separated by thin lines for easy identification.

9. Examples

9.1 [Not included] derivative item in the system input

Let the differential equation of the system be
$$y^{(3)}+2\ddot{y}+5\dot{y}+2y=2u$$ see$n=3$ , then its coefficients can be written immediately:$a_0=2,a_1=5,a_2=2,a_3=1;b_0=2$。See (3):
$$\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -2& -5& -2\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ b_0\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}1& 0& 0\end{array}\right],\mathbf{D}=0$$

9.2 [contains] derivative items in the system input

Let the differential equation of the system be
$$y^{(3)}+2\ddot{y}+5\dot{y}+2y=2\ddot{u}+3\dot{u}+u$$ can see$n=3$ , then its coefficients can be written immediately:$a_0=2,a_1=5,a_2=2,a_3=1;b_0=1,b_1=3,b_2=2,b_3=0$ . Substituting into formula (4) first calculate$h_i$：
$$\begin{gathered} h_i=b_{n-i}-\sum _{j=1}^i{a}_{n-}{h}_{i-j}⟹ \\ {h}_0=b_n=b_3=0, \\ {h}_1=b_{3-1}-\sum _{j=1}^1{a}_{3-j}{h}_{1-j}=b_2-a_2{h}_0=2, \\ {h}_2=b_{3-2}-\sum _{j=1}^2{a}_{3-j}{h}_{2-j}=b_1-a_2{h}_1-a_1{h}_0=3-2\times 2=-1 \\ {h}_3=b_{3-3}-\sum _{j=1}^3{a}_{3-j}{h}_{3-j}=b_0-a_2{h}_2-a_1{h}_1-a_0{h}_0=1+2\times 1-5\times 2=-7 \end{gathered}$$ Substituting into formula (5),
the state space matrix is:
$$\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -2& -5& -2\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}{h}_1\\ h_2\\ h_3\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ -7\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}1& 0& \cdots & 0\end{array}\right],\mathbf{D}=h_0=0$$

9.3 The system input [contains] derivative term, but ${\mathbf{b}}_{\mathbf{n}}=0$

Still taking the differential equation in 9.3 as an example, notice that $b_n=b_3=0$ . Then it can be substituted into formula (6):
$$\mathbf{A}=\left[\begin{array}{ccc}0& 0& -2\\ 1& 0& -5\\ 0& 1& -2\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}{b}_0\\ b_1\\ ⋮\\ b_{n-1}\end{array}\right]=\left[\begin{array}{c}1\\ 3\\ 2\end{array}\right],\mathbf{C}=\left[\begin{array}{cccc}0& 0& \cdots & 1\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(6)}$$

9.4 Controllable standard type

Let the system differential equation be:
$$y^{(3)}+2\ddot{y}+3\dot{y}+3y=2{and}^{\left(3\right)}+\ddot{u}+3\dot{u}+4u$$ can see$n=3,a_0=3,a_1=3,a_2=2,a_3=1;b_0=4,b_1=3,b_2=1,b_3=2$ . Substituting formula (7) to calculate parameter$b_i$：
$$\{\begin{array}{l}{b}_0=b_0-a_0{b}_n=4-3\times 2=-2,\\ b_1=b_1-a_1{b}_n=3-3\times 2=-3,\\ b_2=b_2-a_2{b}_n=1-2\times 2=-3\end{array}\text{\hspace{1em}( 7 )}$$ Substitute into formula (8) to get:
$$\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -3& -3& -2\end{array}\right]$$$$\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}-2& -3& -3\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(8)}$$

9.5 Considerable Standard Form

Since the observable canonical form is the transpose of the controllable canonical form, we can immediately write:
$$\mathbf{A}=\left[\begin{array}{ccc}0& 0& -3\\ 1& 0& -3\\ 0& 1& -2\end{array}\right]$$$$B=\left[\begin{array}{c}-2\\ -3\\ -3\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}0& 0& 1\end{array}\right],\mathbf{D}=0\text{\hspace{1em}(8)}$$