Mathematical basis of vehicle control theory — state equation

Time:2022-5-7

1. Use the equation of state to find the transfer function formula


The equation of state is\(G(s)=\dfrac{Y(s)}{U(s)} = C(sI-A)^{-1}B+D\)

Example 1:\(m-c-k\)System, seeking\(m\overset{··}{x}+c\overset{·}{x}+kx=f\)Transfer function.
Solution:
order\(x_1 = x\)\(x_2=\overset{·}{x}\)Then there are:

\[\begin{cases}
\overset{·}{x_1}=x_2\\
\overset{·}{x_2}=\dfrac{1}{m}[-c x_2-k x_1]+\dfrac{f}{m}
\end{cases}\]

According to the vector expression of the state equation and the vector expression of the output equation

\[\begin{cases}
\overset{·}{X} = AX+Bf \\
Y = CX+Df
\end{cases}\]

You can get:

\[A=\left[\begin{matrix}
0 & 1\\
-\dfrac{k}{m}& -\dfrac{c}{m}
\end{matrix}\right]\qquad B = \left[\begin{matrix}
0\\
\dfrac{1}{m}
\end{matrix}\right]\qquad C=\left[ \begin{matrix}1&0\end{matrix}\right]\qquad D = 0 \qquad X = \left[\begin{matrix}x_1\\x_2\end{matrix}\right]
\]

therefore

\[(sI-A)^{-1} = \left[\begin{matrix}
s&-1\\
\dfrac{k}{m}&s+\dfrac{c}{m}
\end{matrix}\right]^{-1}=\dfrac{1}{\left|\begin{matrix}
s&-1\\
\dfrac{k}{m}&s+\dfrac{c}{m}
\end{matrix}\right|}\left[\begin{matrix}
s+\dfrac{c}{m}&1\\
-\dfrac{k}{m}&s
\end{matrix}\right]
\]

Can get

\[G(s)=\dfrac{1}{s^2+\dfrac{c}{m}s+\dfrac{k}{m}}\left[\begin{matrix}1&0\end{matrix}\right]\left[\begin{matrix}
s+\dfrac{c}{m}&1\\
-\dfrac{k}{m}&s
\end{matrix}\right] \left[\begin{matrix}
0\\
\dfrac{1}{m}
\end{matrix}\right]=\dfrac{1}{ms^2+cs+k}
\]

2. Feedback control

2.1 design of state feedback controller

The control law is\(u = -KX + v\)
The closed-loop system can be obtained by bringing the control law into the state equation\(\begin{cases}
\overset{·}{X}=(A-BK)X+Bv\\
Y = CX
\end{cases}\)

The transfer function is\(G_K(s)=C(sI-A+BK)^{-1}B\)

Example 2:\(\overset{··}{x}=\overset{·}{x}+x+u\), how to design\(u\), make\(x \rarr x_d\quad(t\rarr\infty)\)\(x_d\)Is a constant.

Solution:
Input yes\(u\), the characteristic equation is\(s^2-s-1=0\)

It can be seen that the system is open-loop unstable, because there is a root in the right half plane, which can be designed\(u=-2\overset{·}{x}-2x+x_d\)

Substituting into the above system, we can get\(\overset{··}{x}+\overset{·}{x}+x=x_d\)

For the closed-loop system at this time, the input is\(x_d\), the characteristic equation is\(s^2+s+1=0\), the two equations are negative and the closed-loop is stable.

If the method of equation of state is used, then you can make\(x_1 = x\)\(x_2=\overset{·}{x}\), you can get

\[\begin{cases}
\overset{·}{x_1}=x_2\\
\overset{·}{x_2}=x_1+x_2+u
\end{cases}\]

Then you can get:

\[A=\left[\begin{matrix}
0 & 1\\
1&1
\end{matrix}\right]\qquad B = \left[\begin{matrix}
0\\
1
\end{matrix}\right]\qquad C=\left[ \begin{matrix}1&0\end{matrix}\right]\qquad D = 0 \qquad X = \left[\begin{matrix}x_1\\x_2\end{matrix}\right]
\]

The control law is\(u = -KX + x_d\), where\(K=[k_1, k_2]\), then

\[\begin{aligned}
(sI-A+BK)^{-1} &= \left( \left[ \begin{matrix}
s&0\\
0&s
\end{matrix}\right]-\left[\begin{matrix}
0 & 1\\
1&1
\end{matrix}\right]+\left[\begin{matrix}
0\\
1
\end{matrix}\right][k_1, k_2]\right)^{-1}\\
&=\left[ \begin{matrix}
s&-1\\
-1+k_1&s-1+k_2
\end{matrix}\right]^{-1}\\
&=\dfrac{1}{s^2+(k_2-1)s+k_1-1}\left[\begin{matrix}
s-1+k_2&1\\
1-k_1&s
\end{matrix}\right]
\end{aligned}
\]

The transfer function is

\[\begin{aligned}
G_k(s)&=C(sI-A+BK)^{-1}B\\
&=\dfrac{1}{s^2+(k_2-1)s+k_1-1}\left[ \begin{matrix}1&0\end{matrix}\right]\left[\begin{matrix}
s-1+k_2&1\\
1-k_1&s
\end{matrix}\right]\left[\begin{matrix}
0\\
1
\end{matrix}\right]\\
&=\dfrac{1}{s^2+(k_2-1)s+k_1-1}
\end{aligned}
\]

If order\(k_1=k_2=2\), then the characteristic equation is\(s^2+s+1=0\), same as the above.

The following figure shows the structure of the state feedback system

2.2 design of output feedback controller

The control law is\(u = -HY + v\)
The closed-loop system can be obtained by bringing the control law into the state equation\(\begin{cases}
\overset{·}{X}=(A-BHC)X+Bv\\
Y = CX
\end{cases}\)

The transfer function is\(G_H(s)=C(sI-A+BHC)^{-1}B\)

Output feedback can be regarded as a special case of state feedback. For example, when y = x, C = 1.

The following is the structure diagram of output feedback system

The controller is the design principle of PID controller.

3. Feedback linearization

Example 3:\(\overset{··}{x}=\overset{·}{x}\ ^2+x+u\), how to design\(u\), make\(x\rarr x_d\quad(t\rarr \infty)\), where\(x_d\)Is a constant.

Design\(u=-\overset{·}{x}\ ^2-x+v\)

Substitution system\(\overset{··}{x}=\overset{·}{x}\ ^2+x+u\)

sureRemove nonlinear term, get\(\overset{··}{x}=v\)

The above equation has been linearized, butOpen loop instability。 utilizefeedback controlMethod, design:\(v=-\overset{·}{x}-x+x_d\)

obtainClosed loop stabilitySystem:\(\overset{··}{x}+\overset{·}{x}+x=x_d\)

We can write the tracking error equation of the closed-loop system. order\(\epsilon=x-x_d\), the system can be converted to

\[\overset{··}{\epsilon}+\overset{·}{\epsilon}+\epsilon=0
\]

The equation has two complex roots and can be described as\(s_{1,2}=\alpha\plusmn\beta i\)In which\(\alpha<0\)

The solution can be expressed as:\(\epsilon_{1,2}=e^{\alpha t}(c_1cos\beta t\plusmn c_2isin\beta t)\)

It can be concluded that\(\epsilon_{1,2}\rarr0\), indicating that the system is stable.

Summary of feedback linearization methods:Input state linearizationandInput-output linearization

  • For the former, in this example, you can make\(z=z(x)=\overset{·}{x}\quad \overset{··}{x}=v\), you can get the solution\(u=u(x, v)=-\overset{·}{x}\ ^2-x+v\), the linear system at this time is\(\overset{··}{x}=v\)
  • For the latter, in this example, you can make\(y=h(x)=x\), therefore\(\overset{··}{y}=g(x,u)=\overset{··}{x}=\overset{·}{x}\ ^2+x+u\)So you get the output\(y\)And input\(u\)Relationship, order\(\overset{··}{y}=v\), also get\(u=-\overset{·}{x}\ ^2-x+v\)

It can be seen that the solution process of the two methods in this example is consistent.