超螺旋滑模控制详细介绍(全网独家)

超螺旋滑模控制详细介绍(全网独家)

  • 1. 系统模型
  • 2. 控制量设计
  • 3. 稳定性证明
    • 3.1 李雅普诺夫函数

      V

      0

      V_0

      V0​的求导过程

    • 3.2 关于李雅普诺夫函数导数的结论(必读部分)
    • 3.3 李雅普诺夫函数导数的变换
    • 3.4 矩阵

      Q

      Q

      Q的正定性的保证

    • 3.5 李雅普诺夫函数的更新
    • 3.6 系统各部分总结
  • 4. 总结

关于超螺旋滑模控制(或称超扭滑模控制)的论文有很多,但关于其具体的稳定性证明却少之又少,数学功底不强的人很容易在中间步骤被卡壳。因此,笔者在这里给出详尽的稳定性证明过程,一并将超螺旋滑模控制理论介绍给各位读者,希望能为各位带来一定的参考。

关于该理论的详细证明过程,笔者目前没有找到其他文章,因此本文可以算作是全网第一篇完全详细推导的文章,喜欢的读者可以收藏加点赞。

本文需要读者具有一定的滑模控制理论的知识,可以点击传送门进行学习:滑模控制理论(SMC)概述。强烈建议读者阅读完该文章后再来阅读本文!

1. 系统模型

一般地,对于非线性系统可以建立具有标准柯西形式的微分方程组。令状态量为

x

=

x

1

,

x

2

=

x

˙

1

x = x_1,x_2 = \dot x_1

x=x1​,x2​=x˙1​,则有:

{

x

˙

1

=

x

2

x

˙

2

=

f

+

g

u

\begin{cases} \dot x_1 = x_2 \\ \dot x_2 = f + g \cdot u \end{cases}

{x˙1​=x2​x˙2​=f+g⋅u​与传统的滑模控制相比,超螺旋控制算法使用积分来获得实际控制量,不含高频切换量,因而系统中没有抖振。

令滑模面为

s

s

s,只要满足如下方程:

{

s

˙

=

λ

s

1

2

s

i

g

n

(

s

)

+

ν

ν

˙

=

α

s

i

g

n

(

s

)

(1)

\begin{cases} \dot s = – \lambda \left| s \right| ^{\frac{1}{2}} \cdot sign (s) + \nu \\ \dot \nu = – \alpha \cdot sign(s) \tag{1} \end{cases}

{s˙=−λ∣s∣21​⋅sign(s)+νν˙=−α⋅sign(s)​(1)则系统即为稳定的。

2. 控制量设计

设状态

x

x

x的期望值为

x

d

x_d

xd​,则跟踪误差为

e

1

=

x

1

x

d

e_1 = x_1 – x_d

e1​=x1​−xd​ 。设

e

2

=

e

˙

1

=

x

˙

1

x

˙

d

=

x

2

x

˙

d

e_2 = \dot e_1 = \dot x_1 – \dot x_d = x_2 – \dot x_d

e2​=e˙1​=x˙1​−x˙d​=x2​−x˙d​,并设滑模面为:

s

=

c

1

e

1

+

e

2

(2)

s = c_1 e_1 + e_2 \tag{2}

s=c1​e1​+e2​(2)对其求导

s

˙

=

c

1

e

˙

1

+

e

˙

2

=

c

1

e

2

+

f

+

g

u

x

¨

d

\begin{aligned} \dot s &= c_1 \dot e_1 + \dot e_2 \\ &= c_1 e_2 + f + g \cdot u – \ddot x_d \end{aligned}

s˙​=c1​e˙1​+e˙2​=c1​e2​+f+g⋅u−x¨d​​容易看出,此时如果设

u

=

g

1

(

f

+

x

¨

d

c

1

e

2

λ

s

1

2

s

i

g

n

(

s

)

α

s

i

g

n

(

s

)

)

(3)

u = g^{-1} \left( -f + \ddot x_d – c_1 e_2 – \lambda \left| s \right| ^{\frac{1}{2}} sign (s) – \alpha \cdot sign(s) \right) \tag{3}

u=g−1(−f+x¨d​−c1​e2​−λ∣s∣21​sign(s)−α⋅sign(s))(3)则

s

˙

\dot s

s˙就能具有式(1)的形式。

对于(1)中参数设定为:

λ

˙

=

ω

1

γ

1

2

,

α

=

λ

ε

+

1

2

(

β

+

4

ε

2

)

(4)

\dot \lambda = \omega_1 \sqrt{\frac{\gamma_1}{2}},\\ \alpha = \lambda \varepsilon + \frac{1}{2} \left( \beta + 4 \varepsilon^2 \right) \tag{4}

λ˙=ω1​2γ1​​
​,α=λε+21​(β+4ε2)(4)式中

α

,

β

,

ε

,

λ

,

ω

1

,

γ

1

\alpha, \beta, \varepsilon, \lambda, \omega_1, \gamma_1

α,β,ε,λ,ω1​,γ1​均大于零。

3. 稳定性证明

容易看出,与传统滑模控制不同的是,

u

u

u中含有的不再是滑模面

s

s

s,而是其多项式

s

1

2

s

i

g

n

(

s

)

\left| s \right| ^{\frac{1}{2}} sign(s)

∣s∣21​sign(s)。除此之外,在

s

˙

\dot s

s˙表达式中还出现了另一个参数

ν

\nu

ν(式(1))。不妨把这两者设定为新的状态变量,在此基础上设成李雅普诺夫函数。

{

z

1

=

s

1

2

s

i

g

n

(

s

)

z

2

=

ν

(5)

\begin{cases} z_1 = \left| s \right| ^{\frac{1}{2}} sign(s) \\ z_2 = \nu \end{cases} \tag{5}

{z1​=∣s∣21​sign(s)z2​=ν​(5)则对应的各自导数为

{

z

˙

1

=

1

2

s

1

2

s

˙

=

1

2

s

1

2

(

λ

s

1

2

s

i

g

n

(

s

)

α

s

i

g

n

(

s

)

)

z

˙

2

=

ν

˙

=

α

s

i

g

n

(

s

)

(6)

\begin{cases} \dot z_1 = \frac{1}{2} \left| s \right| ^{-\frac{1}{2}} \dot s = \frac{1}{2} \left| s \right| ^{-\frac{1}{2}} \left( -\lambda \left| s \right| ^{\frac{1}{2}} sign(s) – \alpha \cdot sign(s) \right) \\ \dot z_2 = \dot \nu = – \alpha \cdot sign(s) \end{cases} \tag{6}

{z˙1​=21​∣s∣−21​s˙=21​∣s∣−21​(−λ∣s∣21​sign(s)−α⋅sign(s))z˙2​=ν˙=−α⋅sign(s)​(6)又因为

z

1

=

s

1

2

\left| z_1 \right| = \left| s \right| ^{\frac{1}{2}}

∣z1​∣=∣s∣21​,故

1

z

1

=

s

1

2

\frac{1}{\left| z_1 \right|} = \left| s \right| ^{-\frac{1}{2}}

∣z1​∣1​=∣s∣−21​。故式(6)即为

{

z

˙

1

=

1

2

z

1

(

λ

z

1

+

z

2

)

z

˙

2

=

ν

˙

=

α

s

i

g

n

(

s

)

=

α

s

i

g

n

(

s

)

s

1

2

s

1

2

=

α

z

1

z

1

(7)

\begin{cases} \dot z_1 = \frac{1}{2 \left| z_1 \right| } \left( -\lambda z_1 + z_2 \right) \\ \dot z_2 = \dot \nu = – \alpha \cdot sign(s) = – \alpha \cdot sign(s) \cdot \left| s \right| ^{\frac{1}{2}} \cdot \left| s \right| ^{-\frac{1}{2}} = -\frac{\alpha}{ \left| z_1 \right| } z_1 \end{cases} \tag{7}

{z˙1​=2∣z1​∣1​(−λz1​+z2​)z˙2​=ν˙=−α⋅sign(s)=−α⋅sign(s)⋅∣s∣21​⋅∣s∣−21​=−∣z1​∣α​z1​​(7)即:

{

z

˙

1

=

1

2

z

1

(

λ

z

1

+

z

2

)

z

˙

2

=

α

z

1

z

1

(7)

\begin{cases} \dot z_1 = \frac{1}{2 \left| z_1 \right| } \left( -\lambda z_1 + z_2 \right) \\ \dot z_2 = -\frac{\alpha}{ \left| z_1 \right| } z_1 \end{cases} \tag{7}

{z˙1​=2∣z1​∣1​(−λz1​+z2​)z˙2​=−∣z1​∣α​z1​​(7)设新的状态变量为

Z

=

[

z

1

z

2

]

Z = \left[ \begin{matrix} z_1 \\ z_2 \end{matrix} \right]

Z=[z1​z2​​]并定义李雅普诺夫函数为

V

0

=

(

β

+

4

ε

2

)

z

1

2

+

z

2

2

4

ε

z

1

z

2

=

Z

T

P

Z

(8)

V_0 =\left( \beta + 4 \varepsilon^2 \right) z_1^2 + z_2^2 – 4 \varepsilon z_1 z_2 = Z^T P Z \tag{8}

V0​=(β+4ε2)z12​+z22​−4εz1​z2​=ZTPZ(8)其中

P

=

[

β

+

4

ε

2

2

ε

2

ε

1

]

(9)

P = \left[ \begin{matrix} \beta + 4 \varepsilon^2 & -2 \varepsilon \\ -2 \varepsilon & 1 \end{matrix} \right] \tag{9}

P=[β+4ε2−2ε​−2ε1​](9)

定理1:矩阵

A

A

A正定的充要条件是矩阵

A

A

A的所有特征根均大于零。

根据定理1不难得出矩阵

P

P

P是正定的,因而李雅普诺夫函数

V

0

0

V_0 \geq 0

V0​≥0。

3.1 李雅普诺夫函数

V

0

V_0

V0​的求导过程

直接对(8)求导。

V

˙

0

=

2

(

β

+

4

ε

2

)

z

1

z

˙

1

+

2

z

2

z

˙

2

4

ε

z

2

z

˙

1

4

ε

z

1

z

˙

2

=

2

(

β

+

4

ε

2

)

z

1

1

2

z

1

(

λ

z

1

+

z

2

)

+

2

z

2

(

α

z

1

z

1

)

4

ε

z

2

1

2

z

1

(

λ

z

1

+

z

2

)

4

ε

z

1

(

α

z

1

z

1

)

=

λ

z

1

(

β

+

4

ε

2

)

z

1

2

+

1

z

1

(

β

+

4

ε

2

)

z

1

z

2

2

α

z

1

z

1

z

2

+

2

λ

ε

z

1

z

1

z

2

2

ε

z

1

z

2

2

+

4

α

ε

z

1

z

1

2

=

1

z

1

[

4

α

ε

λ

(

β

+

4

ε

2

)

]

z

1

2

+

1

z

1

[

(

β

+

4

ε

2

)

2

α

+

2

λ

ε

]

z

1

z

2

2

ε

z

1

z

2

2

=

1

z

1

Z

T

[

4

α

ε

λ

(

β

+

4

ε

2

)

1

2

(

β

+

4

ε

2

)

α

+

λ

ε

1

2

(

β

+

4

ε

2

)

α

+

λ

ε

2

ε

]

Z

=

1

z

1

Z

T

Q

Z

(10)

\begin{aligned} \dot V_0 &= 2 \left( \beta + 4 \varepsilon^2 \right) z_1 \dot z_1 +2 z_2 \dot z_2 – 4 \varepsilon z_2 \dot z_1 – 4 \varepsilon z_1 \dot z_2 \\ &= 2 \left( \beta + 4 \varepsilon^2 \right) z_1 \cdot \frac{1}{2 \left| z_1 \right| } \left( -\lambda z_1 + z_2 \right) + 2 z_2 \left( -\frac{\alpha}{ \left| z_1 \right| } z_1 \right) – 4 \varepsilon z_2 \cdot \frac{1}{2 \left| z_1 \right| } \left( -\lambda z_1 + z_2 \right) – 4 \varepsilon z_1 \left( -\frac{\alpha}{ \left| z_1 \right| } z_1 \right) \\ &= – \frac{\lambda}{\left| z_1 \right| } \left( \beta + 4 \varepsilon^2 \right) z_1^2 + \frac{1}{\left| z_1 \right| } \left( \beta + 4 \varepsilon^2 \right) z_1 z_2 – \frac{2 \alpha}{\left| z_1 \right| } z_1 z_2 + \frac{2 \lambda \varepsilon}{\left| z_1 \right| } z_1 z_2 – \frac{2 \varepsilon}{\left| z_1 \right| } z_2^2 + \frac{4 \alpha \varepsilon}{\left| z_1 \right| } z_1^2 \\ &= \frac{1}{\left| z_1 \right| } \left[ 4 \alpha \varepsilon – \lambda \left( \beta + 4 \varepsilon^2 \right) \right] z_1^2 + \frac{1}{\left| z_1 \right| } \left[ \left( \beta + 4 \varepsilon^2 \right) – 2 \alpha + 2 \lambda \varepsilon \right] z_1 z_2 – \frac{2 \varepsilon}{ \left| z_1 \right| } z_2^2 \\ &= \frac{1}{\left| z_1 \right| } Z^T \left[ \begin{matrix} 4 \alpha \varepsilon – \lambda \left( \beta + 4 \varepsilon^2 \right) & \qquad \qquad \quad \frac{1}{2}\left( \beta + 4 \varepsilon^2 \right) – \alpha + \lambda \varepsilon \\ \frac{1}{2}\left( \beta + 4 \varepsilon^2 \right) – \alpha + \lambda \varepsilon & \qquad \qquad \quad -2 \varepsilon \end{matrix} \right] Z \\ &= -\frac{1}{\left| z_1 \right| }Z^TQZ \end{aligned} \tag{10}

V˙0​​=2(β+4ε2)z1​z˙1​+2z2​z˙2​−4εz2​z˙1​−4εz1​z˙2​=2(β+4ε2)z1​⋅2∣z1​∣1​(−λz1​+z2​)+2z2​(−∣z1​∣α​z1​)−4εz2​⋅2∣z1​∣1​(−λz1​+z2​)−4εz1​(−∣z1​∣α​z1​)=−∣z1​∣λ​(β+4ε2)z12​+∣z1​∣1​(β+4ε2)z1​z2​−∣z1​∣2α​z1​z2​+∣z1​∣2λε​z1​z2​−∣z1​∣2ε​z22​+∣z1​∣4αε​z12​=∣z1​∣1​[4αε−λ(β+4ε2)]z12​+∣z1​∣1​[(β+4ε2)−2α+2λε]z1​z2​−∣z1​∣2ε​z22​=∣z1​∣1​ZT[4αε−λ(β+4ε2)21​(β+4ε2)−α+λε​21​(β+4ε2)−α+λε−2ε​]Z=−∣z1​∣1​ZTQZ​(10)注意(10)中最后一个等号前加了负号。这样

Q

Q

Q即为

Q

=

[

4

α

ε

+

λ

(

β

+

4

ε

2

)

1

2

(

β

+

4

ε

2

)

+

α

λ

ε

1

2

(

β

+

4

ε

2

)

+

α

λ

ε

2

ε

]

(11)

Q = \left[ \begin{matrix} -4 \alpha \varepsilon + \lambda \left( \beta + 4 \varepsilon^2 \right) & \qquad -\frac{1}{2}\left( \beta + 4 \varepsilon^2 \right) + \alpha – \lambda \varepsilon \\ -\frac{1}{2}\left( \beta + 4 \varepsilon^2 \right) + \alpha – \lambda \varepsilon & \qquad 2 \varepsilon \end{matrix} \right] \tag{11}

Q=[−4αε+λ(β+4ε2)−21​(β+4ε2)+α−λε​−21​(β+4ε2)+α−λε2ε​](11)这样我们得到李雅普诺夫函数的导数:

V

˙

0

=

1

z

1

Z

T

Q

Z

(12)

\dot V_0 = -\frac{1}{\left| z_1 \right| }Z^TQZ \tag{12}

V˙0​=−∣z1​∣1​ZTQZ(12)

3.2 关于李雅普诺夫函数导数的结论(必读部分)

我们把式(11)所代表的

Q

Q

Q表示为

Q

=

[

A

B

C

D

]

Q = \left[ \begin{matrix} A & B \\ C & D \end{matrix} \right]

Q=[AC​BD​]下面开始求

Q

Q

Q的特征根的一般形式。

p

I

Q

=

p

A

B

C

p

D

=

p

2

(

A

+

D

)

p

+

A

D

B

C

\left| pI – Q \right| = \left| \begin{matrix} p-A & -B \\ -C & p-D \end{matrix} \right| = p^2 – (A+D) p + AD – BC

∣pI−Q∣=∣
∣​p−A−C​−Bp−D​∣
∣​=p2−(A+D)p+AD−BC

Δ

=

b

2

4

a

c

=

(

A

+

D

)

2

4

(

A

D

B

C

)

=

(

A

D

)

2

+

4

B

C

\Delta = b^2 – 4ac = (A+D)^2 – 4(AD – BC) = (A-D)^2 +4BC

Δ=b2−4ac=(A+D)2−4(AD−BC)=(A−D)2+4BC特征根为

p

1

,

2

(

Q

)

=

A

+

D

±

(

A

D

)

2

+

4

B

C

2

p_{1,2} (Q) = \frac{A+D \pm \sqrt{(A-D)^2 +4BC}}{2}

p1,2​(Q)=2A+D±(A−D)2+4BC
​​设两个特征根中大的为

q

max

(

Q

)

q_{\max}(Q)

qmax​(Q),小的为

q

min

(

Q

)

q_{\min}(Q)

qmin​(Q),有

{

p

max

(

Q

)

=

A

+

D

+

(

A

D

)

2

+

4

B

C

2

p

min

(

Q

)

=

A

+

D

(

A

D

)

2

+

4

B

C

2

\begin{cases} p_{\max}(Q) = \frac{A+D + \sqrt{(A-D)^2 +4BC}}{2} \\ p_{\min} (Q)= \frac{A+D – \sqrt{(A-D)^2 +4BC}}{2} \end{cases}



⎧​pmax​(Q)=2A+D+(A−D)2+4BC
​​pmin​(Q)=2A+D−(A−D)2+4BC
​​​为方便表示,把根号部分记为

R

R

R,进而

p

min

(

Q

)

Z

T

Z

=

A

+

D

R

2

(

z

1

2

+

z

2

2

)

(13)

p_{\min} (Q)Z^TZ = \frac{A+D – R}{2} \left( z_1^2 + z_2^2 \right) \tag{13}

pmin​(Q)ZTZ=2A+D−R​(z12​+z22​)(13)另一方面有

Z

T

Q

Z

=

A

z

1

2

+

(

B

+

C

)

z

1

z

2

+

D

z

2

2

(14)

Z^TQZ = A z_1^2 + (B+C)z_1 z_2 + D z_2^2 \tag{14}

ZTQZ=Az12​+(B+C)z1​z2​+Dz22​(14)为比较

p

min

(

Q

)

Z

T

Z

p_{\min} (Q)Z^TZ

pmin​(Q)ZTZ与

Z

T

Q

Z

Z^TQZ

ZTQZ的大小,不妨作差:

2

(

Z

T

Q

Z

p

min

(

Q

)

Z

T

Z

)

=

2

A

z

1

2

+

2

(

B

+

C

)

z

1

z

2

+

2

D

z

2

2

[

A

+

D

R

]

(

z

1

2

+

z

2

2

)

=

(

A

D

+

R

)

z

1

2

+

(

D

A

+

R

)

z

2

2

+

2

(

B

+

C

)

z

1

z

2

=

(

A

D

+

R

)

[

z

1

2

+

D

A

+

R

A

D

+

R

z

2

2

+

2

(

B

+

C

)

A

D

+

R

z

1

z

2

]

=

(

A

D

+

R

)

[

z

1

2

+

(

R

+

D

A

)

2

R

2

(

D

A

)

2

z

2

2

+

2

(

B

+

C

)

(

R

+

D

A

)

R

2

(

D

A

)

2

z

1

z

2

]

=

(

A

D

+

R

)

[

z

1

2

+

(

R

+

D

A

)

2

4

B

C

z

2

2

+

2

(

B

+

C

)

(

R

+

D

A

)

4

B

C

z

1

z

2

]

=

(

A

D

+

R

)

[

(

z

1

+

R

+

D

A

2

B

C

z

2

)

2

+

2

(

B

+

C

)

(

R

+

D

A

)

4

B

C

z

1

z

2

R

+

D

A

B

C

z

1

z

2

]

=

(

A

D

+

R

)

[

(

z

1

+

R

+

D

A

2

B

C

z

2

)

2

+

(

R

+

D

A

)

(

2

B

+

2

C

4

B

C

)

4

B

C

z

1

z

2

]

(15)

\begin{aligned} 2 \left( Z^TQZ – p_{\min} (Q)Z^TZ \right) &= 2A z_1^2 +2(B+C)z_1 z_2 + 2D z_2^2 – \left[ A+D – R \right] (z_1^2 + z_2^2 ) \\ &= \left( A – D + R \right) z_1^2 + \left( D-A+R \right) z_2^2 + 2 (B+C) z_1 z_2 \\ &= \left( A – D + R \right) \left[ z_1^2 + \frac{D-A+R}{A-D+R} z_2^2 + \frac{2(B+C)}{A-D+R}z_1 z_2 \right]\\ &= \left( A – D + R \right) \left[ z_1^2 + \frac{( R + D – A)^2}{R^2 – (D-A)^2} z_2^2 + \frac{2(B+C)(R+D-A)}{R^2 – (D-A)^2}z_1 z_2 \right] \\ &= \left( A – D + R \right) \left[ z_1^2 + \frac{(R+D-A)^2}{4BC}z_2^2 + \frac{2(B+C)(R+D-A)}{4BC}z_1 z_2 \right] \\ &= \left( A – D + R \right) \left[ \left( z_1 + \frac{R+D-A}{2 \sqrt{BC}}z_2 \right)^2 + \frac{2(B+C)(R+D-A)}{4BC}z_1 z_2 – \frac{R+D-A}{\sqrt{BC}}z_1 z_2 \right] \\ &= \left( A – D + R \right) \left[ \left( z_1 + \frac{R+D-A}{2 \sqrt{BC}}z_2 \right)^2 + \frac{(R+D-A)(2B+2C-4\sqrt{BC})}{4BC}z_1 z_2 \right] \tag{15} \end{aligned}

2(ZTQZ−pmin​(Q)ZTZ)​=2Az12​+2(B+C)z1​z2​+2Dz22​−[A+D−R](z12​+z22​)=(A−D+R)z12​+(D−A+R)z22​+2(B+C)z1​z2​=(A−D+R)[z12​+A−D+RD−A+R​z22​+A−D+R2(B+C)​z1​z2​]=(A−D+R)[z12​+R2−(D−A)2(R+D−A)2​z22​+R2−(D−A)22(B+C)(R+D−A)​z1​z2​]=(A−D+R)[z12​+4BC(R+D−A)2​z22​+4BC2(B+C)(R+D−A)​z1​z2​]=(A−D+R)[(z1​+2BC
​R+D−A​z2​)2+4BC2(B+C)(R+D−A)​z1​z2​−BC
​R+D−A​z1​z2​]=(A−D+R)[(z1​+2BC
​R+D−A​z2​)2+4BC(R+D−A)(2B+2C−4BC
​)​z1​z2​]​(15)对于(15),式中

(

A

D

+

R

)

\left( A-D+R \right)

(A−D+R)为常数项;因此最后结果可看成由2部分组成,第一部分为完全平方式,大于等于零;而对于第二部分的分子来说又分为

R

+

D

A

R+D-A

R+D−A和

2

B

+

2

C

4

B

C

2B+2C-4\sqrt{BC}

2B+2C−4BC
​两部分。其中:

R

+

D

A

=

(

A

D

)

2

+

4

B

C

+

D

A

=

(

A

D

)

2

+

4

B

C

(

A

D

)

0

R+D-A = \sqrt{(A-D)^2 +4BC} +D-A = \sqrt{(A-D)^2 +4BC} – (A-D) \geq 0

R+D−A=(A−D)2+4BC
​+D−A=(A−D)2+4BC
​−(A−D)≥0而根据绝对不等式

2

B

+

2

C

4

B

C

4

B

C

4

B

C

=

0

2B+2C-4\sqrt{BC} \geq 4 \sqrt{BC} – 4 \sqrt{BC} = 0

2B+2C−4BC
​≥4BC
​−4BC
​=0故式(15)的第二部分也大于等于零。

到这里我们总结可以得到:

Z

T

Q

Z

p

min

(

Q

)

Z

T

Z

0

Z^TQZ – p_{\min}(Q) Z^TZ \geq 0

ZTQZ−pmin​(Q)ZTZ≥0即

p

min

(

Q

)

Z

T

Z

Z

T

Q

Z

(16)

p_{\min}(Q) Z^TZ \leq Z^TQZ \tag{16}

pmin​(Q)ZTZ≤ZTQZ(16)同理可以得

p

max

(

Q

)

Z

T

Z

Z

T

Q

Z

(17)

p_{\max} (Q)Z^TZ \geq Z^TQZ \tag{17}

pmax​(Q)ZTZ≥ZTQZ(17)

3.3 李雅普诺夫函数导数的变换

式(17)是对

V

˙

0

=

1

z

1

Z

T

Q

Z

\dot V_0 = -\frac{1}{\left| z_1 \right| }Z^TQZ

V˙0​=−∣z1​∣1​ZTQZ作出的,对于

V

0

=

Z

T

P

Z

V_0 = Z^TPZ

V0​=ZTPZ同样根据式(17)有

p

max

(

P

)

Z

T

Z

Z

T

P

Z

(

Z

T

P

Z

)

1

2

p

max

1

2

(

P

)

(

Z

T

Z

)

1

2

=

p

max

1

2

(

P

)

Z

Z

(

Z

T

P

Z

)

1

2

p

max

1

2

(

P

)

=

V

0

1

2

p

max

1

2

(

P

)

(18)

p_{\max} (P)Z^TZ \geq Z^TPZ \Longrightarrow \\ \left( Z^TPZ \right)^{\frac{1}{2}} \leq p_{\max}^{\frac{1}{2}}(P)\left( Z^T Z \right)^{\frac{1}{2}} = p_{\max}^{\frac{1}{2}} (P)\left| \left| Z \right| \right| \Longrightarrow \\ \left| \left| Z \right| \right| \geq \frac{\left( Z^TPZ \right)^{\frac{1}{2}}}{p_{\max}^{\frac{1}{2}}(P)} = \frac{V_0^{\frac{1}{2}}}{p_{\max}^{\frac{1}{2}}(P)} \tag{18}

pmax​(P)ZTZ≥ZTPZ⟹(ZTPZ)21​≤pmax21​​(P)(ZTZ)21​=pmax21​​(P)∣∣Z∣∣⟹∣∣Z∣∣≥pmax21​​(P)(ZTPZ)21​​=pmax21​​(P)V021​​​(18)另一方面

Z

=

z

1

2

+

z

2

2

=

(

s

1

2

s

i

g

n

(

s

)

)

2

+

ν

2

=

s

+

ν

2

s

=

s

1

2

=

z

1

(19)

\left| \left| Z \right| \right| = \sqrt{z_1^2 + z_2^2} = \sqrt{\left( \left| s \right| ^{\frac{1}{2}} sign(s)\right)^2 + \nu^2} = \sqrt{\left| s \right| + \nu^2} \geq \sqrt{\left| s \right|} = \left| s \right| ^{\frac{1}{2}} = \left| z_1 \right| \tag{19}

∣∣Z∣∣=z12​+z22​
​=(∣s∣21​sign(s))2+ν2
​=∣s∣+ν2
​≥∣s∣
​=∣s∣21​=∣z1​∣(19)由(19)推出

z

1

=

s

1

2

Z

1

z

1

1

Z

(20)

\left| z_1 \right| = \left| s \right| ^{\frac{1}{2}} \leq \left| \left| Z \right| \right| \Longrightarrow \\ -\frac{1}{ \left| z_1 \right|} \leq – \frac{1}{\left| \left| Z \right| \right|} \tag{20}

∣z1​∣=∣s∣21​≤∣∣Z∣∣⟹−∣z1​∣1​≤−∣∣Z∣∣1​(20)又根据(16):

V

˙

0

=

1

z

1

Z

T

Q

Z

1

z

1

p

min

(

Q

)

Z

T

Z

=

1

z

1

p

min

(

Q

)

Z

2

1

Z

p

min

(

Q

)

Z

2

=

p

min

(

Q

)

Z

\begin{aligned} \dot V_0 &= -\frac{1}{ \left| z_1 \right|} Z^T Q Z \leq -\frac{1}{ \left| z_1 \right|} p_{\min}(Q) Z^TZ \\ &= -\frac{1}{ \left| z_1 \right|} p_{\min}(Q) \left| \left| Z \right| \right| ^2 \leq – \frac{1}{\left| \left| Z \right| \right| }p_{\min}(Q) \left| \left| Z \right| \right| ^2 \\ &= -p_{\min}(Q) \left| \left| Z \right| \right| \end{aligned}

V˙0​​=−∣z1​∣1​ZTQZ≤−∣z1​∣1​pmin​(Q)ZTZ=−∣z1​∣1​pmin​(Q)∣∣Z∣∣2≤−∣∣Z∣∣1​pmin​(Q)∣∣Z∣∣2=−pmin​(Q)∣∣Z∣∣​再根据(18)

V

˙

0

p

min

(

Q

)

Z

p

min

(

Q

)

V

0

1

2

p

max

1

2

(

P

)

=

r

V

0

1

2

(21)

\dot V_0 \leq -p_{\min}(Q) \left| \left| Z \right| \right| \leq -p_{\min}(Q)\frac{V_0^{\frac{1}{2}}}{p_{\max}^{\frac{1}{2}}(P)} = -r V_0 ^{\frac{1}{2}} \tag{21}

V˙0​≤−pmin​(Q)∣∣Z∣∣≤−pmin​(Q)pmax21​​(P)V021​​​=−rV021​​(21)其中

r

=

p

min

(

Q

)

p

max

1

2

(

P

)

(22)

r = \frac{p_{\min}(Q)}{p_{\max}^{\frac{1}{2}}(P)} \tag{22}

r=pmax21​​(P)pmin​(Q)​(22)

定理2:若系统的李雅普诺夫函数满足

V

˙

r

V

1

2

,

(

r

>

0

)

\dot V \leq – r V ^{\frac{1}{2}}, \qquad \left( r >0 \right)

V˙≤−rV21​,(r>0)则系统具有稳定性。

3.4 矩阵

Q

Q

Q的正定性的保证

根据定理2,式(21)保证了系统具有李雅普诺夫稳定性。读者可能注意到,式(21)只有在

r

0

r \geq 0

r≥0的情况下才能保证系统稳定性,而根据式(22),即需要

p

min

(

Q

)

p_{\min}(Q)

pmin​(Q)和

p

max

1

2

(

P

)

p_{\max}^{\frac{1}{2}}(P)

pmax21​​(P)均大于等于零。由于矩阵

P

P

P为正定的,因此

p

max

1

2

(

P

)

>

0

p_{\max}^{\frac{1}{2}}(P) > 0

pmax21​​(P)>0立即得证;下面需要保证

p

min

(

Q

)

>

0

p_{\min}(Q) > 0

pmin​(Q)>0,即保证矩阵

Q

Q

Q的正定性。

这里再次列出

Q

Q

Q的表达式:

Q

=

[

4

α

ε

+

λ

(

β

+

4

ε

2

)

1

2

(

β

+

4

ε

2

)

+

α

λ

ε

1

2

(

β

+

4

ε

2

)

+

α

λ

ε

2

ε

]

Q = \left[ \begin{matrix} -4 \alpha \varepsilon + \lambda \left( \beta + 4 \varepsilon^2 \right) & \qquad -\frac{1}{2}\left( \beta + 4 \varepsilon^2 \right) + \alpha – \lambda \varepsilon \\ -\frac{1}{2}\left( \beta + 4 \varepsilon^2 \right) + \alpha – \lambda \varepsilon & \qquad 2 \varepsilon \end{matrix} \right]

Q=[−4αε+λ(β+4ε2)−21​(β+4ε2)+α−λε​−21​(β+4ε2)+α−λε2ε​]不妨直接取

α

=

λ

ε

+

1

2

(

β

+

4

ε

2

)

(23)

\alpha = \lambda \varepsilon + \frac{1}{2} \left( \beta + 4 \varepsilon ^2 \right) \tag{23}

α=λε+21​(β+4ε2)(23)这样

Q

Q

Q可以化简为一个对角矩阵

Q

=

[

(

λ

2

ε

)

(

β

+

4

ε

2

)

4

λ

ε

2

0

0

2

ε

]

Q = \left[ \begin{matrix} \left(\lambda – 2 \varepsilon \right) \left( \beta + 4 \varepsilon ^2 \right) – 4 \lambda \varepsilon^2 & \quad 0 \\ 0 & \quad 2 \varepsilon \end{matrix} \right]

Q=[(λ−2ε)(β+4ε2)−4λε20​02ε​]并能够一眼看出

Q

Q

Q的特征根为

p

1

(

Q

)

=

(

λ

2

ε

)

(

β

+

4

ε

2

)

4

λ

ε

2

,

p

2

(

Q

)

=

2

ε

p_1(Q) = \left(\lambda – 2 \varepsilon \right) \left( \beta + 4 \varepsilon ^2 \right) – 4 \lambda \varepsilon^2, \\ p_2 (Q) = 2 \varepsilon

p1​(Q)=(λ−2ε)(β+4ε2)−4λε2,p2​(Q)=2ε其中

p

2

(

Q

)

=

2

ε

>

0

p_2 (Q) = 2 \varepsilon > 0

p2​(Q)=2ε>0立即得证,为保证

p

1

(

Q

)

>

0

p_1(Q) > 0

p1​(Q)>0,需要有

λ

>

2

ε

(

β

+

4

ε

2

)

β

(24)

\lambda > \frac{2 \varepsilon \left( \beta + 4 \varepsilon ^2 \right)}{\beta} \tag{24}

λ>β2ε(β+4ε2)​(24)

3.5 李雅普诺夫函数的更新

在3.4一节中给出了保证矩阵

Q

Q

Q正定性的条件。由于

α

,

λ

\alpha, \lambda

α,λ两参数是人为给出的,因此需要把这两个因素加入到李雅普诺夫函数中,构建新的李雅普诺夫函数:

V

=

V

0

+

1

2

γ

1

(

λ

λ

)

2

+

1

2

γ

2

(

α

α

)

2

(25)

V = V_0 + \frac{1}{2 \gamma_1} \left( \lambda – \lambda^* \right)^2 + \frac{1}{2 \gamma_2} \left( \alpha – \alpha^* \right)^2 \tag{25}

V=V0​+2γ1​1​(λ−λ∗)2+2γ2​1​(α−α∗)2(25)其中

λ

,

α

\lambda^*, \alpha^*

λ∗,α∗为常数(未知)。

对其求导得下式(26):

V

˙

=

V

˙

0

+

1

γ

1

(

λ

λ

)

λ

˙

+

1

γ

2

(

α

α

)

α

˙

r

V

0

1

2

+

1

γ

1

(

λ

λ

)

λ

˙

+

1

γ

2

(

α

α

)

α

˙

=

r

V

0

1

2

+

1

γ

1

(

λ

λ

)

λ

˙

+

1

γ

2

(

α

α

)

α

˙

ω

1

2

γ

1

λ

λ

+

ω

1

2

γ

1

λ

λ

ω

2

2

γ

2

α

α

+

ω

2

2

γ

2

α

α

(26)

\dot V = \dot V_0 + \frac{1}{\gamma_1} \left( \lambda – \lambda^* \right) \dot \lambda + \frac{1}{\gamma_2} \left( \alpha – \alpha^* \right) \dot \alpha \\ \leq -r V_0 ^{\frac{1}{2}} + \frac{1}{\gamma_1} \left( \lambda – \lambda^* \right) \dot \lambda + \frac{1}{\gamma_2} \left( \alpha – \alpha^* \right) \dot \alpha \\ = -r V_0 ^{\frac{1}{2}} + \frac{1}{\gamma_1} \left( \lambda – \lambda^* \right) \dot \lambda + \frac{1}{\gamma_2} \left( \alpha – \alpha^* \right) \dot \alpha – \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| + \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| – \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| + \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \tag{26}

V˙=V˙0​+γ1​1​(λ−λ∗)λ˙+γ2​1​(α−α∗)α˙≤−rV021​​+γ1​1​(λ−λ∗)λ˙+γ2​1​(α−α∗)α˙=−rV021​​+γ1​1​(λ−λ∗)λ˙+γ2​1​(α−α∗)α˙−2γ1​
​ω1​​∣λ−λ∗∣+2γ1​
​ω1​​∣λ−λ∗∣−2γ2​
​ω2​​∣α−α∗∣+2γ2​
​ω2​​∣α−α∗∣(26)根据

(

x

2

+

y

2

+

z

2

)

1

2

x

+

y

+

z

\left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}} \leq \left| x \right| + \left| y \right| + \left| z \right|

(x2+y2+z2)21​≤∣x∣+∣y∣+∣z∣有

r

V

0

1

2

ω

1

2

γ

1

λ

λ

ω

2

2

γ

2

α

α

[

r

2

V

0

+

ω

1

2

2

γ

1

(

λ

λ

)

2

+

ω

2

2

2

γ

2

(

α

α

)

2

]

1

2

-r V_0 ^{\frac{1}{2}} – \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| – \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \leq – \left[ r^2 V_0 + \frac{\omega_1^2}{2 \gamma_1} \left( \lambda – \lambda^* \right)^2 + \frac{\omega_2^2}{2 \gamma_2} \left( \alpha – \alpha^* \right)^2 \right]^{\frac{1}{2}}

−rV021​​−2γ1​
​ω1​​∣λ−λ∗∣−2γ2​
​ω2​​∣α−α∗∣≤−[r2V0​+2γ1​ω12​​(λ−λ∗)2+2γ2​ω22​​(α−α∗)2]21​设

r

,

ω

1

,

ω

2

r, \omega_1, \omega_2

r,ω1​,ω2​中最小的数为

n

=

min

(

r

,

ω

1

,

ω

2

)

n = \min(r, \omega_1, \omega_2)

n=min(r,ω1​,ω2​),则上式为

r

V

0

1

2

ω

1

2

γ

1

λ

λ

ω

2

2

γ

2

α

α

[

r

2

V

0

+

ω

1

2

2

γ

1

(

λ

λ

)

2

+

ω

2

2

2

γ

2

(

α

α

)

2

]

1

2

n

[

V

0

+

1

2

γ

1

(

λ

λ

)

2

+

1

2

γ

2

(

α

α

)

2

]

1

2

=

n

V

1

2

-r V_0 ^{\frac{1}{2}} – \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| – \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \leq – \left[ r^2 V_0 + \frac{\omega_1^2}{2 \gamma_1} \left( \lambda – \lambda^* \right)^2 + \frac{\omega_2^2}{2 \gamma_2} \left( \alpha – \alpha^* \right)^2 \right]^{\frac{1}{2}} \\ \leq – n \left[ V_0 + \frac{1}{2 \gamma_1} \left( \lambda – \lambda^* \right)^2 + \frac{1}{2 \gamma_2} \left( \alpha – \alpha^* \right)^2 \right]^{\frac{1}{2}} \\ = -n V^{\frac{1}{2}}

−rV021​​−2γ1​
​ω1​​∣λ−λ∗∣−2γ2​
​ω2​​∣α−α∗∣≤−[r2V0​+2γ1​ω12​​(λ−λ∗)2+2γ2​ω22​​(α−α∗)2]21​≤−n[V0​+2γ1​1​(λ−λ∗)2+2γ2​1​(α−α∗)2]21​=−nV21​于是代入(26)有

V

˙

r

V

0

1

2

+

1

γ

1

(

λ

λ

)

λ

˙

+

1

γ

2

(

α

α

)

α

˙

ω

1

2

γ

1

λ

λ

+

ω

1

2

γ

1

λ

λ

ω

2

2

γ

2

α

α

+

ω

2

2

γ

2

α

α

n

V

1

2

+

1

γ

1

(

λ

λ

)

λ

˙

+

1

γ

2

(

α

α

)

α

˙

+

ω

1

2

γ

1

λ

λ

+

ω

2

2

γ

2

α

α

(27)

\dot V \leq -r V_0 ^{\frac{1}{2}} + \frac{1}{\gamma_1} \left( \lambda – \lambda^* \right) \dot \lambda + \frac{1}{\gamma_2} \left( \alpha – \alpha^* \right) \dot \alpha – \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| + \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| – \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| + \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \\ \leq -n V^{\frac{1}{2}}+ \frac{1}{\gamma_1} \left( \lambda – \lambda^* \right) \dot \lambda + \frac{1}{\gamma_2} \left( \alpha – \alpha^* \right) \dot \alpha + \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| + \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \tag{27}

V˙≤−rV021​​+γ1​1​(λ−λ∗)λ˙+γ2​1​(α−α∗)α˙−2γ1​
​ω1​​∣λ−λ∗∣+2γ1​
​ω1​​∣λ−λ∗∣−2γ2​
​ω2​​∣α−α∗∣+2γ2​
​ω2​​∣α−α∗∣≤−nV21​+γ1​1​(λ−λ∗)λ˙+γ2​1​(α−α∗)α˙+2γ1​
​ω1​​∣λ−λ∗∣+2γ2​
​ω2​​∣α−α∗∣(27)由于

λ

,

α

\lambda^*, \alpha^*

λ∗,α∗为常数,不妨假设恒有

λ

>

λ

,

α

>

α

\lambda^*>\lambda, \alpha^*>\alpha

λ∗>λ,α∗>α。由于李雅普诺夫稳定性只要证明李雅普诺夫函数存在即可,因此总能找到这样的

λ

,

α

\lambda^*, \alpha^*

λ∗,α∗,该假设是合理的。

此时式(27)为

V

˙

n

V

1

2

+

1

γ

1

(

λ

λ

)

λ

˙

+

1

γ

2

(

α

α

)

α

˙

+

ω

1

2

γ

1

λ

λ

+

ω

2

2

γ

2

α

α

=

n

V

1

2

1

γ

1

λ

λ

λ

˙

1

γ

2

α

α

α

˙

+

ω

1

2

γ

1

λ

λ

+

ω

2

2

γ

2

α

α

=

n

V

1

2

+

λ

λ

(

ω

1

2

γ

1

λ

˙

γ

1

)

+

α

α

(

ω

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2

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˙

γ

2

)

(28)

\dot V \leq -n V^{\frac{1}{2}} + \frac{1}{\gamma_1} \left( \lambda – \lambda^* \right) \dot \lambda + \frac{1}{\gamma_2} \left( \alpha – \alpha^* \right) \dot \alpha + \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| + \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \\ = -n V^{\frac{1}{2}} – \frac{1}{\gamma_1} \left| \lambda – \lambda^* \right| \dot \lambda – \frac{1}{\gamma_2} \left| \alpha – \alpha^* \right| \dot \alpha + \frac{\omega_1}{\sqrt{2 \gamma_1}} \left| \lambda – \lambda^* \right| + \frac{\omega_2}{\sqrt{2 \gamma_2}} \left| \alpha – \alpha^* \right| \\ = -n V^{\frac{1}{2}} + \left| \lambda – \lambda^* \right| \left( \frac{\omega_1}{\sqrt{2 \gamma_1}} – \frac{ \dot \lambda}{\gamma_1} \right) + \left| \alpha – \alpha^* \right| \left( \frac{\omega_2}{\sqrt{2 \gamma_2}} – \frac{ \dot \alpha}{\gamma_2} \right) \tag{28}

V˙≤−nV21​+γ1​1​(λ−λ∗)λ˙+γ2​1​(α−α∗)α˙+2γ1​
​ω1​​∣λ−λ∗∣+2γ2​
​ω2​​∣α−α∗∣=−nV21​−γ1​1​∣λ−λ∗∣λ˙−γ2​1​∣α−α∗∣α˙+2γ1​
​ω1​​∣λ−λ∗∣+2γ2​
​ω2​​∣α−α∗∣=−nV21​+∣λ−λ∗∣(2γ1​
​ω1​​−γ1​λ˙​)+∣α−α∗∣(2γ2​
​ω2​​−γ2​α˙​)(28)

此时若令

λ

˙

=

ω

1

γ

1

2

(29)

\dot \lambda = \omega_1 \sqrt{\frac{\gamma_1}{2}} \tag{29}

λ˙=ω1​2γ1​​
​(29)即可使式(28)变为

V

˙

n

V

1

2

+

α

α

(

ω

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2

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˙

γ

2

)

=

n

V

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η

(30)

\dot V \leq -n V^{\frac{1}{2}} + \left| \alpha – \alpha^* \right| \left( \frac{\omega_2}{\sqrt{2 \gamma_2}} – \frac{ \dot \alpha}{\gamma_2} \right) = -n V^{\frac{1}{2}} + \eta \tag{30}

V˙≤−nV21​+∣α−α∗∣(2γ2​
​ω2​​−γ2​α˙​)=−nV21​+η(30)其中

η

=

α

α

(

ω

2

2

γ

2

α

˙

γ

2

)

(31)

\eta = \left| \alpha – \alpha^* \right| \left( \frac{\omega_2}{\sqrt{2 \gamma_2}} – \frac{ \dot \alpha}{\gamma_2} \right) \tag{31}

η=∣α−α∗∣(2γ2​
​ω2​​−γ2​α˙​)(31)根据定理2,式(30)使得系统具有稳定性。

3.6 系统各部分总结

系统具有如下为标准柯西形式:

{

x

˙

1

=

x

2

x

˙

2

=

f

+

g

u

\begin{cases} \dot x_1 = x_2 \\ \dot x_2 = f + g \cdot u \end{cases}

{x˙1​=x2​x˙2​=f+g⋅u​设计滑模面为

s

=

c

1

e

1

+

e

2

s = c_1 e_1 + e_2

s=c1​e1​+e2​以及控制量

u

u

u:

u

=

g

1

(

f

+

x

¨

d

c

1

e

2

λ

s

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2

s

i

g

n

(

s

)

α

s

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g

n

(

s

)

)

u = g^{-1} \left( -f + \ddot x_d – c_1 e_2 – \lambda \left| s \right| ^{\frac{1}{2}} sign (s) – \alpha \cdot sign(s) \right)

u=g−1(−f+x¨d​−c1​e2​−λ∣s∣21​sign(s)−α⋅sign(s))并设计自适应律为

λ

˙

=

ω

1

γ

1

2

λ

>

2

ε

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β

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4

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β

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=

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\dot \lambda = \omega_1 \sqrt{\frac{\gamma_1}{2}} \\ \lambda > \frac{2 \varepsilon \left( \beta + 4 \varepsilon ^2 \right)}{\beta} \\ \alpha = \lambda \varepsilon + \frac{1}{2} \left( \beta + 4 \varepsilon^2 \right)

λ˙=ω1​2γ1​​
​λ>β2ε(β+4ε2)​α=λε+21​(β+4ε2)则系统具有稳定性:

V

˙

n

V

1

2

+

η

\dot V \leq -n V^{\frac{1}{2}} + \eta

V˙≤−nV21​+η

4. 总结

就笔者而言,超螺旋滑模控制内容的精髓在于巧妙设计了状态量

z

1

=

s

1

2

s

i

g

n

(

s

)

z_1 = \left| s \right| ^{\frac{1}{2}} sign(s)

z1​=∣s∣21​sign(s),使得后续的导数与不等式计算大大简化,很多项可以巧妙消去。此外,尽管在(29)中不等式右边有正数项

η

\eta

η的存在,系统依然可以在一定限度内保持稳定,原因在于我们证明了

V

˙

n

V

1

2

0

\dot V \leq -n V^{\frac{1}{2}} \leq 0

V˙≤−nV21​≤0而非传统的

V

˙

0

\dot V \leq 0

V˙≤0,这更大程度上能够保证系统稳定性。

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