7. Controller Design (Lyapunov and Back-stepping)

7.1 Control Design via Lyapunov Theory

Example

\[ \left\{\begin{aligned} \dot x_1 &= -3 x_1 + 2x_1 x_2^2 + u \\ \dot x_2 &= -x_2^3 - x_2 \end{aligned}\right. \]

Design \(u = k(x_1, x_2)\) such that \(\bar x\) is G.A.S.

\[ V(x) = \frac12 (x_1^2 + x_2^2) \]

\(V(\bar x) \succ 0, \forall \bar x\), and \(V \in \mathbb R^n \to \mathbb R\)

\[ \begin{aligned} \dot V(x) &= x_1\dot x_1 + x_2 \dot x_2 \\ &= -3x_1^2 +2x_1^2x_2^2 + x_1u - x_2^4 - x_2^2 \\ \end{aligned} \]

To make \(\dot V\) ND, one possible choice is \(u = -2x_1x_2^2\), hence,

\[ \dot V(x) = -3x_1^2 -x_2^2 - x_2^4 \]

is globally ND in \(\bar x = 0\). \(\Rightarrow\) \(\bar x = 0\) is a G.A.S. equilibrium for the closed loop system.

7.2 Back-stepping Method

Given the system with the form:

\[ \begin{aligned} \dot x_1(t) &= f(x_1(t)) + g(x_1(t)) x_2(t), x_1 \in \mathbb R^n \\ \dot x_2(t) &= u(t) \end{aligned} \]

\(f\), \(g\) \(\in C^1\) on the neighbor \(D\) of \(\bar x_1 = 0\)
\(f(0) = 0\)

Goal: design \(u = k(x_1, x_2)\) such that \(\bar x = 0\) is an A.S. equilibrium for the closed loop system

7.2.1 Controller Design

Suppose that you know a "fictitious" control law:

\[ x_2 = \Phi(x_1), \Phi(x) \in C^\infty, \Phi(0) = 0 \]

\(\bar x_1 = 0\) is an A.S. equilibrium of the reduced order system:

\[ \dot x_1 = f(x_1) + g(x_1)\Phi(x_1) \]

give \(V_1(x_1) \in C^1, V_1(\bar x) \succ 0\) for \(\bar x = 0\) such that,

\[ \dot V_1(x_1) = \frac{\partial V_1}{\partial x_1}\left[ f(x_1)+g(x_1)\Phi(x_1) \right] \]

\(\dot V_1(x_1) \prec 0\) in \(\bar x = 0\) on \(D\), then the control law is:

\[ u = - \frac{\partial V_1}{\partial x_1}(x_1)g(x_1) - \mu(x_2 - \Phi(x_1)) + \frac{\partial \Phi}{\partial x_1}(x_1)(f(x_1) + g(x_1)x_2), \mu > 0 \]

is A.S. for \(\bar x = 0\).

7.2.2 Proof

\[ V(x_1, x_2) = V_1(x_1) + \frac12 (x_2 - \Phi(x_1))^2 \]

7.2.3 Extension

\[ \begin{aligned} \dot x_1(t) &= f(x_1(t)) + g(x_1(t)) x_2(t), x_1 \in \mathbb R^n \\ \dot x_2(t) &= a(x_1,x_2) + b(x_1,x_2)u(t) \end{aligned} \]

If \(b(x_1,x_2) \neq 0\) is on the domain of interest,

\[ \begin{aligned} u(t) &= -\frac{1}{b(x_1,x_2)}(a(x_1,x_2)-v) \\ \dot x_2 &= v \end{aligned} \]

\(v = k(x_1,x_2)\), by back-stepping

7.3 Feedback Linearization

\[ \begin{aligned} \dot x_1 &= A_1x_1 + A_2x_2 \\ \dot x_2 &= a(x_1,x_2) + b(x_1,x_2)u(t) \end{aligned} \]

\(u = -\frac{1}{b(x_1,x_2)}(a(x_1,x_2)-v)\), \(b(x_1,x_2) \neq 0, \forall (x_1, x_2)\), \(\dot x_2 = v\), \(v = Kx\)