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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Control-Lyapunov function ： ウィキペディア英語版 Control-Lyapunov function In control theory, a control-Lyapunov function〔Freeman (46)〕 is a Lyapunov function $V(x)$ for a system with control inputs. The ordinary Lyapunov function is used to test whether a dynamical system is ''stable'' (more restrictively, ''asymptotically stable''). That is, whether the system starting in a state $x\; \backslash ne\; 0$ in some domain ''D'' will remain in ''D'', or for ''asymptotic stability'' will eventually return to $x\; =\; 0$. The control-Lyapunov function is used to test whether a system is ''feedback stabilizable'', that is whether for any state ''x'' there exists a control $u(x,t)$ such that the system can be brought to the zero state by applying the control ''u''. More formally, suppose we are given an autonomous dynamical system :$$ \dot=f(x,u) where $x\backslash in\backslash mathbf^n$ is the state vector and $u\backslash in\backslash mathbf^m$ is the control vector, and we want to feedback stabilize it to $x=0$ in some domain $D\backslash subset\backslash mathbf^n$. Definition. A control-Lyapunov function is a function $V:D\backslash rightarrow\backslash mathbf$ that is continuously differentiable, positive-definite (that is $V(x)$ is positive except at $x=0$ where it is zero), and such that :$$ \forall x \ne 0, \exists u \qquad \dot(x,u)=\nabla V(x) \cdot f(x,u) < 0. The last condition is the key condition; in words it says that for each state ''x'' we can find a control ''u'' that will reduce the "energy" ''V''. Intuitively, if in each state we can always find a way to reduce the energy, we should eventually be able to bring the energy to zero, that is to bring the system to a stop. This is made rigorous by the following result: Artstein's theorem. The dynamical system has a differentiable control-Lyapunov function if and only if there exists a regular stabilizing feedback ''u''(''x''). It may not be easy to find a control-Lyapunov function for a given system, but if we can find one thanks to some ingenuity and luck, then the feedback stabilization problem simplifies considerably, in fact it reduces to solving a static non-linear programming problem :$$ u^ *(x) = \arg\min_u \nabla V(x) \cdot f(x,u) for each state ''x''. The theory and application of control-Lyapunov functions were developed by Z. Artstein and E. D. Sontag in the 1980s and 1990s. ==Example== Here is a characteristic example of applying a Lyapunov candidate function to a control problem. Consider the non-linear system, which is a mass-spring-damper system with spring hardening and position dependent mass described by :$$ m(1+q^2)\ddot+b\dot+K_0q+K_1q^3=u Now given the desired state, $q\_d$, and actual state, $q$, with error, $e\; =\; q\_d\; -\; q$, define a function $r$ as :$$ r=\dot+\alpha e A Control-Lyapunov candidate is then :$$ V=\fracr^2 which is positive definite for all $q\; \backslash ne\; 0$, $\backslash dot\; \backslash ne\; 0$. Now taking the time derivative of $V$ :$$ \dot=r\dot :$$ \dot=(\dot+\alpha e)(\ddot+\alpha \dot) The goal is to get the time derivative to be :$$ \dot=-\kappa V which is globally exponentially stable if $V$ is globally positive definite (which it is). Hence we want the rightmost bracket of $\backslash dot$, :$$ (\ddot+\alpha \dot)=(\ddot_d-\ddot+\alpha \dot) to fulfill the requirement :$$ (\ddot_d-\ddot+\alpha \dot) = -\frac(\dot+\alpha e) which upon substitution of the dynamics, $\backslash ddot$, gives :$$ (\ddot_d-\frac+\alpha \dot) = -\frac(\dot+\alpha e) Solving for $u$ yields the control law :$$ u= m(1+q^2)(\ddot_d + \alpha \dot+\fracr )+K_0q+K_1q^3+b\dot with $\backslash kappa$ and $\backslash alpha$, both greater than zero, as tunable parameters This control law will guarantee global exponential stability since upon substitution into the time derivative yields, as expected :$$ \dot=-\kappa V which is a linear first order differential equation which has solution :$$ V=V(0)e^ And hence the error and error rate, remembering that $V=\backslash frac(\backslash dot+\backslash alpha\; e)^2$, exponentially decay to zero. If you wish to tune a particular response from this, it is necessary to substitute back into the solution we derived for $V$ and solve for $e$. This is left as an exercise for the reader but the first few steps at the solution are: :$$ r\dot=-\fracr^2 :$$ \dot=-\fracr :$$ r=r(0)e^ t} :$$ \dot+\alpha e= (\dot(0)+\alpha e(0))e^ t} which can then be solved using any linear differential equation methods. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Control-Lyapunov function」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.