2009
DOI: 10.1515/9781400833351
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Stability and Stabilization

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Cited by 51 publications
(9 citation statements)
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“…(3.10)- (3.11). To show that the origin is indeed stable in the expanding phase, one need appeal to the Lyapunov theorem [52] which states:…”
Section: Appendix A: Stability Analysis Of the Equilibriummentioning
confidence: 99%
“…(3.10)- (3.11). To show that the origin is indeed stable in the expanding phase, one need appeal to the Lyapunov theorem [52] which states:…”
Section: Appendix A: Stability Analysis Of the Equilibriummentioning
confidence: 99%
“…which has a solution in _ Xð0Þ ¼ 0, where X ¼ ½X 1 X 2 � T . To demonstrate Lyapunov local Asymptotic Stability by the direct method, a candidate function V must have the following properties [34,35]:…”
Section: Outer Loopmentioning
confidence: 99%
“…{V}_{S}\left(\omega Lcos{X}_{2}-Rsin{X}_{2}\right)-\omega L{V}_{S}\right)$ which has a solution in trueẊ(0)=0 $\dot{X}(0)=0$, where X=X1X2T $X={\left[{X}_{1}{X}_{2}\right]}^{T}$. To demonstrate Lyapunov local Asymptotic Stability by the direct method, a candidate function V must have the following properties [34, 35]: V(0)=0 $V(0)=0$ V(X)>0,X0,XDX $V(X) > 0,X\ne 0,X\in {D}_{X}$ V(X),false‖Xfalse‖ $V(X)\to \infty ,\Vert X\Vert \to \infty $ trueV̇(0)=0 $\dot{V}(0)=0$ trueV̇(X)<0,X0,XDX $\dot{V}(X)< 0,X\ne 0,X\in {D}_{X}$ …”
Section: Resiliency and Stability In A Distributed Control Environmentmentioning
confidence: 99%
“…An important property of the Kalman model to be valid is that the pair (A,C) be detectable [2]. We use the Popov-Belevitch-Hautus (PBH) Lemma [20] to easily discard unacceptable A b values. We then use the Kalman model to obtain an expression of the Kalman filter.…”
Section: Kalman Modelmentioning
confidence: 99%