In the reference ''On stability analysis of active disturbance rejection control for nonlinear time-varying plants with unknow dynamics '',
why when A is Hurwitz, there exists a finite time T1 > 0 such that
Eq.(19) exists?
Even when A is hurwitz, there exists positive elements in the A matrix. How to explain this?
Mohamed Harmouche
$e^{w_0 A t}$ is bounded by and exponential decreasing function in the case of A hurwitz. so after some time $T1$will be less than any constant , in particular 1/(w_0^{n+1}).
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Ibraheem Kasim Ibraheem
when the matrix A is Hurwitz, all of its eigenvalues are in the left-half plane, it is nothing to do with the signs of the matrix elements, got it?
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