Control-Lyapunov function: Difference between revisions

Revision as of 16:02, 22 October 2010

Revision as of 19:13, 8 February 2011

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And hence the error and error rate, remembering that <math>V=\frac{1}{2}(\dot{e}+\alpha e)^2</math>, 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 for <math>V</math> and solve for <math>e</math>. This is left as an exercise for the reader but the first few steps at the solution are shown.:

:<math>

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\dot{e}+\alpha e= (\dot{e}(0)+\alpha e(0))e^{-\frac{\kappa}{2} t}

</math>

Whichwhich can then be solved using any linear differential equation methods.

==Notes==