How come you can always find u(t) such that the integrals come out to be what is needed to drive the state where you want it to go if the reachability condition holds?
Submitted by: waydo Submitted on: October 27, 2004 Identifier: L5.2 The answer to this question is somewhat involved and beyond the scope of this course. For one take on it, check out Dullerud and Paganini, “A Course in Robust Control Theory: A Convex Approach” (Chapter 2). If we consider the discrete-time case, however, the intuition is much easier. Let x(k+1) = A x(k) + B u(k) We can drive x to zero (or to any arbitrary other value) in n steps or less (where as usual n is the dimension of the system) if the reachability condition rank( [B AB A^2B … A^(n-1)B] ) = n holds, just like in the continuous case. The difference is that it in general takes a finite number of steps to do this, rather than an arbitrarily short period of time as in the continuous case. To see this, look at x as k increases: x(k+2) = A x(k+1) + Bu(k+1) = A^2 x(k) + ABu(k) + Bu(k+1) … … x(k+n) = A^n x(k) + A^(n-1)Bu(k) + … + ABu(k+n-2) + Bu(k+n-1) Now all the u’s can be chosen independently, so if the colum
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