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The state model of linear control system is $\Sigma:\dot{x}=Ax+Bu$, where $x$ is the state and $u$ is the input. A state feedback $F:u=Fx+v$ can be treated as a transformation which maps $\Sigma$ to $\Sigma_F:\dot{x}=(A+BF)x+Bv$. Prove the following conclusions:

  • All feedback transformation form a group which is called feedback group;

  • The controllability is invariant under the transformation.

The first question is obviously because the transformations form the linear transformation in $\mathbb R^n$. However what is the controllability in Math? And how can we prove that it is invariant under the transformation.

Any advice is helpful. Thank you.

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gaoxinge
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1 Answers1

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I'm not exactly sure what the group operation is. Since the feedback is a linear function of the state, the collection of $F$ form a subspace, perhaps the group operation is addition?

The second part is more straightforward.

By definition, $(A,B)$ is cc. iff for all $x_0,x_1$ there exists some $u$ such that with input $u$ the solution to the differential equation $\dot{x}=Ax+Bu$ satisfies $x(0) = x_0, x(1) = x_1$ (in other words, $u$ 'steers' the system from $x_0$ at time zero to $x_1$ at time one).

Suppose $(A,B)$ is cc. and consider the system $(A+BF,B)$. Choose $x_0,x_1$ and let $u$ be control that steers $x_0$ to $x_1$ in the system $(A,B)$, and let $x_u$ be the solution of the system $(A,B)$ with input $u$, starting at $x_0$. Then the control $u^* = u-Fx_u$ will steer the system $(A+BF,B)$ from $x_0$ at time zero to $x_1$ at time one. To see this, note that $x_u$ is a solution of the system $\dot{x}=(A+BF)x+Bu$ with input $u^*$ starting at $x_0$. Hence $(A+BF,B)$ is cc.

(Note: Since $A = (A+BF)-BF$, we see that the above is an equivalence, that is, $(A,B)$ is cc. iff $(A+BF,B)$ is cc. for any $F$.)

copper.hat
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