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Controllability Matrix [Control Theory] 
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#1
Nov1410, 09:26 AM

P: 22

I looked at several books as well as internet sources, none of them explain how the controllability matrix is formed.
Given the linear time invariant system x'(t)=Ax(t)+Bu(t) A is an nxn matrix, B is an nx1 matrix, (assuming single input) Then controllability matrix R is given by: R= [B AB A^2B A^3B .....A^(n1)B] System is controllable if Det(R)=/=0 or rank(R)=n Can someone explain me the logic behind how this matrix was formed ? 


#2
Nov1510, 01:27 PM

P: 179

I would begin by taking a very simple system and trying to see how
the C Matrix guarantees that the closed loop poles will be in the Left Half Plane. An interesting question is whether the Matrix can insure not only Global Stability, but also Relative Stability ..... the poles will be near the Real Axis for a stable, non oscillatory, transient response. 


#3
Nov1510, 03:05 PM

P: 196




#4
Nov1510, 04:36 PM

P: 341

Controllability Matrix [Control Theory]
A slightly clearer but somewhat less rigorous connection (only C'bility [itex]\Longrightarrow[/itex] rank condition!) can be made as follows: We can solve the diff. eq. system that you have provided and obtain
[tex] x(t) = \int_0^{\infty}e^{A(t\tau)}Bu(\tau)d\tau + e^{At}x(0) [/tex] Let's assume zero initial conditions for simplicity. Now, since the controllability means that I can reach any x(t), the integral converges to x(t) with some u(t). Let's use the Taylor series of exponential [tex] x(t) = \int_0^{\infty}\left(I+A(t\tau) + \frac{A^2}{2!}(t\tau)^2+\cdots \right)Bu(\tau)d\tau [/tex] You can take the constant terms out and obtain a matrixvector multiplication (though infinite dimensional) [tex] x(t) = \begin{bmatrix}B &AB &A^2B &\cdots\end{bmatrix}\begin{pmatrix}\int_0^{\infty}u(\tau)d\tau \\\int_0^{\infty}(t\tau)u(\tau)d\tau \\ \int_0^{\infty}\frac{1}{2!}(t\tau)^2u(\tau)d\tau\\ \vdots\end{pmatrix} = \mathcal{C}_\infty \mathcal{U} [/tex] I would denote the matrix part as [itex]\mathcal{C}_\infty[/itex] . Now, since we assume controllability, we should be able to obtain any x(t), hence [itex]\mathcal{C}_\infty[/itex] must be full row rank. But from CayleyHamilton theorem we know that the powers of A with degree higher then n1, can be rewritten by the powers of A up to the degree n1. (This is a bad sentence but looking it up is easy so I skip that part.) This means that no extra information about the rank of this matrix can be included after the [itex]A^{n1}B[/itex] since the remaining terms are linear combinations of the first n terms. Thus, [tex]rank(\mathcal{C}_\infty) = rank(\mathcal{C}) = rank(\begin{bmatrix}B &AB &A^2B &\cdots &A^{n1}B\end{bmatrix}[/tex] In case of SISO systems, [itex]\mathcal{C}[/itex] happens to be square so the rank condition equals to the determinant being nonzero. 


#5
Nov1510, 04:56 PM

P: 22

Thanks for the help everybody , after studying the example 1.2.1 @ http://teal.gmu.edu/ececourses/ece52...00000000000000
and trambolin's solution it's crystal clear now. (How can i add "Solved" to the thread title ?) 


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