Given this transfer function

I have two questions:

1) What is the state space model in controllable canonical form?

2) How can you that the system is always controllable; i.e. show that the controllability matrix: $$ \mathscr C = [B\;|\;AB\;|\;A^2B\;...\;A^{n-1}B] $$ always has full rank?

  • $\begingroup$ I've seen many textbooks saying that the matrix has full rank, but I have yet to see one that actually gives any sort of proof. They usually just say, "It is easy to show that the matrix is full rank/the system is controllable." $\endgroup$ – Matt Nov 26 '17 at 23:08

Here is the sketch for the sufficiency part of (2): I assume that you can derive the general solution for a Linear Time Invariant system given by

$$ x(t) = \int_0^{\infty}e^{A(t-\tau)}Bu(\tau)d\tau + e^{At}x(0) $$

We forget about the initial condition since it doesn't affect the controllability property. Now, expand the exponential inside the integral with Taylor expansion

$$ x(t) = \int_0^{\infty}\left(I+A(t-\tau) + \frac{A^2}{2!}(t-\tau)^2+\cdots \right)Bu(\tau)d\tau $$

Then form a infinite dimensional dot product

$$ x(t) = \begin{bmatrix}B &AB &A^2B &\cdots\end{bmatrix}\begin{pmatrix}\int_0^{\infty}u(\tau)d\tau \\\int_0^{\infty}(t-\tau)u(\tau)d\tau \\ \int_0^{\infty}\frac{1}{2!}(t-\tau)^2u(\tau)d\tau\\ \vdots\end{pmatrix} $$

And from the Cayley-Hamilton theorem we can truncate the elements of the product up to the term $A^{n}B$. That's your matrix rank implying controllability part.

Notice that this is not a rigorous proof as we have not touched upon the convergence of the integral or the validity of the truncation. That's why you don't see the proof in the textbooks because it involves some nontrivial and rather irrelevant details.


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