Suppose that the system $$x'(t)=Ax(t)+Bu(t)$$ is controllable in $\mathbb{R}^n$, where $A$ is $n \times n$, $B$ is $ n \times m$ and $u(t)$ is $m \times 1$ Show that the system $$\left \{ \begin{array}{rclccc} x'(t) &=& Ax(t) &+& By(t)& \\ y'(t)&=&&&&u(t) \end{array} \right.$$ with the additional states $y\in\mathbb{R}^m$ and input $u\in\mathbb{R}^m$ is controllable in $\mathbb{R}^{n+m}$.

My attempt

If the first system is controllable, then the matrix $$M = \begin{pmatrix} B & AB & A^2B &\ldots & A^{n-1}B \end{pmatrix}$$ has rank M. Let the second system $$\begin{pmatrix}x' \\ y' \end{pmatrix} = \begin{pmatrix} A & B \\ 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}+\begin{pmatrix}0 \\ I \end{pmatrix}u(t)$$ Then, its Kalman matrix (aka controllability matrix) is $$\bar{M} = \begin{pmatrix} B & AB & A^2B & \ldots & A^{n+m-1}B \\ 0 & 0 & 0 & \ldots & 0\end{pmatrix}$$ The matrix $A$ is $n \times n$ and, by Cayley-Hamilton theorem, $$A^{n+l} = \sum_{k=0}^{n-1} \alpha_{k,l} A^{k}, l=0,1,...$$ I mean, $A^{n+l}$ is a linear combination of the first $n$ powers of $A$. This implies that we have $n$ linearly independent columns in $\bar{M}$ at the the terms until $A^{n-1}B$ (because the first system is controllable). But I don't see how to find the other $m$ columns. What am I doing wrong? Probably my controlability matrix is wrong, since it only can have $n$ l.i rows.

We could solve it for $y$ first and return to a system only in $x'$, substituting $u$ by some $\int u$. The Kalman matrix of first and second system would be the same and, by hypothesis, it would have $\text{rank }n$, correct? But why isn't my first approach correct?


The solution is fairly straightforward.

Short answer

The system is controllable without any modifications. You made a small mistake calculating the new controllability matrix: you are missing the first column of $\bar{M}$, which should be $\begin{bmatrix}0 & I\end{bmatrix}^T$.

Full proof

Let's consider the second system, and define some augmented matrix placeholders:

$\begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix}A & B\\ 0 & 0\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} + \begin{bmatrix}0 \\ I\end{bmatrix} u = \tilde{A}\tilde{x} + \tilde{B}u$

If we want to find out if this system is controllable, we have to prove that its controllability matrix $\tilde{M}$ has full rank (i.e. all its rows or columns are linearly independent). The expression for the controllability matrix is:

$\tilde{M} = \begin{bmatrix}\tilde{B} & \tilde{A}\tilde{B} & \tilde{A}^2\tilde{B}& \dots & \tilde{A}^{n+m-1}\tilde{B}\end{bmatrix}$

If we start to solve for each term in terms of $A$ and $B$, a pattern begins to emerge:

$\tilde{A}\tilde{B} = \begin{bmatrix} B \\ 0 \end{bmatrix} \qquad \tilde{A}^2\tilde{B} = \begin{bmatrix} AB \\ 0 \end{bmatrix} \qquad \tilde{A}^3\tilde{B} = \begin{bmatrix} A^2 B \\ 0 \end{bmatrix} \dots \qquad \tilde{A}^p\tilde{B} = \begin{bmatrix} A^{p-1} B \\ 0 \end{bmatrix}$

Therefore we can rewrite $\tilde{M}$ as:

$\tilde{M} = \begin{bmatrix}0 & B & AB & A^2B & \dots & A^{n-1}B & A^n B & \dots & A^{n+m-2}B\\ I & 0 & 0 & 0 & \dots & 0 & 0 & \dots & 0\end{bmatrix}$

Notice that the middle of the first row is actually $M$, which we can then substitute into $\tilde{M}$:

$\tilde{M} = \begin{bmatrix}0 & M & A^n B & \dots & A^{n+m-2}B\\ I & 0 & 0 & \dots & 0\end{bmatrix}$

Now that we have this matrix in a workable form, let's look at the elements to see if we can make a statement about its rank. Since $M$ occupies the first $n$ rows of $\tilde{M}$, and we know that $M$ has full rank, we can conclude that those rows are linearly independent with each other.

Now we have to ask, are the final $m$ rows linearly independent with respect to each other and the first $n$ rows? Focusing on the first column only, we see the null matrix $0$, followed by the identity matrix $I$. The identity matrix has full rank, and all its rows are also linearly independent of the null matrix. Therefore, without looking at the other columns we can conclude that the last $m$ rows are linearly independent with respect to each other and the rest of the rows in $\tilde{M}$.

In conclusion, all rows of $\tilde{M}$ are linearly independent, therefore it is full rank, therefore the system is controllable.


Couldn't you just define $y(t)$ and be done?

$$ y'(t)=u(t) \\ y(t) = \int{u(t)}dt \\ $$

As long as the system is controllable, as defined by the controllability matrix $M$ being full rank, it doesn't matter what the input signal is, does it?

The only issue I can think of is maybe if $u(t)$ is not able to be integrated, but that seems beyond the scope of the question. $M$ is defined only by $A$ and $B$, so as long as you can define $y(t)$ I don't think it matters what it is. At least, that would be my approach/defense.

  • $\begingroup$ Yes, as I noted I could do this. But someone correctly said that my matrix $\bar{M}$ is wrong. As you can see, $\bar{B}$ is missing $\endgroup$ – Giiovanna Sep 6 '15 at 13:32
  • $\begingroup$ Well, first I would say that your $\bar{M}$ is defined with $A$ and $B$ not $\bar{A}$ and $\bar{B}$. That is, $\bar{A}$ should be the block A matrix for your new system; you are still using the old definition. For instance - why does $I$ not show up in your $\bar{M}$? $\endgroup$ – Chuck Sep 6 '15 at 14:02
  • $\begingroup$ But, that said, I think this is a trick question in that the new system as you have defined it is not controllable - $u(t)$ cannot drive the $x$ state to any value because there is no mapping from u to x. Understanding this, the trick here is that your input is not $u(t)$! The input is $\int{u(t)}$, which can drive x. $\endgroup$ – Chuck Sep 6 '15 at 14:07
  • $\begingroup$ There is no map from u to x but there is from u to y and y to x. $\endgroup$ – Giiovanna Sep 6 '15 at 19:04
  • $\begingroup$ @Giiovanna - No there is not. There is a map from u to the derivative of y and there is a map from y to x. Sorry I didn't respond earlier, I didn't get a notification. Again, as you draw the system, you draw it with u(t) being the input, which it is not. Then you ask, why is the system not controllable with u(t) as the input? Because u(t) is not driving anything. $\endgroup$ – Chuck Sep 7 '15 at 14:18

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