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Here is the state space representation of the system. $$\dot x=\begin{bmatrix}\dot x_1\\\dot x_2\end{bmatrix}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix}u=Ax+Bu$$ From the eigen values we can see that there is a pole at the origin(s=0). The final question to the problem is if this pole is controllable using feedback.

Using feedback, in case I haven't translated correctly is when u has this form $$u=\begin{bmatrix}k_1&k_2\end{bmatrix}x+\begin{bmatrix}0\\1\end{bmatrix}u'$$ This will alter A in the following way $$A'=\begin{bmatrix}0&0\\k_1+1&k_2\end{bmatrix}$$ And for the eigen values $$sI-A'=\begin{bmatrix}s&0\\-k_1-1&s-k_2\end{bmatrix}\\det=s(s-k_2)$$

We can see that the s pole will always be there but how do I explain this? Do I have to say that the pole is uncontrollable or unobservable or simply that the top row of the A matrix won't change? The question gives many points and bearing in mind that I've already found A' in a previous question makes me think there is more to it.

I've read some stuff about the observability and the controllability of a system but not that much to realize if and which poles are unchangeable. I think I can say that x1 is unobservable and uncontrollable because of the top 0 rows.

Update: Finding controllability matrix we see that the system is uncontrollable. Since we can see that k2 moves one pole is it adequate then to say that origin pole is unmovable since it's the only pole left to be uncontrollable?

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  • \$\begingroup\$ Can you define your x vector; is it \$\begin{bmatrix}\dot x_1\\x_1\end{bmatrix}\$ or \$\begin{bmatrix} x_1\\\dot x_1\end{bmatrix}\$? \$\endgroup\$ – Chu Aug 31 '17 at 15:23
  • \$\begingroup\$ Did it @Chu ,check now \$\endgroup\$ – John Katsantas Aug 31 '17 at 15:26
  • \$\begingroup\$ Something wrong. Your equation doesn't define x1. \$\endgroup\$ – Chu Aug 31 '17 at 15:36
  • \$\begingroup\$ This system comes from a linearization process on previous questions.I even checked the solution. Even if it's wrong will it affect what we are trying to do at this point? I could upload the whole problem but it will be a long question. By the way, these are all Δx instead of x if that makes a difference. \$\endgroup\$ – John Katsantas Aug 31 '17 at 15:39
  • \$\begingroup\$ At the moment you have \$ \dot x_1=0\$ and \$ \dot x_2 = x_1+u\$. That doesn't define a system. \$\endgroup\$ – Chu Aug 31 '17 at 15:41
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I would have liked to put this answer as a comment, but I don't have enough reputation to do so.


First a small remark. There is a problem with the dimensions of the matrices in the question: If you plug the expression for \$u\$ in the state equation, then you multiply a \$2\times1\$ matrix with a \$2\times1\$ matrix: $$ \begin{aligned} \dot{x} & = A x + B u \\ & = A x + \begin{bmatrix} 0 \\ 1 \end{bmatrix} (\begin{bmatrix} k_1 & k_2 \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \end{bmatrix} u') \end{aligned} $$ and the problem occurs in the product \$\begin{bmatrix} 0 \\ 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix}\$.

Typically, a state feedback is formulated as \$u = - K x + u'\$ where \$u'\$ is a reference signal. Here, you get \$u = - \begin{bmatrix} k_1 & k_2 \end{bmatrix} x + u'\$, but you can of course take the minus sign into account in the coefficients \$k_1\$ and \$k_2\$. Considering \$u = \begin{bmatrix} k_1 & k_2 \end{bmatrix} x + u'\$, the matrix \$A'\$ is correct.


Your analysis is correct. The open-loop system has two eigenvalues in the origin. With the state-feedback controller, you can only control one pole of this open-loop system, since the system is not controllable. You can easily see this in the \$A\$ and \$B\$ matrix: neither of the states, nor the input affect the evolution of the first state (the top rows of the matrices \$A\$ and \$B\$ are zero). This means that \$\dot{x}_1=0\$ no matter what and so \$x_1\$ stays at its initial value.

If the top element of the matrix \$B\$ was non-zero, then the first state equation would be \$\dot{x}_1=b_1 u\$, which corresponds to an integrator (as mentioned in Chus comment). Nevertheless, since the top element of \$B\$ is zero here, the integrator will start at the initial value of \$x_1\$ and keep integrating zeros.

You can also see that one pole is not moveable from the rank of the controllability matrix. Here, \$rank(\begin{bmatrix}B & AB\end{bmatrix}) = rank(\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}) = 1\$. This means that you can only move one pole. It also means that the states \$x=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\$ that you can reach from a zero initial state live in a one-dimensional space. In fact, those states that you can reach are linear combinations of the columns of the controllability matrix, here \$\alpha \begin{bmatrix} 0 \\ 1 \end{bmatrix}\$, where \$\alpha\$ can be any number you like. And again, you can see that you cannot control the first state.

In conclusion, one pole in the origin is controllable (with \$k_2\$) and the other pole in the origin is not controllable.

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