Chapter H Problem Sheet 8 (Lectures 16–17)

  • 1. Consider the state-output system \(\dot {x}=Ax\), \(y=Cx\) with

    \[ A=\bbm {1&0\\0&-1}\qquad C=\bbm {1&0}. \]

    • (a) Use the Hautus observability matrix to show that this system is detectable and not observable.

    • (b) Show using only the definitions of detectability and observability that this system is detectable and not observable.

    • Solution. (a) The Hautus observability matrix of this system is

      \[ \bbm {sI-A\\C}=\bbm {s-1&0\\0&s+1\\1&0}. \]

      The matrix consisting of the last two rows has determinant \(-(s+1)\) so that for \(s\neq -1\) the Hautus observability matrix has two linearly independent rows and is therefore injective. Since in particular we have injectivity for all \(s\in \mC \) with \(\re (s)\geq 0\), it follows that we have detectability. If \(s=-1\), then the Hautus observability matrix has a zero column and is therefore not injective. Therefore the system is not observable.

      (b) We have

      \[ A-LC=\bbm {1&0\\0&-1}-\bbm {L_1\\L_2}\bbm {1&0} =\bbm {1-L_1&0\\-L_2&-1}, \]

      the characteristic polynomial of which equals \(s^2+L_1s+L_1-1\). This is stable if and only if \(L_1>0\) and \(L_1-1>0\). We can therefore choose \(L_1=2\) and \(L_2=0\), i.e. \(L=\sbm {2\\0}\). Hence we have that the system is detectable.

      The differential equations are

      \[ \dot {x}_1=x_1,\qquad \dot {x}_2=-x_2,\qquad y=x_1. \]

      Let \(x^0=\sbm {0\\1}\). Then \(x_1=0\) and therefore \(y=0\). Therefore \(x^0\) is a nonzero unobservable initial state. Hence the system is not observable. □

  • 2. Consider

    \begin{align*} \ddot {q}_1+2\dot {q}_1+q_1&=0,\\ \ddot {q}_2+q_2&=u. \end{align*} The equations can be written in first order form \(\dot {x}=Ax+Bu\) with

    \[ x=\bbm {q_1\\\dot {q}_1\\q_2\\\dot {q}_2},\qquad A=\bbm {0&1&0&0\\-1&-2&0&0\\ 0&0&0&1\\0&0&-1&0},\qquad B=\bbm {0\\0\\0\\1}. \]

    • (a) Use the Hautus controllability matrix to show that the system is stabilizable and not controllable.

    • (b) Show using only the definitions of stabilizability and controllability that this system is stabilizable and not controllable.

    • Solution. (a) The Hautus controllability matrix is:

      \[ \bbm {sI-A&B}=\bbm {s&-1&0&0&0\\1&s+2&0&0&0\\ 0&0&s&-1&0\\0&0&1&s&1}. \]

      Omitting the third column gives

      \[ \bbm {s&-1&0&0\\1&s+2&0&0\\ 0&0&-1&0\\0&0&s&1}, \]

      which by the block structure has determinant

      \[ -(s^2+2s+1), \]

      the roots of which are stable (in fact, both roots equal \(s=-1\)). Therefore, for all \(s\) with \(\re (s)\geq 0\) we have that this determinant is nonzero so that the matrix is invertible so that the Hautus controllability matrix is surjective. Therefore the system is stabilizable.

      To show that the system is not controllable, we consider the Hautus controllability matrix with \(s=-1\). This is

      \[ \bbm {sI-A&B}=\bbm {-1&-1&0&0&0\\1&1&0&0&0\\ 0&0&-1&-1&0\\0&0&1&-1&1}. \]

      We see that the first two rows are linearly dependent. Therefore the matrix is not surjective and the system is not controllable.

      (b) The idea is that the control \(u=-2\dot {q}_2\) leads to the second equation becoming

      \[ \ddot {q}_2+2\dot {q}_2+q_2=0, \]

      and since this is asymptotically stable, and similarly the first equation above (for \(q_1\)) is asymptotically stable, we obtain asymptotic stability. We formally check this. We have

      \[ F=\bbm {0&0&0&-2}, \]

      so that

      \[ A+BF=\bbm {0&1&0&0\\-1&-2&0&0\\ 0&0&0&1\\0&0&-1&-2}, \]

      which by the block diagonal structure has characteristic polynomial

      \[ (s^2+2s+1)(s^2+2s+1), \]

      and using that both terms in the product are stable, this is stable. Hence the pair \((A,B)\) is stabilizable.

      No matter what \(u\) is, we have that \(q_1\) is fully determined by \(\ddot {q}_1+2\dot {q}_1+q_1=0\) and the initial conditions \(q_1(0)\) and \(\dot {q}_1(0)\). Choose \(x^0\) such that \(x^0_1=x^0_2=0\). Then \(q_1=\dot {q}_1=0\). Choose \(x^1\) such that \(x^1_1=1\). The condition \(x(T)=x^1\) then implies \(q_1(T)=1\) which is never true since \(q_1=0\). Therefore the system is not controllable. □

  • 3. Consider the input-state-output system \(\dot {x}=Ax+Bu\), \(y=Cx\) where

    \[ A=\bbm {-1&0\\1&3},\qquad B=\bbm {0\\1},\qquad C=\bbm {2&1}. \]

    Show that an observer-based stabilizing controller exists without explicitly calculating such a controller.

    • Solution. By Proposition 17.5, it suffices to show that \((A,B)\) is stabilizable and \((A,C)\) is detectable.

      The Hautus controllability matrix is

      \[ \bbm {sI-A&B}=\bbm {s+1&0&0\\-1&s-3&1}. \]

      The matrix obtained by deleting the second column has determinant \(s+1\) and is therefore invertible for \(s\neq -1\). It follows that the Hautus controllability matrix is surjective for \(s\neq -1\) and therefore in particular for all \(s\in \mC \) with \(\re (s)\geq 0\). Hence the system is stabilizable.

      The Hautus observability matrix is

      \[ \bbm {sI-A\\C}=\bbm {s+1&0\\-1&s-3\\2&1}. \]

      The matrix obtained by deleting the second row has determinant \(s+1\) and is therefore invertible for \(s\neq -1\). It follows that the Hautus observability matrix is injective for \(s\neq -1\) and therefore in particular for all \(s\in \mC \) with \(\re (s)\geq 0\). Hence the system is detectable. □