Chapter I Problem Sheet 9 (Lectures 18–19)

• 1. Consider the first order scalar differential equation

  \[ \dot {x}=2x+u, \]

  and the performance output

  \[ y=\bbm {x\\u}, \]

  i.e.

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

  • (a) Determine whether or not \(D\) is injective;

  • (b) Determine whether or not \((A,B)\) is stabilizable ;

  • (c) Determine whether or not the Rosenbrock injectivity condition holds;

  • (d) Determine all symmetric positive semidefinite solutions of the algebraic Riccati equation (18.2), the corresponding state feedback matrices \(F\) and the corresponding closed-loop matrices \(A+BF\).

  • Solution. (a) The second row of \(D\) is an invertible 1-by-1 matrix and therefore \(D\) is injective.

    (b) With \(F=-3\) we have \(A+BF=-1<0\).

    (c) The Rosenbrock matrix is

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

    Omitting the first row we obtain an invertible matrix (the identity matrix) so that the Rosenbrock matrix is injective (for all \(s\in \mC \)).

    (d) The algebraic Riccati equation is (since \(n=1\), this is an equation in a scalar unknown) using that \(C^*D=0\) and \(D^*D=1\):

    \[ 4X+1-X^2=0. \]

    The solutions of this are

    \[ X=2\pm \sqrt {5}. \]

    Since we need \(X\geq 0\), we obtain

    \[ X=2+\sqrt {5}. \]

    We then have

    \[ F=-2-\sqrt {5}, \]

    and therefore

    \[ A+BF=-\sqrt {5}. \]

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• 2. Consider the undamped second order scalar differential equation

  \[ \ddot {q}(t)+q(t)=u(t), \]

  with the state \(x=\sbm {q\\\dot {q}}\) and the performance output

  \[ y=\bbm {q\\\varepsilon u}, \]

  where \(\varepsilon >0\), i.e.

  \[ A=\bbm {0&1\\-1&0},\quad B=\bbm {0\\1},\quad C=\bbm {1&0\\0&0},\qquad D=\bbm {0\\\varepsilon }. \]

  • (a) Determine whether or not \(D\) is injective;

  • (b) Determine whether or not \((A,B)\) is stabilizable ;

  • (c) Determine whether or not the Rosenbrock injectivity condition holds;

  • (d) Determine all symmetric positive semidefinite solutions of the algebraic Riccati equation (18.2) and the corresponding state feedback matrices;

  • (e) Write down the closed-loop system in second order form.

  • Solution. (a) Since \(\varepsilon >0\) we have that \(D\) is injective; the second row is an invertible 1-by-1 matrix.

    (b) Since \(A\) and \(B\) are the same as in Example 19.1 we have as there that \((A,B)\) is stabilizable.

    (c) The Rosenbrock matrix is

    \[ \bbm {s&-1&0\\1&s&-1\\1&0&0\\0&0&\varepsilon }. \]

    If we delete the second row, then we obtain the matrix

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

    which is easily seen to have determinant \(\varepsilon \) and is therefore invertible. It follows that the Rosenbrock matrix is injective (in fact for all \(s\in \mC \)).

    (d) The Riccati equation is (using that \(X\) is symmetric and that \(C^*D=0\) and \(D^*D=\varepsilon ^2\))

    \(\seteqnumber{0}{I.}{0}\)

    \begin{multline*} \bbm {0&-1\\1&0}\bbm {X_1&X_0\\X_0&X_2} +\bbm {X_1&X_0\\X_0&X_2}\bbm {0&1\\-1&0} +\bbm {1&0\\0&0}\bbm {1&0\\0&0} \\ -\varepsilon ^{-2}\bbm {X_1&X_0\\X_0&X_2}\bbm {0\\1}\bbm {0&1}\bbm {X_1&X_0\\X_0&X_2}=\bbm {0&0\\0&0}, \end{multline*} which is

    \[ \bbm {-2X_0&X_1-X_2\\X_1-X_2&2X_0} +\bbm {1&0\\0&0} -\varepsilon ^{-2}\bbm {X_0^2&X_0X_2\\X_0X_2&X_2^2}=\bbm {0&0\\0&0}, \]

    which is

    \[ \bbm { -2X_0+1-\varepsilon ^{-2}X_0^2 &X_1-X_2-\varepsilon ^{-2}X_0X_2 \\X_1-X_2-\varepsilon ^{-2}X_0X_2 &2X_0-\varepsilon ^{-2}X_2^2 }=\bbm {0&0\\0&0}. \]

    We note that from the bottom-right corner we have that \(X_0\geq 0\). The top-left corner gives (using that \(X_0\geq 0\) so that we can exclude the negative solution):

    \[ X_0=\frac {-1+\sqrt {1+\varepsilon ^{-2}}}{\varepsilon ^{-2}} =-\varepsilon ^2+\varepsilon \sqrt {\varepsilon ^2+1}. \]

    The bottom-right corner then gives (using that \(X_2\geq 0\))

    \[ X_2=\varepsilon \sqrt {-2\varepsilon ^2+2\varepsilon \sqrt {\varepsilon ^2+1}}. \]

    The off-diagonal entry gives

    \[ X_1=\sqrt {\varepsilon ^2+1}~\sqrt {-2\varepsilon ^2+2\varepsilon \sqrt {\varepsilon ^2+1}}. \]

    It follows that

    \[ X=\bbm { \sqrt {\varepsilon ^2+1}~\sqrt {-2\varepsilon ^2+2\varepsilon \sqrt {\varepsilon ^2+1}}. & -\varepsilon ^2+\varepsilon \sqrt {\varepsilon ^2+1} \\ -\varepsilon ^2+\varepsilon \sqrt {\varepsilon ^2+1} &\varepsilon \sqrt {-2\varepsilon ^2+2\varepsilon \sqrt {\varepsilon ^2+1}} }. \]

    This is the unique symmetric positive semidefinite solution. The optimal feedback then is (again using that \(C^*D=0\) and \(D^*D=\varepsilon ^2\)):

    \[ F=-\varepsilon ^{-2}\bbm {0&1}X =\bbm { 1-\sqrt {1+\varepsilon ^{-2}} &-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} }. \]

    (e) The closed-loop system then is (in second order form)

    \[ \ddot {q}(t)+\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}}~\dot {q}(t)+\sqrt {1+\varepsilon ^{-2}}~q(t)=0. \]

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