Chapter J Problem Sheet 10 (Lectures 20–21)

  • 1. Consider the first order scalar differential equation

    \[ \dot {x}=2x+w_1+u, \]

    the performance output

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

    and the measurement

    \[ y=x+w_2, \]

    i.e.

    \[ A=2,~~B_1=\bbm {1&0},~~B_2=1,~~C_1=\bbm {1\\0},~~D_{12}=\bbm {0\\1},~~C_2=1,~~D_{21}=\bbm {0&1}. \]

    • (a) Determine whether or not \(D_{21}\) is surjective;

    • (b) Determine whether or not \((A,C_2)\) is detectable;

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

    • (d) Determine the stabilizing solution of the algebraic Riccati equation (21.4);

    • (e) Determine the unique controller which solves the \(H^2\) measurement feedback problem (you may use the solution of the other Riccati equation from Problem Sheet 9).

    • Solution. (a) Since the second column of \(D_{21}\) is an invertible 1-by-1 matrix, \(D_{21}\) is surjective.

      (b) With \(L=3\) we have \(A-LC_2=-1\) which is stable, so that \((A,C_2)\) is detectable.

      (c) The relevant Rosenbrock matrix is

      \[ \bbm {sI-A&-B_1\\C_2&D_{21}}=\bbm {s-2&-1&0\\1&0&1}. \]

      Omitting the first column gives an invertible matrix so that the Rosenbrock matrix is surjective for all \(s\in \mC \) (in particular, for those with \(\re (s)=0\)).

      (d) The relevant algebraic Riccati equation is (using that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=1\))

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

      The solutions of this are

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

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

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

      We then have

      \[ L=2+\sqrt {5}, \]

      (e) From Problem Sheet 9 we have \(F=-2-\sqrt {5}\) (the additional zero column in \(B_1\) doesn’t change the Riccati equation), which combined with the above computed \(L\) gives the controller matrices

      \[ A_c=A+B_2F-LC_2=-2-2\sqrt {5},\qquad B_c=L=2+\sqrt {5},\qquad C_c=F=-2-\sqrt {5}. \]

      □

  • 2. Consider the undamped second order scalar differential equation

    \[ \ddot {q}(t)+q(t)=w_1(t)+u(t), \]

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

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

    where \(\varepsilon >0\), and with \(\delta >0\) the measured output

    \[ y=q+\delta w_2. \]

    i.e.

    \begin{gather*} A=\bbm {0&1\\-1&0},\quad B_1=\bbm {0&0\\1&0},\quad B_2=\bbm {0\\1},\quad C_1=\bbm {1&0\\0&0},\quad D_{12}=\bbm {0\\\varepsilon }, \\ C_2=\bbm {1&0},\quad D_{21}=\bbm {0&\delta }. \end{gather*}

    • (a) Determine whether or not \(D_{21}\) is surjective;

    • (b) Determine whether or not \((A,C_2)\) is detectable;

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

    • (d) Determine the stabilizing solution of the algebraic Riccati equation (21.4);

    • (e) Determine the unique controller which solves the \(H^2\) measurement feedback problem (you may use the solution of the other Riccati equation from Problem Sheet 9).

    • Solution. (a) Since \(\delta >0\) we have that \(D_{21}\) is surjective (its second column is an invertible 1-by-1 matrix).

      (b) From Problem Sheet 7 we have that \((A,C_2)\) is observable and therefore detectable.

      (c) The relevant Rosenbrock matrix is

      \[ \bbm {sI-A&-B_1\\C_2&D_{21}} =\bbm {s&-1&0&0\\1&s&-1&0\\1&0&0&\delta }. \]

      The final three columns form a lower-triangular matrix with nonzero entries on the diagonal and therefore has nonzero determinant. Therefore the final three columns are linearly independent. Therefore the Rosenbrock matrix has rank 3 and is therefore surjective (for all \(s\in \mC \) in fact).

      (d) The Riccati equation (21.4) is (using that \(Y\) is symmetric, that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\))

      \begin{multline*} \bbm {0&1\\-1&0}\bbm {Y_1&Y_0\\Y_0&Y_2} +\bbm {Y_1&Y_0\\Y_0&Y_2}\bbm {0&-1\\1&0} +\bbm {0&0\\1&0}\bbm {0&1\\0&0} \\-\delta ^{-2}\bbm {Y_1&Y_0\\Y_0&Y_2}\bbm {1\\0}\bbm {1&0}\bbm {Y_1&Y_0\\Y_0&Y_2} =\bbm {0&0\\0&0}, \end{multline*} which is

      \[ \bbm {2Y_0&Y_2-Y_1\\Y_2-Y_1&-2Y_0} +\bbm {0&0\\0&1} -\delta ^{-2}\bbm {Y_1^2&Y_0Y_1\\Y_0Y_1&Y_0^2} =\bbm {0&0\\0&0}, \]

      which is

      \[ \bbm {2Y_0-\delta ^{-2}Y_1^2 &Y_2-Y_1-\delta ^{-2}Y_0Y_1 \\Y_2-Y_1-\delta ^{-2}Y_0Y_1 &-2Y_0+1-\delta ^{-2}Y_0^2 }=\bbm {0&0\\0&0}. \]

      From the top-left corner we obtain that \(Y_0\geq 0\). From the bottom-right corner we then obtain (using that \(Y_0\geq 0\) to pick the correct solution):

      \[ Y_0=-\delta ^2+\delta ^2\sqrt {1+\delta ^{-2}}. \]

      The top-left corner then gives (using that \(Y_1\geq 0\) to pick the correct sign)

      \[ Y_1=\delta ^2\sqrt {-2+2\sqrt {1+\delta ^{-2}}}. \]

      The off-diagonal entry then gives

      \[ Y_2=\sqrt {1+\delta ^{-2}}~\delta ^2\sqrt {-2+2\sqrt {1+\delta ^{-2}}}. \]

      Hence

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

      We then have (using again that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\)):

      \begin{align*} L&=\delta ^{-2} \delta ^2\bbm { \sqrt {-2+2\sqrt {1+\delta ^{-2}}}& -1+\sqrt {1+\delta ^{-2}}\\ -1+\sqrt {1+\delta ^{-2}}& \sqrt {1+\delta ^{-2}}~\sqrt {-2+2\sqrt {1+\delta ^{-2}}} } \bbm {1\\0} \\&=\bbm { \sqrt {-2+2\sqrt {1+\delta ^{-2}}}\\ -1+\sqrt {1+\delta ^{-2}}}. \end{align*} (e) On Problem Sheet 9 we already solved the Riccati equation (21.3) and obtained

      \begin{gather*} X=\varepsilon ^2\bbm { \sqrt {1+\varepsilon ^{-2}}~ \sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} & -1+\sqrt {1+\varepsilon ^{-2}} \\ -1+\sqrt {1+\varepsilon ^{-2}} &\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} }, \\ F=\bbm { 1-\sqrt {1+\varepsilon ^{-2}} &-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} }. \end{gather*} The controller then is

      \begin{gather*} \dot {x}_c=\bbm {-\sqrt {-2+2\sqrt {1+\delta ^{-2}}}&1\\1-\sqrt {1+\varepsilon ^{-2}}-\sqrt {1+\delta ^{-2}}&-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}}}x_c +\bbm {\sqrt {-2+2\sqrt {1+\delta ^{-2}}}\\ -1+\sqrt {1+\delta ^{-2}}}y,\\ u=\bbm { 1-\sqrt {1+\varepsilon ^{-2}} &-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}}}x_c. \end{gather*} □

Bibliography

The suspension system model is based on [Hrovat, 1997] (authored by an employee of Ford Research). The fixed structure optimization is based on [Scheibe and Smith, 2009]. The tape drive model is based on [Cherubini, 2022] (authored by an employee of IBM Research) and other articles by that author.

Further background on material in the first seven chapters can for example be found in [Bolton, 2015] and on the material in the later chapters in [Trentelman et al., 2002] (for the material on Controllability and Observability also [Logemann and Ryan, 2014]).

Bibliography

  • [Bolton, 2015]  Bolton, W. (2015). Mechatonics. sixth edition.

  • [Cherubini, 2022]  Cherubini, G. (2022). Advanced control systems for data storage on magnetic tape: A long-lasting success story. IEEE Control Systems Magazine, 42(4):8–11.

  • [Hrovat, 1997]  Hrovat, D. (1997). Survey of advanced suspension developments and related optimal control applications. Automatica, 33:1781–1817.

  • [Logemann and Ryan, 2014]  Logemann, H. and Ryan, E. P. (2014). Ordinary Differential Equations: Analysis, Qualitative Theory and Control.

  • [Scheibe and Smith, 2009]  Scheibe, F. and Smith, M. C. (2009). Analytical solutions for optimal ride comfort and tyre grip for passive vehicle suspensions. Vehicle Systems Dynamics, 47:1229–1252.

  • [Trentelman et al., 2002]  Trentelman, H. L., Stoorvogel, A. A., and Hautus, M. (2002). Control Theory for Linear Systems.