Chapter E Problem Sheet 5 (Lectures 10–11)

  • 1. Consider the first order scalar differential equation \(\dot {x}=w+u\), \(z=x\), with exo-system \(\dot {x}_e=0\), \(w=x_e\), \(x_e(0)=d\in \mR \), i.e.

    \[ A=0,~~B_1=1,~~B_2=1,~~C_1=1,~~D_{11}=0,~~A_e=0,~~C_e=1. \]

    • (a) Show that the Rosenbrock matrix \(\sbm {sI-A&-B_2\\C_1&0}\) is surjective for \(s=0\) (the only eigenvalue of \(A_e\)).

    • (b) Show that \((A,B_2)\) is stabilizable.

    • (c) What does (a) allow us to conclude about the solvability of the regulator equations and, in combination with (b), about solvability of the full-information disturbance rejection problem?

    • (d) Why can we not use the transfer function \(C_1(sI-A)^{-1}B_2\) to draw the conclusions in (c)?

    • (e) Find a control \(u\) which solves the full-information disturbance rejection problem.

    • Solution. (a) The Rosenbrock matrix is

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

      This has determinant 1 for all \(s\in \mC \) and is therefore invertible and therefore surjective for all \(s\in \mC \) and in particular for \(s=0\).

      (b) We can choose \(F_1=-1\) to obtain \(A+B_2F_1=-1\) which is asymptotically stable. Therefore \((A,B_2)\) is stabilizable.

      (c) From Proposition 10.4 we obtain that the regulator equations are solvable and from Theorem 8.6 we then obtain that the full-information disturbance rejection problem is solvable.

      (d) Since 0 is an eigenvalue of \(A\), the transfer function \(C_1(sI-A)^{-1}B_2\) is not defined in \(s=0\) (the eigenvalue of \(A_e\)). Therefore we cannot use the transfer function.

      (e) The regulator equations are

      \[ 1+V=0,\qquad \Pi =0. \]

      This gives \(V=-1\). With the above choice \(F_1=-1\) we then obtain

      \[ u=-x-d. \]

      □

  • 2. Consider the undamped second order scalar differential equation

    \[ \ddot {q}+q=u. \]

    Let \(x:=\sbm {q\\\dot {q}}\). The exo-system is \(\dot {x}_e=A_ex_e\), \(w=C_ex_e\) with

    \[ A_e=\bbm {0&1\\-1&0},\qquad C_e=\bbm {1&0}. \]

    Define \(z:=q-w\), so that

    \[ A=\bbm {0&1\\-1&0},\quad B_1=\bbm {0\\0},\quad B_2=\bbm {0\\1},\quad C_1=\bbm {1&0},\quad D_{11}=-1. \]

    • (a) Show that the Rosenbrock matrix \(\sbm {sI-A&-B_2\\C_1&0}\) is surjective for all \(s\in \mC \).

    • (b) Show that \((A,B_2)\) is stabilizable.

    • (c) What does (a) allow us to conclude about the solvability of the regulator equations and, in combination with (b), about solvability of the full-information disturbance rejection problem?

    • (d) Why can we not use the transfer function \(C_1(sI-A)^{-1}B_2\) to draw the conclusions in (c)?

    • (e) Find a control \(u\) which solves the full-information output regulation problem.

    • Solution. (a) The Rosenbrock matrix is

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

      Developing by the third row, we see that the determinant of this matrix is \(1\), so that it is invertible and therefore surjective for all \(s\in \mC \).

      (b) We can choose \(F_1=\bbm {0&-1}\) to obtain

      \[ A+B_2F_1=\bbm {0&1\\-1&-1}, \]

      which has characteristic polynomial \(s^2+s+1\) which is stable by Routh–Hurwitz (degree two and all coefficients positive).

      (c) From Proposition 10.4 we obtain that the regulator equations are solvable and from Theorem 8.6 we then obtain that the full-information disturbance rejection problem is solvable.

      (d) \(A=A_e\) the eigenvalues of \(A\) are also eigenvalues of \(A_e\). Therefore the transfer function is undefined for \(s\) equal to the eigenvalues of \(A_e\) and therefore we cannot use this transfer function.

      (e) The regulator equations are

      \begin{gather*} \bbm {0&1\\-1&0}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}} +\bbm {0\\1}\bbm {V_1&V_2} =\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}\bbm {0&1\\-1&0}, \\ \bbm {1&0}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}-\bbm {1&0}=\bbm {0&0}, \end{gather*} which we can re-write as

      \[ \bbm {\Pi _{21}&\Pi _{22}\\-\Pi _{11}+V_1&-\Pi _{12}+V_2} =\bbm {-\Pi _{12}&\Pi _{11}\\-\Pi _{22}&\Pi _{21}},\qquad \bbm {\Pi _{11}-1&\Pi _{12}}=\bbm {0&0}. \]

      From the second equation we obtain \(\Pi _{11}=1\), \(\Pi _{12}=0\). Substituting this into the first equation gives

      \[ \bbm {\Pi _{21}&\Pi _{22}\\-1+V_1&V_2} =\bbm {0&1\\-\Pi _{22}&\Pi _{21}}. \]

      The first row gives \(\Pi _{21}=0\), \(\Pi _{22}=1\). Substituting this into the second row gives \(V_2=0\) and \(V_1=0\). Therefore we obtain

      \[ V=\bbm {0&0},\qquad \Pi =\bbm {1&0\\0&1}. \]

      With the above \(F_1=\bbm {0&-1}\) we obtain \(V-F_1\Pi =\bbm {0&1}\) so that the control is (using that \(x_{e,2}=\dot {x}_{e,1}=\dot {w}\))

      \[ u=-\dot {q}+\dot {w}. \]

      □

  • 3. Consider the second order scalar differential equation \(\ddot {q}+2\dot {q}+q=w+u\), \(z=\dot {q}\) and \(x:=\sbm {q\\\dot {q}}\). Further consider the exo-system \(\dot {x}_e=0\), \(w=x_e\), \(x_e(0)=d\in \mR \). This means that

    \[ A=\bbm {0&1\\-1&-2},\quad B_1=\bbm {0\\1},\quad B_2=\bbm {0\\1},\quad C_1=\bbm {0&1},\quad D_{11}=0,\quad A_e=0,\quad C_e=1. \]

    • (a) Show that the Rosenbrock matrix \(\sbm {sI-A&-B_2\\C_1&0}\) is not surjective for \(s=0\).

    • (b) Show that the regulator equations have a solution.

    • (c) Why does the combination of (a) and (b) not contradict Proposition 10.4?

    • (d) Find a control \(u\) which solves the full-information disturbance rejection problem.

    • Solution. (a) The Rosenbrock matrix is

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

      Developing by the third row we see that the determinant equals \(-s\), so that for \(s=0\) the Rosenbrock matrix is not invertible and therefore not surjective.

      Note that we could also have used the transfer function \(C_1(sI-A)^{-1}B_2\) which equals \(G(s)=\frac {s}{s^2+2s+1}\), so that \(G(0)=0\) which is not surjective so that the Rosenbrock matrix is not surjective for \(s=0\).

      (b) The regulator equations are

      \[ \bbm {0&1\\-1&-2}\bbm {\Pi _1\\\Pi _2}+\bbm {0\\1}+\bbm {0\\1}V=\bbm {0\\0},\qquad \bbm {0&1}\bbm {\Pi _1\\\Pi _2}=0. \]

      From the second equation we obtain that \(\Pi _2=0\). The first row in the first equation then is \(0=0\) whereas the second row in the first equation is \(-\Pi _1+1+V=0\). We are therefore free to pick \(V\in \mR \) and obtain

      \[ \Pi =\bbm {1+V\\0}. \]

      (c) Proposition 10.4 says that surjectivity of the Rosenbrock matrix for all eigenvalues of \(A_e\) is sufficient, not that it is necessary.

      (d) Since \(A\) is stable, we can pick \(F_1=0\) and we obtain the control

      \[ u=Vd. \]

      [As an aside, note that this control gives that \(\ddot {q}+2\dot {q}+q\) is constant. The solutions of this are a constant plus something that goes to zero as \(t\) goes to infinity by stability. Therefore \(z=\dot {q}\) equals something that goes to zero as \(t\) goes to infinity, as desired.] □

  • 4. Consider the first order scalar differential equation

    \[ \dot {x}+2x=3w+u,\qquad z=x,\qquad y=x, \]

    with exo-system

    \[ \dot {x}_e=0,\qquad x_e(0)=d\in \mR , \qquad w=x_e. \]

    Solve the measurement feedback disturbance rejection problem.

    • Solution. This is of the standard form with

      \[ A=-2,~~B_1=3,~~B_2=1,~~C_1=1,~~D_{11}=0,~~C_2=1,~~D_{21}=0,~~A_e=0,~~C_e=1. \]

      The regulator equations are

      \[ -2\Pi +3+V=0,\qquad \Pi =0, \]

      which gives \(\Pi =0\), \(V=-3\). Since \(A\) is stable, we can take \(F_1=0\). We have

      \[ \bbm {A&B_1C_e\\0&A_e}-\bbm {L_1\\L_2}\bbm {C_1&D_{21}C_e}=\bbm {-2&3\\0&0}-\bbm {L_1\\L_2}\bbm {1&0} =\bbm {-2-L_1&3\\-L_2&0}, \]

      which has characteristic polynomial \(s^2+(2+L_1)s+3L_2\), which is stable if and only if \(2+L_1>0\) and \(L_2>0\). We can therefore choose \(L_1=0\) and \(L_2=1\). The controller then is

      \[ A_c=\bbm {-2&0\\-1&0},\qquad B_c=\bbm {0\\1},\qquad C_c=\bbm {0&-3},\qquad D_c=0. \]

      □

  • 5. Consider the first order scalar differential equation

    \[ \dot {x}+x=u,\qquad z=x-w, \]

    with exo-system

    \[ \dot {x}_e=0,\qquad w=x_e,\qquad x_e(0)=r\in \mR , \]

    and measurement

    \[ y=x. \]

    Note that this gives

    \[ A=-1,~~B_1=0,~~B_2=1,~~C_1=1,~~D_{11}=-1,~~C_2=1,~~D_{21}=0,~~A_e=0,~~C_e=1. \]

    • (a) Show that \(\left (\sbm {A&B_1C_e\\0&A_e},\bbm {C_2&D_{21}C_e}\right )\) is not detectable.

    • (b) By considering the closed-loop system

      \[ \bbm {\dot {x}\\\dot {x}_c}=\bbm {A&B_2C_c\\B_cC_2&A_c}\bbm {x\\x_c}+\bbm {B_1\\B_cD_{21}}w,\qquad z=\bbm {C_1&0}\bbm {x\\x_c}+D_{11}w, \]

      with \(A\), \(B_1\), \(B_2\), \(C_1\), \(D_{11}\), \(C_2\) and \(D_{21}\) from above and arbitrary \(A_c\in \mR ^{n_c\times n_c}\), \(B_c\in \mR ^{n_c\times 1}\) and \(C_c\in \mR ^{1\times n_c}\), show that the regulation requirement in the measurement feedback output regulation problem cannot be satisfied, i.e. show that there exist \(x(0)\), \(x_c(0)\) and \(x_e(0)\) such that \(\lim _{t\to \infty }z(t)=0\) does not hold.

    • Solution. (a) Detectability means the existence of an \(L\) such that

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

      is asymptotically stable. Since this matrix has a zero column no matter what \(L\) is, it always has zero as an eigenvalue and is therefore not stable. Alternatively, the characteristic polynomial is \(s^2+(1+L_1)s\), which is not stable.

      (b) The closed-loop system is

      \[ \bbm {\dot {x}\\\dot {x}_c}=\bbm {-1&C_c\\B_c&A_c}\bbm {x\\x_c}+\bbm {0\\0}w,\qquad z=\bbm {1&0}\bbm {x\\x_c}-w, \]

      For \(x(0)=x_c(0)=0\) we have \(x(t)=x_c(t)=0\) for all \(t\geq 0\). Hence \(z(t)=-w\) for all \(t\geq 0\). So if we choose \(x_e(0)=1\) so that \(x_e=w=1\) we have \(z(t)=-1\) for all \(t\geq 0\). In particular we do not have \(\lim _{t\to \infty }z(t)=0\) for these particular initial conditions.

      [As an aside: the issue here is that the measurement \(y\) provides no information about \(w\), also not indirectly though the state \(x\).] □