Chapter D Problem Sheet 4 (Lectures 8–9)

  • 1. Consider the first order scalar differential equation \(\dot {x}(t)+6x(t)=2u(t)\). Find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }x(t)=r\) no matter what \(x(0)\) is.

    • Solution. We define the performance output and exo-system

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

      so that the regulation requirement will give what is desired. We have

      \[ A=-6,~~B_1=0,~~B_2=2,~~C_1=1,~~D_{11}=-1,~~A_e=0,~~C_e=1. \]

      The regulator equations are

      \[ -6\Pi +2V=0,\qquad \Pi -1=0, \]

      which gives \(\Pi =1\) and \(V=3\). Since \(A\) is asymptotically stable, we can take \(F_1=0\). We then have that

      \[ u=3r, \]

      solves our problem.

  • 2. Consider the first order scalar differential equation \(\dot {x}(t)=x(t)+u(t)\). Find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }x(t)=r\) no matter what \(x(0)\) is.

    • Solution. We define the performance output and exo-system

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

      so that the regulation requirement will give what is desired. We have

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

      The regulator equations are

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

      which gives \(\Pi =1\) and \(V=-1\). We choose \(F_1=-2\) so that \(A+B_2F_1=-1\) is asymptotically stable. We then have \(V-F_1\Pi =-1+2=1\), so that

      \[ u=-2x+r, \]

      solves our problem.

  • 3. Consider the first order scalar differential equation \(\dot {x}(t)+x(t)=w(t)+u(t)\). Find a control \(u\) such that \(\lim _{t\to \infty }x(t)=0\) no matter what \(x(0)\) is when \(w(t)=r\), where \(r\in \mR \).

    • Solution. We define the performance output and exo-system

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

      so that the regulation requirement will give what is desired. We have

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

      The regulator equations are

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

      which gives \(\Pi =0\) and \(V=-1\). Since \(A\) is asymptotically stable, we can take \(F_1=0\) and we obtain that

      \[ u=-r, \]

      solves our problem.

  • 4. Consider the second order scalar differential equation

    \[ \ddot {q}+2\zeta \dot {q}+q=u, \]

    where \(\zeta >0\). Find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }q(t)=r\) no matter what the initial conditions are.

    • Solution. We define the state \(x:=\sbm {q\\\dot {q}}\), the performance output and exo-system

      \[ z=x_1-w,\qquad \dot {x}_e=0,\qquad w=x_e,\qquad x_e(0)=r, \]

      so that the regulation requirement will give what is desired. We have

      \begin{gather*} A=\bbm {0&1\\-1&-2\zeta },\qquad B_1=\bbm {0\\0},\qquad B_2=\bbm {0\\1},\qquad C_1=\bbm {1&0},\qquad D_{11}=-1,\\ A_e=0,\qquad C_e=1. \end{gather*} The regulator equations are

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

      From the second equation we obtain \(\Pi _1=1\). The first row in the first equation gives \(\Pi _2=0\) and the second row then gives \(-\Pi _1+V=0\) which gives \(V=1\). Since \(A\) is asymptotically stable, we can take \(F_1=0\) and we obtain that

      \[ u=r, \]

      solves our problem.