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\(\)
Chapter D Problem Sheet 4 (Lectures 8–9)
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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.
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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. □
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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. □
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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
\(\seteqnumber{0}{D.}{0}\)
\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. □