Chapter F Problem Sheet 6 (Lectures 12–13)
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1. Consider the first order scalar differential equation
\[ \dot {x}+2x=4u. \]
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(a) Write this in the standard form \(\dot {x}=Ax+Bu\).
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(b) Determine the Kalman controllability matrix and use this to show that the system is controllable.
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(c) Determine the Hautus controllability matrix and use this to show that the system is controllable.
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(d) Determine the controllability Gramian \(Q_T\) and use this to show that the system is controllable. Moreover, determine a control which steers the system from a given \(x^0\in \mR \) to a given \(x^1\in \mR \) in a given time \(T>0\).
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(e) Determine the infinite-time controllability Gramian \(Q\) by solving the control Lyapunov equation and use this to show that the system is controllable.
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Solution. (a) We have
\[ A=-2,\qquad B=4. \]
(b) Since \(n=1\), the Kalman controllability matrix is simply \(B\). Since a \(1\times 1\) matrix is surjective if and only if it is nonzero, we have that the Kalman controllability matrix is surjective and therefore the system is controllable.
(c) The Hautus controllability matrix is
\[ \bbm {s+2&4}. \]
This matrix is surjective no matter what \(s\) is since the second column gives an invertible 1-by-1 matrix.
(d) The controllability Gramian is
\[ Q_T=\int _0^T 16\e ^{-4t}\,dt =\left [-4\e ^{-4t}\right ]_{t=0}^T =4\left (1-\e ^{-4T}\right ). \]
From Remark 12.8 we then see that a control which steers the system from state \(x^0\) to state \(x^1\) in time \(T\) is given by
\(\seteqnumber{0}{F.}{0}\)\begin{align*} u(t)&=\e ^{-2(T-t)}~\frac {1}{1-\e ^{-4T}}(x^1-\e ^{-2T}x^0). \end{align*}
(e) The control Lyapunov equation is
\[ -4Q+16=0, \]
which gives \(Q=4\) as the infinite-time controllability Gramian. Since this is invertible, the system is controllable. □
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2. Consider the second order scalar differential equation
\[ \ddot {q}+q=u. \]
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(a) Write this in the standard form \(\dot {x}=Ax+Bu\) with \(x:=\sbm {q\\\dot {q}}\).
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(b) Determine the Hautus controllability matrix and use this to show that the system is controllable.
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Solution. (a) We have
\[ A=\bbm {0&1\\-1&0},\qquad B=\bbm {0\\1}. \]
(b) The Hautus controllability matrix is
\[ \bbm {sI-A&B}=\bbm {s&-1&0\\1&s&1}. \]
Selecting the last two columns, we obtain a matrix with determinant \(-1\). So we have that the Hautus controllability matrix is surjective for all \(s\in \mC \). Therefore, the system is controllable. □
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3. For the input-state system \(\dot {x}=Ax+Bu\) with
\[ A=\bbm {5&4\\-3&-2},\qquad B=\bbm {1\\-1}, \]
determine the reachable subspace.
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4. Determine whether or not \(\dot {x}=Ax+Bu\) is controllable when
\[ A=\bbm {0&1&-1\\2&1&0\\-3&5&1},\qquad B=\bbm {1\\0\\1}. \]
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5. Determine whether or not \(\dot {x}=Ax+Bu\) is controllable when
\[ A=\bbm {0&1&2\\0&1&5\\0&0&1},\qquad B=\bbm {1&0\\0&1\\0&1}. \]
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Solution. The Kalman controllability matrix is \(\bbm {B&AB&A^2B}\). We calculate only the first four columns (as it turns out this suffices)
\[ \bbm {B&AB}=\bbm {1&0&0&3\\0&1&0&6\\0&1&0&1}. \]
Omitting the zero column gives the matrix
\[ \bbm {1&0&3\\0&1&6\\0&1&1}. \]
Developing by the first column, we see that its determinant equals \(-5\). Therefore the first, second and fourth column of the Kalman controllability matrix are linearly independent so that the Kalman controllability matrix is surjective and therefore the system is controllable. □
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6. Solve the control Lyapunov equation for
\[ A=\bbm {-2&0\\0&-3},\qquad B=\bbm {1\\2}. \]
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Solution. The control Lyapunov equation is
\[ \bbm {-2&0\\0&-3}\bbm {Q_1&Q_0\\Q_0&Q_2}+\bbm {Q_1&Q_0\\Q_0&Q_2}\bbm {-2&0\\0&-3} +\bbm {1\\2}\bbm {1&2}=\bbm {0&0\\0&0}, \]
which is
\[ \bbm {-4Q_1+1&-5Q_0+2\\-5Q_0+2&-6Q_2+4}=\bbm {0&0\\0&0}. \]
Solving this gives \(Q_1=\frac {1}{4}\), \(Q_2=\frac {2}{3}\), \(Q_0=\frac {2}{5}\) so that
\[ Q=\bbm {\frac {1}{4}&\frac {2}{5}\\\frac {2}{5}&\frac {2}{3}}. \]
□
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7. Show using only the definition of controllability that \(\dot {x}=Ax+Bu\) with
\[ A=\bbm {1&0\\0&1},\qquad B=\bbm {1\\2}, \]
is not controllable, i.e. show that for all \(T>0\) there exist \(x^0,x^1\in \mR ^2\) such that there does not exist a control \(u:[0,T]\to \mR \) for which we have \(x(0)=x^0\) and \(x(T)=x^1\).
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Solution. The differential equations are
\[ \dot {x}_1=x_1+u,\qquad \dot {x}_2=x_2+2u. \]
We therefore have
\[ 2\dot {x}_1-\dot {x}_2=2(x_1+u)-(x_2+2u)=2x_1-x_2. \]
If we define \(z:=2x_1-x_2\), then this is \(\dot {z}=z\). Choose \(x^0=\sbm {0\\0}\) and \(x^1=\sbm {1\\0}\). Then \(z(0)=0\), which combined with \(\dot {z}=z\) gives that \(z(t)=0\) for all \(t\geq 0\). Therefore \(2x_1(t)=x_2(t)\) for all \(t\geq 0\) no matter what \(u\) is. If \(x(T)=x^1\), then we obtain the contradiction \(2=2x_1(T)=x_2(T)=0\). Therefore a control \(u:[0,T]\to \mR \) such that \(x(0)=x^0\) and \(x(T)=x^1\) does not exist. □
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