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\(\)
Chapter G Problem Sheet 7 (Lectures 14–15)
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1. Consider the undamped second order scalar differential equation
\[ \ddot {q}+q=0,\qquad y=q. \]
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(a) Write this in the standard form \(\dot {x}=Ax\), \(y=Cx\) with \(x=\sbm {q\\\dot {q}}\).
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(b) Use the Kalman observability matrix to show that this system is observable.
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(c) Use the Hautus observability matrix to show that this system is observable.
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Solution. (a) We have
\[ A=\bbm {0&1\\-1&0},\qquad C=\bbm {1&0}. \]
(b) The Kalman observability matrix is
\[ \bbm {C\\CA}=\bbm {1&0\\0&1}. \]
Since this is the identity matrix, it is injective. It follows that the system is observable.
(c) The Hautus observability matrix is
\[ \bbm {sI-A\\C}=\bbm {s&-1\\1&s\\1&0}. \]
Selecting the first and last row gives a matrix with determinant 1. In particular, that matrix is invertible so that the Hautus observability matrix is injective for all \(s\in \mC \). Hence the system is observable. □
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2. Consider the state-output system \(\dot {x}=Ax\), \(y=Cx\) with
\[ A=\bbm {1&0&0\\1&1&1\\0&0&1},\qquad C=\bbm {0&1&0}. \]
Determine whether or not this system is observable.
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Solution. The Kalman observability matrix is
\[ \bbm {C\\CA\\CA^2}=\bbm {0&1&0\\1&1&1\\2&1&2}. \]
Since this matrix has two identical columns, the columns are not linearly independent and therefore the matrix is not invertible and therefore not injective. Therefore the system is not observable. □
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3. Consider the state-output system \(\dot {x}=Ax\), \(y=Cx\) with
\[ A=\bbm {1&0\\1&1},\qquad C=\bbm {1&0}. \]
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(a) Use the Kalman observability matrix to show that this system is not observable.
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(b) Use the Hautus observability matrix to show that this system is not observable.
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(c) Show using only the definition of observability that this system is not observable, i.e. show that there exists a nonzero \(x^0\in \mR ^2\) so that with the initial condition \(x(0)=x^0\) the unique solution satisfies \(y(t)=0\) for all \(t\geq 0\).
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Solution. (a) The Kalman observability matrix is
\[ \bbm {C\\CA}=\bbm {1&0\\1&0}. \]
This matrix is not injective (for example the vector \(\sbm {0\\1}\) is in its nullspace). Therefore the system is not observable.
(b) The Hautus observability matrix is
\[ \bbm {sI-A\\C}=\bbm {s-1&0\\-1&s-1\\1&0}. \]
For \(s=1\) this matrix equals
\[ \bbm {0&0\\-1&0\\1&0}. \]
Because of the zero column, this matrix is not injective (for example the vector \(\sbm {0\\1}\) is in its nullspace). Therefore the system is not observable.
(c) Let \(x^0=\sbm {0\\1}\). The differential equations are
\[ \dot {x}_1=x_1,\qquad \dot {x}_2=x_1+x_2. \]
With the initial condition \(x_1(0)=x^0_1(0)=0\) we obtain \(x_1=0\). This gives \(y(t)=x_1(t)=0\) for all \(t\geq 0\). Clearly \(x^0\) is nonzero. Therefore we have shown that the system is not observable. □
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4. Consider the second order scalar differential equation
\[ \ddot {q}+4\dot {q}+3q=0,\qquad y=q. \]
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(a) Write this in the standard form \(\dot {x}=Ax\), \(y=Cx\) with \(x=\sbm {q\\\dot {q}}\).
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(b) Determine the infinite-time observability Gramian \(S\) by solving the observation Lyapunov equation and use this to show that the system is observable.
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(c) Use the infinite-time observability Gramian to determine \(\int _0^\infty |h(t)|^2\,dt\) where \(h\) is the impulse response of
\[ \ddot {q}(t)+4\dot {q}(t)+3q(t)=u(t),\qquad y(t)=q(t). \]
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Solution. (a) We have
\[ A=\bbm {0&1\\-3&-4},\qquad C=\bbm {1&0}. \]
(this is asymptotically stable by knowledge of second order systems).
(b) The observation Lyapunov equation \(A^*S+SA+C^*C=0\) is (using that \(S\) is symmetric)
\[ \bbm {0&-3\\1&-4}\bbm {S_1&S_0\\S_0&S_2} +\bbm {S_1&S_0\\S_0&S_2}\bbm {0&1\\-3&-4} +\bbm {1&0}\bbm {1\\0} =\bbm {0&0\\0&0}. \]
This is
\[ \bbm {-6S_0+1&-3S_2+S_1-4S_0\\-3S_2+S_1-4S_0&2S_0-8S_2}=\bbm {0&0\\0&0}. \]
From the top-left corner we obtain \(S_0=\frac {1}{6}\). The bottom-right corner then gives \(S_2=\frac {1}{24}\) and the off-diagonal entry then gives \(S_1=\frac {19}{24}\). Hence
\[ S=\frac {1}{24}\bbm {19&4\\4&1}. \]
The determinant of \(24S\) equals 3, so that \(S\) is invertible. Hence the system is observable.
(c) For the input-state-output system we have \(B=\bbm {0\\1}\), so the
\[ \int _0^\infty |h(t)|^2\,dt=\ipd {SB}{B}=\frac {1}{24}. \]
□