Chapter 16 Stabilizability and detectability
16.1 Stabilizability
We consider the input-state system
\(\seteqnumber{0}{16.}{0}\)\begin{equation} \label {eq:is2} \dot {x}(t)=Ax(t)+Bu(t). \end{equation}
Here \(A\in \mR ^{n\times n}\) and \(B\in \mR ^{n\times m}\).
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Definition 16.1. The input-state system (16.1) (or equivalently, the pair of matrices \((A,B)\)) is said to be stabilizable if there exists a \(F\in \mR ^{m\times n}\) such that \(A+BF\) is asymptotically stable.
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Theorem 16.2. The following are equivalent:
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1. The input-state system is stabilizable;
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2. The Hautus controllability matrix is surjective for all \(s\in \mC \) with \(\re (s)\geq 0\);
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3. The Hautus controllability matrix is surjective for all eigenvalues \(s\) of \(A\) with \(\re (s)\geq 0\);
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4. For all \(x^0\in \mR ^n\) there exists a control \(u\) such that the solution of (16.1) with \(x(0)=x^0\) satisfies \(\int _0^\infty \|x(t)\|^2+\|u(t)\|^2\,dt<\infty \).
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16.2 Detectability
We consider the state-output system
\(\seteqnumber{0}{16.}{1}\)\begin{equation} \label {eq:so2} \dot {x}(t)=Ax(t),\qquad y(t)=Cx(t). \end{equation}
Here \(A\in \mR ^{n\times n}\) and \(C\in \mR ^{p\times n}\).
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Definition 16.4. The state-output system (16.2) (or equivalently, the pair of matrices \((A,C)\)) is said to be detectable if there exists a \(L\in \mR ^{n\times p}\) such that \(A-LC\) is asymptotically stable.
16.3 Examples
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Example 16.8. Consider the input-state system
\[ \dot {x}_1=x_1+u,\qquad \dot {x}_2=-x_2. \]
This can be written in standard form with
\[ A=\bbm {1&0\\0&-1},\qquad B=\bbm {1\\0}. \]
We will show in various ways that this system is stabilizable, but not controllable.
We first only use the definitions. Choosing \(u=-2x_1\) we obtain
\[ \dot {x}_1=-x_1,\qquad \dot {x}_2=-x_2, \]
which is an asymptotically stable system: all solutions are \(x_1(t)=x_1^0\e ^{-t}\), \(x_2(t)=x_2^0\e ^{-t}\), which converge to zero as \(t\to \infty \). Therefore \(F=\bbm {-2&0}\) (for which \(u=Fx\) gives \(u=-2x_1\)) is stabilizing. The system is not controllable, since \(x_2(t)=x_2^0\e ^{-t}\) no matter what \(u\) is. In more detail: choosing \(x^0\) with \(x^0_2=1\) gives \(x_2(t)=\e ^{-t}\) (no matter what \(u\) is) and choosing \(x^1_2=-1\) then means that there does not exist a \(T\) with \(x(T)=x^1\) since \(\e ^{-T}=-1\) is never true.
We now use the Hautus controllability matrix. We have
\[ \bbm {sI-A&B}=\bbm {s-1&0&1\\0&s+1&0}. \]
Selecting the last two columns gives a matrix with determinant \(-(s+1)\). Therefore this matrix is invertible for \(s\neq -1\). In particular, it is invertible for all \(s\in \mC \) with \(\re (s)\geq 0\). It follows that the Hautus controllability matrix is surjective for all \(s\in \mC \) with \(\re (s)\geq 0\). This shows stabilizability. If \(s=-1\), then the Hautus controllability matrix has a zero row and is therefore not surjective. This shows that controllability does not hold.
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Example 16.9. For the measurement feedback equivalent of Example 9.2 with \(\zeta =0\) and with measurement \(y=x_2\) we are interested in detectability of the pair
\[ A=\bbm {0&1&0&0\\-1&0&1&0\\0&0&0&1\\0&0&-\omega _e^2&0},\qquad C=\bbm {0&1&0&0}, \]
where \(\omega _e>0\).
The Kalman observability matrix is
\[ \bbm {C\\CA\\CA^2\\CA^3} =\bbm {0&1&0&0\\-1&0&1&0\\0&-1&0&1\\1&0&-1-\omega _e^2&0}. \]
Developing by the last column, the determinant equals
\[ -\det \bbm {0&1&0\\-1&0&1\\1&0&-1-\omega _e^2}, \]
and developing by the first row shows that this equals \(\omega _e^2\). Since \(\omega _e>0\) we have that the Kalman observability matrix is invertible and therefore injective and therefore the system is observable and therefore detectable.
The Hautus observability matrix is
\[ \bbm {sI-A\\C} =\bbm {s&-1&0&0\\1&s&-1&0\\0&0&s&-1\\0&0&\omega _e^2&s\\0&1&0&0}. \]
Omitting the fourth row gives the submatrix
\[ \bbm {s&-1&0&0\\1&s&-1&0\\0&0&s&-1\\0&1&0&0}, \]
whose determinant is (develop by the last row and then by the first row)
\[ -\det \bbm {s&0&0\\1&-1&0\\0&s&-1}=-s. \]
So for \(s\neq 0\) we have found four linearly independent rows. For \(s=0\) we consider instead the first four rows (develop by the first row and then by the first column):
\[ \det \bbm {0&-1&0&0\\1&0&-1&0\\0&0&0&-1\\0&0&\omega _e^2&0} =\det \bbm {1&-1&0\\0&0&-1\\0&\omega _e^2&0}=\omega _e^2. \]
Since \(\omega _e>0\), we obtain that when \(s=0\) we have also found four linearly independent rows. Therefore for all \(s\in \mC \) the Hautus observability matrix is injective so that the system is observable and therefore detectable.