Chapter 14 Observability I
We consider the state-output system
\(\seteqnumber{0}{14.}{0}\)\begin{equation} \label {eq:so} \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 14.2. The unobservable subspace consists of all initial state \(x^0\) for which the solution of (14.1) with \(x(0)=x^0\) satisfies \(y(t)=0\) for all \(t\geq 0\).
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Definition 14.3. The Kalman observability matrix is the \(pn\times n\) matrix
\[ \bbm {C\\CA\\CA^2\\\vdots \\CA^{n-1}}. \]
The observability Gramian, indexed by \(T>0\), is the \(n\times n\) matrix
\[ S_T=\int _0^T \e ^{A^*t}C^*C\e ^{At}\,dt. \]
The Hautus observability matrix, indexed by \(s\in \mC \), is the \((n+p)\times n\) matrix
\[ \bbm {sI-A\\C}. \]
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Theorem 14.4. The following are equivalent:
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1. The state-output system is observable;
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2. The Kalman observability matrix is injective;
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3. The Hautus observability matrix is injective for all \(s\in \mathbb {C}\);
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4. The Hautus observability matrix is injective for all eigenvalues \(s\) of \(A\);
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5. The observability Gramian \(S_T\) is invertible for some \(T>0\);
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6. The observability Gramian \(S_T\) is invertible for all \(T>0\);
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7. For every monic polynomial \(r\) of degree \(n\) there exists a matrix \(L\in \mR ^{n\times p}\) such that the characteristic polynomial of \(A-LC\) equals \(r\).
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8. The reachable subspace equals \(\{0\}\).
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Proposition 14.8. Let \(A\) be asymptotically stable.
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• The state-output system is observable if and only if the infinite-time observability gramian \(S\) is invertible.
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• The unobservable subspace equals the nullspace of the infinite-time observability gramian \(S\).
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• The infinite-time observability gramian \(S\) is the unique solution of the observation Lyapunov equation
\[ A^*S+SA+C^*C=0. \]
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Remark 14.9. The interpretation of the observability Gramian is that with \(y\) the output of the state-output system (14.1) with initial condition \(x^0\)
\[ \ipd {S_Tx^0}{x^0}=\int _0^T \|y(t)\|^2\,dt,\qquad \ipd {Sx^0}{x^0}=\int _0^\infty \|y(t)\|^2\,dt. \]
This is easily seen as follows:
\(\seteqnumber{0}{14.}{1}\)\begin{multline*} \ipd {S_Tx^0}{x^0} =\Ipd {\int _0^T \e ^{A^*t}C^*C\e ^{At}x^0\,dt}{x^0} =\int _0^T \ipd {\e ^{A^*t}C^*C\e ^{At}x^0}{x^0}\,dt \\ =\int _0^T \ipd {C\e ^{At}x^0}{C\e ^{At}x^0}\,dt =\int _0^T \|y(t)\|^2\,dt. \end{multline*} If we have an input-state-output system \(\dot {x}=Ax+Bu\), \(y=Cx\) with a one-dimensional input (i.e. \(m=1\)), then we can consider the corresponding state-output system with initial condition \(x^0=B\). The above interpretation then gives
\[ \ipd {SB}{B}=\int _0^\infty \|C\e ^{At}B\|^2\,dt, \]
i.e.
\[ \ipd {SB}{B}=\int _0^\infty \|h(t)\|^2\,dt, \]
so we obtain the norm squared of the impulse response of the original input-state-output system.
14.1 Examples
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Example 14.11. Consider the first order scalar differential equation
\[ \dot {x}+x=0,\qquad y=x, \]
i.e.
\[ A=-1,\qquad C=1. \]
It is clear from the definition that this system is observable: if \(y(t)=0\) for all \(t\geq 0\), then \(x(t)=0\) for all \(t\geq 0\) and in particular \(x(0)=0\). Hence the only state which is unobservable is zero.
The Kalman observability matrix in this case is the \(1\times 1\) matrix \(1\). Since this is nonzero, it is injective and we also see from this that the system is observable.
The infinite-time observability Gramian is the solution of
\[ -2S+1=0, \]
i.e. \(S=\frac {1}{2}\). Since this is nonzero, it is invertible and we see from this as well that the system is observable.
The observability Gramian \(S_T\) equals
\[ S_T=\int _0^T \e ^{-2t}\,dt=\frac {1-\e ^{-2T}}{2}. \]
The formula from Remark 14.5 then gives
\[ x(0)=\frac {2}{1-\e ^{-2T}}\int _0^T \e ^{-t}y(t)\,dt. \]
Using that \(y(t)=x(t)=\e ^{-t}x(0)\), we see by direct computation that the right-hand side indeed equals \(x(0)\). Note that in principle we could use \(x(0)=y(0)\), but the formula from Remark 14.5 uses the observed values of \(y\) on the whole interval \([0,T]\) and will therefore by more “robust” (to for example measurement error).