Chapter 12 Observability
We consider the state-output system
\(\seteqnumber{0}{12.}{0}\)\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 12.1.
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• Let \(T>0\) and \(x^0\in \mR ^n\). The initial state \(x^0\) is said to be unobservable in time \(T\) if the solution of (12.1) with \(x(0)=x^0\) satisfies \(y(t)=0\) for all \(t\in [0,T]\).
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• Let \(x^0\in \mR ^n\). The initial state \(x^0\) is said to be unobservable if the solution of (12.1) with \(x(0)=x^0\) satisfies \(y(t)=0\) for all \(t\geq 0\).
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• Let \(T>0\). The system (12.1) (or equivalently the pair of matrices \((A,C)\)) is said to be observable in time \(T\) if the only unobservable state in time \(T\) is zero.
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• The system (12.1) (or equivalently the pair of matrices \((A,C)\)) is said to be observable if the only unobservable state is zero.
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Definition 12.2. 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 12.3. The following are equivalent:
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1. The state-output system is observable in time \(T\) for some \(T>0\)
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2. The state-output system is observable
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3. The Kalman observability matrix is injective
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4. The Hautus observability matrix is injective for all \(s\in \mathbb {C}\)
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5. The Hautus observability matrix is injective for all eigenvalues \(s\) of \(A\)
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6. The observability Gramian \(S_T\) is invertible for some \(T>0\)
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7. The observability Gramian \(S_T\) is invertible for all \(T>0\)
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Definition 12.5.
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• Let \(T>0\). The set of states which are unobservable in time \(T>0\) is denoted by \(\mathcal {U}_T\).
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• The set of states which are unobservable is denoted \(\mathcal {U}\) and is called the unobservable subspace.
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Proposition 12.6. The unobservable subspace equals the nullspace of the Kalman observability matrix and also equals the nullspace of the observability Gramian \(S_T\) (for any \(T>0\)).
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Proposition 12.8. 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 12.9. The interpretation of the observability Gramian is that with \(y\) the output of the state-output system (12.1) 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:
\(\seteqnumber{0}{12.}{1}\)\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, 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.
12.1 Examples
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Example 12.12. Consider the first order system
\[ \dot {x}+x=0,\qquad y=x, \]
i.e.
\[ A=-1,\qquad C=1. \]
It is clear from the definition that this system is observable in any time \(T>0\): if \(y(t)=0\) for all \(t\in [0,T]\), then \(x(t)=0\) for all \(t\in [0,T]\) and in particular \(x(0)=0\). Hence the only state which is unobservable in time \(T\) 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 12.4 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 12.4 uses the observed values of \(y\) on the whole interval \([0,T]\) and will therefore by more “robust” (to for example measurement error).
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Example 12.13. Consider the second order (low-pass) system
\[ \ddot {q}(t)+2\zeta \dot {q}(t)+q(t)=0, \]
where \(\zeta \geq 0\), the state is \(x=\sbm {q\\\dot {q}}\) and the output is \(y=q\). Hence
\[ A=\bbm {0&1\\-1&-2\zeta },\qquad C=\bbm {1&0}. \]
We can see from the definition that this system is observable. If \(q(t)=0\) for all \(t\in [0,T]\), then by differentiating we have that \(\dot {q}(t)=0\) for all \(t\in [0,T]\) and therefore \(x(t)=0\) for all \(t\in [0,T]\). In particular \(x(0)=0\) and therefore the only unobservable state (in time \(T\)) is zero.
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Example 12.14. Consider the second order (band-pass) system
\[ \ddot {q}(t)+2\zeta \dot {q}(t)+q(t)=0, \]
where \(\zeta \geq 0\), the state is \(x=\sbm {q\\\dot {q}}\) and the output is \(y=\dot {q}\). Hence
\[ A=\bbm {0&1\\-1&-2\zeta },\qquad C=\bbm {0&1}. \]
We can show from the definition that this system is observable, but this is slightly more complicated than it was for Example 12.13. Assume that \(y(t)=0\) for all \(t\in [0,T]\), i.e. that \(\dot {q}(t)=0\) for all \(t\in [0,T]\). Then \(\ddot {q}(t)=0\) for all \(t\in [0,T]\). From the differential equation we then obtain that \(q(t)=0\) for all \(t\in [0,T]\). Hence \(x(t)=0\) for all \(t\in [0,T]\) and in particular \(x(0)=0\). We see that zero is the only state which is unobservable in time \(T\) and therefore that the system is observable.
We can also consider the Kalman observability matrix. We have
\[ CA=\bbm {-1&-2\zeta }, \]
so that
\[ \bbm {C\\CA}=\bbm {0&1\\-1&-2\zeta }, \]
which has determinant \(1\), which is nonzero, and therefore is invertible and therefore is injective. Hence we have observability.
The Hautus observability matrix is
\[ \bbm {s&-1\\1&s+2\zeta \\0&1}. \]
Picking the second and last rows gives the matrix
\[ \bbm {1&s+2\zeta \\0&1}, \]
which is upper-triangular with nonzero elements on the diagonal and is therefore invertible (no matter what \(s\in \mC \) is). It follows that the Hautus observability matrix is injective for all \(s\in \mC \) and from this we see that the system is observable.
If \(\zeta >0\), then \(A\) is asymptotically stable and we can consider the infinite-time observability Gramian. The observation Lyapunov equation is (using that \(S\) is symmetric)
\[ \bbm {0&-1\\1&-2\zeta } \bbm {S_1&S_0\\S_0&S_2} +\bbm {S_1&S_0\\S_0&S_2} \bbm {0&1\\-1&-2\zeta } +\bbm {0\\1}\bbm {0&1}=\bbm {0&0\\0&0}, \]
which is
\[ \bbm {-2S_0&-S_2+S_1-2\zeta S_0\\ -S_2+S_1-2\zeta S_0&2S_0-4\zeta S_2+1}=\bbm {0&0\\0&0}. \]
The top-left corner gives that \(S_0=0\). The bottom-right corner then gives that \(S_2=\frac {1}{4\zeta }\). The off-diagonal entry then gives \(S_1=\frac {1}{4\zeta }\). Hence
\[ S=\frac {1}{4\zeta }\bbm {1&0\\0&1}. \]
Since \(S\) is invertible, we see that the system is observable.
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Example 12.15. We can use the observability Gramian from Example 12.14 to compute \(\int _0^\infty |h(t)|^2\,dt\) where \(h\) is the impulse response of the system
\[ \ddot {q}(t)+2\zeta \dot {q}(t)+q(t)=u(t),\qquad y(t)=\dot {q}(t), \]
where \(\zeta >0\). By Remark 12.9 we have
\[ \int _0^\infty |h(t)|^2\,dt=\ipd {SB}{B}, \]
where
\[ S=\frac {1}{4\zeta }\bbm {1&0\\0&1},\qquad B=\bbm {0\\1}, \]
so we have
\[ \int _0^\infty |h(t)|^2\,dt=\frac {1}{4\zeta }. \]
In particular, we see that this gets smaller the larger \(\zeta \) is. See also Figure 7.2b.
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Example 12.16. We use the Kalman observability matrix to show that the state-output system with
\[ A=\bbm {1&1&1\\0&1&1\\0&0&1},\qquad C=\bbm {1&0&0}, \]
is observable. We have that the Kalman observability matrix is
\[ \bbm {C\\CA\\CA^2}=\bbm {1&0&0\\1&1&1\\1&2&3}. \]
Developing by the first row, we see that the determinant of this matrix is 1. Therefore the Kalman observability matrix is invertible and therefore injective. It follows that the system is observable.
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Example 12.17. We use the Kalman observability matrix to show that the state-output system with
\[ A=\bbm {0&0\\0&0},\qquad C=\bbm {1&0\\0&1}, \]
is observable. We have that the Kalman observability matrix is
\[ \bbm {C\\CA}=\bbm {1&0\\0&1\\0&0\\0&0}. \]
The first two rows are linearly independent (as they form the 2-by-2 identity matrix) and therefore the Kalman observability matrix is injective. It follows that the system is observable.
12.2 Case study: control of a tape drive
Since we measure the whole state, it is clear that the state is observable from the measurement. To make tape drives smaller, manufacturers are looking into removing the sensor which measure the tension in the tape. This would leave only the two velocity measurements. We then have
\[ A=\bbm { -\frac {d_1}{M_1}&0&\frac {1}{M_1}\\[1mm] 0&-\frac {d_2}{M_2}&-\frac {1}{M_2}\\ -k&k&0 },\qquad C=\bbm {1&0&0\\0&1&0}. \]
The Kalman observability matrix is (since \(n=3\))
\[ \bbm {C\\CA\\CA^2}. \]
We have
\[ CA=\bbm {-\frac {d_1}{M_1}&0&\frac {1}{M_1}\\[1mm] 0&-\frac {d_2}{M_2}&-\frac {1}{M_2}}, \]
so that
\[ \bbm {C\\CA}=\bbm { 1&0&0\\0&1&0\\ -\frac {d_1}{M_1}&0&\frac {1}{M_1}\\[1mm] 0&-\frac {d_2}{M_2}&-\frac {1}{M_2} }, \]
which has rank 3 (the first three rows form a lower triangular matrix with nonzero elements on the diagonal). Therefore the Kalman observability matrix is injective (note that we didn’t even need the \(CA^2\) rows) and the system is observable. So we have observability even in the absence of measurement of the tension in the tape.
If we measure only the tension in the tape, then
\[ C=\bbm {0&0&1}, \]
and we have
\[ \bbm {C\\CA\\CA^2} =\bbm {0&0&1\\-k&k&0\\k\frac {d_1}{M_1}&-k\frac {d_2}{M_1}&\frac {-k}{M_1}-\frac {k}{M_2}}, \]
whose determinant (e.g. developing by the first row) is \(k^2\left (\frac {d_2}{M_2}-\frac {d_1}{M_1}\right )\). Therefore if \(\frac {d_2}{M_2}=\frac {d_1}{M_1}\) (in particular if \(d_1=d_2\) and \(M_1=M_2\)), then we do not have observability. More precisely we see that in this case
\[ \bbm {1\\1\\0}, \]
is in the nullspace of the Kalman observability matrix and therefore is in the unobservable subspace. This means that if the two mass-damper systems are identical, then if the velocities of the two reels are the same, we cannot determine what that common velocity is from measurement of the tension in the tape.
In concordance with Theorem 10.4 we consider observability of \((\sbm {A&B_1C_e\\0&A_e},\bbm {C_2&D_{21}C_e})\). We have
\(\seteqnumber{0}{12.}{1}\)\begin{gather*} \widetilde {C}:=\bbm {C_2&D_{21}C_e}= \bbm { 1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&1&0&0 }, \\ \widetilde {A}:=\bbm {A&B_1C_e\\0&A_e} = \bbm { -\frac {d_1}{M_1}&0&\frac {1}{M_1}&0&0&0&0\\ 0&-\frac {d_2}{M_2}&-\frac {1}{M_2}&0&0&0&0\\ -k&k&0&0&0&k&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&1\\ 0&0&0&0&0&-\omega ^2&0 }. \end{gather*} We see that
\[ \widetilde {C}\widetilde {A} =\bbm { -\frac {d_1}{M_1}&0&\frac {1}{M_1}&0&0&0&0\\ 0&-\frac {d_2}{M_2}&-\frac {1}{M_2}&0&0&0&0\\ -k&k&0&0&0&k&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0 }. \]
The third row of \(\widetilde {C}\widetilde {A}\) times \(\widetilde {A}\) equals
\[ \bbm {k\frac {d_1}{M_1}&-k\frac {d_2}{M_2}&-\frac {k}{M_1}-\frac {k}{M_2}&0&0&0&k}. \]
Therefore the Kalman observability matrix includes all rows in the following matrix (we pick all rows from \(C\), the third row from \(\widetilde {C}\widetilde {A}\) and the above computed row from \(\widetilde {C}\widetilde {A}^2\)):
\[ \bbm { 1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&1&0&0\\ -k&k&0&0&0&k&0\\ k\frac {d_1}{M_1}&-k\frac {d_2}{M_2}&-\frac {k}{M_1}-\frac {k}{M_2}&0&0&0&k }, \]
and since this is lower-triangular with nonzero entries on the diagonal this matrix is invertible and therefore the Kalman observability matrix is injective and the pair \((\widetilde {A},\widetilde {C})\) is observable.
12.3 Case study: a suspension system
If we measure the whole state, then we certainly have observability. We consider the case where we only measure the suspension stroke and the difference of the velocity of the two masses. This gives
\[ C=\bbm {0&0&1&0\\0&1&0&-1},\qquad A=\bbm {0&k_{us}&0&0\\ \frac {-1}{m_{us}}&0&0&0\\ 0&1&0&-1\\ 0&0&0&0 }. \]
We then have
\[ CA=\bbm { 0&1&0&-1\\ \frac {-1}{m_{us}}&0&0&0 }, \]
and
\[ CA^2=\bbm { \frac {-1}{m_{us}}&0&0&0\\ 0&\frac {-k_{us}}{m_{us}}&0&0 }, \]
and
\[ CA^3=\bbm { 0&\frac {-k_{us}}{m_{us}}&0&0\\ \frac {k_{us}}{m_{us}^2}&0&0&0 }. \]
The Kalman observability matrix therefore is
\[ \bbm { 0&0&1&0\\ 0&1&0&-1\\ 0&1&0&-1\\ \frac {-1}{m_{us}}&0&0&0\\ \frac {-1}{m_{us}}&0&0&0\\ 0&\frac {-k_{us}}{m_{us}}&0&0\\ 0&\frac {-k_{us}}{m_{us}}&0&0\\ \frac {k_{us}}{m_{us}^2}&0&0&0 }. \]
Omitting duplicate rows gives the matrix
\[ \bbm { 0&0&1&0\\ 0&1&0&-1\\ \frac {-1}{m_{us}}&0&0&0\\ 0&\frac {-k_{us}}{m_{us}}&0&0\\ \frac {k_{us}}{m_{us}^2}&0&0&0 }. \]
Omitting the last row gives the matrix
\[ \bbm { 0&0&1&0\\ 0&1&0&-1\\ \frac {-1}{m_{us}}&0&0&0\\ 0&\frac {-k_{us}}{m_{us}}&0&0 }. \]
We develop the determinant by the first row:
\[ -\det \bbm { 0&1&-1\\ \frac {-1}{m_{us}}&0&0\\ 0&\frac {-k_{us}}{m_{us}}&0 }, \]
and the last row
\[ -\frac {k_{us}}{m_{us}}\det \bbm { 0&-1\\ \frac {-1}{m_{us}}&0 }=\frac {k_{us}}{m_{us}^2}. \]
We therefore see that the Kalman observability matrix is injective and therefore we obtain observability with only these two measurements.
12.4 Case study: a suspension system (optimal fixed structure)
We consider the fixed structure suspension system from Section 1.2.1. One way of choosing the constants \(d\) and \(k_s\) is by choosing those which minimize
\[ \int _0^\infty \|h(t)\|^2\,dt, \]
where \(h\) is the impulse response (from \(w\) to \(z\)). By Plancherel’s Theorem this also minimizes
\[ \int _{-\infty }^\infty \|F(\omega )\|^2\,d\omega , \]
where \(F\) is the frequency response (from \(w\) to \(z\)). From Remark 12.9 we can calculate the above integral in terms of the infinite-time observability Gramian. It is possible but very tedious to do this by hand, so we use computer algebra instead. The result is that
\[ \int _0^\infty \|h(t)\|^2\,dt= a_1d+a_2(k_s)d^{-1}, \]
with
\[ \begin {aligned} a_1&=r_1^2\frac {(m_s+m_{us})^2}{2k_{us}m_s^2} +\frac {k_{us}}{2m_s^2}, \\ a_2(k_s)&=r_1^2\frac {(m_s+m_{us})^3k_s^2-2k_{us}m_sm_{us}(m_s+m_{us})k_s+k_{us}^2m_s^2m_{us}}{2k_{us}^2m_s^2} \\&\qquad \qquad +r_2^2\frac {m_s+m_{us}}{2}+\frac {(m_s+m_{us})k_s^2}{2m_s^2}. \end {aligned} \]
Defining \(f(d,k_s)=a_1d+a_2(k_s)d^{-1}\), we have a function of two variables of which we want to find the minimum. Setting the partial derivative with respect to \(d\) equal to zero gives
\[ d=\sqrt {\frac {a_2(k_s)}{a_1}}. \]
Setting the partial derivative with respect to \(k_s\) equal to zero gives \(a_2'(k_s)=0\), which is
\[ r_1^2\frac {(m_s+m_{us})^3k_s-k_{us}m_sm_{us}(m_s+m_{us})}{k_{us}^2m_s^2} +\frac {(m_s+m_{us})k_s}{m_s^2}=0. \]
Solving this gives
\[ k_s=\frac {r_1^2k_{us}m_sm_{us}}{r_1^2(m_s+m_{us})^2+k_{us}^2}. \]
Substituting this back into the equation for \(d\) gives
\[ d=\frac {m_s\sqrt {r_1^4(m_s+m_{us})k_{us}m_{us}m_s+r_1^2m_{us}k_{us}^3+r_2^2(m_s+m_{us})k_{us}^3+r_1^2r_2^2(m_s+m_{us})^3k_{us}}}{r_1^2(m_s+m_{us})^2+k_{us}^2}. \]
The minimum value of \(f\) (i.e. of \(\int _0^\infty \|h(t)\|^2\,dt\)) is given by
\[ \frac {\sqrt {r_1^4(m_s+m_{us})k_{us}m_{us}m_s+r_1^2m_{us}k_{us}^3+r_2^2(m_s+m_{us})k_{us}^3+r_1^2r_2^2(m_s+m_{us})^3k_{us}}}{k_{us}m_s}. \]
We give the impulse responses in Figure 12.1a and the Bode magnitude diagrams in Figure 12.1b. Comparing these to Figures 7.3 and 6.9, we see that the output is far less oscillatory and the peaks in the Bode diagram are far less pronounced.
12.5 Problems
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(a) Consider the second order system
\[ \ddot {q}+q=0,\qquad y=q. \]
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(i) Write this in the standard form \(\dot {x}=Ax\), \(y=Cx\) with \(x=\sbm {q\\\dot {q}}\).
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(ii) Use the Kalman observability matrix to show that this system is observable.
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(iii) Use the Hautus observability matrix to show that this system is observable.
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Solution. We have
\[ A=\bbm {0&1\\-1&0},\qquad C=\bbm {1&0}. \]
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.
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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(b) Consider the second order system
\[ \ddot {q}+4\dot {q}+3q=0,\qquad y=q. \]
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(i) Write this in the standard form \(\dot {x}=Ax\), \(y=Cx\) with \(x=\sbm {q\\\dot {q}}\).
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(ii) 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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(iii) 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. We have
\[ A=\bbm {0&1\\-3&-4},\qquad C=\bbm {1&0}. \]
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.
For the input-state-output system we have \(B=\bbm {0\\1}\), so the
\[ \int _0^\infty |h(t)|^2\,dt=\ipd {SB}{B}=\frac {1}{24}. \]
□
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12.6 Further Problems
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• Consider two first order systems
\[ T_1\dot {x}_1+x_1=0,\qquad T_2\dot {x}_2+x_2=0, \]
where \(T_1,T_2>0\), with the output
\[ y=x_1+x_2. \]
Determine for which parameter values this system is observable. For those parameter values for which it is not observable, determine the unobservable subspace.
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Solution. We have
\[ A=\bbm {\frac {-1}{T_1}&0\\0&\frac {-1}{T_2}},\qquad C=\bbm {1&1}. \]
The Kalman observability matrix is
\[ \bbm {C\\CA}=\bbm {1&1\\\frac {-1}{T_1}&\frac {-1}{T_2}}, \]
which has determinant \(\left (\frac {1}{T_1}-\frac {1}{T_2}\right )\). We see that the Kalman observability matrix is injective if and only if \(T_1\neq T_2\). Therefore we have observability if and only if \(T_1\neq T_2\). If \(T_1=T_2\), then the nullspace of the Kalman observability matrix is spanned by
\[ \bbm {1\\-1}, \]
and this is the unobservable subspace. □