We consider the input-state system
\(\seteqnumber{0}{13.}{0}\)\begin{equation} \label {eq:is2} \dot {x}(t)=Ax(t)+Bu(t). \end{equation}
Here \(A\in \mR ^{n\times n}\) and \(B\in \mR ^{n\times m}\).
Definition 13.1. The input-state system (13.1) (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.
Theorem 13.2. The following are equivalent:
1. The input-state system is stabilizable
2. The Hautus controllability matrix is surjective for all \(s\in \mC \) with \(\re (s)\geq 0\)
3. The Hautus controllability matrix is surjective for all eigenvalues \(s\) of \(A\) with \(\re (s)\geq 0\)
4. For all \(x^0\in \mR ^n\) there exists a control \(u\) such that the solution of (13.1) with \(x(0)=x^0\) satisfies \(\int _0^\infty \|x(t)\|^2+\|u(t)\|^2\,dt<\infty \).
Proposition 13.3. If an input-state system is controllable, then it is stabilizable.
We consider the state-output system
\(\seteqnumber{0}{13.}{1}\)\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 \(C\in \mR ^{p\times n}\).
Definition 13.4. The state-output system (13.2) (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.
Proposition 13.5. The pair \((A,C)\) is detectable if and only if the pair \((A^T,C^T)\) is stabilizable. Furthermore \(F=-L^T\) relates stabilizing feedbacks and output injections.
Theorem 13.6. The following are equivalent:
1. The state-output system is detectable
2. The Hautus observability matrix is injective for all \(s\in \mC \) with \(\re (s)\geq 0\)
3. The Hautus observability matrix is injective for all eigenvalues \(s\) of \(A\) with \(\re (s)\geq 0\)
Proposition 13.7. If a state-output system is observable, then it is detectable.
Definition 13.8. An observer for the state-output system
\[ \dot {x}(t)=Ax(t),\qquad y(t)=Cx(t), \]
where \(A\in \mR ^{n\times n}\) and \(C\in \mR ^{p\times n}\) is an input-state system
\[ \dot {x}_c(t)=A_cx_c(t)+B_cy(t), \]
where \(A_c\in \mR ^{n\times n}\) and \(B_c\in \mR ^{n\times p}\) such that for all \(x(0)\) and \(x_c(0)\) we have \(\lim _{t\to \infty }x(t)-x_c(t)=0\).
Let \(L\in \mR ^{n\times p}\) be such that \(A-LC\) is asymptotically stable. Then
\[ A_c=A-LC,\qquad B_c=L, \]
gives an observer. This is because we have
\[ \dot {x}_c=(A-LC)x_c+LCx, \]
so that
\[ \dot {x}-\dot {x}_c=Ax-(A-LC)x_c-LCx=(A-LC)(x-x_c), \]
so that \(z:=x-x_c\) satisfies \(\dot {z}=(A-LC)z\). Since \(A-LC\) is asymptotically stable, we have that \(\lim _{t\to \infty }z(t)=0\) for all \(z(0)\). It follows that \(\lim _{t\to \infty }x(t)-x_c(t)=0\) for all \(x(0)\) and \(x_c(0)\).
Remark 13.9. If we have the input-state-output system
\[ \dot {x}(t)=Ax(t)+Bu(t),\qquad y(t)=Cx(t), \]
then with the same reasoning we have that
\[ \dot {x}_c(t)=(A-LC)x_c(t)+Ly(t)+Bu(t), \]
is such that \(\lim _{t\to \infty }x(t)-x_c(t)=0\) for all \(x(0)\) and \(x_c(0)\) and all \(u\) (since \(Bu\) simply cancels out).
Remark 13.10. The idea is that the state \(x_c\) of the observer is an estimate of the state \(x\) based only on the measured output \(y\). Of course, by the above definition this is only necessarily a good estimate as \(t\to \infty \). However, by appropriately choosing the output injection \(L\), we can obtain a good estimate for all positive time.
Remark. The controller in Theorem 10.4 has (we use tildes for the matrices as in this section to distinguish them from those in Theorem 10.4)
\[ \widetilde {A}=\bbm {A&B_1C_e\\0&A_e},\qquad \widetilde {B}=\bbm {B_2\\0},\qquad \widetilde {C}=\bbm {C_2&D_{21}C_e}. \]
Hence it is an observer for
\[ \bbm {x\\x_e}'=\bbm {A&B_1C_e\\0&A_e}\bbm {x\\x_e}+\bbm {B_2\\0}u,\qquad y=\bbm {C_2&D_{21}C_e}\bbm {x\\x_e}, \]
which is the combination of the original system with the exo-system (with \(w\) eliminated). The control from Theorem 10.4 is
\[ u=\bbm {F_2&V-F_2\Pi }x_c, \]
rather than the full-information control
\[ u=\bbm {F_2&V-F_2\Pi }\bbm {x\\x_e}, \]
(i.e. the unknown \(\bbm {x\\x_e}\) is replaced by its estimate \(x_c\) obtained from the observer).
Example 13.11. 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.
Example 13.12. Consider two second order systems: one damped on which no control acts and one undamped on which a control acts:
\(\seteqnumber{0}{13.}{2}\)\begin{align*} \ddot {q}_1+2\zeta \dot {q}_1+q_1&=0,\\ \ddot {q}_2+q_2&=u, \end{align*} where \(\zeta >0\). We have
\[ x=\bbm {q_1\\\dot {q}_1\\q_2\\\dot {q}_2},\qquad A=\bbm {0&1&0&0\\-1&-2\zeta &0&0\\ 0&0&0&1\\0&0&-1&0},\qquad B=\bbm {0\\0\\0\\1}. \]
The control \(u=-2\zeta _2\dot {q}_2\) leads to the second equation becoming
\[ \ddot {q}_2+2\zeta _2\dot {q}_2+q_2=0, \]
and since this is asymptotically stable for \(\zeta _2>0\), and similarly the first equation above (for \(q_1\)) is asymptotically stable, we obtain asymptotic stability. This means that
\[ F=\bbm {0&0&0&-2\zeta _2}, \]
is stabilizing. We can also check that as follows; we have
\[ A+BF=\bbm {0&1&0&0\\-1&-2\zeta &0&0\\ 0&0&0&1\\0&0&-1&-2\zeta _2}, \]
which by the block diagonal structure has characteristic polynomial
\[ (s^2+2\zeta s+1)(s^2+2\zeta _2s+1), \]
and using that both terms in the product are stable, this is stable. Hence the pair \((A,B)\) is stabilizable whereas it is easy to see due to the block structure that it is not controllable.
We can also use the Hautus controllability matrix. We have that \(A\) has two stable eigenvalues (the roots of \(s^2+2\zeta s+1\)) and the two unstable eigenvalues \(\pm i\). We only have to consider the Hautus controllability matrix for the unstable eigenvalues and in fact only for one of the complex conjugate pair. The Hautus controllability matrix is:
\[ \bbm {sI-A&B}=\bbm {s&-1&0&0&0\\1&s+2\zeta &0&0&0\\ 0&0&s&-1&0\\0&0&1&s&1}. \]
Omitting the third column gives
\[ \bbm {s&-1&0&0\\1&s+2\zeta &0&0\\ 0&0&-1&0\\0&0&s&1}, \]
which by the block structure has determinant
\[ -(s^2+2\zeta s+1), \]
the roots of which are stable. Therefore, for all \(s\) with \(\re (s)\geq 0\) we have that this determinant is nonzero so that the matrix is invertible so that the Hautus controllability matrix is surjective. From Theorem 13.2 we then obtain that the system is stabilizable.
(a) Consider the input-state system \(\dot {x}=Ax+Bu\) with
\[ A=\bbm {-1&2\\0&1},\qquad B=\bbm {1\\1}. \]
Use the Hautus controllability matrix to show that this system is stabilizable, but not controllable.
Solution. The Hautus controllability matrix is
\[ \bbm {sI-A&B}=\bbm {s+1&-2&1\\0&s-1&1} \]
Choosing first and last columns gives a matrix with determinant \(s+1\). Therefore that matrix is invertible as long as \(s\neq -1\). It follows that the Hautus controllability matrix is surjective for \(s\neq -1\). In particular, it is surjective for all \(s\in \mC \) with \(\re (s)\geq 0\), so that the system is stabilizable.
When \(s=-1\) the Hautus controllability matrix is
\[ \bbm {0&-2&1\\0&-2&1}, \]
which identical rows and is therefore not surjective. It follows that the system is not controllable. □
(b) Consider the state-output system
\[ \dot {x}_1=x_1,\qquad \dot {x}_2=-x_2,\qquad y=x_1. \]
(i) Using only the definitions, show that this system is detectable, but not observable.
(ii) Use the Hautus observability matrix to show that this system is detectable, but not observable.
Solution. We have
\[ A=\bbm {1&0\\0&-1},\qquad C=\bbm {1&0}, \]
so that with \(L=\bbm {2\\0}\) we have
\[ A-LC=\bbm {-1&0\\0&-1}, \]
which is asymptotically stable (we immediately see that its eigenvalues are both \(-1\)). Therefore, the system is detectable.
We have \(x(t)=x^0\bbm {\e ^t\\\e ^{-t}}\), so that \(y(t)=x^0_1\e ^t\). Therefore the initial condition \(x^0=\sbm {0\\1}\) leads to \(y=0\). Hence the initial condition \(\sbm {0\\1}\) is unobservable. Hence there is a nonzero unobservable initial state and therefore the system is not observable.
The Hautus observability matrix is
\[ \bbm {sI-A\\C}=\bbm {s-1&0\\0&s+1\\1&0}. \]
The determinant of the matrix formed by the last two rows equals \(-(s+1)\). It follows that this matrix is invertible for all \(s\neq -1\). Therefore the Hautus observability matrix is injective for all \(s\neq -1\) and in particular for all \(s\in \mC \) with \(\re (s)\geq 0\). Hence the system is detectable. For \(s=-1\) the Hautus observability matrix has a zero column and is therefore not injective. Therefore the system is not observable. □
• Determine \(A\in \mR ^{1\times 1}\) and \(B\in \mR ^{1\times 1}\) such that \((A,B)\) is stabilizable, but not controllable.
Solution. Let \(A=-1\) and \(B=0\). Then \(A\) is asymptotically stable and therefore \((A,B)\) is stabilizable (\(F=0\) will do). However, since \(B=0\) the system is clearly not controllable (as the control doesn’t affect the state at all). □
• Determine \(A\in \mR ^{2\times 2}\) and \(B\in \mR ^{2\times 1}\) with \(B\) nonzero such that \((A,B)\) is stabilizable, but not controllable.
Solution. Let
\[ A=\bbm {-1&0\\0&1},\qquad B=\bbm {0\\1}. \]
Then \((A,B)\) is stabilizable since \(F=\bbm {0&-3}\) gives
\[ A+BF=\bbm {-1&0\\0&-2}, \]
which is asymptotically stable. However, the system is clearly not controllable since the control doesn’t influence the first component of the state (either directly or indirectly). □