Chapter 10 The Rosenbrock matrix and solvability of the regulator equations
-
Proof. Assume that \(G(s)\) is surjective. Let \(q\in \mR ^n\) and \(r\in \mR ^p\) be arbitrary. We then have
\[ \bbm {q\\r}=\bbm {sI-A&-B\\C&D}\bbm {x\\u}, \]
where \(u\) is a solution of
\[ G(s)u=r-C(sI-A)^{-1}q, \]
(which exists by our surjectivity assumption) and
\[ x=(sI-A)^{-1}q+(sI-A)^{-1}Bu, \]
since
\[ \bbm {sI-A&-B\\C&D}\bbm {(sI-A)^{-1}q+(sI-A)^{-1}Bu\\u} =\bbm {q+Bu-Bu\\C(sI-A)^{-1}q+C(sI-A)^{-1}Bu+Du}=\bbm {q\\r}. \]
Assume that the Rosenbrock matrix is surjective. Let \(r\in \mR ^p\). Then there exist \(x\in \mR ^n\) and \(u\in \mR ^m\) such that
\[ \bbm {0\\r}=\bbm {sI-A&-B\\C&D}\bbm {x\\u}. \]
It follows that \(x=(sI-A)^{-1}Bu\) and subsequently that \(r=Cx+Du=C(sI-A)^{-1}Bu+Du=G(s)u\). Hence \(G(s)\) is surjective.
Assume that \(G(s)\) is injective. Let \(x\in \mR ^n\) and \(u\in \mR ^m\) be such that
\[ \bbm {sI-A&-B\\C&D}\bbm {x\\u}=\bbm {0\\0}. \]
This means
\[ (sI-A)x-Bu=0,\qquad Cx+Du=0. \]
From the first of these equations we obtain \(x=(sI-A)^{-1}Bu\), which substituted in the second equation gives \(C(sI-A)^{-1}Bu+Du=0\), i.e. \(G(s)u=0\). By the injectivity assumption we then have \(u=0\). From the first of the above equations we then have \((sI-A)x=0\) which since \(s\) is not an eigenvalue of \(A\) gives \(x=0\). Therefore the Rosenbrock matrix for this \(s\) is injective.
Assume that the Rosenbrock matrix is injective. Let \(G(s)u=0\). Then
\[ \bbm {sI-A&-B\\C&D}\bbm {(sI-A)^{-1}Bu\\u}=\bbm {0\\0}. \]
By the injectivity assumption it follows that \(\sbm {(sI-A)^{-1}Bu\\u}=0\) and in particular that \(u=0\). Hence \(G(s)\) is injective. □
-
Remark 10.3. We can re-write the regulator equations as
\[ \bbm {I&0\\0&0}\bbm {\Pi \\V}A_e-\bbm {A&B_2\\C_1&0}\bbm {\Pi \\V}=\bbm {B_1\\D_{11}}C_e, \]
which shows that the regulator equations are a generalized Sylvester equation.
Using the Kronecker product and the vectorization of a matrix, this can in turn be written as the standard linear system
\(\seteqnumber{0}{10.}{0}\)\begin{equation} \label {eq:Kronecker} \left (A_e^T\otimes \bbm {I_n&0_{n\times m_2}\\0_{p_1\times n}&0_{p_1\times m_2}} -I_{n_e}\otimes \bbm {A&B_2\\C_1&0_{p_1\times m_2}}\right ) \vecc \left (\bbm {\Pi \\V}\right ) =\vecc \left (\bbm {B_1C_e\\D_{11}C_e}\right ). \end{equation}
-
Remark 10.5. A partial converse of Proposition 10.4 is also true in case \(A_e\) has only eigenvalues with nonnegative real part: if the regulator equations are solvable for all \(B_1\), \(D_{11}\) and \(C_e\), then the surjectivity condition on the Rosenbrock matrix \(\sbm {sI-A&-B_2\\C_1&0}\) must hold. This relates to the Kronecker form (10.1) where \(B_1\), \(D_{11}\) and \(C_e\) appear on the right-hand side.
10.1 Examples
-
Example 10.7. Consider the second order scalar differential equation
\[ \ddot {q}(t)+2\zeta \dot {q}(t)+q(t)=u(t), \]
where \(\zeta >0\). Let \(x:=\sbm {q\\\dot {q}}\). The objective is to find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }\dot {q}(t)=r\) no matter what \(x(0)\) is.
We initially define the performance output and exo-system
\[ z=x_2-w,\qquad \dot {x}_e=0,\qquad w=x_e,\qquad x_e(0)=r, \]
since then the regulation requirement becomes what we desire.
We have \(n=2\), \(m_1=m_2=p_1=1\) and
\[ x=\bbm {q\\\dot {q}},\quad A=\bbm {0&1\\-1&-2\zeta },\quad B_1=\bbm {0\\0},\quad B_2=\bbm {0\\1},\quad C_1=\bbm {0&1},\quad D_{11}=-1. \]
We note that the transfer function \(C_1(sI-A)^{-1}B_2\) is (by Chapter 5)
\[ G(s)=\frac {s}{s^2+2\zeta s+1}, \]
which has a zero at \(s=0\) which is an eigenvalue of \(A_e\). Therefore by Corollary 10.6 we expect that there might be an issue with solvability of the problem.
The regulator equations are
\[ \bbm {0&1\\-1&-2\zeta }\bbm {\Pi _1\\\Pi _2} +\bbm {0\\1}V =0,\qquad \bbm {0&1}\bbm {\Pi _1\\\Pi _2}-1=0. \]
It follows from the second equation that \(\Pi _2=1\) so that the first equation is
\[ \bbm {1\\-\Pi _1-2\zeta +V}=\bbm {0\\0}. \]
We see that the first component of this gives the contradiction \(1=0\). Therefore the regulator equations do not have a solution (as we expected might be the case).
We instead consider
\[ n_e=2,\qquad A_e=\bbm {0&1\\0&0},\qquad C_e=\bbm {0&1}. \]
The regulator equations are
\(\seteqnumber{0}{10.}{1}\)\begin{gather*} \bbm {0&1\\-1&-2\zeta }\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}} +\bbm {0\\1}\bbm {V_1&V_2} =\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}\bbm {0&1\\0&0}, \\ \bbm {0&1}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}-\bbm {0&1}=0. \end{gather*} The second of these equations give \(\Pi _{21}=0\), \(\Pi _{22}=1\). The first equation then is
\[ \bbm {0&1\\-1&-2\zeta }\bbm {\Pi _{11}&\Pi _{12}\\0&1} +\bbm {0&0\\V_1&V_2} =\bbm {0&\Pi _{11}\\0&0}, \]
which is
\[ \bbm {0&1\\-\Pi _{11}&-\Pi _{12}-2\zeta } +\bbm {0&0\\V_1&V_2} =\bbm {0&\Pi _{11}\\0&0}, \]
from which we deduce \(\Pi _{11}=1\), \(V_1=1\) and \(\Pi _{12}=V_2-2\zeta \). We see that now the regulator equations have a solution, but that this solution is not unique (as \(V_2\in \mR \) is arbitrary). We have
\[ \Pi =\bbm {1&V_2-2\zeta \\0&1},\qquad V=\bbm {1&V_2}. \]
Since \(A\) is asymptotically stable, we can take \(F_1=0\) and we obtain the control
\[ u=x_{e,1}+V_2x_{2,e}. \]
With the initial condition \(x_e^0=\sbm {0\\r}\) we obtain \(x_e(t)=\sbm {rt\\r}\) and \(w(t)=r\), so that the above control is
\[ u(t)=rt+V_2r, \]
where \(V_2\in \mR \) is an arbitrary constant. We note that this arbitrary constant arises because the transfer function from \(u\) to \(z\) has a zero at \(s=0\), so that constant inputs result in a zero output (therefore the arbitrary term \(V_2r\) in \(u\) does not affect the output and hence for the purposes of output regulation is indeed irrelevant).
-
\[ \dot {x}=u. \]
The objective is to find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }x(t)=r\) no matter what \(x(0)\) is. We have \(z=x-w\)
\[ n=1,\quad A=0,\quad B_1=0,\quad B_2=1,\quad C_1=1,\quad D_{11}=-1. \]
We note that zero is an eigenvalue of \(A\) which will also be an eigenvalue of \(A_e\). Therefore Corollary 10.6 will not be applicable, but Proposition 10.4 is.
We choose the exo-system
\[ n_e=1,\quad A_e=0,\qquad C_e=1. \]
The Rosenbrock matrix is
\[ \bbm {sI-A&-B_2\\C_1&0}=\bbm {s&-1\\1&0}, \]
which has determinant 1 and therefore is invertible and therefore surjective for all \(s\in \mC \) (in particular: for all eigenvalues of \(A_e\)). Therefore by Proposition 10.4 we have solvability of the regulator equations (and therefore of the output regulation problem).
The regulator equations are
\[ V=0,\quad \Pi -1=0, \]
which gives \(\Pi =1\) and \(V=0\). Since \(A\) is not asymptotically stable, we have to choose a nonzero \(F_1\); we have \(A+B_2F_1=F_1\) so any \(F_1<0\) will work. We choose \(F_1=-1\). The feedback then is
\[ u=-x+x_e. \]
We check directly that this indeed solves the problem. We have
\[ \dot {x}=-x+x_e,\quad \dot {x}_e=0,\quad x(0)=r,\quad w=x_e. \]
This gives \(w=x_e=r\) and therefore
\[ \dot {x}=-x+r. \]
The solution of this with \(x(0)=x^0\) is \(x(t)=r+(x^0-r)\e ^{-t}\), which indeed satisfies \(\lim _{t\to \infty }x(t)=r\) no matter what \(x(0)\) is.