Chapter 21 \(H^2\) control: the measurement feedback case
We consider input-state-output systems with a state \(x:[0,\infty )\to \mR ^n\), external input \(w:[0,\infty )\to \mR ^{m_1}\), control input \(u:[0,\infty )\to \mR ^{m_2}\), performance output \(z:[0,\infty )\to \mR ^{p_1}\) and measured output \(y:[0,\infty )\to \mR ^{p_2}\) described by
\(\seteqnumber{0}{21.}{0}\)\begin{equation} \label {eq:H2:xzy} \dot {x}=Ax+B_1w+B_2u,\qquad z=C_1x+D_{12}u,\qquad y=C_2x+D_{21}w, \end{equation}
with the initial condition \(x(0)=x^0\) where \(x^0\in \mR ^n\) and
\[ A\in \mR ^{n\times n},~ B_1\in \mR ^{n\times m_1},~ B_2\in \mR ^{n\times m_2},~ C_1\in \mR ^{p_1\times n},~ D_{12}\in \mR ^{p_1\times m_2},~ C_2\in \mR ^{p_2\times n},~ D_{21}\in \mR ^{p_2\times m_1}. \]
-
Definition 21.1 (\(H^2\) measurement feedback problem). The objective is to find matrices
\[ A_c\in \mR ^{n_c\times n_c},~ B_c\in \mR ^{n_c\times p_2},~ C_c\in \mR ^{m_2\times n_c},~ D_c\in \mR ^{m_2\times p_2}, \]
such that with the control
\(\seteqnumber{0}{21.}{1}\)\begin{equation} \label {eq:H2:controller} \dot {x}_c=A_cx_c+B_cy,\quad u=C_cx_c+D_cy, \end{equation}
where \(x_c(0)=x_c^0\), we have
-
1. \(\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega \) is minimized amongst all such matrices,
-
2. for all \(x^0\) and \(x_c^0\) and \(w=0\) we have \(\lim _{t\to \infty }x(t)=0\) and \(\lim _{t\to \infty }x_c(t)=0\),
where \(F_{zw}\) is the frequency response of the combination of (21.1) and (21.2).
-
-
Theorem 21.3. Assume that \(D_{12}\) is injective, that \((A,B_2)\) is stabilizable, that the Rosenbrock matrix
\[ \bbm {sI-A&-B_2\\C_1&D_{12}}, \]
is injective for all \(s\in \mC \) with \(\re (s)=0\), that \(D_{21}\) is surjective, that \((A,C_2)\) is detectable and that the Rosenbrock matrix
\[ \bbm {sI-A&-B_1\\C_2&D_{21}}, \]
is surjective for all \(s\in \mC \) with \(\re (s)=0\).
Then there exists a unique solution \(X\) of the algebraic Riccati equation
\(\seteqnumber{0}{21.}{2}\)\begin{equation} \label {eq:RiccatiH2XX} A^*X+XA+C_1^*C_1-(XB_2+C_1^*D_{12})(D_{12}^*D_{12})^{-1}(B_2^*X+D_{12}^*C_1)=0, \end{equation}
such that \(A+BF\) is asymptotically stable where
\[ F=-(D_{12}^*D_{12})^{-1}\left (D_{12}^*C_1+B_2^*X\right ), \]
there exists a unique solution \(Y\) of the algebraic Riccati equation
\(\seteqnumber{0}{21.}{3}\)\begin{equation} \label {eq:RiccatiH2Y} AY+YA^*+B_1B_1^*-(YC_2^*+B_1D_{21}^*)(D_{21}D_{21}^*)^{-1}(C_2Y+D_{21}B_1^*)=0, \end{equation}
such that \(A-LC_2\) is asymptotically stable where
\[ L=(B_1D_{21}^*+YC_2^*)(D_{21}D_{21}^*)^{-1}. \]
These solutions \(X\) and \(Y\) are symmetric positive semidefinite. Furthermore,
\[ A_c=A+B_2F-LC_2,\qquad B_c=L,\qquad C_c=F,\qquad D_c=0, \]
is the unique solution of the \(H^2\) measurement feedback problem and
\[ \trace (B_1^*XB_1)+\trace (B_2^*XYXB_2) =\frac {1}{2\pi }\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega =\int _0^\infty \|h_{zw}(t)\|_{\HS }^2\,dt. \]
-
Remark 21.4. Note that the controller from Theorem 21.3 is an observer-based stabilizing controller.
21.1 Examples
-
Example 21.5. Consider the first order scalar differential equation
\[ \dot {x}(t)+x(t)=w_1(t)+u(t). \]
The measurement is
\[ y(t)=x_1(t)+\delta w_2(t), \]
where \(\delta >0\) and \(w_2\) has the interpretation of measurement error or unmodelled dynamics. The performance output is
\[ z=\bbm {x\\\varepsilon u}, \]
where \(\varepsilon >0\).
We can put this into the standard form (21.1) with \(n=1\), \(m_1=2\), \(m_2=1\), \(p_1=2\), \(p_2=1\),
\(\seteqnumber{0}{21.}{4}\)\begin{gather*} A=-1\qquad B_1=\bbm { 1&0},\qquad B_2=1,\qquad \\ C_1=\bbm {1\\0},\qquad D_{12}=\bbm {0\\\varepsilon },\qquad C_2=1,\qquad D_{21}=\bbm {0&\delta }. \end{gather*} The Riccati equation (21.3) we already solved in Example 20.7:
\[ X=\varepsilon ^2+\varepsilon \sqrt {\varepsilon ^2+1}, \qquad F=1-\sqrt {1+\varepsilon ^{-2}}. \]
The Riccati equation (21.4) is
\[ 2Y-1+Y^2\delta ^{-2}=0, \]
with symmetric positive semidefinite solution
\[ Y=\frac {-1+\sqrt {1+\delta ^{-2}}}{\delta ^{-2}} =-\delta ^2+\delta \sqrt {\delta ^2+1}. \]
We then have
\[ L=-1+\sqrt {1+\delta ^{-2}}. \]
The controller then is
\(\seteqnumber{0}{21.}{4}\)\begin{gather*} \dot {x}_c= \left (1-\sqrt {1+\varepsilon ^{-2}}-\sqrt {1+\delta ^{-2}}\right )x_c +\left (1-\sqrt {1+\delta ^{-2}}\right )y, \\ u=\left (-1+\sqrt {1+\varepsilon ^{-2}}\right )x_c. \end{gather*}
-
Example 21.6. Consider the undamped second order scalar differential equation
\[ \ddot {q}(t)+q(t)=w_1(t)+u(t), \]
with the state \(x=\sbm {q\\\dot {q}}\) and the performance output
\[ z=\bbm {\dot {q}\\\varepsilon u}, \]
where \(\varepsilon >0\), and with \(\delta >0\) the measured output
\[ y=\dot {q}+\delta w_2. \]
To put this into the \(H^2\) measurement feedback framework, we have \(n=2\), \(m_2=2\), \(m_2=1\), \(p_1=2\), \(p_2=1\) and
\(\seteqnumber{0}{21.}{4}\)\begin{gather*} A=\bbm {0&1\\-1&0},\quad B_1=\bbm {0&0\\1&0},\quad B_2=\bbm {0\\1},\quad C_1=\bbm {0&1\\0&0},\quad D_{12}=\bbm {0\\\varepsilon }, \\ C_2=\bbm {0&1},\quad D_{21}=\bbm {0&\delta }. \end{gather*} The Riccati equation (21.3) we already solved in Example 20.8, so that
\[ X=\varepsilon \bbm {1&0\\0&1},\qquad F=\bbm {0&-\varepsilon ^{-1}}. \]
The Riccati equation (21.4) is (using that \(Y\) is symmetric, that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\))
\(\seteqnumber{0}{21.}{4}\)\begin{multline*} \bbm {0&1\\-1&0}\bbm {Y_1&Y_0\\Y_0&Y_2} +\bbm {Y_1&Y_0\\Y_0&Y_2}\bbm {0&-1\\1&0} +\bbm {0&0\\1&0}\bbm {0&1\\0&0} \\-\delta ^{-2}\bbm {Y_1&Y_0\\Y_0&Y_2}\bbm {0\\1}\bbm {0&1}\bbm {Y_1&Y_0\\Y_0&Y_2} =\bbm {0&0\\0&0}, \end{multline*} which is
\[ \bbm {2Y_0&Y_2-Y_1\\Y_2-Y_1&-2Y_0} +\bbm {0&0\\0&1} -\delta ^{-2}\bbm {Y_0^2&Y_0Y_2\\Y_0Y_2&Y_2^2} =\bbm {0&0\\0&0}, \]
which is
\[ \bbm {2Y_0-\delta ^{-2}Y_0^2 &Y_2-Y_1-\delta ^{-2}Y_0Y_2 \\Y_2-Y_1-\delta ^{-2}Y_0Y_2 &-2Y_0+1-\delta ^{-2}Y_2^2 }=\bbm {0&0\\0&0}. \]
From the top-left corner we obtain \(Y_0=0\) or \(Y_0=2\delta ^2\).
We first consider the case \(Y_0=2\delta ^2\). The off-diagonal entry then gives
\[ -Y_1-Y_2=0, \]
which since \(Y_1,Y_2\geq 0\) (since \(Y\) is symmetric positive semidefinite) gives \(Y_1=Y_2=0\). The determinant of \(Y\) then equals \(-Y_0^2=-4\delta ^4<0\) which contradicts that \(Y\) is symmetric positive semidefinite. We can therefore ignore this case.
We now consider the case \(Y_0=0\). The bottom-right corner then gives \(Y_2=\delta \) and the off-diagonal entry gives \(Y_1=\delta \). Hence
\[ Y=\delta \bbm {1&0\\0&1}. \]
We then have (using again that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\)):
\[ L=\delta ^{-2}\delta \bbm {1&0\\0&1}\bbm {0\\1} =\bbm {0\\\delta ^{-1}}. \]
The controller then is
\[ \dot {x}_c=\bbm {0&1\\-1&-\varepsilon ^{-1}-\delta ^{-1}}x_c+\bbm {0\\\delta ^{-1}}y,\qquad u=\bbm {0&-\varepsilon ^{-1}}x_c. \]