Chapter 16 \(H^2\) control: measurement feedback
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
\(\seteqnumber{0}{16.}{0}\)\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
\[ A\in \mR ^{n\times n},~ B_1\in \mR ^{n\times m_1},~ B_2\in \mR ^{n\times m_2},~ C_1\in \mR ^{p_1\times n},~ D_{12}\in \mR ^{p_1\times m_2},~ C_2\in \mR ^{p_2\times n},~ D_{21}\in \mR ^{p_2\times m_1}. \]
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Definition 16.1 (\(H^2\) 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
\(\seteqnumber{0}{16.}{1}\)\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
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1. \(\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega \) is minimized amongst all such matrices,
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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 (16.1) and (16.2)
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Theorem 16.2. 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
\(\seteqnumber{0}{16.}{2}\)\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
\(\seteqnumber{0}{16.}{3}\)\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
\[ \trace (B_1^*XB_1)+\trace (B_2^*XYXB_2) =\frac {1}{2\pi }\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega =\int _0^\infty \|h_{zw}(t)\|_{\HS }^2\,dt. \]
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Remark 16.3. The Riccati equation (16.4) can be rewritten as the system of equations (these are called Lur’e equations) in the unknowns \(Y\), \(K\) and \(J\)
\(\seteqnumber{0}{16.}{4}\)\begin{align*} AY+YA^*+B_1B_1^*&=KK^*,\\ C_2Y+D_{21}B_1^*&=JK^*\\ D_{21}D_{21}^*&=JJ^*, \end{align*} and the differential equation (16.2) for the controller can be re-written as the differential-algebraic equation
\[ \dot {x}_c=(A+B_2F)x_c-Kq_c,\qquad Jq_c=C_2x_c-y. \]
If \(D_{21}\) is surjective, then we can eliminate \(J\) and \(K\) through
\[ J=(D_{21}D_{21}^*)^{1/2}, \qquad K=(B_1D_{21}^*+YC_2^*)(D_{21}D_{21}^*)^{-1/2}, \]
and the first Lur’e equation then becomes the Riccati equation.
The Lur’e equations continue to be well-defined even if \(D_{21}\) is not surjective. It is possible to use the Lur’e equations to formulate a theorem similar to Theorem 16.2 valid also for the case where \(D_{21}\) may not be surjective, but the precise additional conditions become technical.
16.1 Examples
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Example 16.5. Consider the first order system
\[ \dot {x}(t)+x(t)=w_1(t)+u(t). \]
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 (16.1) with \(n=1\), \(m_1=2\), \(m_2=1\), \(p_1=2\), \(p_2=1\),
\(\seteqnumber{0}{16.}{4}\)\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 (16.3) we already solved in Example 15.10:
\[ X=\varepsilon ^2+\varepsilon \sqrt {\varepsilon ^2+1}, \qquad F=1-\sqrt {1+\varepsilon ^{-2}}. \]
The Riccati equation (16.4) 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
\(\seteqnumber{0}{16.}{4}\)\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*} In Figure 16.1 we give the Bode magnitude diagram of the system with this controller (for various values of \(\varepsilon \) and \(\delta \)).
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Example 16.6. Consider the undamped second order system
\[ \ddot {q}(t)+q(t)=w_1(t)+u(t), \]
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
\(\seteqnumber{0}{16.}{4}\)\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 (16.3) we already solved in Example 15.11, so that
\[ X=\varepsilon \bbm {1&0\\0&1},\qquad F=\bbm {0&-\varepsilon ^{-1}}. \]
The Riccati equation (16.4) is (using that \(Y\) is symmetric, that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\))
\(\seteqnumber{0}{16.}{4}\)\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 semi-definite) 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 semi-definite. 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
\[ \dot {x}_c=\bbm {0&1\\-1&-\varepsilon ^{-1}-\delta ^{-1}}x_c+\bbm {0\\\delta ^{-1}}y,\qquad u=\bbm {0&-\varepsilon ^{-1}}x_c. \]
16.2 Problems
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(a) Consider the undamped second order system
\[ \ddot {q}(t)+q(t)=w_1(t)+u(t), \]
with the state \(x=\sbm {q\\\dot {q}}\) and the performance output
\[ z=\bbm {q\\\varepsilon u}, \]
where \(\varepsilon >0\), and with \(\delta >0\) the measured output
\[ y=q+\delta w_2. \]
i.e.
\(\seteqnumber{0}{16.}{4}\)\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 {1&0\\0&0},\quad D_{12}=\bbm {0\\\varepsilon }, \\ C_2=\bbm {1&0},\quad D_{21}=\bbm {0&\delta }. \end{gather*}
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(i) Determine whether or not \(D_{21}\) is surjective.
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(ii) Determine whether or not \((A,C_2)\) is detectable.
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(iii) Determine whether or not the Rosenbrock surjectivity condition holds.
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(iv) Solve the \(H^2\) measurement feedback problem for this system.
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Solution. (i) Since \(\delta >0\) we have that \(D_{21}\) is surjective (picking the second column given a 1-by-1 invertible matrix).
(ii) From a problem in Section 12.5 we have that \((A,C_2)\) is observable and therefore detectable.
(iii) The relevant Rosenbrock matrix is
\[ \bbm {sI-A&-B_1\\C_2&D_{21}} =\bbm {s&-1&0&0\\1&s&-1&0\\1&0&0&\delta }. \]
The final three columns form a lower-triangular matrix with nonzero entries on the diagonal and therefore has nonzero determinant. Therefore the final three columns are linearly independent. Therefore the Rosenbrock matrix has rank 3 and is therefore surjective (for all \(s\in \mC \) in fact).
(iv) In Section 14.2 we already solved the Riccati equation (16.3) and obtained
\(\seteqnumber{0}{16.}{4}\)\begin{gather*} X=\varepsilon ^2\bbm { \sqrt {1+\varepsilon ^{-2}}~ \sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} & -1+\sqrt {1+\varepsilon ^{-2}} \\ -1+\sqrt {1+\varepsilon ^{-2}} &\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} }, \\ F=\bbm { 1-\sqrt {1+\varepsilon ^{-2}} &-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}} }. \end{gather*} The Riccati equation (16.4) is (using that \(Y\) is symmetric, that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\))
\(\seteqnumber{0}{16.}{4}\)\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 {1\\0}\bbm {1&0}\bbm {Y_1&Y_0\\Y_0&Y_2} =\bbm {0&0\\0&0}, \end{multline*} which is
\[ \bbm {2Y_0&Y_2-Y_1\\Y_2-Y_1&-2Y_0} +\bbm {0&0\\0&1} -\delta ^{-2}\bbm {Y_1^2&Y_0Y_1\\Y_0Y_1&Y_0^2} =\bbm {0&0\\0&0}, \]
which is
\[ \bbm {2Y_0-\delta ^{-2}Y_1^2 &Y_2-Y_1-\delta ^{-2}Y_0Y_1 \\Y_2-Y_1-\delta ^{-2}Y_0Y_1 &-2Y_0+1-\delta ^{-2}Y_0^2 }=\bbm {0&0\\0&0}. \]
From the top-left corner we obtain that \(Y_0\geq 0\). From the bottom-right corner we then obtain (using that \(Y_0\geq 0\) to pick the correct solution):
\[ Y_0=-\delta ^2+\delta ^2\sqrt {1+\delta ^{-2}}. \]
The top-left corner then gives (using that \(Y_1\geq 0\) to pick the correct sign)
\[ Y_1=\delta ^2\sqrt {-2+2\sqrt {1+\delta ^{-2}}}. \]
The off-diagonal entry then gives
\[ Y_2=\sqrt {1+\delta ^{-2}}~\delta ^2\sqrt {-2+2\sqrt {1+\delta ^{-2}}}. \]
Hence
\[ Y=\delta ^2\bbm { \sqrt {-2+2\sqrt {1+\delta ^{-2}}}& -1+\sqrt {1+\delta ^{-2}}\\ -1+\sqrt {1+\delta ^{-2}}& \sqrt {1+\delta ^{-2}}~\sqrt {-2+2\sqrt {1+\delta ^{-2}}} }. \]
We then have (using again that \(B_1D_{21}^*=0\) and \(D_{21}D_{21}^*=\delta ^2\)):
\(\seteqnumber{0}{16.}{4}\)\begin{align*} L&=\delta ^{-2} \delta ^2\bbm { \sqrt {-2+2\sqrt {1+\delta ^{-2}}}& -1+\sqrt {1+\delta ^{-2}}\\ -1+\sqrt {1+\delta ^{-2}}& \sqrt {1+\delta ^{-2}}~\sqrt {-2+2\sqrt {1+\delta ^{-2}}} } \bbm {1\\0} \\&=\bbm { \sqrt {-2+2\sqrt {1+\delta ^{-2}}}\\ -1+\sqrt {1+\delta ^{-2}}}. \end{align*} The controller then is
\(\seteqnumber{0}{16.}{4}\)\begin{gather*} \dot {x}_c=\bbm {-\sqrt {2+2\sqrt {1+\delta ^{-2}}}&1\\1-\sqrt {1+\varepsilon ^{-2}}-\sqrt {1+\delta ^{-2}}&-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}}}x_c +\bbm {\sqrt {-2+2\sqrt {1+\delta ^{-2}}}\\ -1+\sqrt {1+\delta ^{-2}}}y,\\ u=\bbm { 1-\sqrt {1+\varepsilon ^{-2}} &-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}}}}x_c. \end{gather*} □