Chapter 19 \(H^\infty \) control: state 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 \(u:[0,\infty )\to \mR ^{m_2}\) and performance output \(z:[0,\infty )\to \mR ^{p_1}\) described by
\(\seteqnumber{0}{19.}{0}\)\begin{equation} \label {eq:Hinf:xz} \dot {x}=Ax+B_1w+B_2u,\qquad z=C_1x+D_{12}u, \end{equation}
with the initial condition \(x(0)=x^0\) where \(x^0\in \mR ^n\) and
\(\seteqnumber{0}{19.}{1}\)\begin{equation} \label {eq:Hinf:xzmatrices} 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}. \end{equation}
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Definition 19.1 (\(H^\infty \) state feedback problem). The objective is to, for a given \(\gamma >0\), find a matrix
\[ F\in \mR ^{m_2\times n}, \]
such that with the control \(u=Fx\) we have
\[ \sup _{\omega \in \mR } \|F_{zw}(\omega )\|<\gamma , \]
where \(F_{zw}\) is the frequency response of the closed-loop system
\(\seteqnumber{0}{19.}{2}\)\begin{equation} \label {eq:closedHinfty} \dot {x}=(A+B_2F)x+B_1w,\qquad z=(C_1+D_{12}F)x, \end{equation}
and such that for all \(x^0\in \mR ^n\) and \(w=0\) we have \(\lim _{t\to \infty }x(t)=0\).
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Remark 19.2. The matrix norm considered in Definition 19.1 differs from that considered in the \(H^2\) problem. Here we consider the operator or spectral norm. For a matrix \(M\) this is given by
\[ \max _{\|v\|=1}\|Mv\|, \]
where the vector norms are the Euclidean norms. This is also equal to the square root of the largest eigenvalue of \(M^*M\).
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Theorem 19.3. Assume that \(D_{12}\) is injective and that the Rosenbrock matrix
\[ \bbm {sI-A&-B_2\\C_1&D_{12}} \]
is injective for all \(s\in \mC \) with \(\re (s)=0\). Let \(\gamma >0\). The following are equivalent:
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1. The \(H^\infty \) state feedback problem as given in Definition 19.1 is solvable for this \(\gamma \).
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2. There exists a symmetric positive semi-definite \(X\in \mR ^{n\times n}\) which solves the algebraic Riccati equation
\(\seteqnumber{0}{19.}{3}\)\begin{equation} \label {eq:X:Hinfty} A^*X+XA+C_1^*C_1+\gamma ^{-2}XB_1B_1^*X-(XB_2+C_1^*D_{12})(D_{12}^*D_{12})^{-1}(B_2^*X+D_{12}^*C_1)=0, \end{equation}
and is such that \(A+B_2F+\gamma ^{-2}B_1B_1^*X\) is asymptotically stable, where
\(\seteqnumber{0}{19.}{4}\)\begin{equation} \label {eq:F:Hinfty} F=-(D_{12}^*D_{12})^{-1}\left (D_{12}^*C_1+B_2^*X\right ). \end{equation}
If \(X\) is as in (2.), then a solution \(F\) of the \(H^\infty \) state feedback problem is given by (19.5). Moreover, we have that \(A+B_2F\) is asymptotically stable.
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Remark 19.4. Careful consideration of Theorem 19.3 shows that if for a given \(\gamma _0>0\) the Riccati equation (19.4) has a symmetric positive semi-definite solution such that \(A+B_2F+\gamma _0^{-2}B_1B_1^*X\) is asymptotically stable, then it has such a solution for all \(\gamma >\gamma _0\). On the other hand, if for a given \(\gamma _0\) such a solution does not not exist, then it does not exist for any \(\gamma <\gamma _0\) either.
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Remark 19.5. At first sight it might seem strange that the asymptotic stability condition in Theorem 19.3 is on \(A+B_2F+\gamma ^{-2}B_1B_1^*X\) rather than on the closed-loop matrix \(A+B_2F\). There is a good explanation for this. The worst case disturbance equals \(w=\gamma ^{-2}B_1^*Xx\) and the asymptotic stability condition therefore corresponds to \(u\) being optimal and \(w\) being worst case.
19.1 Examples
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Example 19.6. Consider the first order system
\[ \dot {x}(t)+x(t)=w(t)+u(t). \]
For the \(H^\infty \) state feedback problem, we consider the performance output
\[ z(t)=\bbm {x(t)\\\varepsilon u(t)}, \]
where \(\varepsilon >0\). To put this into the \(H^\infty \) state feedback framework, we have \(n=m_1=m_2=1\), \(p_1=2\) and
\[ A=-1,\quad B_1=1,\quad B_2=1,\quad C_1=\bbm {1\\0},\quad D_{12}=\bbm {0\\\varepsilon }. \]
The Riccati equation (19.4) is (using that \(C_1^*D_{12}=0\) and \(D_{12}^*D_{12}=\varepsilon ^2\))
\[ -2X+1+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X^2=0. \]
This quadratic equation has the two (possible complex) solutions
\[ X=\frac {-1\pm \sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}}{\varepsilon ^{-2}-\gamma ^{-2}} =\frac {-\varepsilon ^2\gamma ^2\pm \varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2}. \]
To obtain a real solution we need
\[ \varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2\geq 0, \]
i.e.
\[ \gamma \geq \frac {\varepsilon }{\sqrt {\varepsilon ^2+1}}. \]
There is the special case where \(\varepsilon =\gamma \) which gives the one solution
\[ X=\frac {1}{2}. \]
If \(\gamma >\varepsilon \) then there is only one symmetric positive semi-definite solution:
\[ \frac {-\varepsilon ^2\gamma ^2+\varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2}. \]
If \(\gamma <\varepsilon \) then both solutions are symmetric positive semi-definite; we have
\[ F=-\varepsilon ^{-2}~ \frac {-\varepsilon ^2\gamma ^2\pm \varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2} \]
and
\(\seteqnumber{0}{19.}{5}\)\begin{align*} A+B_2F+\gamma ^{-2}B_1B_1^*X &=-1 +\left (\gamma ^{-2}-\varepsilon ^{-2}\right )~ \frac {-\varepsilon ^2\gamma ^2\pm \varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2} \\&=-1 -\frac {-\varepsilon ^2\gamma ^2\pm \varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\varepsilon ^2\gamma ^2} \\&=-\pm \frac {\sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\varepsilon \gamma }, \end{align*} so that we need the plus sign for asymptotic stability of \(A+B_2F+\gamma ^{-2}B_1B_1^*X\) and we also need \(\gamma >\frac {\varepsilon }{\sqrt {\varepsilon ^2+1}}\) rather than just \(\gamma \geq \frac {\varepsilon }{\sqrt {\varepsilon ^2+1}}\) as established above. Therefore we have
\[ X= \frac {-\varepsilon ^2\gamma ^2+\varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2}. \]
The optimal feedback is
\[ F= -\varepsilon ^{-2}~\frac {-\varepsilon ^2\gamma ^2+\varepsilon \gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2} =\frac {\gamma ^2-\varepsilon ^{-1}\gamma \sqrt {\varepsilon ^2\gamma ^2+\gamma ^2-\varepsilon ^2}}{\gamma ^2-\varepsilon ^2} \]
For the special case \(\gamma =\varepsilon \) we instead have \(X=\frac {1}{2}\), \(F=-\frac {1}{2\varepsilon ^2}\).
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Example 19.7. Consider the undamped second order system
\[ \ddot {q}(t)+q(t)=w(t)+u(t), \]
with the state \(x=\sbm {q\\\dot {q}}\) and the performance output
\[ z=\bbm {\dot {q}\\\varepsilon u}, \]
where \(\varepsilon >0\). To put this into the \(H^\infty \) state feedback framework, we have \(n=2\), \(m_1=m_2=1\), \(p_1=2\) and
\[ A=\bbm {0&1\\-1&0},\quad B_1=\bbm {0\\1},\quad B_2=\bbm {0\\1},\quad C_1=\bbm {0&1\\0&0},\qquad D_{12}=\bbm {0\\\varepsilon }. \]
The Riccati equation is (using that \(X\) is symmetric and that \(C_1^*D_{12}=0\) and \(D_{12}^*D_{12}=\varepsilon ^2\))
\(\seteqnumber{0}{19.}{5}\)\begin{multline*} \bbm {0&-1\\1&0}\bbm {X_1&X_0\\X_0&X_2} +\bbm {X_1&X_0\\X_0&X_2}\bbm {0&1\\-1&0} +\bbm {0&0\\1&0}\bbm {0&1\\0&0} \\+\gamma ^{-2}\bbm {X_1&X_0\\X_0&X_2}\bbm {0\\1}\bbm {0&1}\bbm {X_1&X_0\\X_0&X_2} -\varepsilon ^{-2}\bbm {X_1&X_0\\X_0&X_2}\bbm {0\\1}\bbm {0&1}\bbm {X_1&X_0\\X_0&X_2}=\bbm {0&0\\0&0}, \end{multline*} which is
\[ \bbm {-2X_0&X_1-X_2\\X_1-X_2&2X_0} +\bbm {0&0\\0&1} +\left (\gamma ^{-2}-\varepsilon ^{-2}\right )\bbm {X_0^2&X_0X_2\\X_0X_2&X_2^2}=\bbm {0&0\\0&0}, \]
which is
\[ \bbm { -2X_0+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_0^2 &X_1-X_2+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_0X_2 \\X_1-X_2+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_0X_2 &2X_0+1+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_2^2 }=\bbm {0&0\\0&0}. \]
The top-left corner gives \(X_0=0\) or \(X_0=2\left (\gamma ^{-2}-\varepsilon ^{-2}\right )^{-1}\).
We first consider the case \(X_0=2\left (\gamma ^{-2}-\varepsilon ^{-2}\right )^{-1}\). The off-diagonal entry then gives
\[ X_1+X_2=0. \]
Since \(X_1,X_2\geq 0\) (because \(X\) is symmetric positive semi-definite), this implies \(X_1=X_2=0\). The determinant of \(X\) then equals \(-X_0^2\) which is negative which contradicts that \(X\) is symmetric positive semi-definite. Therefore this case must be excluded.
We now consider the case \(X_0=0\). Then the bottom-right corner gives that we must have
\[ \gamma >\varepsilon , \]
and that \(X_2=\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{-1/2}\) and finally the off-diagonal entry gives that \(X_1=\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{-1/2}\). It follows that
\[ X=\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{-1/2}\bbm {1&0\\0&1}. \]
The feedback then is (again using that \(C_1^*D_{12}=0\) and \(D_{12}^*D_{12}=\varepsilon ^2\)):
\[ F=-\varepsilon ^{-2}\bbm {0&1}\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{-1/2}\bbm {1&0\\0&1} =\bbm {0&-\varepsilon ^{-2}\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{-1/2}}. \]
We have
\(\seteqnumber{0}{19.}{5}\)\begin{align*} A+B_2F+\gamma ^{-2}B_1B_1^*X &=\bbm {0&1\\-1&0} +\bbm {0&0\\0& \left (\gamma ^{-2}-\varepsilon ^{-2}\right )\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{-1/2}} \\ &=\bbm {0&1\\-1& -\left (\varepsilon ^{-2}-\gamma ^{-2}\right )^{1/2}}, \end{align*} which is stable under the condition \(\gamma >\varepsilon \) which we already saw above.
In Figure 19.1 we plot the Bode magnitude diagrams for both the \(H^2\) and the \(H^\infty \) closed-loop systems. We see that the maximum is lower in the \(H^\infty \) case (because this is what it is designed to do) whereas “on average” the \(H^2\) closed-loop system is better (because that is what it is designed to do).
19.2 Problems
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(a) Consider the undamped second order system
\[ \ddot {q}(t)+q(t)=w(t)+u(t), \]
with the state \(x=\sbm {q\\\dot {q}}\) and the performance output
\[ z=\bbm {q\\\varepsilon u}, \]
where \(\varepsilon >0\).
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(i) Write this in the standard form (15.1).
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(ii) Determine whether or not \(D_{12}\) is injective.
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(iii) Determine whether or not \((A,B_2)\) is stabilizable.
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(iv) Determine whether or not the Rosenbrock injectivity condition holds.
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(v) Solve the \(H^\infty \) state feedback problem for this system.
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(vi) Interpret the closed-loop system as a second order system and compare it to the uncontrolled system.
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Solution. We have
\(\seteqnumber{0}{19.}{5}\)\begin{gather*} A=\bbm {0&1\\-1&0},\quad B_1=\bbm {0\\1},\quad B_2=\bbm {0\\1},\quad C_2=\bbm {1&0\\0&0},\quad D_{21}=\bbm {0\\\varepsilon }. \end{gather*}
(ii), (iii), (iv) These conditions were already checked in Section 14.2.
(v) The Riccati equation is (using that \(X\) is symmetric and that \(C_1^*D_{12}=0\) and \(D_{12}^*D_{12}=\varepsilon ^2\))
\(\seteqnumber{0}{19.}{5}\)\begin{multline*} \bbm {0&-1\\1&0}\bbm {X_1&X_0\\X_0&X_2} +\bbm {X_1&X_0\\X_0&X_2}\bbm {0&1\\-1&0} +\bbm {1&0\\0&0}\bbm {1&0\\0&0} \\ +\gamma ^{-2}\bbm {X_1&X_0\\X_0&X_2}\bbm {0\\1}\bbm {0&1}\bbm {X_1&X_0\\X_0&X_2} -\varepsilon ^{-2}\bbm {X_1&X_0\\X_0&X_2}\bbm {0\\1}\bbm {0&1}\bbm {X_1&X_0\\X_0&X_2}=\bbm {0&0\\0&0}, \end{multline*} which is
\[ \bbm {-2X_0&X_1-X_2\\X_1-X_2&2X_0} +\bbm {1&0\\0&0} +\left (\gamma ^{-2}-\varepsilon ^{-2}\right )\bbm {X_0^2&X_0X_2\\X_0X_2&X_2^2}=\bbm {0&0\\0&0}, \]
which is
\[ \bbm { -2X_0+1+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_0^2 &X_1-X_2+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_0X_2 \\X_1-X_2+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_0X_2 &2X_0+\left (\gamma ^{-2}-\varepsilon ^{-2}\right )X_2^2 }=\bbm {0&0\\0&0}. \]
We separately consider the special case where \(\gamma ^{-2}-\varepsilon ^{-2}=0\), i.e. \(\gamma =\varepsilon \). The above then is
\[ \bbm { -2X_0+1 &X_1-X_2 \\X_1-X_2 &2X_0 }=\bbm {0&0\\0&0}; \]
from the bottom-right we obtain \(X_0=0\), but this contradicts the top-left. Remark 19.4 then gives that we must have \(\gamma >\varepsilon \), i.e. \(\gamma ^{-2}-\varepsilon ^{-2}<0\).
We now return to the general case. Given the above, the bottom-right corner shows that \(X_0\geq 0\). The top-left corner then gives (noting that we take the squareroot of a positive number and we pick the plus sign to satisy \(X_0\geq 0\)):
\[ X_0=\frac {-1+\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}}{\varepsilon ^{-2}-\gamma ^{-2}}. \]
The bottom-right corner then gives
\[ X_2=\frac {\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}}}{\left (\varepsilon ^{-2}-\gamma ^{-2}\right )}. \]
The off-diagonal entry gives
\[ X_1=\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}~ \frac {\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}}}{\varepsilon ^{-2}-\gamma ^{-2}}. \]
It follows that
\[ X=\frac {1}{\varepsilon ^{-2}-\gamma ^{-2}}\bbm { \sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}~ \sqrt {-2+2\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}} & -1+\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}} \\ -1+\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}} &\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}} }. \]
The optimal feedback then is (again using that \(C_1^*D_{12}=0\) and \(D_{12}^*D_{12}=\varepsilon ^2\)):
\[ F=-\varepsilon ^{-2}\bbm {0&1}X =\frac {\varepsilon ^{-2}}{\varepsilon ^{-2}-\gamma ^{-2}}\bbm { 1-\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}} &-\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}} }. \]
(vi) The closed-loop system then is (in second order form)
\[ \ddot {q}(t)+\frac {1}{1-\gamma ^{-2}\varepsilon ^2}\sqrt {-2+2\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}}~\dot {q}(t)+\frac {1}{1-\gamma ^{-2}\varepsilon ^2}\sqrt {1+\varepsilon ^{-2}-\gamma ^{-2}}~q(t)=w(t). \]
We see that a damping term is introduced (the coefficient of \(\dot {q}\)) and that the coefficient of \(q\) is changed. □