Chapter 15 \(H^2\) 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
\(\seteqnumber{0}{15.}{0}\)\begin{equation} \label {eq:H2: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
\(\seteqnumber{0}{15.}{1}\)\begin{equation} \label {eq:control: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 15.1 (\(H^2\) state feedback problem). The objective is to find a matrix
\[ F\in \mR ^{m_2\times n}, \]
such that with the control \(u=Fx\) we have
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1. \(\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega \) is minimized amongst all such \(F\),
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2. for all \(x^0\) and \(w=0\) we have \(\lim _{t\to \infty }x(t)=0\),
where \(F_{zw}\) is the frequency response of the closed-loop system
\(\seteqnumber{0}{15.}{2}\)\begin{equation} \label {eq:closedH2} \dot {x}=(A+B_2F)x+B_1w,\qquad z=(C_1+D_{12}F)x. \end{equation}
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Remark 15.2. For a matrix \(M\in \mR ^{p\times m}\) we have the Hilbert-Schmidt or Frobenius norm
\[ \|M\|_{\HS }=\sqrt {\sum _{i=1}^p\sum _{j=1}^mM_{ij}^2}. \]
When either \(p=1\) or \(m=1\) (so that \(M\) is either a row or a column vector), this norm coincides with the Euclidean vector norm.
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Remark 15.3. By Plancherel’s equality we have
\[ \frac {1}{2\pi }\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega =\int _0^\infty \|h_{zw}(t)\|_{\HS }^2\,dt, \]
where \(h_{zw}\) is the impulse response, so that equivalently to (1) in Definition 15.1 we could have:
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• \(\int _0^\infty \|h_{zw}(t)\|_{\HS }^2\,dt\) is minimized amongst all such \(F\), where \(h_{zw}\) is the impulse response of the closed-loop system (15.3)
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Remark 15.4. More generally we could have \(z=C_1x+D_{11}w+D_{12}u\), which would result in the closed-loop \(z=(C_1+D_{12}F)x+D_{11}w\). However, if \(D_{11}\neq 0\), then the integral that we wish to minimize will be infinite for all \(F\). Therefore, we have only considered the case \(D_{11}=0\).
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Theorem 15.6. Assume that \(D_{12}\) is injective, that \((A,B_2)\) is stabilizable and that the Rosenbrock matrix
\[ \bbm {sI-A&-B_2\\C_1&D_{12}}, \]
is injective for all \(s\in \mC \) with \(\re (s)=0\). Then there exists a unique solution \(X\) of the algebraic Riccati equation
\(\seteqnumber{0}{15.}{3}\)\begin{equation} \label {eq:RiccatiH2} 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
\[ F=-(D_{12}^*D_{12})^{-1}\left (D_{12}^*C_1+B_2^*X\right ). \]
This \(X\) is symmetric positive semidefinite. Furthermore, \(F\) is the unique solution of the \(H^2\) state feedback problem and
\[ \trace (B_1^*XB_1) =\frac {1}{2\pi }\int _{-\infty }^\infty \|F_{zw}(\omega )\|_{\HS }^2\,d\omega =\int _0^\infty \|h_{zw}(t)\|_{\HS }^2\,dt. \]
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Remark 15.7. Recall that the trace of a square matrix is the sum of its diagonal elements.
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Remark 15.8. Definition 15.1 and Theorem 15.6 consider the \(H^2\) state feedback problem with stability; similarly as in Chapter 14, the \(H^2\) state feedback problem without stability can be formulated and solved: this involves the minimal symmetric positive semidefinite solution of the algebraic Riccati equation and does not involve the stabilizability and Rosenbrock conditions.
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Remark 15.9. The Riccati equation (15.4) can be rewritten as the system of equations (these are called Lur’e equations) in the unknowns \(X\), \(K\) and \(J\)
\(\seteqnumber{0}{15.}{4}\)\begin{align*} A^*X+XA+C_1^*C_1&=K^*K\\ B_2^*X+D_{12}^*C_1&=J^*K\\ D_{12}^*D_{12}&=J^*J, \end{align*} and the feedback \(u=Fx\) can be re-written as
\[ Ju+Kx=0. \]
If \(D_{12}\) is injective, then we can eliminate \(J\) and \(K\) through
\[ J=(D_{12}^*D_{12})^{1/2}, \qquad K=(D_{12}^*D_{12})^{-1/2}(B_2^*X+D_{12}^*C_1), \]
and the first Lur’e equation then becomes the Riccati equation.
The Lur’e equations continue to be well-defined even if \(D_{12}\) is not injective. It is possible to use the Lur’e equations to formulate a theorem similar to Theorem 15.6 valid also for the case where \(D_{12}\) may not be injective, but the precise additional conditions become technical.
15.1 Examples
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Example 15.10. Consider the first order system
\[ \dot {x}(t)+x(t)=w(t)+u(t). \]
For the \(H^2\) state feedback problem, we consider the performance output
\[ z(t)=\bbm {x(t)\\\varepsilon u(t)}, \]
where \(\varepsilon >0\). To put this into the \(H^2\) 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 is the same as in Example 14.5, so that
\[ X=-\varepsilon ^2+\varepsilon \sqrt {\varepsilon ^2+1}, \qquad F=1-\sqrt {1+\varepsilon ^{-2}}. \]
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Example 15.11. 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 \(H^2\) 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 the same as in Example 14.6, so that
\[ X=\varepsilon \bbm {1&0\\0&1},\qquad F=\bbm {0&-\varepsilon ^{-1}}. \]
15.2 Case study: a suspension system
We consider the state feedback \(H^2\) problem for the suspension system design. By Section 11.3, the system is controllable and therefore stabilizable. We clearly have that \(D_{12}\) is injective. We now consider the Rosenbrock condition from Theorem 15.6.
The Rosenbrock matrix is
\[ \bbm {sI-A&-B_2\\C_1&D_{12}}= \bbm { s&-k_{us}&0&0&0\\ \frac {1}{m_{us}}&s&0&0&\frac {-1}{m_{us}}\\ 0&-1&s&1&0\\ 0&0&0&s&\frac {1}{m_s}\\ \frac {r_1}{k_{us}}&0&0&0&0\\ 0&0&r_2&0&0\\ 0&0&0&0&\frac {-1}{m_s} }. \]
To show that this is injective, we need to identify 5 independent rows; we pick rows 1,3,5,6 and 7. That gives the submatrix
\[ \bbm { s&-k_{us}&0&0&0\\ 0&-1&s&1&0\\ \frac {r_1}{k_{us}}&0&0&0&0\\ 0&0&r_2&0&0\\ 0&0&0&0&\frac {-1}{m_s} }. \]
To show that this is indeed injective, we calculate its determinant. We develop by the last row to obtain
\[ \frac {-1}{m_s}\det \bbm { s&-k_{us}&0&0\\ 0&-1&s&1\\ \frac {r_1}{k_{us}}&0&0&0\\ 0&0&r_2&0\\ }, \]
we develop this by the last column to obtain
\[ \frac {-1}{m_s}\det \bbm { s&-k_{us}&0\\ \frac {r_1}{k_{us}}&0&0\\ 0&0&r_2\\ }, \]
developing this by the last row gives
\[ \frac {-r_2}{m_s}\det \bbm { s&-k_{us}\\ \frac {r_1}{k_{us}}&0\\ }=\frac {-r_1r_2}{m_s}, \]
which is nonzero when \(r_1,r_2>0\).
We investigate what happens when \(r_1\) or \(r_2\) equals zero. If \(r_2=0\), then when \(s=0\) the third column of the Rosenbrock matrix equals zero; therefore the Rosenbrock condition does not hold. If \(r_1=0\), then we consider rows 1,3,4,6 and 7 of the Rosenbrock matrix:
\[ \bbm { s&-k_{us}&0&0&0\\ 0&-1&s&1&0\\ 0&0&0&s&\frac {1}{m_s}\\ 0&0&r_2&0&0\\ 0&0&0&0&\frac {-1}{m_s} }, \]
whose determinant we develop by the last row to obtain
\[ \frac {-1}{m_s} \det \bbm { s&-k_{us}&0&0\\ 0&-1&s&1\\ 0&0&0&s\\ 0&0&r_2&0\\ }, \]
and develop by the last row to obtain
\[ \frac {r_2}{m_s} \det \bbm { s&-k_{us}&0\\ 0&-1&1\\ 0&0&s\\ }, \]
and since this matrix is upper triangular, this equals
\[ \frac {-s^2r_2}{m_s}. \]
Therefore, as long as \(s\neq 0\), these rows are linearly independent. If \(s=0\), then we instead consider rows 1,2,3,6 and 7 and obtain
\[ \bbm { 0&-k_{us}&0&0&0\\ \frac {1}{m_{us}}&0&0&0&\frac {-1}{m_{us}}\\ 0&-1&0&1&0\\ 0&0&r_2&0&0\\ 0&0&0&0&\frac {-1}{m_s} }. \]
Developing the determinant by the last row gives
\[ \frac {-1}{m_s}\det \bbm { 0&-k_{us}&0&0\\ \frac {1}{m_{us}}&0&0&0\\ 0&-1&0&1\\ 0&0&r_2&0 }, \]
developing by the last row again gives
\[ \frac {r_2}{m_s}\det \bbm { 0&-k_{us}&0\\ \frac {1}{m_{us}}&0&0\\ 0&-1&1 }, \]
developing by the last column gives
\[ \frac {r_2}{m_s}\det \bbm { 0&-k_{us}\\ \frac {1}{m_{us}}&0 }=\frac {r_2k_{us}}{m_sm_{us}}. \]
Therefore, for \(s=0\), these rows are linearly independent. So if \(r_1=0\), then the Rosenbrock condition is still satisfied. Therefore the suspension stroke term needs to be present in the cost function for the Rosenbrock condition to be satisfied, but the tyre deflection (“handling”) term could be absent and the conditions from Theorem 15.6 would still be satisfied.
The algebraic Riccati equation is too complicated to solve by hand. We use the specific parameter values from the introduction to solve it numerically. This gives the state feedback matrix
\[ F=\bbm {-0.2944&-1908&-56125&6027}. \]
We give the impulse responses in Figure 15.1a and the Bode diagrams in Figure 15.1b. Comparing these to Figures 12.1a and 12.1b (the optimal fixed structure), we see that the output is less oscillatory and the peaks in the Bode diagram are slightly less pronounced.
