Remark 17.2.  An issue with the problem in Definition 17.1 is that the integral typically only has an infimum which is not a minimum. Therefore we will relax the minimality to \(\int _{-\infty }^\infty \|F_{\zt \wt }(\omega )\|_{\HS }^2\,d\omega \) being “close to minimal”.

Example 17.3.  Consider similarly as in Example 10.7 the first order system

\[ T\dot {x}(t)+x(t)=K_1w(t)+K_2u(t)+\widetilde {K}_1\wt _1(t), \]

where \(T,K_1,\widetilde {K}_1,K_2>0\) and the second-order exosystem

\[ \ddot {w}=-\omega ^2 w. \]

The objective is that \(\lim _{t\to \infty }x(t)=0\) for all \(w\) as above if \(\wt =0\), i.e. the performance output for regulation is \(z=x\). The measurement equals \(y=x+\delta \wt _2\), where \(\delta >0\). For the \(H^2\) optimization, the performance output is

\[ \zt =\bbm {x\\\varepsilon u}, \]

where \(\varepsilon >0\) and (as usual) the first component means that we want \(x\) to be small (consistent with the regulation requirement) and the second component means that \(u\) should not be too large.

Since the measurement equals the performance output for regulation plus something depending on \(\wt \), this is a problem of the special form we considered above.

We have

\begin{gather*} A=\frac {-1}{T},\quad B_1=\frac {K_1}{T},\quad \widetilde {B}_1=\bbm {\dfrac {\widetilde {K}_1}{T}&0},\quad B_2=\frac {K_2}{T}, \\ C_1=1,\quad D_{11}=0,\quad \widetilde {C}_1=\bbm {1\\0},\quad \widetilde {D}_{11}=\bbm {0\\0},\quad \widetilde {D}_{12}=\bbm {0\\\varepsilon },\quad C_2=1,\quad D_{21}=0,\quad \widetilde {D}_{21}=\delta , \\ A_e=\bbm {0&1\\-\omega ^2&0},\qquad C_e=\bbm {1&0}. \end{gather*} In Example 10.7 we solved the regulator equations and obtained \(\Pi =0\) and \(V=\bbm {-\frac {K_1}{K_2}&0}\). In Example 16.5 we solve the \(H^2\) problem for the transformed system (with notation differing by tildes). This gives

\[ F_2=\frac {1-\sqrt {1+K_2^2\varepsilon ^{-2}}}{K_2},\qquad L_1=\frac {-1+\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T}. \]

We now determine an additional output injection \(\widetilde {L}\) (by doing pole placement “by hand”). Writing

\[ \widetilde {L}=\bbm {\widetilde {L}_1\\\widetilde {L}_2\\\widetilde {L}_3}, \]

we have

\[ \bbm {A-L_1C_2&B_1C_e-L_1D_{21}C_e\\0&A_e}-\widetilde {L}\bbm {C_2&D_{21}C_e} =\bbm { -\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T}-\widetilde {L}_1&\frac {K_1}{T}&0\\ -\widetilde {L}_2&0&1\\ -\widetilde {L}_3&-\omega ^2&0 }, \]

the characteristic polynomial of which is

\[ s^3 +\left (\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T}+\widetilde {L}_1\right )s^2 +\left (\frac {K_1}{T}\widetilde {L}_2+\omega ^2\right )s +\omega ^2\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T} +\omega ^2\widetilde {L}_1 +\frac {K_1}{T}\widetilde {L}_3. \]

We desire the characteristic polynomial

\[ (s+s_0)(s^2+2\zeta \omega s+\omega ^2) =s^3+\left (s_0+2\zeta \omega \right )s^2+\left (\omega ^2+2s_0\zeta \omega \right )s+s_0\omega ^2. \]

Comparing coefficients, we obtain the equations

\begin{gather*} s_0+2\zeta \omega =\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T}+\widetilde {L}_1,\\ \omega ^2+2s_0\zeta \omega =\frac {K_1}{T}\widetilde {L}_2+\omega ^2\\ s_0\omega ^2=\omega ^2\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T} +\omega ^2\widetilde {L}_1 +\frac {K_1}{T}\widetilde {L}_3. \end{gather*} Solving this (solving the third equation last) gives

\begin{align*} \widetilde {L}_1&=s_0+2\zeta \omega -\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T}\\ \widetilde {L}_2&=2s_0\zeta \omega \frac {T}{K_1}\\ \widetilde {L}_3&=\frac {-T}{K_1}2\zeta \omega ^3. \end{align*} The overall output injection \(L=\sbm {L_1\\0}+\widetilde {L}\) then is

\[ L=\bbm { \frac {-1}{T}+s_0+2\zeta \omega \\ 2s_0\zeta \omega \frac {T}{K_1}\\ \frac {-T}{K_1}2\zeta \omega ^3 }. \]

The controller then is

\begin{gather*} A_c=\bbm { -s_0-2\zeta \omega +\frac {\sqrt {1+K_2^2\varepsilon ^{-2}}}{T} &0&0\\ -2s_0\zeta \omega \frac {T}{K_1}&0&1\\ \frac {T}{K_1}2\zeta \omega ^3&-\omega ^2&0 },\qquad B_c=\bbm { \frac {-1}{T}+s_0+2\zeta \omega \\ 2s_0\zeta \omega \frac {T}{K_1}\\ \frac {-T}{K_1}2\zeta \omega ^3 },\\ C_c=\bbm {\frac {1-\sqrt {1+K_2^2\varepsilon ^{-2}}}{K_2}&-\frac {K_1}{K_2}&0},\qquad D_c=0. \end{gather*} As mentioned, we would ideally want to keep the eigenvalues of \(A-L_1C_2\) and move those of \(A_e\) slightly to the left. This amounts to choosing \(s_0=\frac {\sqrt {1+\widetilde {K}_1^2\delta ^{-2}}}{T}\) and \(\zeta >0\) “small”. This results in a frequency response from \(w\) to \(z\) as in Figure 17.1. In the figure we see the regulation requirement (the frequency response is zero at the given frequency \(\omega \)) and the \(H^2\) requirement (the frequency response is a lot “lower” on average over the whole frequency range compared to without control).

Example 17.4.  Consider the undamped second order system

\[ \ddot {q}(t)+\omega _0^2q(t)=K_1\omega _0^2w(t)+K_2\omega _0^2u(t)+\widetilde {K}_1\wt _1(t), \]

where \(\omega _0,K_1,K_2,\widetilde {K}_1>0\) and with the state \(x=\sbm {q\\\dot {q}}\). We consider the performance output for regulation \(z=\dot {q}\), the measurement \(y=\dot {q}+\delta \wt _2\) where \(\delta >0\) (so that we have the special form considered above) and the \(H^2\) performance output

\[ z=\bbm {\dot {q}\\\varepsilon u}, \]

where \(\varepsilon >0\).

This gives

\begin{gather*} A=\bbm {0&1\\-\omega _0^2&0},\quad B_1=\bbm {0\\K_1\omega _0^2},\quad B_2=\bbm {0\\K_2\omega _0^2},\quad \widetilde {B}_1=\bbm {0&0\\\widetilde {K}_1\omega _0^2&0},\quad \\ C_1=\bbm {0&1},\quad D_{12}=0,\quad \widetilde {C}_1=\bbm {0&1\\0&0},\quad \widetilde {D}_{12}=\bbm {0\\\varepsilon }, \\ C_2=\bbm {0&1},\quad D_{21}=0,\quad \widetilde {D}_{21}=\bbm {0&\delta } \end{gather*} We assume that the exo-system is

\[ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}, \]

where \(\omega _e>0\).

We solved the regulator equations in Example 9.15, where we obtained

\[ \Pi =\bbm {0&0\\0&0},\qquad V=\bbm {-\dfrac {K_1}{K_2}&0}. \]

In Example 16.6 we solve the \(H^2\) problem for the transformed system (with notation differing by tildes). This gives

\[ F_2=\bbm {0&-\varepsilon ^{-1}},\qquad L_1=\bbm {0\\\delta ^{-1}\widetilde {K}_1\omega _0^2}. \]

We now determine an additional output injection \(\widetilde {L}\) (by doing pole placement “by hand”). Writing

\[ \widetilde {L}=\bbm {\widetilde {L}_1\\\widetilde {L}_2\\\widetilde {L}_3\\\widetilde {L}_4}, \]

we have

\[ \bbm {A-L_1C_2&B_1C_e-L_1D_{21}C_e\\0&A_e}-\widetilde {L}\bbm {C_2&D_{21}C_e} =\bbm { 0&1-\widetilde {L}_1&0&0\\ -\omega _0^2&-\delta ^{-1}\widetilde {K}_1\omega _0^2-\widetilde {L}_2&K_1\omega _0^2&0\\ 0&-\widetilde {L}_3&0&1\\ 0&-\widetilde {L}_4&-\omega _e^2&0 }, \]

the characteristic polynomial of which is

\[ s^4 +\left (\delta ^{-1}\widetilde {K}_1\omega _0^2+\widetilde {L}_2\right )s^3 +\left (\omega _e^2+K_1\omega _0^2\widetilde {L}_3+\omega _0^2[1-\widetilde {L}_1]\right )s^2 +\left (\delta ^{-1}\widetilde {K}_1\omega _0^2\omega _e^2+\widetilde {L}_2\omega _e^2+K_1\omega _0^2\widetilde {L}_4\right )s +\omega _0^2\omega _e^2[1-\widetilde {L}_1]. \]

We desire the characteristic polynomial

\begin{multline*} (s^2+2\zeta _0\omega _0s+\omega _0^2)(s^2+2\zeta _e\omega _e s+\omega _e^2) \\= s^4 +\left (2\zeta _0\omega _0+2\zeta _e\omega _e \right )s^3 +\left (\omega _0^2+\omega _e^2+4\zeta _0\omega _0\zeta _e\omega _e\right )s^2 +\left (2\zeta _0\omega _0\omega _e^2+2\zeta _e\omega _e\omega _0^2\right )s +\omega _0^2\omega _e^2. \end{multline*} Comparing coefficients, we obtain the equations

\begin{align*} \delta ^{-1}\widetilde {K}_1\omega _0^2+\widetilde {L}_2 &=2\zeta _0\omega _0+2\zeta _e\omega _e\\ \omega _e^2+K_1\omega _0^2\widetilde {L}_3+\omega _0^2[1-\widetilde {L}_1] &=\omega _0^2+\omega _e^2+4\zeta _0\omega _0\zeta _e\omega _e\\ \delta ^{-1}\widetilde {K}_1\omega _0^2\omega _e^2+\widetilde {L}_2\omega _e^2+K_1\omega _0^2\widetilde {L}_4 &=2\zeta _0\omega _0\omega _e^2+2\zeta _e\omega _e\omega _0^2\\ \omega _0^2\omega _e^2[1-\widetilde {L}_1] &=\omega _0^2\omega _e^2. \end{align*} From the last equation we obtain \(\widetilde {L}_1=0\). We then solve the first, second and third equation in order to obtain

\begin{align*} \widetilde {L}_1&=0\\ \widetilde {L}_2&=2\zeta _0\omega _0+2\zeta _e\omega _e-\delta ^{-1}\widetilde {K}_1\omega _0^2\\ \widetilde {L}_3&=\frac {4\zeta _0\zeta _e\omega _e}{\omega _0K_1}\\ \widetilde {L}_4&=\frac {2\zeta _e\omega _e}{K_1}\left (1-\frac {\omega _e^2}{\omega _0^2}\right ). \end{align*} The overall output injection \(L=\sbm {L_1\\0}+\widetilde {L}\) then is

\[ L=\bbm { 0\\ 2\zeta _0\omega _0+2\zeta _e\omega _e\\ \frac {4\zeta _0\zeta _e\omega _e}{\omega _0K_1}\\ \frac {2\zeta _e\omega _e}{K_1}\left (1-\frac {\omega _e^2}{\omega _0^2}\right ) }. \]

With the above \(F_2\), \(L\) and \(V\), we can then write down the controller.

The characteristic polynomial of \(A-L_1C_2\) equals \(s^2+\delta ^{-1}\widetilde {K}_1\omega _0^2s+\omega _0^2\), so we would choose \(\zeta _0=\frac {1}{2}\delta ^{-1}\widetilde {K}_1\omega _0\). We want to move the eigenvalues of \(A_e\) slightly to the left, so we would choose \(\zeta _e>0\) “small”.