Chapter 10 Output regulation and disturbance rejection: 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
\(\seteqnumber{0}{10.}{0}\)\begin{equation} \label {eq:OR:xzy} \dot {x}=Ax+B_1w+B_2u,\qquad z=C_1x+D_{11}w,\qquad y=C_2x+D_{21}w, \end{equation}
with the initial condition \(x(0)=x^0\) where \(x^0\in \mR ^n\) and
\[ A\in \mR ^{n\times n},~ B_1\in \mR ^{n\times m_1},~ B_2\in \mR ^{n\times m_2},~ C_1\in \mR ^{p_1\times n},~ D_{11}\in \mR ^{p_1\times m_1},~ C_2\in \mR ^{p_2\times n},~ D_{21}\in \mR ^{p_2\times m_1}. \]
We assume that the external input \(w\) is generated by an exo-system
\[ \dot {x}_e=A_ex_e,\qquad w=C_ex_e, \]
with the initial condition \(x_e(0)=x_e^0\) where \(x_e^0\in \mR ^{n_e}\) and
\[ A_e\in \mR ^{n_e\times n_e},\qquad C_e\in \mR ^{m_1\times n_e}. \]
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Definition 10.1 (Measurement feedback output regulation and disturbance rejection 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
\(\seteqnumber{0}{10.}{1}\)\begin{equation} \label {eq:OR:controller} \dot {x}_c=A_cx_x+B_cy,\quad u=C_cx_c+D_cy, \end{equation}
where \(x_c(0)=x_c^0\), the following two conditions are both satisfied:
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1. for all \(x^0\), \(x_c^0\) and \(x_e^0\) we have \(\lim _{t\to \infty }z(t)=0\);
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2. for \(x_e^0=0\) and all \(x^0\) and \(x_c^0\) we have \(\lim _{t\to \infty }x(t)=0\) and \(\lim _{t\to \infty }x_c(t)=0\).
The first of these is the regulation requirement, the second is the stability requirement.
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Remark 10.2. If \(y=\sbm {x_e\\x}\), then with \(n_c=0\) (i.e. the state \(x_c\) is trivial) and \(D_c=F\) the above reduces to the full information problem from Definition 9.2.
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Theorem 10.4. Assume that \((A,B_2)\) is stabilizable, that \(\left (\sbm {A&B_1C_e\\0&A_e},\bbm {C_2&D_{21}C_e}\right )\) is detectable and that the regulator equations have a solution \(\Pi \), \(V\). Let \(F_2\in \mR ^{m_2\times n}\) be such that \(A+B_2F_2\) is asymptotically stable and let \(L\in \mR ^{(n+n_e)\times p_2}\) be such that \(\sbm {A&B_1C_e\\0&A_e}-L\bbm {C_2&D_{21}C_e}\) is asymptotically stable. Then
\(\seteqnumber{0}{10.}{2}\)\begin{gather*} A_c=\bbm {A&B_1C_e\\0&A_e}-L\bbm {C_2&D_{21}C_e}+\bbm {B_2\\0}\bbm {F_2&V-F_2\Pi },\\ B_c=L,\qquad C_c=\bbm {F_2&V-F_2\Pi },\qquad D_c=0. \end{gather*} solves the measurement feedback output regulation and disturbance rejection problem.
Moreover, for all \(x^0\), \(x_c^0\) and \(x_e^0\) we have \(\lim _{t\to \infty } x(t)-\Pi x_e(t)=0\).
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Remark 10.5. If \(D_c=0\), then we can write the combination of (10.1) and (10.2) as
\[ \bbm {\dot {x}\\\dot {x}_c}=\bbm {A&B_2C_c\\B_cC_2&A_c}\bbm {x\\x_c}+\bbm {B_1\\B_cD_{21}}w,\qquad z=\bbm {C_1&0}\bbm {x\\x_c}+D_{11}w, \]
which we can view as an input/state/output system with input \(w\) and output \(z\).
If \(D_c\neq 0\), then the formulas become more complicated. We do not need that case.
10.1 Examples
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Example 10.6. Consider the first order system
\[ \dot {x}+x=w+u. \]
The objective is to find a control \(u\) such that \(\lim _{t\to \infty }x(t)=0\) no matter what \(x^0\) is when \(w(t)=d\) where \(d\in \mR \) is unknown. Assume that \(y=x\) is the measurement.
This is of the standard form with
\[ A=-1,\quad B_1=1,\quad B_2=1,\quad C_1=1,\quad D_{11}=0,\quad C_2=1,\quad D_{21}=0, \]
and (so that \(w=x_e=x_e^0=d\))
\[ A_e=0,\qquad C_e=1. \]
The regulator equations are
\[ -\Pi +1+V=0,\qquad \Pi =0, \]
which gives \(\Pi =0\), \(V=-1\). Since \(A\) is stable, we can take \(F_2=0\). In the full-information case, this would give \(u=-d\) as control (which cancels the disturbance \(w\) completely); however, here we assume that \(u\) can only depend on the measurement \(y=x\) (and therefore not directly on \(d\)).
We have
\[ \bbm {A&B_1C_e\\0&A_e}-\bbm {L_1\\L_2}\bbm {C_1&D_{21}C_e}= \bbm {-1&1\\0&0}-\bbm {L_1\\L_2}\bbm {1&0} =\bbm {-1-L_1&1\\-L_2&0}. \]
The characteristic polynomial of this matrix is \(s^2+(1+L_1)s+L_2\), which is stable if and only if \(1+L_1>0\) and \(L_2>0\). We can therefore choose \(L_1=0\) and \(L_2=1\). We see that the detectability condition from Theorem 10.4 is satisfied. The controller from that theorem is
\[ A_c=\bbm {-1&0\\-1&0},\quad B_c=\bbm {0\\1},\quad C_c=\bbm {0&-1},\quad D_c=0. \]
Theorem 10.4 tells us that this solves our problem, but to see this more explicitly, we investigate further. The differential equations for the controller are
\[ \dot {x}_{c,1}=-x_{c,1},\quad \dot {x}_{c,2}=x-x_{c,1},\quad u=-x_{c,2}. \]
From this we can get a differential equation for the control \(u\)
\[ \ddot {u}+\dot {u}+u=-d, \]
since
\[ \ddot {u}+\dot {u}+u=-\ddot {x}_{c,2}-\dot {x}_{c,2}+u =-(\dot {x}-\dot {x}_{c,1})-(x-x_{c,1})+u=-\dot {x}-x+u=-d. \]
From this we see that \(u\) equals \(-d\) plus a term which decays to zero; so “asymptotically” we get back the full-information control \(-d\).
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Example 10.7. Consider the first order system
\[ T\dot {x}(t)+x(t)=K_1w(t)+K_2u(t), \]
where \(T,K_1,K_2>0\). The objective is to find a control \(u\) such that \(\lim _{t\to \infty }x(t)=0\) no matter what \(x^0\) is when \(w(t)=R\sin (\omega _e t-\varphi )\) where \(\omega _e>0\) is known and \(R>0\) and \(\varphi \in \mR \) are unknown. Assume that \(y=x\) is the measurement.
This is of the standard form with
\[ A=\frac {-1}{T},\quad B_1=\frac {K_1}{T},\quad B_2=\frac {K_2}{T},\quad C_1=1,\quad D_{11}=0,\quad C_2=1,\quad D_{21}=0, \]
and
\[ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}. \]
We solved the regulator equations in Example 9.14, which gave
\[ \Pi =0,\qquad V=\bbm {-\dfrac {K_1}{K_2}&0}. \]
Since \(A\) is asymptotically stable we can choose \(F_2=0\). Since \(A_e\) is not asymptotically stable, we do have to choose a nonzero \(L\). We want
\[ \bbm { \frac {-1}{T}&\frac {K_1}{T}&0\\ 0&0&1\\ 0&-\omega _e^2&0 } -\bbm {L_1\\L_2\\L_3} \bbm {1&0&0} =\bbm { \frac {-1}{T}-L_1&\frac {K_1}{T}&0\\ -L_2&0&1\\ -L_3&-\omega _e^2&0 } \]
to be asymptotically stable. The characteristic polynomial is
\[ s^3+\left (\frac {1}{T}+L_1\right )s^2+\left (\omega _e^2+\frac {K_1}{T}L_2\right )s+\frac {\omega _e^2}{T}+\omega _e^2L_1+\frac {K_1}{T}L_3. \]
We desire the characteristic polynomial (since we know that for \(s_0,\zeta >0\) this is stable)
\[ (s+s_0)(s^2+2\zeta \omega _e s+\omega _e^2) =s^3+(s_0+2\zeta \omega _e)s^2+\left (2s_0\zeta \omega _e+\omega _e^2\right )s+s_0\omega _e^2. \]
Comparing coefficients, we obtain the equations
\(\seteqnumber{0}{10.}{2}\)\begin{align*} \frac {1}{T}+L_1&=s_0+2\zeta \omega _e \\ \omega _e^2+\frac {K_1}{T}L_2&=2s_0\zeta \omega _e+\omega _e^2 \\ \frac {\omega _e^2}{T}+\omega _e^2L_1+\frac {K_1}{T}L_3&=s_0\omega _e^2. \end{align*} We choose \(s_0=\frac {1}{T}\) (to leave the stable eigenvalue unaffected). The first equation then gives \(L_1=2\zeta \omega _e\). The second equation gives \(L_2=\frac {2\zeta }{K_1}\). The third equation gives \(L_3=-\frac {2\zeta \omega _e^3T}{K_1}\). Hence
\[ L=\bbm { 2\zeta \omega _e\\ \frac {2\zeta }{K_1}\\ -\frac {2\zeta \omega _e^3T}{K_1} } \]
The Bode magnitude diagram of the closed-loop system (as in Remark 10.5) with the above choices is given in Figure 10.1 where it is compared to the frequency response from \(w\) to \(z\) when \(u=0\). We see that the Bode magnitude diagrams are very similar, but that the controlled one has a zero at the given frequency \(\omega _e\).
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Example 10.8. Consider the undamped second order system
\[ \ddot {q}(t)+\omega _0^2q(t)=K_1\omega _0^2w(t)+K_2\omega _0^2u(t), \]
where \(\omega _0,K_1,K_2>0\). Assume that \(z=\dot {q}\) is the performance output and that \(y=\dot {q}\) is the measurement. This is of the standard form with
\(\seteqnumber{0}{10.}{2}\)\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},\\ C_1=\bbm {0&1},\quad D_{11}=0,\quad C_2=\bbm {0&1},\quad D_{21}=0,\quad \end{gather*} We consider the exosystem
\[ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}. \]
In Example 9.15 we solved the regulator equations and obtained
\[ \Pi =\bbm {0&0\\0&0},\qquad V=\bbm {-\dfrac {K_1}{K_2}&0}. \]
There we chose the state feedback
\[ F_2=\bbm {0&\dfrac {-2\zeta _0}{K_2\omega _0}}. \]
We want to find a \(L\) so that
\(\seteqnumber{0}{10.}{2}\)\begin{align*} \bbm {A&B_1C_e\\0&A_e}-L\bbm {C_2&D_{21}C_e} &=\bbm { 0&1&0&0\\ -\omega _0^2&0&K_1\omega _0^2&0\\ 0&0&0&1\\ 0&0&-\omega _e^2&0 } -\bbm {L_1\\L_2\\L_3\\L_4} \bbm {0&1&0&0} \\ &=\bbm { 0&1-L_1&0&0\\ -\omega _0^2&-L_2&K_1\omega _0^2&0\\ 0&-L_3&0&1\\ 0&-L_4&-\omega _e^2&0 } \end{align*} is asymptotically stable. The characteristic polynomial of this matrix is
\[ s^4+L_2s^3+\left (K_1\omega _0^2L_3+\omega _e^2+\omega _0^2[1-L_1]\right )s^2 +\left (\omega _e^2L_2+K\omega _0^2L_4\right )s+\omega _0^2\omega _e^2[1-L_1]. \]
We desire the characteristic polynomial (which for \(\zeta _0,\zeta _e>0\) is stable)
\(\seteqnumber{0}{10.}{2}\)\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
\(\seteqnumber{0}{10.}{2}\)\begin{align*} L_2 &=2\zeta _0\omega _0+2\zeta _e\omega _e \\K_1\omega _0^2L_3+\omega _e^2+\omega _0^2[1-L_1] &=\omega _0^2+\omega _e^2+4\zeta _0\omega _0\zeta _e\omega _e \\\omega _e^2L_2+K\omega _0^2L_4 &=2\zeta _0\omega _0\omega _e^2+2\zeta _e\omega _e\omega _0^2 \\\omega _0^2\omega _e^2[1-L_1] &=\omega _0^2\omega _e^2 \end{align*} This gives
\[ L=\bbm { 0\\ 2\zeta _0\omega _0+2\zeta _e\omega _e\\ \frac {4\zeta _0\zeta _e\omega _e}{K_1\omega _1}\\ \frac {2\zeta _e\omega _e(\omega _0^2-\omega _e^2)}{K\omega _0^2} }. \]
We then have
\[ A_c=\bbm { 0&1&0&0\\ -\omega _0^2&-4\zeta _0\omega _0+2\zeta _e\omega _e&0&0\\ 0&-\frac {4\zeta _0\zeta _e\omega _e}{K_1\omega _1}&0&1\\ 0&-\frac {2\zeta _e\omega _e(\omega _0^2-\omega _e^2)}{K\omega _0^2}&-\omega _e^2&0 }, \quad B_c=\bbm { 0\\ 2\zeta _0\omega _0+2\zeta _e\omega _e\\ \frac {4\zeta _0\zeta _e\omega _e}{K_1\omega _1}\\ \frac {2\zeta _e\omega _e(\omega _0^2-\omega _e^2)}{K\omega _0^2} },\quad C_c=\bbm {0&\dfrac {-2\zeta _0}{K_2\omega _0}&-\dfrac {K_1}{K_2}&0}. \]
10.2 Case study: control of a tape drive
We consider the measurement feedback output regulation and disturbance rejection problem for the tape drive system. In addition to what we did for the full information feedback problem in Section 9.2, we need to find an output injection matrix \(L\) which makes \(\sbm {A&B_1C_e\\0&A_e}-L\bbm {C_2&D_{21}C_e}\) asymptotically stable. As this is a 7-by-3 matrix, this is non-trivial to do by hand. Therefore, we do this numerically using pole placement (see Remark 12.11). The eigenvalues of \(\sbm {A&B_1C_e\\0&A_e}\) are the union of those of \(A\), which are stable by Section 2.3 and which we want to leave as they are, and those of \(A_e\), which are \(0\) (with multiplity 2) and \(\pm i\omega _e\), which are unstable and which we want to move slightly to the left. The resulting Bode magnitude plots from \(v_e\) to \(v_1-r_v\) and \(T-r_T\) are given in Figure 10.2 (and are compared to those without any control). We can clearly see the effect of the disturbance rejection at the frequency \(\omega _0\). The choice of the desired characteristic polynomial does not uniquely determine the matrix \(L\) (this is because \(p_2>1\)). The algorithms utilized by matlab and octave produce different \(L\)’s with the one produced by octave being clearly superior (though both achieve the given objective).
10.3 Problems
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• Consider the first order system
\[ \dot {x}+2x=3w+u. \]
Find a measurement based control \(u\) such that \(\lim _{t\to \infty }x(t)=0\) no matter what \(x^0\) is when \(w(t)=d\) where \(d\in \mR \) is unknown. Assume that \(y=x\) is the measurement.
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Solution. This is similar to Example 10.6 (just different numbers).
This is of the standard form with
\[ A=-2,\quad B_1=3,\quad B_2=1,\quad C_1=1,\quad D_{11}=0,\quad C_2=1,\quad D_{21}=0, \]
and (so that \(w=x_e=x_e^0=d\))
\[ A_e=0,\qquad C_e=1. \]
The regulator equations are
\[ -2\Pi +3+V=0,\qquad \Pi =0, \]
which gives \(\Pi =0\), \(V=-3\). Since \(A\) is stable, we can take \(F_2=0\).
We have
\[ \bbm {A&B_1C_e\\0&A_e}-\bbm {L_1\\L_2}\bbm {C_1&D_{21}C_e}= \bbm {-2&3\\0&0}-\bbm {L_1\\L_2}\bbm {1&0} =\bbm {-2-L_1&3\\-L_2&0}. \]
The characteristic polynomial of this matrix is \(s^2+(2+L_1)s+3L_2\), which is stable if and only if \(2+L_1>0\) and \(L_2>0\). We can therefore choose \(L_1=0\) and \(L_2=1\). We see that the detectability condition from Theorem 10.4 is satisfied. The controller from that theorem is
\[ A_c=\bbm {-2&0\\-1&0},\quad B_c=\bbm {0\\1},\quad C_c=\bbm {0&-3},\quad D_c=0. \]
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10.4 Further Problems
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• We consider the situation as in Section 1.4 but without the external force \(F_e\) so that \(w=v_e\). Assume that \(v_e\) is a sinusoid with given frequency \(\omega _e>0\), i.e. consider the exosystem
\[ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}. \]
Consider the performance output \(z=q\) and the measured output \(y=q\). Solve the measurement feedback disturbance rejection problem for this situation. Assume that \(d=0\).
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Solution. We have similarly as in Section 1.4
\(\seteqnumber{0}{10.}{2}\)\begin{gather*} A=\bbm {0&1\\\frac {-k}{m}&\frac {-d}{m}},\quad B_1=\bbm {-1\\0},\quad B_2=\bbm {0\\\frac {1}{m}},\\ C_1=\bbm {1&0},\quad D_{11}=0,\quad D_{12}=0,\quad C_2=\bbm {1&0},\quad D_{21}=0,\quad D_{22}=0. \end{gather*} We solved the regulator equations in Section 9.3 and obtained
\[ \Pi =\bbm {0&0\\1&0},\qquad V=\bbm {0&m}. \]
We want to find a \(L\) such that
\(\seteqnumber{0}{10.}{2}\)\begin{align*} \bbm {A&B_1C_e\\0&A_e}-L\bbm {C_2&D_{21}C_e} &=\bbm { 0&1&-1&0\\ \frac {-k}{m}&0&0&0\\ 0&0&0&1\\ 0&0&-\omega _e^2&0} -\bbm {L_1\\L_2\\L_3\\L_4}\bbm {1&0&0&0} \\&=\bbm { -L_1&1&-1&0\\ \frac {-k}{m}-L_2&0&0&0\\ -L_3&0&0&1\\ -L_4&0&-\omega _e^2&0} \end{align*} is asymptotically stable. The characteristic polynomial of this matrix is
\[ s^4 +L_1s^3 +\left (\frac {k}{m}+\omega _e^2-L_3+L_2\right )s^2 +\left (-L_4+L_1\omega _e^2\right )s +\omega _e^2\left [\frac {k}{m}+L_2\right ]. \]
We desire the characteristic polynomial (which for \(\zeta _0,\zeta _e>0\) is stable) with \(\omega _0^2=\frac {k}{m}\))
\(\seteqnumber{0}{10.}{2}\)\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
\(\seteqnumber{0}{10.}{2}\)\begin{align*} L_1 &=2\zeta _0\omega _0+2\zeta _e\omega _e \\\frac {k}{m}+\omega _e^2-L_3+L_2 &=\omega _0^2+\omega _e^2+4\zeta _0\omega _0\zeta _e\omega _e \\-L_4+L_1\omega _e^2 &=2\zeta _0\omega _0\omega _e^2+2\zeta _e\omega _e\omega _0^2 \\\omega _e^2\left [\frac {k}{m}+L_2\right ] &=\omega _0^2\omega _e^2 \end{align*} This gives
\[ L=\bbm { 2\zeta _0\sqrt {\frac {k}{m}}+2\zeta _e\omega _e\\ 0\\ -4\zeta _0\sqrt {\frac {k}{m}}\zeta _e\omega _e\\ 2\zeta _e\omega _e\left [\omega _e^2-\frac {k}{m}\right ] }. \]
Since \(A\) is not stable, we need a stabilizing feedback; as in Section 9.3, we pick \(F_2=\bbm {0&-d_0}\) where \(d_0>0\). We then have
\[ A_c=\bbm { -2\zeta _0\sqrt {\frac {k}{m}}-2\zeta _e\omega _e&1&-1&0\\ \frac {-k}{m}&\frac {-d_0}{m}&\frac {d_0}{m}&1\\ 4\zeta _0\sqrt {\frac {k}{m}}\zeta _e\omega _e&0&0&1\\ 2\zeta _e\omega _e\left [\frac {k}{m}-\omega _e^2\right ]&0&-\omega _e^2&0}, \quad B_c=\bbm { 2\zeta _0\sqrt {\frac {k}{m}}+2\zeta _e\omega _e\\ 0\\ -4\zeta _0\sqrt {\frac {k}{m}}\zeta _e\omega _e\\ 2\zeta _e\omega _e\left [\omega _e^2-\frac {k}{m}\right ] },\quad C_c=\bbm {0&-d_0&d_0&m}. \]
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• Consider the undamped second order system
\[ \ddot {q}(t)+\omega _0^2q(t)=K_1\omega _0^2w(t)+K_2\omega _0^2u(t), \]
where \(\omega _0,K_1,K_2>0\). Assume that \(w\) is a sinusoid with given frequency \(\omega _e>0\), i.e. consider the exosystem
\[ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}. \]
Consider the performance output \(z=q\) and the measured output \(y=q\). Solve the measurement feedback disturbance rejection problem for this situation.
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Solution. We have
\(\seteqnumber{0}{10.}{2}\)\begin{gather*} A=\bbm {0&1\\-\omega _0^2&0},\qquad B_1=\bbm {0\\K_1\omega _0^2},\qquad B_2=\bbm {0\\K_2\omega _0^2},\\ C_1=\bbm {1&0},\qquad D_{11}=0,\qquad C_2=\bbm {1&0},\qquad D_{21}=0, \\ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}. \end{gather*} We solved the regulator equations in Section 9.3 (this part of the answer was valid for \(\zeta =0\) as well) and obtained
\[ \Pi =\bbm {0&0\\0&0},\qquad V=\bbm {-\dfrac {K_1}{K_2}&0}. \]
We want to find a \(L\) such that
\(\seteqnumber{0}{10.}{2}\)\begin{align*} \bbm {A&B_1C_e\\0&A_e}-L\bbm {C_2&D_{21}C_e} &=\bbm { 0&1&0&0\\ -\omega _0^2&0&K_1\omega _0^2&0\\ 0&0&0&1\\ 0&0&-\omega _e^2&0} -\bbm {L_1\\L_2\\L_3\\L_4}\bbm {1&0&0&0} \\&=\bbm { -L_1&1&0&0\\ -\omega _0^2-L_2&0&K_1\omega _0^2&0\\ -L_3&0&0&1\\ -L_4&0&-\omega _e^2&0} \end{align*} is asymptotically stable. The characteristic polynomial of this matrix is
\[ s^4 +L_1s^3 +\left (L_2+\omega _0^2+\omega _e^2\right )s^2 +\left (L_1\omega _e^2+L_3K_1\omega _0^2\right )s +L_2\omega _e^2+L_4K_1\omega _0^2+\omega _e^2\omega _0^2. \]
We desire the characteristic polynomial (which for \(\zeta _0,\zeta _e>0\) is stable)
\(\seteqnumber{0}{10.}{2}\)\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
\(\seteqnumber{0}{10.}{2}\)\begin{align*} L_1 &=2\zeta _0\omega _0+2\zeta _e\omega _e \\L_2+\omega _0^2+\omega _e^2 &=\omega _0^2+\omega _e^2+4\zeta _0\omega _0\zeta _e\omega _e \\L_1\omega _e^2+L_3K_1\omega _0^2 &=2\zeta _0\omega _0\omega _e^2+2\zeta _e\omega _e\omega _0^2 \\L_2\omega _e^2+L_4K_1\omega _0^2+\omega _e^2\omega _0^2 &=\omega _0^2\omega _e^2 \end{align*} This gives
\[ L=\bbm { 2\zeta _0\omega _0+2\zeta _e\omega _e\\ 4\zeta _0\omega _0\zeta _e\omega _e\\ \frac {2\zeta _e\omega _e(\omega _0^2-\omega _e^2)}{K_1\omega _0^2}\\ -\frac {4\zeta _0\zeta _e\omega _e^3}{K_1\omega _0} }. \]
Since \(A\) is not stable, we need a stabilizing feedback; as in Example 10.8 we pick (here \(\zeta _0>0\))
\[ F_2=\bbm {0&\frac {-2\zeta _0}{K_2\omega _0}}. \]
We then have
\[ A_c=\bbm { -2\zeta _0\omega _0+2\zeta _e\omega _e&1&0&0\\ -\omega _0^2-4\zeta _0\omega _0\zeta _e\omega _e&-2\zeta _0\omega _0&0&0\\ -\frac {2\zeta _e\omega _e(\omega _0^2-\omega _e^2)}{K_1\omega _0^2}&0&0&1\\ -\frac {4\zeta _0\zeta _e\omega _e^3}{K_1\omega _0}&0&-\omega _e^2&0 }, \quad B_c=\bbm { 2\zeta _0\omega _0+2\zeta _e\omega _e\\ 4\zeta _0\omega _0\zeta _e\omega _e\\ \frac {2\eta _e\omega _e(\omega _0^2-\omega _e^2)}{K_1\omega _0^2}\\ -\frac {4\zeta _0\zeta _e\omega _e^3}{K_1\omega _0} },\quad C_c=\bbm {0&\frac {-2\zeta _0}{K_2\omega _0}&-\dfrac {K_1}{K_2}&0}. \]
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