Remark 9.1.  The two must common forms of an exosystem are the following. The first is:

\[ n_e=m_1=1,\qquad A_e=0,\quad C_e=1, \]

in which case we obtain \(w=x_e^0\), i.e. the external input \(w\) is constant. The second is:

\[ n_e=2,~m_1=1,\qquad A_e=\bbm {0&1\\-\omega _e^2&0},\quad C_e=\bbm {1&0}, \]

where \(\omega _e>0\) is given, which is the first order form of the second order equation \(\ddot {w}=-\omega _e^2w\) which means that the external input \(w\) is a sinusoid of frequency \(\omega _e\), i.e. \(w(t)=R\sin (\omega _e t-\varphi )\) for \(R\geq 0\) and \(\varphi \in \mR \).

Remark 9.4.  We can re-write the regulator equations as

\[ \bbm {I&0\\0&0}\bbm {\Pi \\V}A_e-\bbm {A&B_2\\C_1&0}\bbm {\Pi \\V}=\bbm {B_1\\D_{11}}C_e, \]

which shows that the regulator equations are a generalized Sylvester equation.

Using the Kronecker product and the vectorization of a matrix, this can in turn be written as the standard linear system

\[ \left (A_e^T\otimes \bbm {I_n&0_{n\times m_2}\\0_{p_1\times n}&0_{p_1\times m_2}} -I_{n_e}\otimes \bbm {A&B_2\\C_1&0_{p_1\times m_2}}\right ) \vecc \left (\bbm {\Pi \\V}\right ) =\vecc \left (\bbm {B_1C_e\\D_{11}C_e}\right ). \]

Remark 9.7.  Because of the stability requirement, stabilizability of \((A,B_2)\) is necessary for the full information output regulation and disturbance rejection problem to be solvable. If \(A_e\) has only eigenvalues with nonnegative real part, then solvability of the regulator equations is also necessary for the full information output regulation and disturbance rejection problem to be solvable.

Remark 9.9.  A partial converse of Proposition 9.8 is also true in case \(A_e\) has only eigenvalues with nonnegative real part: if the regulator equations are solvable for all \(B_1\), \(D_{11}\) and \(C_e\), then the surjectivity condition on \(\sbm {sI-A&-B_2\\C_1&0}\) must hold.

Remark 9.10.  If the eigenvalues of \(A_e\) are not eigenvalues of \(A\) and the transfer function \(C_1(sI-A)^{-1}B_2\) is surjective for all eigenvalues \(s\) of \(A_e\), then the surjectivity condition from Proposition 9.8 holds. This can be seen as follows. Let \(q\in \mR ^n\) and \(z\in \mR ^{p_1}\) be arbitrary and let \(s\) be an eigenvalue of \(A_e\). We then have

\[ \bbm {q\\z}=\bbm {sI-A&-B_2\\C_1&0}\bbm {x\\u}, \]

where \(u\) is a solution of

\[ C_1(sI-A)^{-1}B_2u=z-C_1(sI-A)^{-1}q, \]

(which exists by our surjectivity assumption) and

\[ x=(sI-A)^{-1}q+(sI-A)^{-1}B_2u. \]

When \(m_2=p_1=1\), then surjectivity of \(C_1(sI-A)^{-1}B_2\) simply means that it is nonzero. Therefore the condition becomes that eigenvalues of \(A_e\) should not be eigenvalues of \(A\) and should not be zeros of the transfer function \(C_1(sI-A)^{-1}B_2\). If this condition is satisfied, then the regulator equations are solvable (and the full information output regulation and disturbance rejection problem is therefore solvable) whereas if this condition is not satisfied, then we should expect problems with solvability.

Remark 9.11.  The regulator equations do not necessarily have a unique solution (but any solution will give a solution of the full information output regulation and disturbance rejection problem).

Example 9.12.  Consider the first order system

\[ T\dot {x}(t)+x(t)=Ku(t), \]

where \(T,K>0\). The objective is to find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }x(t)=r\) no matter what \(x(0)\) is.

To put this in the framework of the full information output regulation and disturbance rejection problem, define the performance output and exosystem

\[ z=x-w,\qquad \dot {x}_e=0,\qquad w=x_e,\qquad x_e(0)=r. \]

We then have \(w(t)=x_e(t)=r\) for all \(t\) so that \(z(t)=x(t)-r\) and the regulation requirement implies that \(\lim _{t\to \infty }x(t)=r\) no matter what \(x(0)\) is.

We have \(n=n_e=m_1=m_2=p_1=1\) and

\[ A=-\frac {1}{T},\quad B_1=0,\quad B_2=\frac {K}{T},\quad C_1=1,\quad D_{11}=-1,\quad A_e=0,\quad C_e=1. \]

The regulator equations are

\[ -\frac {1}{T}\Pi +\frac {K}{T}V=0,\qquad \Pi -1=0. \]

Hence \(\Pi =1\) and \(V=\frac {1}{K}\). Since \(A\) is asymptotically stable we can take \(F_2=0\). From Theorem 9.6 we have that (using that \(x_e=r\))

\[ u=\frac {r}{K}, \]

solves our problem.

We explicitly check that this indeed works. Substituting \(u=\frac {r}{K}\) into the differential equation gives

\[ T\dot {x}(t)+x(t)=r, \qquad x(0)=x^0, \]

which has solution \(x(t)=r+(x^0-r)\e ^{-t/T}\), which we see indeed satisfies \(\lim _{t\to \infty }x(t)=r\) no matter what \(x^0\) is.

We note that with a general \(F_2\) we obtain

\[ u=\left (\frac {1}{K}-F_2\right )r+F_2x, \]

and substituting this into the differential equation gives

\[ T\dot {x}(t)+\left (1-KF_2\right )x(t)=\left (1-KF_2\right )r, \]

which has general solution \(x(t)=r+(x^0-r)\e ^{-t(1-KF_2)/T}\). Therefore if \(F_2\) is such that \(1-KF_2>0\) (i.e. \(A+B_2F_2\) is asymptotically stable), then we again have \(\lim _{t\to \infty }x(t)=r\) no matter what \(x^0\) is. We now have the freedom to choose \(F_2\) in such as a way as to make this convergence faster.

Example 9.13.  Consider the second order system

\[ \ddot {q}(t)+2\zeta \omega _0\dot {q}(t)+\omega _0^2q(t)=K\omega _0^2u(t), \]

where \(\omega _0,K>0\) and \(\zeta >0\). The objective is to find a control \(u\) such that for a given \(r\in \mR \) we have \(\lim _{t\to \infty }\dot {q}(t)=r\) no matter what \(x(0)\) is.

We have \(n=2\), \(m_1=m_2=p_1=1\) and

\[ x=\bbm {q\\\dot {q}},\quad A=\bbm {0&1\\-\omega _0^2&-2\zeta \omega _0},\quad B_1=\bbm {0\\0},\quad B_2=\bbm {0\\K\omega _0^2},\quad C_1=\bbm {0&1},\quad D_{11}=-1. \]

We note that the transfer function \(C_1(sI-A)^{-1}B_2\) is (by Section 5.1)

\[ G(s)=\frac {K\omega _0^2s}{s^2+2\zeta \omega _0s+\omega _0^2}, \]

which has a zero at \(s=0\) which by Remark 9.1 will be an eigenvalue of \(A_e\). Therefore by Remark 9.10 we expect there to be an issue with solvability of the problem.

Consistent with Remark 9.1, we choose

\[ n_e=1,\quad A_e=0,\qquad C_e=1. \]

The regulator equations are

\[ \bbm {0&1\\-\omega _0^2&-2\zeta \omega _0}\bbm {\Pi _1\\\Pi _2} +\bbm {0\\K\omega _0^2}V =0,\qquad \bbm {0&1}\bbm {\Pi _1\\\Pi _2}-1=0. \]

It follows from the second equation that \(\Pi _2=1\) so that the first equation is

\[ \bbm {1\\-\omega _0^2\Pi _1-2\zeta \omega _0+K\omega _0^2V}=\bbm {0\\0}. \]

We see that the first component of this gives the contradiction \(1=0\). Therefore the regulator equations do not have a solution (as we expected might be the case).

We instead consider

\[ n_e=2,\qquad A_e=\bbm {0&1\\0&0},\qquad C_e=\bbm {0&1}. \]

The regulator equations are

\begin{gather*} \bbm {0&1\\-\omega _0^2&-2\zeta \omega _0}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}} +\bbm {0\\K\omega _0^2}\bbm {V_1&V_2} =\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}\bbm {0&1\\0&0}, \\ \bbm {0&1}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}-\bbm {0&1}=0. \end{gather*} The second of these equations give \(\Pi _{21}=0\), \(\Pi _{22}=1\). The first equation then is

\[ \bbm {0&1\\-\omega _0^2&-2\zeta \omega _0}\bbm {\Pi _{11}&\Pi _{12}\\0&1} +\bbm {0&0\\K\omega _0^2V_1&K\omega _0^2V_2} =\bbm {0&\Pi _{11}\\0&0}, \]

which is

\[ \bbm {0&1\\-\omega _0^2\Pi _{11}&-\omega _0^2\Pi _{12}-2\zeta \omega _0} +\bbm {0&0\\K\omega _0^2V_1&K\omega _0^2V_2} =\bbm {0&\Pi _{11}\\0&0}, \]

from which we deduce \(\Pi _{11}=1\), \(V_1=\frac {1}{K}\) and \(\Pi _{12}=KV_2-2\zeta \omega _0^{-1}\). We see that now the regulator equations have a solution, but that this solution is not unique (as \(V_2\in \mR \) is arbitrary). We have

\[ \Pi =\bbm {1&KV_2-2\zeta \omega _0^{-1}\\0&1},\qquad V=\bbm {\frac {1}{K}&V_2}. \]

Since \(A\) is asymptotically stable, we can take \(F_2=0\) and we obtain the control

\[ u=\frac {1}{K}x_{e,1}+V_2x_{2,e}. \]

With the initial condition \(x_e^0=\sbm {0\\r}\) we obtain \(x_e(t)=\sbm {rt\\r}\) and \(w(t)=r\), so that the above control is

\[ u(t)=\frac {rt}{K}+V_2r, \]

where \(V_2\in \mR \) is an arbitrary constant. We note that this arbitrary constant arises because the transfer function from \(u\) to \(z\) has a zero at \(s=0\), so that constant inputs result in a zero output (therefore the arbitrary term \(V_2r\) in \(u\) does not affect the output and hence for the purposes of output regulation is indeed irrelevant).

Example 9.14.  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 \(\omega _e,R>0\) and \(\varphi \in \mR \). Note that this problem has an obvious solution: \(u=-\frac {K_1}{K_2}w\) cancels out the disturbance \(w\) completely (regardless of whether it is a sinusoid or not).

This is of the standard form with

\[ A=\frac {-1}{T},\quad B_1=\frac {K_1}{T},\quad B_2=\frac {K_2}{T},\quad C_1=1,\quad D_{11}=0. \]

Consistent with Remark 9.1, we introduce the exosystem

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

The regulator equations are

\[ \frac {-1}{T}\bbm {\Pi _{11}&\Pi _{12}} +\frac {K_1}{T}\bbm {1&0} +\frac {K_2}{T}\bbm {V_1&V_2} =\bbm {\Pi _{11}&\Pi _{12}}\bbm {0&1\\-\omega _e^2&0}, \qquad \bbm {\Pi _{11}&\Pi _{12}}=0, \]

which has solution \(\Pi =0\) and \(V_1=-\frac {K_1}{K_2}\) and \(V_2=0\). Since \(A\) is asymptotically stable we can choose \(F_2=0\). The control therefore is

\[ u=-\frac {K_1}{K_2}w, \]

which is consistent with what we obtained above.

Example 9.15.  Consider the second order system

\[ \ddot {q}(t)+2\zeta \omega _0\dot {q}(t)+\omega _0^2q(t)=K_1\omega _0^2w(t)+K_2\omega _0^2u(t), \]

where \(\omega _0,K_1,K_2>0\) and \(\zeta \geq 0\). The objective is to find a control \(u\) such that \(\lim _{t\to \infty }\dot {q}(t)=0\) for all initial conditions and all \(w\) which are known sinusoids with a given frequency \(\omega _e>0\).

Note that this problem has an obvious solution: \(u=-\frac {K_1}{K_2}w\) cancels out the disturbance \(w\) completely (regardless of whether it is a sinusoid or not).

We have

\begin{gather*} A=\bbm {0&1\\-\omega _0^2&-2\zeta \omega _0},\qquad B_1=\bbm {0\\K_1\omega _0^2},\qquad B_2=\bbm {0\\K_2\omega _0^2},\qquad C_1=\bbm {0&1},\qquad D_{11}=0, \\ A_e=\bbm {0&1\\-\omega _e^2&0},\qquad C_e=\bbm {1&0}. \end{gather*} The regulator equations are

\begin{gather*} \bbm {0&1\\-\omega _0^2&-2\zeta \omega _0}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}} +\bbm {0\\K_1\omega _0^2}\bbm {1&0} +\bbm {0\\K_2\omega _0^2}\bbm {V_1&V_2} =\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}\bbm {0&1\\-\omega _e^2&0}, \\ \bbm {0&1}\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}}=\bbm {0&0}. \end{gather*} From the second equation we obtain \(\Pi _{21}=\Pi _{22}=0\). The first equation then is

\[ \bbm {0&0\\-\omega _0^2\Pi _{11}&-\omega _0^2\Pi _{12}} +\bbm {0&0\\K_1\omega _0^2&0} +\bbm {0&0\\K_2\omega _0^2V_1&K_2\omega _0^2V_2} =\bbm {-\omega _e^2\Pi _{12}&\Pi _{11}\\0&0}. \]

The top row gives \(\Pi _{11}=\Pi _{12}=0\). The bottom row then gives \(V_1=-\frac {K_1}{K_2}\) and \(V_2=0\). If \(\zeta >0\), then \(A\) is stable and we can therefore choose \(F_2=0\). We then obtain the control \(u=Vx_e\), which is

\[ u=-\frac {K_1}{K_2}w, \]

which is the one we already saw above. If \(\zeta =0\), then we can for example choose \(F_2=\bbm {0&\frac {-2\zeta _0}{K_2\omega _0}}\) for \(\zeta _0>0\) (so that \(A+B_2F_2\) is the matrix corresponding to a second order system with natural frequency \(\omega _0\) and damping ratio \(\zeta _0\), which we know is stable) and obtain the feedback

\[ u=-\frac {K_1}{K_2}w-\frac {2\zeta _0}{K_2\omega _0}\dot {q}. \]

Remark 9.16.  The solutions in Example 9.14 and Example 9.15 are obvious. However, this solution assumes that we know the disturbance \(w\), which is unrealistic.

Remark 9.17.  If we ignore the disturbance, then

\[ A_e=\bbm {0&0\\0&0},\qquad C_e=\bbm {1&0\\0&1}, \]

and the solution of the regulator equations is

\[ \Pi =\bbm {1&0\\1&0\\0&1},\qquad V=\bbm {d_1&-1\\d_2&1}. \]

With \(F_2=0\), this gives the control

\[ u_1=d_1r_v-r_T,\qquad u_2=d_2r_v+r_T. \]

With this control the equations become

\begin{align*} M_1\frac {d}{dt}(v_1-r_v)+d_1(v_1-r_v)-(T-r_T)&=0, \\ M_2\frac {d}{dt}(v_2-r_v)+d_2(v_2-r_v)+(T-r_T)&=0, \\ \frac {d}{dt}(T-r_T)+k\left [(v_1-r_v)-(v_2-r_v)\right ]=kv_e. \end{align*} This is the same system that we started with, but with the replacements

\[ v_1\leftrightarrow v_1-r_v,\qquad v_2\leftrightarrow v_2-r_v,\qquad T\leftrightarrow T-r_T. \]

In particular, the Bode plot from Section 6.5 is still relevant, but should now be interpreted as being from the disturbance \(v_e\) to the errors \(v_1-r_v\), \(v_2-r_v\) and \(T-r_T\).