Example 9.1. Consider the first order scalar differential equation

\[ \dot {x}(t)+x(t)=w(t)+u(t). \]

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: \(u=-w\) cancels out the disturbance \(w\) completely (regardless of whether it is a sinusoid or not).

Since the objective is \(\lim _{t\to \infty }x(t)=0\) no matter what \(x(0)\) is, the performance output is

\[ z=x. \]

Consistent with Remark 8.1, we introduce the exo-system

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

since this gives precisely \(w(t)=R\sin (\omega _e t-\varphi )\) as solutions of \(\dot {x}_e=A_ex_e\), \(w=C_ex_e\).

With the above we obtain

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

The regulator equations are

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

which has solution \(\Pi =0\) and \(V_1=-1\) and \(V_2=0\). Since \(A\) is asymptotically stable we can choose \(F_1=0\). The control therefore is

\[ u=-w, \]

which is consistent with what we obtained above.

Example 9.2. Consider the second order scalar differential equation

\[ \ddot {q}(t)+2\zeta \dot {q}(t)+q(t)=w(t)+u(t), \]

where \(\zeta \geq 0\). 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: \(u=-w\) cancels out the disturbance \(w\) completely (regardless of whether it is a sinusoid or not). If \(\zeta =0\) this does not satisfy the stability requirement though.

We write our second order scalar differential equation in first order form with state \(x:=\sbm {q\\\dot {q}}\). Since the regulation requirement is \(\lim _{t\to \infty }\dot {q}(t)=0\), we choose \(z=\dot {q}=x_2\). Therefore we have

\begin{gather*} A=\bbm {0&1\\-1&-2\zeta },\qquad B_1=\bbm {0\\1},\qquad B_2=\bbm {0\\1},\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*} where we have used the same exo-system as in Example 9.1 since the disturbances \(w\) are the same as there.

The regulator equations are

\begin{gather*} \bbm {0&1\\-1&-2\zeta }\bbm {\Pi _{11}&\Pi _{12}\\\Pi _{21}&\Pi _{22}} +\bbm {0\\1}\bbm {1&0} +\bbm {0\\1}\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

\[ \bbm {0&0\\-\Pi _{11}&-\Pi _{12}} +\bbm {0&0\\1&0} +\bbm {0&0\\V_1&V_2} =\bbm {-\omega _e^2\Pi _{12}&\Pi _{11}\\0&0}. \]

The top row gives \(\Pi _{11}=\Pi _{12}=0\). The bottom row then gives \(V_1=-1\) and \(V_2=0\). If \(\zeta >0\), then \(A\) is stable and we can therefore choose \(F_1=0\). We then obtain the control \(u=Vx_e\), which is

\[ u=-w, \]

which is the one we already saw above. If \(\zeta =0\), then we can for example choose \(F_1=\bbm {0&-2\zeta _0}\) for \(\zeta _0>0\) (so that \(A+B_2F_1\) is the matrix corresponding to a second order scalar differential equatin with damping ratio \(\zeta _0\), which we know is stable) and obtain the feedback

\[ u=-w-2\zeta _0\dot {q}. \]

Substituting this gives \(\ddot {q}+2\zeta _0\dot {q}+q=0\) which indeed satisfies both requirements.

Remark 9.3. The solutions in Example 9.1 and Example 9.2 are obvious. However, this solution assumes that we know the disturbance \(w\), which is unrealistic. In Chapter 11 we solve the case where we do not know the disturbance \(w\) and that solution will build on the calculations in this chapter.