\section{Combined Controller-Estimator Compensator}
We start with combining a control law with a full-order estimator. We begin with writing the equations that govern the behavior of the dynamic system consisting of the plant model and the full-order estimator. The resulting system's order is $2n$, and the equations that describe the system have the form
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{\b x}(t)\\
\dot{\tilde{\b x}}(t)
\end{array}\right]&=&\left[\begin{array}{cc}
\b A & \b 0\\
\b L\b C & \b A-\b L\b C
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\tilde{\b x}(t)
\end{array}\right]+\left[\begin{array}{c}
\b B\\
\b B
\end{array}\right]\b u(t)\\
\b y(t) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\tilde{\b x}(t)
\end{array}\right].
\end{eqnarray*}
Suppose that we now apply the control law
\[
\b u(t)=-\b K\tilde{\b x}(t)+\b v(t)
\]
instead of the actual state-feedback control law. In the above, the vector $\b v$ denotes an external input signal. Combining the above two equations together yields a model of the closed-loop system driven by the combined controller-estimator compensator. We have
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{\b x}(t)\\
\dot{\tilde{\b x}}(t)
\end{array}\right]&=&\left[\begin{array}{cc}
\b A &  -\b B\b K\\
\b L\b C & \b A-\b L\b C-\b B\b K
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\tilde{\b x}(t)
\end{array}\right]+\left[\begin{array}{c}
\b B\\
\b B
\end{array}\right]\b v(t)\\
\b y(t) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\tilde{\b x}(t)
\end{array}\right].
\end{eqnarray*}
To analyze the closed-loop system it is convenient to perform a change of state variables using the following transformation matrix
\[
\left[\begin{array}{cc}
\b I_n & \b 0\\
-\b I_n & \b I_n
\end{array}\right].
\]
Note that
\[
\left[\begin{array}{cc}
\b I_n & \b 0\\
-\b I_n & \b I_n
\end{array}\right]^{-1}=\left[\begin{array}{cc}
\b I_n & \b 0\\
\b I_n & \b I_n
\end{array}\right].
\]
In the new coordinates the equations describing the closed-loop system are
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{\b x}(t)\\
\dot{\tilde{\b x}}(t)-\dot{\b x}(t)
\end{array}\right]&=&\left[\begin{array}{cc}
\b A -\b B\b K & -\b B\b K\\
\b 0 & \b A-\b L\b C
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\tilde{\b x}(t)-\b x(t)
\end{array}\right]+\left[\begin{array}{c}
\b B\\
\b 0
\end{array}\right]\b v(t)\\
\b y(t) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\tilde{\b x}(t)
\end{array}\right].
\end{eqnarray*}
Note that in the above system the subsystem corresponding to the error component $\b e(t)=\tilde{\b x}(t)-\b x(t)$ is uncontrollable. Furthermore, the $2n$ poles of the closed-loop system are equal to the individual eigenvalues of both $\b A-\b B\b K$ and $\b A-\b L\b C$, which means that the design of the control law is separated from the construction of the estimator. This is what is known as the {\em separation principle\/}. The closed-loop transfer function relating $\b Y(s)$ and $\b V(s)$ is
\begin{eqnarray*}
\b Y(s) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{cc}
s\b I_n -\b A +\b B\b K & \b B\b K\\
\b 0 & s\b I_n -\b A+\b L\b C
\end{array}\right]^{-1}\left[\begin{array}{c}
\b B\\
\b 0
\end{array}\right]\b V(s)\\
&=& \b C\left(s\b I_n -\b A +\b B\b K \right)^{-1}\b B\b V(s).
\end{eqnarray*}
The above expression is identical to that for the closed-loop system if we applied the state-feedback control law
\[
\b u(t)=-\b K\b x(t)+\b v(t).
\]
Thus, the combined estimator-controller compensator yields the same closed-loop transfer function as the actual state-feedback control law.

We now briefly discuss the issue of the estimator pole selection. Franklin, Powell, and Emami-Naeini~\cite{franklin86} recommend that the estimator poles, that is, the eigenvalues of the matrix $\b A-\b L\b C$, to be a factor of 2 to 6 times faster than the controller poles. Such a choice ensures a faster decay of the estimator error $\b e(t)=\tilde{\b x}(t)-\b x(t)$ compared with the desired controller dynamics. This in turn causes the controller poles to dominate the closed-loop system response. Since the estimator poles represent a measure of the speed with which the error $\b e(t)=\tilde{\b x}(t)-\b x(t)$ decays to zero, one would tend to assign estimator poles deep in the left-hand plane. However, fast decay requires large gains which may lead to saturation of some signals, and unpredictable nonlinear effects. If the estimator poles were slower than the controller poles, the closed-loop system response would be dominated by the estimator, an undesirable situation. As it is usual in engineering practice, the term compromise can be used to describe the process of constructing the final controller structure.

We now analyze the combined estimator-controller compensator that uses the reduced-order estimator. The closed-loop system driven by this type of compensator enjoys the same properties as in the case of the full-order estimator. To prove the above statement we begin with writing the equations describing the closed-loop system. We have
\begin{eqnarray*}
\dot{\b x}(t) &=& \b A\b x(t) +\b B\b u(t)\\
\dot{\b z}(t) &=& \b D\b z(t)+\b T\b B\b u(t)+\b G\b C\b x(t)\\
\b y(t) &=& \b C\b x(t)\\
\b u(t) &=& -\b K\tilde{\b x}(t)+\b v(t),
\end{eqnarray*}
where
\[
\tilde{\b x}=\left[\begin{array}{c}
\b T\\
\b C
\end{array}\right]^{-1}\left[\begin{array}{c}
\b z\\
\b y
\end{array}\right].
\]
Let
\[
\left[\begin{array}{c}
\b T\\
\b C
\end{array}\right]^{-1}=\left[\begin{array}{cc}
\b M,&\b N
\end{array}\right],
\]
where $\b M\in\mb{\Bb R}^{n \times (n-p)}$ and $\b N\in\mb{\Bb R}^{n \times p}$. Note that
\[
\b M\b T+\b N\b C=\b I_n.
\]
With the above notation we can represent the control input in the form
\begin{eqnarray*}
\b u(t) &=& -\b K\left[\begin{array}{cc}
\b M, & \b N
\end{array}\right]\left[\begin{array}{c}
\b z(t)\\
\b y(t)
\end{array}\right]+\b v(t)\\
&=& -\b K\b M\b z(t)-\b K\b N\b y(t)+\b v(t).
\end{eqnarray*}
Using the above equations, and noting that $\b y(t)=\b C\b x(t)$, we represent equations that describe the closed-loop system as
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{\b x}(t)\\
\dot{\b z}(t)
\end{array}\right]&=&\left[\begin{array}{cc}
\b A & \b 0\\
\b G\b C & \b D
\end{array}\right]\left[\begin{array}{c}
{\b x}(t)\\
{\b z}(t)
\end{array}\right]+\left[\begin{array}{c}
\b B\\
\b T\b B
\end{array}\right]\left(-\b K\b M\b z(t)-\b K\b N\b y(t)+\b v(t) \right)\\
&=& \left[\begin{array}{cc}
\b A-\b B\b K\b N\b C & -\b B\b K\b M\\
\b G\b C-\b T\b B\b K\b N\b C & \b D-\b T\b B\b K\b M
\end{array}\right]\left[\begin{array}{c}
{\b x}(t)\\
{\b z}(t)
\end{array}\right]+\left[\begin{array}{c}
\b B\\
\b T\b B
\end{array}\right]\b v(t)\\
\b y(t) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\b z(t)
\end{array}\right].
\end{eqnarray*}
To proceed further we perform a transformation of the state variable of the above system. We use the following transformation matrix
\[
\left[\begin{array}{cc}
\b I_n & \b 0\\
-\b T & \b I_{n-p}
\end{array}\right].
\]
Note that
\[
\left[\begin{array}{cc}
\b I_n & \b 0\\
-\b T & \b I_{n-p}
\end{array}\right]^{-1}=\left[\begin{array}{cc}
\b I_n & \b 0\\
\b T & \b I_{n-p}
\end{array}\right].
\]
In the new coordinates the equations describing the closed-loop system become
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{\b x}(t)\\
\dot{\b z}(t)-\b T\dot{\b x}(t)
\end{array}\right]&=&\left[\begin{array}{cc}
\b A-\b B\b K\b N\b C-\b B\b K\b M\b T & -\b B\b K\b M\\
-\b T\b A+\b G\b C+\b D\b T & \b D
\end{array}\right]\left[\begin{array}{c}
{\b x}(t)\\
{\b z}(t)-\b T\b x(t)
\end{array}\right]\\
&{}&+\left[\begin{array}{c}
\b B\\
\b 0
\end{array}\right]\b v(t)\\
\b y(t) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\b z(t)-\b T\b x(t)
\end{array}\right].
\end{eqnarray*}
Taking into account the equations
\begin{eqnarray*}
\b T\b A-\b D\b T-\b G\b C &=& \b 0\\
\b M\b T &=& \b I_n -\b N\b C,
\end{eqnarray*}
we obtain
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{\b x}(t)\\
\dot{\b z}(t)-\b T\dot{\b x}(t)
\end{array}\right]&=&\left[\begin{array}{cc}
\b A-\b B\b K & -\b B\b K\b M\\
\b 0 & \b D
\end{array}\right]\left[\begin{array}{c}
{\b x}(t)\\
{\b z}(t)-\b T\b x(t)
\end{array}\right]+\left[\begin{array}{c}
\b B\\
\b 0
\end{array}\right]\b v(t)\\
\b y(t) &=& \left[\begin{array}{cc}
\b C & \b 0
\end{array}\right]\left[\begin{array}{c}
\b x(t)\\
\b z(t)-\b T\b x(t)
\end{array}\right].
\end{eqnarray*}
It is now apparent that the separation principle also holds if we use the reduced-order estimator in constructing the compensator, that is, the selection of the gain matrix $\b K$ that assigns closed-loop poles is independent, separate, from the selection of the estimator gain matrix $\b L$ that assigns the estimator poles. Furthermore, the transfer function matrix relating $\b Y(s)$ and $\b V(s)$ in the closed-loop system with the reduced-order estimator is exactly the same as the one that would be obtained using actual state-feedback control law $\b u(t)=-\b K\b x(t)+\b v(t)$.
\begin{example}\rm
Consider the following diagram of an armature controlled DC motor, where the system parameters are: $R_a=5\,\Omega$, $L_a=200\,mH$, $K_b=0.1\,V/rad/sec$, $K_i=0.1\,Nm/A$, the gear ratio $N_1/N_2=1/50$. The armature inertia is $2\times10^{-3}\,kgm^2$. The load of $10\,kg$ is located at an effective radius of $0.2\,m$. The gear inertia and friction are negligible.
\begin{figure}
\vspace{4.5in}
\special{bmp:c:/mybook/hw2fig2.bmp x=6in y=4.5in}
\caption{Block diagram of a DC motor system}
\label{fig:open_dc}
\end{figure}
Let $x_1=i_a$, $x_2=\theta$, $x_3=\dot{\theta}=\omega$, $u=e_a$, and $y=\theta_l$.

We first construct the state space model of the DC motor system. We start with writing the equation relating torque to angular acceleration. We have
\[
T_m=J_{eq}\ddot{\theta},
\]
where
\begin{eqnarray*}
J_{eq}&=&J_{armature}+\left(N_1/N_2 \right)^2J_{load}\\
&=&2\times10^{-3}\,kgm^2+(1/50)^2\times10\times(0.2)^2\,kgm^2\\
&=&2.16\times10^{-3}.
\end{eqnarray*}
Kirchhoff's voltage law applied to the armature circuit gives
\[
R_ai_a+L_a\frac{di_a}{dt}+K_b\frac{d\theta}{dt}=e_a.
\]
We also have
\begin{eqnarray*}
T_m &=& K_ii_a\\
&=& J_{eq}\ddot{\theta}.
\end{eqnarray*}
Taking into account the definition of the state variables, we represent the above equations in state-space format
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{x}_1\\
\dot{x}_2\\
\dot{x}_3
\end{array}\right] &=& \left[\begin{array}{ccc}
-\frac{R_a}{L_a} & 0 & -\frac{K_b}{L_a}\\
0 & 0 & 1\\
\frac{K_i}{J_{eq}} & 0 & 0
\end{array}\right]\left[\begin{array}{c}
{x}_1\\
{x}_2\\
{x}_3
\end{array}\right]+\left[\begin{array}{c}
\frac{1}{L_a}\\
0\\
0
\end{array}\right]u\\
y &=& \left[\begin{array}{ccc}
0 & \frac{N_1}{N_2} & 0
\end{array}\right]\left[\begin{array}{c}
{x}_1\\
{x}_2\\
{x}_3
\end{array}\right].
\end{eqnarray*}
Substituting the given parameter values we get
\begin{eqnarray*}
\left[\begin{array}{c}
\dot{x}_1\\
\dot{x}_2\\
\dot{x}_3
\end{array}\right] &=& \left[\begin{array}{ccc}
-25 & 0 & -0.5\\
0 & 0 & 1\\
46.296 & 0 & 0
\end{array}\right]\left[\begin{array}{c}
{x}_1\\
{x}_2\\
{x}_3
\end{array}\right]+\left[\begin{array}{c}
5\\
0\\
0
\end{array}\right]u\\
y &=& \left[\begin{array}{ccc}
0 & 0.02 & 0
\end{array}\right]\left[\begin{array}{c}
{x}_1\\
{x}_2\\
{x}_3
\end{array}\right].
\end{eqnarray*}


Our next step is to design a state-feedback controller $u=-\b k\b x + v$, such that the eigenvalues of $\b A-\b b\b k$ are
\[
-1+j,-1-j,-10.
\]
To allocate the poles into the desired locations we first transform the system model into the controller companion form. We first form the controllability matrix
\[
\left[\begin{array}{ccc}
\b b, & \b A\b b, & \b A^2\b b
\end{array}\right]=\left[\begin{array}{ccc}
5 & -125 & 3009.3\\
0 & 0 & 231.5\\
0 & 231.5 &-5787
\end{array}\right]
\]
The system is controllable. The last row of the inverse of the controllability matrix is
\[
\b q_1=\left[\begin{array}{ccc}
0 & 0.0043 & 0
\end{array}\right],
\]
and hence the transformation matrix we are seeking has the form
\begin{eqnarray*}
\b Q &=& \left[\begin{array}{c}
\b q_1\\
\b q_1\b A\\
\b q_1\b A^2
\end{array}\right]\\
&=& \left[\begin{array}{ccc}
0 & 0.0043 & 0\\
0 & 0 & 0.0043\\
0.2 & 0 & 0
\end{array}\right].
\end{eqnarray*}
The system's matrices in the new coordinates are
\[
\tilde{\b A} = \b Q\b A\b Q^{-1}
= \left[\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & -23.148 & -25
\end{array}\right],
\]
\[
\tilde{\b b} = \b Q\b b
= \left[\begin{array}{c}
0\\
0\\
1
\end{array}\right],
\]
and
\[
\tilde{\b c} = \b c\b Q^{-1}
= \left[\begin{array}{ccc}
4.63 & 0 & 0
\end{array}\right].
\]
The desired closed-loop characteristic polynomial $\alpha_c(s)$ is
\begin{eqnarray*}
\alpha_s(s) &=& (s+1-j)(s+1+j)(s+10)\\
&=& s^3+12s^2+22s+20.
\end{eqnarray*}
We now want to find $\tilde{\b k}$ so that
\[
\tilde{\b A}-\tilde{\b b}\tilde{\b k}=\left[\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
-20 & -22 & -12
\end{array}\right].
\]
We have
\[
\tilde{\b k}=\left[\begin{array}{ccc}
20 & -1.148 & -13
\end{array}\right],
\]
and
\begin{eqnarray*}
\b k &=& \tilde{\b k}\b Q\\
&=&\left[\begin{array}{ccc}
-2.6 & 0.0864 & -0.005
\end{array}\right].
\end{eqnarray*}
Hence
\[
\b A-\b b \b k = \left[\begin{array}{ccc}
-12 & -0.432 & -0.4752\\
0 & 0 & 1\\
46.294 & 0 & 0
\end{array}\right].
\]
The transfer function of the closed-loop system is
\[
\frac{Y(s)}{V(s)}=\frac{4.6296}{s^3+12s^2+22s+20}.
\]
The step response of the closed-loop system is shown in Figure~\ref{fig:motcl}.
\begin{figure}
\vspace{4in}
\hskip.5in\special{bmp:c:/ee680/motcl.bmp x=5.33in y=4in}
\caption{Step response of the closed-loop DC motor system.}
\label{fig:motcl}
\end{figure}


Next, we design the full-order estimator placing the estimator poles at $\{-4, -5+2j, -5-2j\}$, and then synthesize the combined controller-estimator compensator.
To compute the estimator gain vector $\b l$ we first transform the pair $(\b A, \b c)$ into the observer companion form. This is equivalent to transforming the pair $(\b A^T, \b c^T)$ into the controller companion form. The transformation matrix that brings the pair $(\b A, \b c)$ into its observer companion form is
\[
\hat{\b Q}= \left[\begin{array}{ccc}
1.1 & -27 & 650\\
0 & 0 & 50\\
0 & 50 & -1250
\end{array}\right].
\]
The matrices $\b A$ and $\b c$ in the new coordinates have the form:
\[
\hat{\b A}=\hat{\b Q}^{-1}\b A\hat{\b Q}=\left[\begin{array}{ccc}
0 & 0 & 0\\
1 & 0 & -23.1481\\
0 & 1 & -25
\end{array}\right],
\]
and
\[
\hat{\b c}=\b c\hat{\b Q}=\left[\begin{array}{ccc}
0 & 0 & 1
\end{array}\right].
\]
The desired characteristic polynomial of the estimator is
\begin{eqnarray*}
\det\left(\b I_3-\b A+\b l\b c\right) &=& (s+4)(s+5-2j)(s+5+2j)\\
&=& s^3+14s^2+69s+116.
\end{eqnarray*}
We now find $\hat{\b l}$ so that
\[
\hat{\b A}-\hat{\b l}\hat{\b c}=\left[\begin{array}{ccc}
0 & 0 & -116\\
1 & 0 & -69\\
0 & 1 & -14
\end{array}\right].
\]
We have
\[
\hat{\b l}=\left[\begin{array}{c}
116\\
45.8519\\
-11
\end{array}\right],
\]
and
\[
\b l=\hat{\b Q}\hat{\b l}=\left[\begin{array}{c}
-8263\\
-550\\
16043
\end{array}\right].
\]
The dynamics of the full order state estimator are described by
\[
\dot{\tilde{\b x}}=(\b A-\b l\b c)\tilde{\b x}+\b b u+\b l y,
\]
that is,
\[
\dot{\tilde{\b x}}=\left[\begin{array}{ccc}
-25 & 165.2544 & -0.5\\
0 & 11 & 1\\
46.2963 & -320.8519 & 0
\end{array}\right]\tilde{\b x} + \left[\begin{array}{c}
5\\
0\\
0
\end{array}\right]u+\left[\begin{array}{c}
-8263\\
-550\\
16043
\end{array}\right]y.
\]
We then connect the full-order estimator to the DC motor system thus obtaining a combined controller-full-order-estimator compensator. In Figure~\ref{fig:mobser} we show a plot of $x_1$ and its estimate for the closed-loop system with the estimator in the loop, where $v=0$, $\b x(0)=\left[\begin{array}{ccc}
4 & 2 & -1
\end{array}\right]^T$, and $\tilde{\b x}(0)=\b 0$.
\begin{figure}
\vspace{4in}
\hskip.5in\special{bmp:c:/ee680/mobser.bmp x=5.33in y=4in}
\caption{A plot of the first state, marked by 1, and its estimate, marked by 2, versus time for the closed-loop system with the full-order estimator in the loop.}
\label{fig:mobser}
\end{figure}


Our last part of this example is to design the reduced-order state estimator. We choose the reduced-order estimator poles at $\{-5+2j, -5-2j\}$. The reduced-order estimator dynamics are given by
\[
\dot{\b z}=\b D \b z+\b E u+ \b G y.
\]
Let
\[
\b G=\left[\begin{array}{c}
100\\
100
\end{array}\right].
\]
The estimator's poles are supposed to be $-5+2j$ and $-5-2j$. Thus, the desired characteristic polynomial of $\b D$ should be
\[
\det(s\b I_2-\b D)=s^2+10s+29.
\]
Let
\[
\b D=\left[\begin{array}{cc}
0 & 1\\
-29 & -10
\end{array}\right].
\]
We now solve the following matrix equation for $\b T$:
\[
-\b D\b T+\b T\b A=\b G\b c.
\]
Let $\b G_c=\b G\b c$, then using MATLAB we can solve the above equation using the command
\begin{verbatim}
t=(-kron(eye(3),D)+kron(A',eye(2)))\Gc(:);
T=[t(1:2) t(3:4) t(5:6)]
\end{verbatim}
to get
\[
\b T=\left[\begin{array}{ccc}
-0.5493 & 0.7586 & -0.2507\\
2.1271 & -2.0000 & 1.0332
\end{array}\right].
\]
Note that
\[
\det\left[\begin{array}{c}
\b T\\
\b c
\end{array}\right]=6.8691e-04 \ne 0,
\]
and
\[
\left[\begin{array}{c}
\b T\\
\b c
\end{array}\right]^{-1}=\left[\begin{array}{ccc}
-30.0840 & -7.2980 & 411.3200\\
0 & 0 & 50\\
61.9338 & 15.9921 & -750
\end{array}\right].
\]
We also have
\begin{eqnarray*}
\b E &=& \b T\b b\\
&=& \left[\begin{array}{c}
-2.7463\\
10.6357
\end{array}\right].
\end{eqnarray*}
Hence, the dynamics of the reduced order estimator are described by
\[
\dot{\b z}=\left[\begin{array}{cc}
0 & 1\\
-29 & -10
\end{array}\right]\b z+\left[\begin{array}{c}
-2.7463\\
10.6357
\end{array}\right]u+\left[\begin{array}{c}
100\\
100
\end{array}\right]y.
\]
An estimate of the entire state can be obtained from
\begin{eqnarray*}
\hat{\b x} &=& \left[\begin{array}{c}
\b T\\
\b c
\end{array}\right]^{-1}\left[\begin{array}{c}
\b z\\
y
\end{array}\right]\\
&=& \left[\begin{array}{ccc}
-30.0840 & -7.2980 & 411.3200\\
0 & 0 & 50\\
61.9338 & 15.9921 & -750
\end{array}\right]\left[\begin{array}{c}
\b z\\
y
\end{array}\right].
\end{eqnarray*}
A plot of $x_1$ and its estimate versus time for the closed-loop system with the combined controller-reduced-order-estimator compensator is depicted in Figure~\ref{fig:mrobs}, where $v=0$, $\b x(0)=\left[\begin{array}{ccc}
4 & 2 & -1
\end{array}\right]^T$, and $\tilde{\b z}(0)=\b 0$.
\begin{figure}
\vspace{4in}
\hskip.5in\special{bmp:c:/ee680/mrobs.bmp x=5.33in y=4in}
\caption{A plot of the first state, marked by 1, and its estimate, marked by 2, versus time for the closed-loop system with the reduced-order estimator in the loop.}
\label{fig:mrobs}
\end{figure}
The step responses of the closed-loop systems with the full-order, as well as with the reduced-order estimators in the loop, are the same as the step response of the closed-loop system with the state-feedback controller $u=-\b k\b x+v$, which nicely illustrates our analysis of the combined controller-estimator compensators.
\end{example}