I have a DC motor system with three states. The states are:

  • $x_1$ = position (degrees),
  • $x_2$ = velocity (degrees per second)
  • $x_3$ = current (Amps).

My plant input $u(t)$ is measured in volts.

The dynamics are such that:

$$\begin{gather} \frac{dX}{dt} = AX + Bu(t) \\ Y = CX + Du(t) \end{gather}$$ And: $$\begin{align} A &= \left[\begin{matrix} 0 & 1 & 0 \\ 0 & -\dfrac{b_m}{J_m} & \dfrac{K_m}{J_m} \\ 0 & -\dfrac{K_m}{L} & -\dfrac{R}{L} \end{matrix}\right] \\ B &= \left[\begin{matrix} 0 \\ 0 \\ \dfrac{1}{L} \end{matrix}\right] \\ C &= \left[\begin{matrix} 1 & 0 & 0 \end{matrix}\right] \\ D &= [0] \end{align}$$

Where $J_m$ is the rotor inertia, $b_m$ is the viscous damping, $R$ is the armature resistance, $k_m$ is the motor constant, and $L$ is the inductance.

My objective is to control the position of my DC motor.

The rank of my observability matrix is equal to the number of states in my system, so the system is fully observable.

$$Q = \text{rank} \left(\left[\begin{matrix} C \\ CA \\ CA^2 \end{matrix}\right]\right)$$

I am measuring the position using an encoder and hope to employ a reduced order observer to do this.

Partitioning the A matrix into quadrants yields the measured and observed dynamics:

$$A = \left[\begin{matrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix}\right]$$

In my case: $$\begin{align} A11 &= [0] \\ A12 &= \left[\begin{matrix} 1 & 0 \end{matrix}\right] \\ A21 &= \left[\begin{matrix} 0 \\ 0 \end{matrix}\right] \\ A22 &= \left[\begin{matrix} -\dfrac{b_m}{J_m} & \dfrac{K_m}{J_m} \\ -\dfrac{K_m}{L} & -\dfrac{R}{L} \end{matrix}\right] \end{align}$$

Now the Observability of my reduced state observer partition is:

$$Q_{Obs} = \text{rank} \left(\left[\begin{matrix} A_{12} \\ A_{12}A_{22} \end{matrix}\right]\right)$$

The rank of this is 2 (full rank).

I've seen in other projects that the following condition must be satisfied:

$$\begin{align} Q_{Obs} &\approx Q-(\text{number of states}-\text{number of measured states}) \\ 2 &\approx 3-(3-1) \\ 2 &\approx 1 \end{align}$$

  1. How many states need to be measured in a reduced order observer?
  2. Why is it that my system is fully observable, but I cannot measure all the states of my system?
  3. If I measure both position and velocity it seems that $Q_{Obs}=1$. Making the equation:

$$Q_{Obs} \approx Q-(\text{number of states}-\text{number of measured states})$$

simplify to:

$$\begin{align} 1 &\approx 3-(3-2) \\ 1 &\approx 2 \end{align}$$

Does this mean the system is not observable without measuring all three states?

4. If I assume the mechanical time constant is much slower than the electrical time constant (poles are closer to 0 in the real axis):

  • Can I measure only position and estimate velocity while neglecting the electrical dynamics?

  • Will this provide a robust estimator when implemented?

  • 1
    $\begingroup$ Like Suba Thomas said, it turns out you can measure or observe any combination of the states in a reduced order system. The confusion for me came while I was implementing the system. I was getting tripped up because the matrices in the error dynamics were often multiplied by scalars (due to odd number of states). The world makes sense once again. The expression in my 3rd question is bogus. $\endgroup$
    – Bret
    Commented Aug 20, 2016 at 2:08

1 Answer 1

  1. It is not always states that are measured. The measurements $y$ are combinations of states in general or a subset. In the reduced order observer, in this case, you have one measurement.
  2. The system being fully observable means that using the available measurement, all the states can be observed or indirectly measured.
  3. The equation does not make sense. But as I mention above you don't have to directly measure all the states.
  4. You can measure just the position and velocity and not neglect the electrical dynamics. Use the complete dynamics to estimate the current. This is the purpose of the observer.


I think the observability rank $Q_{Obs}$ is tripping you. Observability is a property of the system not the observer. 'Observability of my reduced state observer' is nonsense speak.

With one measurement the system is observable. That means using the knowledge of the system behavior ($A$, $B$, $C$, and $D$) together with the input signal $u$ and the measurement $y$ all the states can be estimated. The reduced state observer goes one step ahead: because one of the states is already measured that need not be estimated, only the remaining two need to estimated. You wouldn't use a reduced state observer if the measurement was very noisy.

If the system is observable with one measurement then it is definitely so with two. Again, if you have enough confidence in those measurements, they can be used directly and a reduced state observer can be used to estimate just the remaining state.

  • 1
    $\begingroup$ Thanks. Regarding 3: The equation should be an inequality, not the approximate sign. Since I have a single-input-single-output system, how can I estimate two states when only one can be directly measured from the output? For example, my error dynamics on my estimation should be two dimensional, but dependent on a one-dimensional output. But if I measure two states, and estimate one, my error dynamics are one dimensional, but it is dependent on a two dimensional output. I don't understand how this can be implemented. Is there some heuristic that must be satisfied as the equation in 3? $\endgroup$
    – Bret
    Commented Aug 19, 2016 at 20:36
  • 1
    $\begingroup$ I will expand the answer. A comment would be too long. $\endgroup$ Commented Aug 19, 2016 at 21:03

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