Why would I use three state variables for a second order system? [state space modeling]

Question

I'm currently being asked to write a state space model of a falling object (equation attached).

I had taken a linear systems course five years ago in which I did a lot of conversion between high order differential equations and systems of first order linear equations. However, this task seems odd for a few reasons:

1. Most nth ordered differential equations that I had converted only required n state variables. For example, if I had a second order differential equation, I needed two state variables. In this problem, it's asking for 3 state variable which seems odd. Is the third state variable even necessary?

2. This differential equation directly includes the independent variable (time) itself. It also includes an initial condition p0 which seems odd.

I'm able to derive a solution to this problem, however it just seems very odd in how I'm doing it. This is my solution:

Answer 1

I'll point out that your solution: $$ \dot{X} = \left[\begin{matrix} \dot{p} \\ \ddot{p} \\ \dddot{p}\end{matrix}\right] = \left[\begin{matrix} x_2 \\ x_3 \\ 0 \end{matrix}\right]= \left[\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix}\right]\\ $$

multiplies out to:

$$ \dot{p} = \dot{p} \\ \ddot{p} = \ddot{p} \\ $$

which is a trivial statement and doesn't express the system you presented.

I am not a fan of this question (not you, but the question you were asked to solve.) As others have mentioned, there's no reason to have acceleration $\ddot{p}$ as a state, because you get acceleration when you take the derivative of your state vector if your state vector includes speed. That is, if $x = \left[\begin{matrix} p \\ \dot{p} \end{matrix}\right]$, then $\dot{x} = \left[\begin{matrix} \dot{p} \\ \ddot{p}\end{matrix}\right]$.

Further, in order to get the solution here, I believe the question is wanting you to rearrange the equation in the form of $\ddot{x} = \text{<stuff>}$, but you wind up with an expression that implies that acceleration is a function of position and speed, which is not the case. Position is a function of speed and acceleration.

Answer 2

Initial conditions are important, let's assume this is a rigid body falling from 10m height at time 't_0', then to know its position at any time 't' I must know its initial position. Also, the third state (acceleration) is redundant information as it has been already described that body is falling at a constant acceleration, which simply means jerk (rate of change of accel.) is zero.

P.S: I have not seen any real system so far where we need to have acceleration as a system state. In system dynamics we are mostly concerned with forces/torques, which lead to accelerations, so typical state vectors include positions and velocities as system states.