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

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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:

enter image description here

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    $\begingroup$ Homework assignments rarely have to do with real systems. It's simply an assignment to practice. You assume right in this case, it's odd to use 3 states. Generally you have as many state variables as you have energy systems. This 2nd order system has potential and kinetic, thus you shouldn't need anything more than 2 state variables that you correctly identify. $\endgroup$
    – DrMrstheMonarch
    Sep 10, 2019 at 23:31
  • $\begingroup$ @morbo one could argue that expressing this as a third order allows one to adaptively obtain the acceleration due to gravity (or any other constant external force) when used as a model for an observer. Such observer would then work in any gravity, so for example both on Earth and the moon. $\endgroup$
    – fibonatic
    Oct 7, 2020 at 22:38

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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.

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  • $\begingroup$ So you're suggesting just take the time derivative of the equation to figure out how to represent p triple dot? $\endgroup$
    – Izzo
    Sep 12, 2019 at 15:20
  • $\begingroup$ @Izzo - That's the point everyone's making about not needing that state. No, ordinarily you should not do that, and also if you do do that then you're left with no position term in your equation. Either way you're going to wind up with a row of zeros. I would just solve for $\ddot{p}$, write your equation out that way, use the $\dot{p} = \dot{p}$ assignment, and have the $\dddot{p}$ term be a row of zeros. As I mentioned, I'm not a fan of this question. You're using a state vector that's too large and it's having you make acceleration be a function of position and speed, which it's not. $\endgroup$
    – Chuck ♦
    Sep 12, 2019 at 15:39
  • $\begingroup$ Well if I take the derivative of the equation to yield an equation that features p triple dot, I don't believe it would yield a row of zeros. However, it would yield a relationship in which p triple dot is not a linear combination of the states. Thus this question makes absolutely not sense to me anymore. $\endgroup$
    – Izzo
    Sep 12, 2019 at 16:36
  • $\begingroup$ @Izzo - Ah yeah, sorry I meant column of zeros. If you take the time derivative it's still a linear combination of states, $d/dt\left(p = p_0 + \dot{p}t + \frac{1}{2}\ddot{p}t^2\right) \rightarrow \dot{p} = 0 + \dot{p} + t\ddot{p} + \frac{1}{2}\left(2\ddot{p}t + t^2\dddot{p}\right)$. Assuming I did that correctly lol. Time's not a state here, so treat it like a parameter/coefficient when you convert to the matrix. $\endgroup$
    – Chuck ♦
    Sep 12, 2019 at 16:46
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    $\begingroup$ @Izzo - Yeah, I think what morbo said is correct, in that this seems more like a (poor) homework question for the sake of going through the motions. Your problem statement states that the object is under "constant acceleration," which means $\ddot{p}$ is not a function of time, so $d/dt(\ddot{p}) = 0$. Everything about the question is upsetting to me haha. $\endgroup$
    – Chuck ♦
    Sep 12, 2019 at 17:02
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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.

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  • $\begingroup$ The thing that confuses me about this is that the equation given seems to be a "solution" to a differential equation rather than a differential equation itself. Why would we ever model a solution in state space? $\endgroup$
    – Izzo
    Sep 11, 2019 at 13:08
  • $\begingroup$ No, the given equation is a linear ODE, it is NOT the solution of an ODE. AN ODE of order 'n' has 'n' linearly independent solutions. $\endgroup$
    – Franky
    Sep 12, 2019 at 8:01

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