In the case of a free fall object, if $h$ denotes distance from floor and $v$ the speed, then, given an input $U = g$, and $X=[h \quad v]^T$ and $\dot{X}=[\dot{h} \quad \dot{v}]^T$, I found that the state space equation of $B$ contains $t$, that is $B=[t \quad 1]^T$. I used $\frac{dh}{dt}=v + gt$ to get that result.

Does it mean that this system is not time-invariant but in fact, time variant? Am I correct, or did I do something wrong?

This is the model that I follow.


1 Answer 1


In your notation you are mixing up the integration of v. For the state space representation one would write: $$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$ This expresses the equations of motion of a free point mass $m$ in one dimension $h$ with external force $F_\text{ext}=mg$. The change of position / height is $\dot h(t) = v(t)$ and the change of velocity depends on the force: $\dot v = \frac{1}{m} F_\text{ext} = g.$ In the state-space terminology we have the state vector $x=(h,v)$ and the force is the input of the system. (Even for time-dependent $F_\text{ext}(t)$ the system is time-invariant.)

Writing $\dot h = v + g t$ is not correct. We have $\dot v(t) = g$ which integrates to $v(t) = v_0 + g t$ with integration constant $v_0$. Integrating again gives $\dot h = h_0 + v_0 t + \tfrac 1 2 g t^2$ with integration constant $h(t=0)=h_0$.

Supplement to answer comment: For a drag force that is proportional to velocity $F_\text{drag}=-b\, v(t)$, the velocity's equation of motion changes to $\dot v=g - \frac{b}{m} v$. To express this, one would add a constant matrix element to the system function, so the system is still time-invariant: $$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & \frac{-b}{m} \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$

  • $\begingroup$ But, how we write the drag (negative force) as function of v if F itself is an input? $\endgroup$ Apr 26, 2021 at 3:56
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    $\begingroup$ I amended the answer to include this specific drag force. This holds only for drag proportional to v. For any other force that is a function of v, but not proportional to v, the system function is not linear time-invariant anymore. $\endgroup$
    – Hannes
    Apr 26, 2021 at 14:55
  • $\begingroup$ Than the $F_{ext}$ means exclude from (proportional to velocity) drag force? While the gravity force go to $F_{ext}$? Thank you, $\endgroup$ Apr 27, 2021 at 8:43
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    $\begingroup$ Yes. Gravitation is independent of the system variables h and v (because it is the same everywhere at every time), so it is part of the input and not part of the system function. Drag forces do depend on system variables so are part of the system function. $\endgroup$
    – Hannes
    Apr 27, 2021 at 9:46

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