# State space of free falling object is time variant?

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.

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: $$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & \frac{-b}{m} \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$
• Than the $F_{ext}$ means exclude from (proportional to velocity) drag force? While the gravity force go to $F_{ext}$? Thank you, Apr 27, 2021 at 8:43