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I am learning about control theory.
Let's consider this system.

$$ m a(t) + b v(t) + k x(t) = f(t) $$

$a$ is acceleration
$v$ is velocity
$x$ is displacement
$f$ is external force

In my textbook, in chapter about "state space model" , two state variables ($x$ and $v$) are necessary
to completely describe this system.
However, I think $x$ can completely describe this mass's motion by itself.
If we know $x$, $v$ can derived from $x$.
So I have two questions.

  1. Why two variables are needed.
  2. Why "$x$ and $v$" ? Can "$v$ and $a$" or "$x$ and $a$" also describe this system ?
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3 Answers 3

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Let's rewrite your system as

$$ m \ddot{x}(t) + b \dot{x}(t) + k x(t) = f(t) $$

then you can see what you're saying: Everything I need to know is in $x(t)$!

But is it really?

State space systems are predicated on having a single, first order differential equation to solve. As you can see in the rewritten equation above, there is a double-derivative term.

In order to write that equation as a first order equation, we need to have two state variables: $$ X(t) = \left[ \begin{array}{c} x\\ \dot{x} \end{array} \right] $$ and so $$ A\dot{X}(t) + B X(t) = f(t) $$ where $A=[b\ m]$ and $B=[k\ 0]$.

Let's try to do the same thing using $x$ and $\ddot{x}$.

$$ X'(t) = \left[ \begin{array}{c} x\\ \ddot{x} \end{array} \right] $$ and so $$ A'\dot{X'}(t) + B' X'(t) = f(t) $$ where $A'=[b\ 0]$ and $B'=[k\ m]$.

So I suppose you could chose the state variables to be $x$ and $\ddot{x}$... but it would mean an unused $\dddot{x}$ variable.

Some examples of first order systems can be found in this document.

enter image description here

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  • $\begingroup$ It seems very strange for me that degree of freedom is 1, but number of state is 2. I think "state" means "how is this system now", and "how is this system now" means "where is the mass", and this is "x". Could you explain about this point? and one more question. Are there a system having only one state variable? $\endgroup$
    – user15151
    Sep 24, 2015 at 14:59
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    $\begingroup$ Any system described by a first order scalar differential equation will have only one state variable: $b\dot{x}(t) + kx(t) = f(t)$. Do not conflate mechanical degrees of freedom with system state. They are very different things. $\endgroup$
    – Peter K.
    Sep 24, 2015 at 15:02
  • $\begingroup$ Can you give an example of one state system? using real objects or something. $\endgroup$
    – user15151
    Sep 24, 2015 at 15:12
  • $\begingroup$ @ShuS Surely you can Google yourself? See my edit. $\endgroup$
    – Peter K.
    Sep 24, 2015 at 15:15
  • $\begingroup$ In upper left figure, state variable is Vm , right? How to get output (displacement) from it using output matrix? $\endgroup$
    – user15151
    Sep 24, 2015 at 15:24
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$v$ cannot be derived from a single $x$ at $t_0$. Once you use require more information than one $x$ at $t_0$ to compute $v$, and thus the momentum, you've added to what you need to describe the state of the system.

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It's useful to think in the definition of a Markov state: all information about the state at $t+1$ can be derived from state at time $t$ and control at time $t$, ONLY. No information older than $t$ allowed.

Sure you can derive $v$ from $x$, but that would mean that you need $x$ at time $t$ and at $t-1$, which violates the Markov state assumption. To solve that you need to include $v$ in the state, so that you don't need to track the "past".

This is why we have as many variables as there are energy absorbing components in electric circuits: we could track the whole history of a capacitor and infer its behaviour but that would violate the Markov assumption.

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