3 Added extra info for first order systems based on comments below
source | link

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

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.

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

2 added 1 character in body
source | link

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]$$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.

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.

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.

1
source | link

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.