3 Added extra info for first order systems based on comments below edited Sep 24 '15 at 15:15 Peter K.♦ 17.4k88 gold badges3131 silver badges6363 bronze badges 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. 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. 2 added 1 character in body edited Sep 24 '15 at 14:44 Peter K.♦ 17.4k88 gold badges3131 silver badges6363 bronze badges 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 answered Sep 24 '15 at 14:13 Peter K.♦ 17.4k88 gold badges3131 silver badges6363 bronze badges 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.