# Time responses (position and speed) of system

This is a basic question regarding state space representation and differential equations. I want to find the time response of states $x_{1} = x$ and $x_{2} = \dot{x}$ of the following system:

$$m\ddot{x} + k\dot{x} = 0$$

The state space is

$$\begin{pmatrix} \dot{x_{1}} \\ \dot{x_{2}} \end{pmatrix}= \begin{pmatrix} 0 & 1 \\ -\frac{k}{m} & 0 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}$$

What is the fastet and simplest way to calculate the time response of the two states $x_{1}(t)$ and $x_{2}(t)$?

1. Just solve the differential equation for $x$ which equals $x_{1}(t)$ and then get $x_{2}(t)$ using the derivative $\dot{x_{1}}(t) = x_{2}(t)$?
2. Using Laplace, transform the differential equation, solve for $X(s)$ and back transform to the time domain to get $x_{1}(t)$. Then again get $x_{2}(t)$ using the derivative?
3. Or is there another possibility to get the time responses of the states directly from the state space representation?
4. Or is there even another approach?
• In case of this simple linear time invariant system you can use the eigenvalues, $\lambda_1$ and $\lambda_2$ of the matrix, such that $x_1(t)=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}$, where $C_1$ and $C_2$ depend on the initial conditions. – fibonatic May 2 '15 at 0:47

fibonatic above solved your problem by inspection; I'm just detailing why it was so easy: essentially, diagonalization of the matrix.

For starters, define $k/m\equiv \omega^2$, and you are expected to reflexively recognize the 2x2 matrix involved has paired eigenvalues $\pm i\omega$ with celebrated (un-normalized) eigenvectors $v_1 (1,i\omega)/2$ and $v_2 (1,-i\omega)/2$, respectively; that is, $x_1=v_1+v_2$ and $x_2=i\omega(v_1-v_2)$. So your system collapses to two decoupled differential equations, $$\begin{pmatrix} \dot{v_{1}} \\ \dot{v_{2}} \end{pmatrix}= \begin{pmatrix} i\omega & 0 \\ 0 & -i\omega \end{pmatrix} \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}$$ The solutions are obviously $v_1=C_1 e^{i\omega t}$ and $v_2=C_2 e^{-i\omega t}$, and you never had to solve a 2nd order differential equation, options 1 and 2---the central idea behind the exercise. This, then, amounts to your option 3: $x_1=C_1 e^{i\omega t}+ C_2 e^{-i\omega t}$, and the velocity, $x_2=i\omega (C_1 e^{i\omega t}- C_2 e^{-i\omega t})$.