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I am having a hard time understanding how a differential equation based on a spring mass damper system $$ m\ddot{x} + b\dot{x} + kx = 0$$ can be described as an second order transfer function for an inpulse response, which looks something like this $$\frac{(\omega_n)^2}{s^2+2\zeta\omega_n + (\omega_n)^2}$$

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If you want to derive the transfer function out of a differential equation, first you need to select "input" and "output" of the system. In your system I believe the equation is

$$ m \ddot{x} + b\dot{x} + kx = ku $$

where $u$ is the input and $x$ is the output. If you select all initial conditions as $0$, then you can obtain the transfer function given, which is the relation $X(s)/U(s)$. From this, you can calculate output of the system for any given input as

$$x(t) = \mathcal{L}^{-1} \left\{ \frac{k}{ms^2+bs + k} U(s) \right\}$$

In particular if you select $u=\delta(t)$ (Dirac delta) you can obtain the transfer function itself.

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    $\begingroup$ Well that's more the form i was looking for, but if you should apply units for x(t) and U(t) could you then clarify what they express? $\endgroup$ – user25778 Jun 5 '14 at 0:01
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    $\begingroup$ In this particular system (spring mass damper) $x$ is the position of the mass in meters and $u$ is the force that is applied to the mass in Newtons. Of course there are other physical systems with the same transfer function. The units may change for these systems but the underlying mathematics is essentially same. $\endgroup$ – obareey Jun 5 '14 at 6:56
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    $\begingroup$ You are assuming that the initial conditions $ x (0)=x'(0)=0 $, aren't you? $\endgroup$ – Américo Tavares Jun 5 '14 at 11:11
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    $\begingroup$ Yes. This comes from the definition of transfer function. See en.wikibooks.org/wiki/Control_Systems/Transfer_Functions $\endgroup$ – obareey Jun 5 '14 at 13:59
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    $\begingroup$ @LawrenceKesteloot You can also write the equations as $m \ddot{x} + b\dot{x} + kx = f$ where $f$ is the force. But then the system gain would be $1/k$ instead of $1$. Generally, a $k$ is added to the input to normalize the system output. You can just think it as a mathematical trick to simplify calculations. $\endgroup$ – obareey Aug 19 '15 at 6:28

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