Deriving Second Order Transfer Function from Spring Mass Damper System. .

Deriving Second Order Transfer Function from Spring Mass Damper System. .
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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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1 Answer

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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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James H. Sterling
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James H. Sterling

James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.