Transfer Function of ODE using Laplace Transform

Hi.

Hi. I'm wondering if anyone can help me with the process of finding the Transfer functions of some differential equations.

I just don't get it tbh! My lecturer gave us the following very ambiguous example:

$\displaystyle a_{2} \frac{d^{2}q_0}{dt^2} + a_{1}\frac{dq_{0}}{dt} + a_{0}q_{0} = b_{0} q_{i}$

$\displaystyle \frac{d^{2}q_0}{dt^2} + \frac{a_{1}}{a_{2}}\frac{dq_{0}}{dt} + \frac{a_{0}}{a_{2}}q_{0} = \frac{b_{0}}{a_{2}} q_{i}$

$\displaystyle \mathcal{L}[\frac{d^{2}q_0}{dt^2} + \frac{a_{1}}{a_{2}}\frac{dq_{0}}{dt} + \frac{a_{0}}{a_{2}}q_{0}] = \mathcal{L}[\frac{b_{0}}{a_{2}} q_{i}]$

$\displaystyle [s^{2} Q_{0}(s)-sq_{0}(0)-q_{0}'(0)] + \frac{a_{1}}{a_{2}}[sQ_{0}(s)-q_{0}(0)] + \frac{a_0}{a_2}Q_{0}(s) = \frac{b_0}{a_2}Q_{i}(s)$

Assuming 0 initial conditions:

$\displaystyle s^{2} Q_{0}(s) + \frac{a_{1}}{a_{2}}sQ_{0}(s) + \frac{a_0}{a_2}Q_{0}(s) = \frac{b_0}{a_2}Q_{i}(s) $

$\displaystyle [s^2 + \frac{a_{1}}{a_{2}}s + \frac{a_0}{a_2}]Q_{0}(s) = \frac{b_0}{a_2}Q_{i}(s) $

$\displaystyle Q_{0}(s) = \frac{b_0}{a_2}Q_{i}(s) \times \frac{1}{s^2 + \frac{a_{1}}{a_{2}}s + \frac{a_0}{a_2}}$

And apparently the transfer function is given by: $\displaystyle \frac{Q_{0}(s)}{Q_{i}(s)}$

Apparently, this is a representation of a time domain dynamical system, with forcing function $\displaystyle q_i$, and response function $\displaystyle q_0 $.

Which is fair enough!

But how on earth does that apply to finding the transfer function of, for example, these two:

$\displaystyle \frac{d^{2}x}{dt^{2}} + 2 \sigma \omega_{n} \frac{dx}{dt} + \omega_{n}^{2}x = y sin(wt)$

$\displaystyle \frac{d^{2}x}{dt^{2}} + 2 \sigma \omega_{n} \frac{dx}{dt} + \omega_{n}^{2}x = K \omega_{n}^{2} $

I'm told that sigma and omega are constants in this case...