i want to find the transfer function of a differential equation (given below)
$\ddot\theta = a [ ([b\times Xin] - bk\dot\theta) - \ddot\theta] - c\phi $
(where $\phi$ and $\theta$ are time dependent) taking the laplce transform yeilds
$s^2\theta = a[(\frac bs \times Xin) - bks\theta) - s^2\theta] - \frac {c}{s^2}$
However, i fail to relate $\theta$ to $Xin$ to give a TF in the form ($\theta /Xin = g(s)$)
Can some one help with this, a step by step would really help
I really wonder why you put two pair of unnecessary brackets in your formula? Furthermore $c\phi$ is a constant or is $\phi$ time dependent? If the latter, you can set it to zero to determine the behavior of $X_{in}$ to $\theta$. But I will asume you mean $c\theta$?
$$\ddot\theta(t) = a (b X_{in}(t) - bk\dot\theta(t) - \ddot\theta(t)) - c\theta(t) $$
Becomes
$$\ddot\theta(t) = a b X_{in}(t) - abk\dot\theta(t) - a\ddot\theta(t) - c\theta(t) $$
$$(1 + a)\ddot\theta(t) + abk\dot\theta(t) + c\theta(t) = a b X_{in}(t)$$
Taking the laplace transform
$$(1 + a)s^2\theta(s) + abks\theta(s) + c\theta(s) = a b X_{in}(s)$$
$$((1 + a)s^2 + abks + c)\theta(s) = a b X_{in}(s)$$
$$\theta(s) = \frac{a b}{(1 + a)s^2 + abks + c} X_{in}(s)$$
$$G(s) = \frac{\theta(s)}{X_{in}(s)} = \frac{a b}{(1 + a)s^2 + abks + c}$$
-edit- since $\phi$ is actually time dependent. Your solution becomes
$$\theta(s) = \frac{a b}{(1 + a)s^2 + abks} X_{in}(s) + \frac{c}{(1 + a)s^2 + abks} \phi(s)$$
Now to determine $G_1(s) = \frac{\theta(s)}{X_{in}(s)}$, you simply set $\phi(s)$ to zero. And to determine $G_2(s) = \frac{\theta(s)}{\phi(s)}$, you set $X_{in}(s)$ to zero.