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 $c\theta$?

$$\ddot\theta(t) = a (b X_{in}(t) - bk\dot\theta(t) - \ddot\theta(t)) - c\theta(t) $$


$$\ddot\theta(t) = a b X_{in}(t) - abk\dot\theta(t) - a\ddot\theta(t) - c\theta(t) $$

$$(1 + a)\ddot\theta(t) + abk\dot\theta(t) + c\theta(t) = a b X_{in}(t)$$

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.

  • $\begingroup$ thanks for your answer. $\phi$ is time dependent, sorry for leaving this out $\endgroup$ – introVertice May 8 '16 at 23:14
  • $\begingroup$ then to determine the transfer function, $\frac{\theta(s)}{X_{in}(s)}$, you can set $\phi(s)$ to zero. Your tf becomes; $$G(s) = \frac{\theta(s)}{X_{in}(s)} = \frac{a b}{(1 + a)s^2 + abks}$$. You can determine also the behavior from $\frac{\theta(s)}{\phi(s)}$, et cetera. $\endgroup$ – WG- May 8 '16 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.