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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

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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.

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  • $\begingroup$ 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

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