# Transfer Function of ODE using Laplace Transform

• December 11th 2008, 02:47 AM
Mush
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:

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

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

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

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

Assuming 0 initial conditions:

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

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

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

And apparently the transfer function is given by: $\frac{Q_{0}(s)}{Q_{i}(s)}$
Apparently, this is a representation of a time domain dynamical system, with forcing function $q_i$, and response function $q_0$.
Which is fair enough!

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

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

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

I'm told that sigma and omega are constants in this case...
• December 11th 2008, 03:55 AM
CaptainBlack
Quote:

Originally Posted by Mush
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:

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

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

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

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

Assuming 0 initial conditions:

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

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

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

And apparently the transfer function is given by: $\frac{Q_{0}(s)}{Q_{i}(s)}$
Apparently, this is a representation of a time domain dynamical system, with forcing function $q_i$, and response function $q_0$.
Which is fair enough!

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

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

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

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

It does not, you have a specific input here rather than an arbitary input. The idea of a transfer function does not apply to these, you can find the output corresponding to these inputs by multiplying the Laplace transform of the input by the transfer function of the appropriate system and then taking the inverse LT.

Oh-yes, what is $y$?

CB
• December 11th 2008, 04:08 AM
Mush
Quote:

Originally Posted by CaptainBlack
It does not, you have a specific input here rather than an arbitary input. The idea of a transfer function does not apply to these, you can find the output corresponding to these inputs by multiplying the Laplace transform of the input by the transfer function of the appropriate system and then taking the inverse LT.

Oh-yes, what is $y$?

CB

I think it's an arbitrary function of t.

$\mathcal L (y(t)) = Y(s)$

Quoting straight from the exam paper that these questions were taken from:

"For the time domain differential equation :

$\frac{d^2x}{dt^2}+2\sigma \omega_{n} \frac{dx}{dt}+\omega^2_{n}x=y.sin({\omega t})$

and assuming zero initial conditions, find the transfer function:

$F(s) = \frac{Y(s)}{X(s)}$"

And also:

"For the Differential Equation:

$\ddot{x} + 2\sigma\omega_n\dot{x} + \omega_n^2 x = K\omega_n^2$

find the transfer function when: $x(0) = \dot{x}(0)=0$"

I'm still not sure how to proceed! Especially in the 2nd question as there seems to be NO function of y.
• December 11th 2008, 06:08 AM
Mush
Anyone?!