Derivation of second order system transfer function

[ryan88]

Hi,

I am trying to derive the general transfer function for a second order dynamic system, shown below:

$$\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$

In order to do this I am considering a mass-spring-damper system, with an input force of f(t) that satisfies the following second-order differential equation:

$$m\frac{d^2y}{\dt^2}+c\frac{dy}{dt}+ky=f(t)$$

Using the following two relationships:

$$c=2\zeta\omega_nm$$

$$\frac{k}{m}=\omega_n^2$$

I get this:

$$\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=\frac{f(t)}{m}$$

$$\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=\frac{1}{m}\mathcal{L}\left\{f(t)\right\}$$

$$Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=\frac{F(s)}{m}$$

$$\frac{Y(s)}{F(s)}=\frac{1}{m(s^2+2\zeta\omega_ns+\omega_n^2)}$$

Wheras my lecturer has the following in his notes:

$$\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=K\omega_n^2x(t)$$

$$\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=K\omega_n^2\mathcal{L}\{x(t)\}$$

$$Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=K\omega_n^2X(s)$$

$$\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$

This obvisously gives the correct transfer function. So, from the two approaches, I have come to the conclusion that:

$$\frac{f(t)}{m}=K\omega_n^2x(t)$$

But I do not understand the physical reasoning behind this. Can anyone offer any help with this?

Thanks,

Ryan

 

[viscousflow]

That is standard notation. The "trick" is to multiply the right hand side by $$\frac{k}{k}$$. As for physical intuition. Perform a unit analysis. You should be able to draw a clear conclusion from that.

 

[ryan88]

Ah yes, I completely missed that. Although substituting $$\frac{k}{m}=\omega_n^2$$ leaves the gain of the system as $$\frac{1}{k}$$ which is then not dimensionless. I thought this transfer function was supposed to be dimensionless?

 

[viscousflow]

No transfer functions are hardly dimensionless. Transfer functions are the ratio of system $$\frac{output}{input}$$. Thus you can see that the transfer function can hold any units as long as it contains the output-input relationship you are looking for.

 

[ryan88]

Ok, thanks for your help viscousflow. It is very much appreciated.

Ryan