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