Derivation of second order system transfer function

AI Thread Summary
The discussion focuses on deriving the transfer function for a second-order dynamic system, specifically a mass-spring-damper system. The user, Ryan, compares his derivation with that of his lecturer and identifies a discrepancy related to the input force and its relationship to the system's parameters. He seeks clarification on the physical reasoning behind the equation f(t)/m = Kω_n²x(t) and the dimensionality of transfer functions. Participants emphasize that transfer functions are not dimensionless and can have units as long as they represent the output-input relationship. The conversation concludes with Ryan expressing gratitude for the insights provided.
ryan88
Messages
41
Reaction score
0
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
 
Engineering news on Phys.org
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.
 
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?
 
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.
 
Ok, thanks for your help viscousflow. It is very much appreciated.

Ryan
 
Hi all, i have some questions about the tesla turbine: is a tesla turbine more efficient than a steam engine or a stirling engine ? about the discs of the tesla turbine warping because of the high speed rotations; does running the engine on a lower speed solve that or will the discs warp anyway after time ? what is the difference in efficiency between the tesla turbine running at high speed and running it at a lower speed ( as fast as possible but low enough to not warp de discs) and: i...
Back
Top