Derivation of second order system transfer function

  • Thread starter ryan88
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  • #1
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Main Question or Discussion Point

Hi,

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

[tex]\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}[/tex]

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:

[tex]m\frac{d^2y}{\dt^2}+c\frac{dy}{dt}+ky=f(t)[/tex]

Using the following two relationships:

[tex]c=2\zeta\omega_nm[/tex]

[tex]\frac{k}{m}=\omega_n^2[/tex]

I get this:

[tex]\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=\frac{f(t)}{m}[/tex]

[tex]\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\}[/tex]

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

[tex]\frac{Y(s)}{F(s)}=\frac{1}{m(s^2+2\zeta\omega_ns+\omega_n^2)}[/tex]

Wheras my lecturer has the following in his notes:

[tex]\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=K\omega_n^2x(t)[/tex]

[tex]\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)\}[/tex]

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

[tex]\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}[/tex]

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

[tex]\frac{f(t)}{m}=K\omega_n^2x(t)[/tex]

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

Thanks,

Ryan
 

Answers and Replies

  • #2
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That is standard notation. The "trick" is to multiply the right hand side by [tex]\frac{k}{k}[/tex]. As for physical intuition. Perform a unit analysis. You should be able to draw a clear conclusion from that.
 
  • #3
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Ah yes, I completely missed that. Although substituting [tex]\frac{k}{m}=\omega_n^2[/tex] leaves the gain of the system as [tex]\frac{1}{k}[/tex] which is then not dimensionless. I thought this transfer function was supposed to be dimensionless?
 
  • #4
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No transfer functions are hardly dimensionless. Transfer functions are the ratio of system [tex]\frac{output}{input}[/tex]. 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.
 
  • #5
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Ok, thanks for your help viscousflow. It is very much appreciated.

Ryan
 

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