Derivation of second order system transfer function

  1. 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
     
  2. jcsd
  3. 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.
     
  4. 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?
     
  5. 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.
     
  6. Ok, thanks for your help viscousflow. It is very much appreciated.

    Ryan
     
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