Magnitude and Phase of Transfer Function

  1. 1. The problem statement, all variables and given/known data

    I am given the transfer function H(s) = 10/(s(s^2 + 80s +400)) where s = jω [j is the imaginary unit i] and I am trying to get it into its magnitude and phase components.

    3. The attempt at a solution

    I rearranged it to 1/(40jω(1+ 4jω/20 + (jω/20)^2)) which is the standard form for transfer functions. I am wanting to plot the bode plots so I took the 20 * log base 10 of the entire transfer function and got for the magnitude: -20log(40) - 20log(1/(jω))- 20log(1+ 4jω/20 + (jω/20)^2)
    and for the phase I got: -90° - tan^-1((ω/5)/(1-ω^2/400))

    but in my textbook when they show the bode plots for the phase it has - tan^-1(ω/(1-ω^2/400)) for one of the phase factors and I am not sure why my numerator is ω/5 and there one is just ω.

    If you don't study electrical engineering and are not sure of some of the stuff I said then basically H(s) = 10/(s(s^2 + 80s +400)) where s = jω is a complex number and j is the same as the imaginary unit i, so I need help to put this complex number into its magnitude and phase, I am pretty sure i got the magnitude part right.

    Please can anyone help me by at least showing me how to get the phase part of this complex number
  2. jcsd
  3. gneill

    Staff: Mentor

    Since the numerator of your transfer function is purely real we can concentrate on the denominator; it's phase will be the negative of (180° shifted) the transfer function's phase.

    ## s ( s^2 + 80 s + 400 ) → \omega j (-\omega^2 + 80 j \omega + 400) ##

    ## -80 \omega^2 + j (400 \omega - \omega^3) ##

    Divide through by 400ω:

    ## -\frac{\omega}{5} + j \left( 1 - \frac{\omega^2}{400} \right) ##

    Looks like the book's solution is not correct.
  4. rude man

    rude man 6,082
    Homework Helper
    Gold Member

    The phase part is just tan-1{(imaginary part)/(real part)}.

    As gneill pointed out, your transfer function phase angle will be the negative of the numerator phase angle, i.e.

    ψ = - tan-1{(imaginary part)/(real part)} of the denominator.

    {However, -ψ is not the same as ψ shifted by 180. E.g if ψ = 30 deg, -ψ is the same as 360 - 30 = 330 deg, not 30 - 180 or 30 + 180.}

    Also, be careful to preserve numerator and denominator signs in the arc tan expression. So arc tan(-a/b) is not the same angle as arc tan(a/-b).
  5. gneill

    Staff: Mentor

    Good catch, rude man; That was a brain fart on my part :redface: I hate when that happens :smile:
  6. rude man

    rude man 6,082
    Homework Helper
    Gold Member

    Don't feel bad! After writing that I wasn't so sure myself that what I wrote was really right! It's really something to think about - after all -V does = V at 180! Subtle business, electricity!

    No to mention that what I wrote wasn't quite right anyway - the angle of the transfer function, not that of the numerator, is the negative of the angle of the denominator, as you of course had already pointed out. Oy vey!
    Last edited: Feb 27, 2012
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?