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Phase angle

  1. Jul 6, 2008 #1
    Can someone show me how they got this:

    [tex]\frac{\omega_n^2}{-\omega^2+j2 \zeta \omega_n \omega}[/tex]

    Has this phase:

    [tex] -90-tan^{-1} \frac{ \omega}{2 \zeta \omega_n} [/tex]

    Why isnt it simply:

    [tex] 0 -tan^{-1} \frac{- 2 \zeta \omega_n}{ \omega} [/tex]
  2. jcsd
  3. Jul 6, 2008 #2


    User Avatar
    Science Advisor

    Hi Cyrus. Look at the denominator, it's phase is clearly a "second quadrant" angle. Inverse tan however only covers quadrands I and IV, so you want to express it as 180 - invtan(2 zeta omega_n / omega) right.

    Now you're right to do what you did next, subtract the phase of the denom from the phase of the num to get :

    -180 + invtan(2 zeta omega_n / omega).

    Now just use invtan(x) = 90 - invtan(1/x) to get the desired expression.

    BTW. Sorry that I'm too lazy to latex today, I hope you can follow it anyway.
    Last edited: Jul 6, 2008
  4. Jul 6, 2008 #3
    Ah, that makes sense. Thanks uart!

    Im so used to using arctan2() command in matlab that I forgot all this stuff.
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