Why is the phase angle not just -90 degrees?

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SUMMARY

The discussion focuses on the calculation of the phase angle for the expression \(\frac{\omega_n^2}{-\omega^2+j2 \zeta \omega_n \omega}\). The correct phase angle is derived as \(-90 - \tan^{-1} \left(\frac{\omega}{2 \zeta \omega_n}\right)\), rather than simply \(-\tan^{-1} \left(\frac{-2 \zeta \omega_n}{\omega}\right)\). The reasoning involves recognizing that the denominator represents a second quadrant angle, necessitating the adjustment of the inverse tangent function to account for this. The final expression is simplified using the identity \(\tan^{-1}(x) = 90 - \tan^{-1}(1/x)\).

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Cyrus
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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 isn't it simply:

[tex]0 -tan^{-1} \frac{- 2 \zeta \omega_n}{ \omega}[/tex]
 
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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:
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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