Phase angle

1. Jul 6, 2008

Cyrus

Can someone show me how they got this:

$$\frac{\omega_n^2}{-\omega^2+j2 \zeta \omega_n \omega}$$

Has this phase:

$$-90-tan^{-1} \frac{ \omega}{2 \zeta \omega_n}$$

Why isnt it simply:

$$0 -tan^{-1} \frac{- 2 \zeta \omega_n}{ \omega}$$

2. Jul 6, 2008

uart

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
3. Jul 6, 2008

Cyrus

Ah, that makes sense. Thanks uart!

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