Calculate phase shift using transfer function

[z.gary]

Homework Statement

Given that you have found the following transfer function for a circuit, H(jω), what is the phase at very low frequencies?
(jωRC-ω2LC)/((1-ω2LC) + jωRC)
a:∏
b:-∏/2
c:+∏/2
d:0

The Attempt at a Solution

I understand that the phase angle is ψ=arc tan(img(ω)/real(ω)), so i eliminated the terms in denominate with ω since the frequency is low, and i got: -ω^2LC+jωRC, then i put it into the equation: arc tan(ωRC/ω^2LC) and have no idea what to do with it. can someone tell me how to solve this arc tan equation? thanks.

 

[NascentOxygen]

Hi z.gary, welcome to Physics Forums.

You usually can treat the numerator and denominator separately. Find their separate angles for w << 1, then subtract the denominator angle from that of the numerator.

 

[rude man]

Homework Statement

Given that you have found the following transfer function for a circuit, H(jω), what is the phase at very low frequencies?
(jωRC-ω2LC)/((1-ω2LC) + jωRC)
a:∏
b:-∏/2
c:+∏/2
d:0

The Attempt at a Solution

I understand that the phase angle is ψ=arc tan(img(ω)/real(ω)), so i eliminated the terms in denominate with ω since the frequency is low, and i got: -ω^2LC+jωRC, then i put it into the equation: arc tan(ωRC/ω^2LC) and have no idea what to do with it. can someone tell me how to solve this arc tan equation? thanks.

Your approach is good but not 100%.

The denominator is indeed 1. So that phase angle is zero.

But the phase of the numerator is actually arc tan wRC/(-w^2 LC) = arc tan R/(-wL).
So as w → 0 what is arc tan R/(-wL)?

[Note carefully how I wrote the last equation. In general you have to respect the sign of the arc tan argument including whether the sign belongs in the numerator or denominator of the argument. In this case it doesn't matter but in general it does: arc tan (-x/y) is generally not the same as arc tan (x/-y).]

Draw the angle in polar coordinates to get your answer.