# Phase angle and transfer function for low pass RL circuit

I am trying to do some calculations for a low pass LR circuit. (V out is across inductor). My input waveform is a square wave which has a period equivalent to L/R. I am having trouble making sense of my answers - any help will be much appreciated.

I worked out the transfer function T out to be 1/(1+jwL/R) which substituting the period in resulted in T = 1/(1+j*2Pi). If I work out the magnitude of T by taking the modulus I end up with T = 1/(1+(2Pi)^2) = 0.157. I.e around 16% which seems a bit low? The bit which has me really lost is the phase angle. Phi = atan(Im(T)/Re(T)) which ends up as Phi = atan(-2Pi). Now if I do this in radians I get =-1.41 radians which is -80 degrees. If I do atan(-360) degrees I get -90 degrees?

These both suggest that the frequency of Vin is already starting to be filtered out by circuit (Because of low |T| and maximum phase angle lag) however I think there should be a pole at R/L which is the frequency of Vin currently. If my understanding is correct the pole should be the point on the bode plot when the phase angle is half and is the corner frequency for the transfer function. This would have a loss of 3dB which is 1/Sqrt(2) which is more like |T| = 70 not 15.

I hope someone can make sense of my problem and show me the flaw in my logic. Many thanks!

The natural frequency is defined as: $$\omega_c = \frac{R}{L} = 2 \pi f_c$$ $\omega$ in the transfer function and $\omega_c$ are both in rad/s. So the same frequency means $\omega = \omega_c$. There is no $2 \pi$ in there.
So overall, $$\vert{T}\vert = \sqrt{\frac{1}{1 + {(\frac{\omega}{\omega_c})}^2}}$$ when $\omega = \omega_c$, $$= \sqrt{ \frac{1}{2} } = 0.707$$