text: # Could somebody clarify the phase angle and gain of this transfer function?
  
  Suppose a circuit with a transfer function of the form $$\A(\omega)=\displaystyle\frac{A_0}{1-\frac{j\omega}{\omega_0}}\$$ has $$\A_0=-10\; V/V\$$ and $$\\omega_0=100\;rad/sec.\$$
  
  We were asked for the gain when $$\\omega = \omega_0\$$. My formula for the gain is: $$A_{v},dB = 20\log\left|\frac{A_{0}}{ \sqrt{1+(-\frac{ω}{w_{0}})^2}}\right|$$
  
  Hence, at $$\\omega = \omega_0\$$, the gain should be 16.98dB since
  
  $$A_{v},dB = 20\log\left|\frac{-10}{ \sqrt{2}}\right|$$
  
  Is that right?
  
  Then next we have to determine the phase at:
  
  1. $$\\omega = \omega_0\$$
  2. $$\\omega = 1\$$; $$\\omega_0=100\; rad/s\$$
  3. $$\\omega = 100,000\$$; $$\\omega_0=100\; rad/s\$$
  
  My solution:
  
  $$phase = \tan^{-1}\left(\frac{\omega}{\omega_{0}}\right)$$
  
  So number 1 would have been 45, number 2 is approximately 0, then number 3 should be 90.
  
  I wasn't able to confirm the other numbers but number 3's answer is 270. Why is that? So that means the right phase angle is -90?
  
  The phase calculations come from algebra rules:
  
  $$Phase(\frac{Num}{Den}) = Phase(Num) - Phase(Den)$$
  
  That, coupled with the fact that there are 360 degrees in a complete cycle should allow you to figure out the your error.
  
  • That is how i came up with my formula for phase but i guess I forgot to add 180 to the phase angles.
  – user266967
  Commented Dec 10, 2020 at 7:03
  • You are missing the negative sign. Your solution should be -tan^-1(w/wo). Commented Dec 10, 2020 at 20:58