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
explanation: The text extract contains a detailed math question related to the phase angle and gain of a transfer function in a circuit. The user provides a specific transfer function and asks for clarification on the gain and phase angle calculations at different frequencies. The user also provides their own calculations and solutions, and seeks confirmation and correction for their answers.
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distilabel_metadata.content: # 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract contains a detailed math question related to the phase angle and gain of a transfer function in a circuit. The user provides a specific transfer function and asks for clarification on the gain and phase angle calculations at different frequencies. The user also provides their own calculations and solutions, and seeks confirmation and correction for their answers.
  
  Label: question-and-answer