# 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?

$$Phase(\frac{Num}{Den}) = Phase(Num) - Phase(Den)$$