Could somebody clarify the phase angle and gain of this transfer function?

Question

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?

Answer 1

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.