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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?

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1 Answer 1

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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.

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    \$\begingroup\$ That is how i came up with my formula for phase but i guess I forgot to add 180 to the phase angles. \$\endgroup\$
    – user266967
    Dec 10, 2020 at 7:03
  • \$\begingroup\$ You are missing the negative sign. Your solution should be -tan^-1(w/wo). \$\endgroup\$
    – Michael
    Dec 10, 2020 at 20:58

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