How do I determine the phase response of a high pass filter?

Question

I am stuck trying to determine the phase response of a high pass filter. I was able to find the transfer function for the high pass filter and the magnitude but I am stuck finding the phase. I found the transfer function of a high pass filter as $$\frac{V_{out}(j\omega)}{V_{in}(j\omega)}=\frac{j\omega}{j\omega+\frac{1}{RC}}$$

and I calculated the magnitude of the high pass filter as

$$|V_{out}(j\omega)|=\frac{\omega}{\sqrt{\omega^{2}+(\frac{1}{RC})^{2}}}$$

I found online that the phase response of a high pass filter is $$\frac{\pi}{2}-tan^{-1}(\omega RC)$$ But i cannot figure out how the derivation got there. I would really like to understand their logic. Thank you for anyone who can teach me the rest of this derivation

Answer 1

The phase response is just the argument of the transfer function (just as the magnitude response is the absolute value).

The argument of a quotient is the argument of the numerator minus the argument of the denominator, i.e.,

$$ \phi = \operatorname{arg} \frac{j\omega}{j\omega+\frac{1}{RC}} = \operatorname{arg}(j\omega) - \operatorname{arg}(j\omega+\frac{1}{RC}) = \frac{\pi}{2} - \operatorname{atan2}(\omega, 1/RC ) $$ which is essentially just what you wrote.

Note that I did not write \$\operatorname{tan}^{-1}(y/x)\$ but \$\operatorname{atan2(y,x)}\$ because the former is only correct if \$x\$ is positive. It is incorrect when \$x\$ is zero or negative. In your case here, when \$\omega\$ is always positive that makes no difference, but I think it is good practice to use atan2.