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

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

$$\phi = \operatorname{arg} \frac{j\omega}{j\omega+\frac{1}{RC}} = \operatorname{arg}(j\omega) - \operatorname{arg}(j\omega+\frac{1}{RC}) = \frac{\pi}{2} - \operatorname{atan2}(\omega, 1/RC )$$ 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.
• Thank you for responding. How does $$arg(j\omega)$$ become $$\frac{\pi}{2}$$ and $$arg(j\omega+\frac{1}{RC})$$ become $$atan2(\omega,(1/RC))$$? I have never seen this notation before on this Commented Sep 5, 2013 at 3:04
• Well, the argument is just the angle of the complex number in the complex plane. So $jw$ with positive $w$ has argument 90 degrees or $\pi/2$. For complex numbers with both real and imaginary part ($x + jy$) this is given by the arcustangent $tan^{-1}(y/x)$, when x is positive. In the general case one has to make slight adjustments and hence the atan2 function. The name is because it is named like this in many programming languages (see eg en.wikipedia.org/wiki/Atan2). See also en.wikipedia.org/wiki/Arg_%28mathematics%29 for general information. Commented Sep 5, 2013 at 3:13