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


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


1 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.

  • \$\begingroup\$ 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 \$\endgroup\$ Commented Sep 5, 2013 at 3:04
  • \$\begingroup\$ 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. \$\endgroup\$
    – Andreas H.
    Commented Sep 5, 2013 at 3:13
  • \$\begingroup\$ Ahhh yes this is slowly coming back to me now. I remember learning this in my system dynamics and controls class as an undergrad but that was 3 years ago :/ I am a mechanical engineer so it's good to see the electrical engineering people help me out! \$\endgroup\$ Commented Sep 5, 2013 at 3:17
  • \$\begingroup\$ sure, no problem \$\endgroup\$
    – Andreas H.
    Commented Sep 5, 2013 at 3:37

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