I’m trying to brush up on some things I left behind, let’s consider the usual RC parallel circuit, I’m trying to get the bode plot of its impedance.

The equivalent impedance is obviously \$ Z = \frac{R}{1+sCR} \$ which has a pole at \$ s = -\frac{1}{RC} \$.

Now I want to use the frequency domain for this, so I use \$s = j\omega \$ and \$f = \frac{\omega}{ 2\pi}\$ and I get:

\$ \omega = -\frac{1}{jRC} \rightarrow \omega = \frac{j}{RC} \rightarrow f = \frac{j}{2\pi RC} \$

I know the frequency should just be \$f = \frac{1}{2\pi RC}\$, but I can’t get what I did wrong.

Any help?

  • \$\begingroup\$ Can you draw your circuit? \$\endgroup\$
    – Miss Mulan
    Aug 17, 2022 at 17:37
  • \$\begingroup\$ Remember what is the definition of this frequency... \$\endgroup\$
    – Antonio51
    Aug 17, 2022 at 17:41

1 Answer 1


From the impedance transfer function start with replacing \$s=j\omega\$, then work out the magnitude:

$$\begin{align} Z(s)&=\dfrac{R}{sRC+1} \\ Z(j\omega)&=\dfrac{R}{j\omega RC+1} \\ |Z(j\omega)|&=\dfrac{R}{\sqrt{(\omega RC)^2+1}} \tag{1} \end{align}$$

Can you take it from here?

  • \$\begingroup\$ Ah got it, this made me feel a bit dumb, thank you. If I wanted to convert the pole directly from s to f should I just ignore j? \$\endgroup\$
    – ale_zec
    Aug 17, 2022 at 18:13
  • \$\begingroup\$ @ale_zec wWell, you can't, really, since \$s\$ is just the Laplace operator, it doesn't mean a specific pole or zero. It needs to be used in the context of the transfer function. Or, if you insist on doing it, be sure to convert the complex to magnitude/phase. \$\endgroup\$ Aug 17, 2022 at 20:11
  • \$\begingroup\$ Thank you, appreciate the help \$\endgroup\$
    – ale_zec
    Aug 17, 2022 at 20:14
  • \$\begingroup\$ @ale_zec You're welcome. ALso, don't forget that this site is not your average forum, it's a Q&A site. If unsure, take the tour to familiarize yourself with it. \$\endgroup\$ Aug 17, 2022 at 20:21
  • 1
    \$\begingroup\$ @ale_zec Oh no, nothing weird, and contribution is not mandatory. It's just that the way your reply affected my retina was in a similar way to other's who are new to the site (aka in the lines of "wham! bam! thank you ma'am!"). Fortunately that's not the case. \$\endgroup\$ Aug 18, 2022 at 5:49

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