# Unclear relationship between s domain and frequency domain

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?

• Can you draw your circuit? Aug 17, 2022 at 17:37
• Remember what is the definition of this frequency... Aug 17, 2022 at 17:41

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}
• @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. Aug 17, 2022 at 20:11