# Bode plot from s-domain singularities

Considering any transfer function H(s), is there any meaningful way to use the s-variable singularities to obtain the f-variable ones to trace the Bode plot?

I know direct conversion with $$\ s = \alpha + j\omega\$$ makes no sense, but why does it work at times? I assume it's just coincidence, can anyone think of an example in which it doesn't?

Examples

1. $$\ H(s) = \frac{R}{1+sCR} \;\$$:

$$\s_p = -\frac{1}{RC} \rightarrow f_p = \frac{1}{2\pi RC}\$$

1. $$\ H(s) = \frac{1+sRC_1}{s(C_1+C_2)} \frac{1}{1+sRC_{eq}} \;\$$ where $$\ \; C_{eq} = \frac{C_1C_2}{C_1+C_2} \;\$$:

$$\ \quad s_p = -\frac{1}{RC_{eq}} \rightarrow f_p = \frac{1}{2\pi RC_{eq}}\$$

$$\ \quad s_z = -\frac{1}{RC_1} \rightarrow f_z = \frac{1}{2\pi RC_1}\$$

Edit: my bad Tim, english is my 2nd language and I'm not extremely familiar with these topics, I'll try to provide more context.

I was given a certain transfer function, of which I need to find zeros, poles and Bode plot.
Finding the zeros and poles is easy using the Laplace variable, and I know I can find the corner frequencies of the Bode plot by using $$\ H(j\omega)\$$.

My question is, why do the corner frequencies of the Bode plot match the Laplace variable singularities when there is no direct correlation between the 2?

E.g. why do these have the same 'form'

$$\s = -\frac{1}{RC} \rightarrow f = \frac{1}{2\pi RC}\$$

When this obviously doesn't work?

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

• I think the coincidence is caused by the format of the transfer function. Say, the first $H(s)$ has (1+sCR) as the denominator. The pole is to make sCR = -1, while the cutoff frequency is to make sCR = j (wRC = 1). The -1 and +1 makes the difference only appears in the sign. But I'm not sure if the format as $\frac{constant}{1+sRC}$ always holds. Aug 24 at 14:56
• It's not clear from the terminology of your question what you're actually asking. Are you trying to find out why the corner frequencies of a Bode plot (which I think are your $f_p$) can be derived from the real-valued poles of the transfer function? I keep erasing "edit your question to use more common terminology" because you're obviously a newbie and that would be cruel because how do you know what's common? But maybe if you include a picture of a Bode plot with the areas of your concern pointed out it would help us understand your misunderstanding. Aug 24 at 15:05
• @TimWescott sorry about that, tried to make myself clearer Aug 24 at 15:35
• @concernedmiddleageman s is a complex number .Where did the imaginary part go from your first equation? Aug 24 at 15:51
• @MissMulan I think that's sort of the point. The complex number isn't in the f equation because it isn't derived from the s equation, see the edit at the bottom. I'm asking if there is any way to derive it, and if there isn't I'm asking why they are so similar (at least in most cases I encountered in my rather short experience with the topic). Aug 24 at 16:33

Considering any transfer function H(s), is there any meaningful way to use the s-variable singularities to obtain the f-variable ones to trace the Bode plot?

The s-plane pole zero diagram and the bode plot are totally linked mathematically. Maybe this picture from my basic web-site will help realize that: -

If you viewed the diagram from the top you would see the standard pole-zero diagram: -

And if you viewed the diagram from the right you would see the bode plot: -

All are totally mathematically linked. The example used in the diagrams above is a 2nd-order low-pass filter if that helps.

My question is, why do the corner frequencies of the Bode plot match the Laplace variable singularities when there is no direct correlation between the 2

There is a direct correlation between the two as I hopefully have demonstrated.

• Thank you Andy, this helps a bit. So basically, what we consider the Bode plot's corner frequency is the projection of the pole onto the jw axis? Aug 24 at 17:44
• S-plane maps to bode-plot maps to pole zero diagram @concernedmiddleageman . I can't answer your specific comment directly because how we define terms and what the s plane transfer function is dictates how we describe those points. If you can be specific? Aug 24 at 18:17
• I believe you mean the cut-off frequency where amplitude drops by 3 dB yes? Aug 24 at 18:19
• I do yes, but I think I'll need a much deeper dive in this to fully understand, I'm not sure if i can get it just from your comments, although I'll appreciate the effort if you'll try. I will accept your answer since it has proven helpful, thank you again Aug 24 at 18:42
• @GeorgeGuo thanks for giving help but the 2nd-order filter is not specifically tied to any type of circuit topology. $\omega_n$ is the natural resonant frequency. Aug 24 at 19:14

Consider the simple case where $$\H(s) = \frac{1}{s + \omega_0}\$$. You generate the Bode plot by calculating $$\H(j \omega)\$$, sweeping $$\\omega\$$ along the real axis.

The magnitude of $$\H(s)\$$ is given by $$\\left | H(j\omega) \right | = \frac{1}{\sqrt{\omega^2 + \omega_0^2}}\$$.

When $$\\omega \ll \omega_0\$$, $$\\left | H(j\omega) \right |\$$ is mostly constant, because $$\\sqrt{\omega^2 + \omega_0^2}\$$ is dominated by $$\\omega_0\$$. When $$\\omega \gg \omega_0\$$, then the $$\\omega\$$ term dominates, and $$\\left | H(j\omega) \right | \simeq \frac{1}{\omega}\$$. The transition happens around $$\\frac{\omega_0}{4} < \omega < 4 \omega_0\$$ -- beyond this there's some effect, but not much.

You can actually do a pretty good sketch of the magnitude response of a transfer function with real poles and zeros just by drawing straight lines between the poles' and zeros' frequencies, of the correct slope and amplitude, then rounding off the magnitude.