Bode plot from s-domain singularities

Question

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}\$

2. \$ 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} \$

Answer 1

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.

Answer 2

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.