I studied Bode plots and the author of my book is somehow relating zeros of transfer functions to Bode plots. What is the relation between zeros or poles of a Bode plot and transfer function? We usually take zeros at \$s=-a\$ (let's say) but how will we show it in a Bode plot since the log of negative values doesn't exist?


1 Answer 1


Bode plot

Take a typical 2nd order low pass filter amplitude response bode plot: -

enter image description here

3-D picture introducing pole-zero diagram

Here's the bigger picture of that response when combined with a pole zero diagram: -

enter image description here

Traditional pole zero diagram

Looking down from above onto the 3-D picture shows the traditional pole zero diagram: -

enter image description here

Pole zero geometry and |H(s)|

What is the relation between zeros or poles of bode plot and transfer function?

If you know the pole positions (or the zero positions) you can predict the bode plot magnitude by calculating the distance from each pole (or zero) to any particular point on the bode plot. This reveals the magnitude along the \$j\omega\$ axis. Note that the red dot below is a variable point on the \$j\omega\$ axis that is required to be calculated. The geometry of the pole zero diagram is examined: -

enter image description here

The reciprocal of \$d_1\cdot d_2\$ gives you the magnitude of the bode plot at any point on the \$j\omega\$ axis. If there was a zero involved (at a distance \$n_1\$ to the particular point on the \$j\omega\$ axis), the magnitude would be the reciprocal of: -

$$\dfrac{d_1\cdot d_2}{n_1}$$

Some pictures from here.

Pole zero diagram and phase response

If you want to know how the phase response is derived from the pole zero diagram, here it is for the conjugate pole example from above: -

enter image description here

But, of course, you could just take the transfer function and mathematically derive phase and magnitude response without reference to the geometry of the pole zero diagram. This is shown further down on this page.

  • 3
    \$\begingroup\$ Holy smokes... 5 years of studying control systems and nobody ever showed me the Bode plot and the S-plane in a 3D plot. I kind of knew how they were related, on an intuitive level and through calculations, but those diagrams make everything so much clearer! \$\endgroup\$
    – craq
    Apr 15, 2020 at 5:25
  • 1
    \$\begingroup\$ I'd have killed for an explanation like this when I was still studying electronics. Very well done. \$\endgroup\$
    – Mast
    Apr 15, 2020 at 12:34
  • 1
    \$\begingroup\$ @craq same here! 3D makes engineering so much better! \$\endgroup\$
    – JoeyB
    Apr 15, 2020 at 19:32
  • \$\begingroup\$ In the pictures you linked, on the website under the header "PEAKING FREQ AND AMPLITUDE" where it says \$|H(j\omega)| = \frac{1}{sqrt(...)}\$ Is that correct? I can't see how that is the magnitude. It looks like they just calculated the magnitude of the denominator, and then reciprocated that magnitude which as far as I know is not valid. \$\endgroup\$
    – DKNguyen
    Oct 26, 2020 at 5:33
  • \$\begingroup\$ @DKNguyen it looks good to me. Rael and complex terms are individually squared then the sum square rooted. \$\endgroup\$
    – Andy aka
    Oct 26, 2020 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.