I am currently looking for a way to derive the anti-resonance frequency of a transfer function (TF).

I seems that the resonance frequency can be derived from:

$\left\lvert G(jw) \right\rvert_{max} = \frac{\partial \left\lvert G(jw) \right\rvert}{\partial w} = 0$

since it is the maximum amplitude of the magnitude plot in the bode diagram.

An easy way for an approximation would be to calculate the zeros of the TF (similar to the pole frequencies and their approximation to the resonance freq. which are closely located, see also Resonance from bode plot)

A TF can be for example:

$G(s) = \frac{1}{s^2(J_m+J_l)} \frac{J_l s^2 + ds + c}{\frac{J_m J_l}{J_m+J_l}s^2 + d s + c}$

the bode plot for $J_m = 0.002, J_l = 0.02, d = 0.5, c = 50$ is

Bode Plot


Well, first of all we can write:

$$\left|\text{G}\left(\omega\text{j}\right)\right|=\left|\frac{1}{\left(\omega\text{j}\right)^2\cdot\left(0.002+0.02\right)}\cdot\frac{\left(\omega\text{j}\right)^2\cdot0.02+\omega\text{j}\cdot0.5+50}{\frac{0.002\cdot0.02}{0.002+0.02}\cdot\left(\omega\text{j}\right)^2+\omega\text{j}\cdot0.5+50}\right|=$$ $$\frac{1}{\left|-\omega^2\cdot0.022\right|}\cdot\frac{\left|-\omega^2\cdot0.02+\omega\text{j}\cdot0.5+50\right|}{\left|-\frac{0.00004}{0.022}\cdot\omega^2+\omega\text{j}\cdot0.5+50\right|}=$$ $$\frac{1}{0.022\omega^2}\cdot\frac{\sqrt{\left(50-0.02\omega^2\right)^2+\left(0.5\omega\right)^2}}{\sqrt{\left(50-\frac{0.00004}{0.022}\cdot\omega^2\right)^2+\left(0.5\omega\right)^2}}\tag1$$

Where $\text{j}^2=-1$.

Finding the point in the Bodeplot we need to solve:


  • $\begingroup$ Thank you for your answer. Can you please explain how you obtained the solution in (2)? If I try to solve it with MATLAB I get nothing useful. clear all; clc; syms Jl Jm d c w; Gs_abs = (1/((Jm+Jl)*ww))*(sqrt((c-Jlw^2)^2+(wd)^2)/sqrt((c-((JmJlw^2‌​)/(Jm+Jl)))^2+(wd)^2‌​)); Gs_abs_diff = diff(Gs_abs,w); solve(Gs_abs_diff,w) $\endgroup$ – Zeit Jan 17 '18 at 8:29
  • $\begingroup$ @Zeit You're welcome, I'm glad that I could help you. I used Mathematica 10.0 to find the result I got in $\left(2\right)$. $\endgroup$ – Jan Jan 17 '18 at 8:38
  • $\begingroup$ @Earland, can you maybe post the general solution for (2)? $\endgroup$ – Zeit Jan 17 '18 at 8:44
  • $\begingroup$ @Zeit What do you define as a general solution? Because in your question the values are given. $\endgroup$ – Jan Jan 17 '18 at 8:46
  • 1
    $\begingroup$ @ Eerland, can you pls send me the solution as mentioned ;) $\endgroup$ – Zeit Jan 27 '18 at 17:22

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