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I have an exercise that asks to create a second order Transfer Function (TF) of this form:

$$ \frac{a}{s^2+bs+a} $$

The TF should have both poles be critically damped, and a bandwidth of 100Hz.


So, what did I try?

Well for starters, I used the following form of the TF (not sure how it is called): $$ \frac{ω_n^2}{s^2+2ω_nζs+ω_n^2} $$

  • ζ is critically damped when equal to 1, so that is what I did.
  • Now for the bandwidth, I tried at first to set $ω_n=100*2 \pi$ but it did not work. I googled it but was not able to find what is the relationship between the natural frequency ($ω_n$) and the bandwidth, of the TF.

So that is what I want help with. How do I create this second order TF with (critical damping and) bandwidth 100 Hz?

p.s. it is not important, but all calculations are done with MATLAB.

p.s2 when saying bandwidth, I'm going with the -3dB definition

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    $\begingroup$ You forgot a factor 2 in the linear term in $s$ of your transfer function. $\endgroup$ – Kwin van der Veen Nov 15 '17 at 18:21
  • $\begingroup$ Read on Bode plots. $\endgroup$ – Preston Roy Nov 16 '17 at 14:37
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Notice that the frequency response of a system can be obtained from its transfer function by putting $s = j w$, where $j^2=-1$. Consequently, $$H(w) = \frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}\Bigg|_{s=jw,\zeta=1}=\frac{\omega_n^2}{\omega_n^2-\omega^2 + 2j\omega_n \omega}.$$ Therefore, $$|H(w)| = \frac{\omega_n^2}{\sqrt{(\omega_n^2-\omega^2)^2+4\omega_n^2\omega^2}}=\frac{\omega_n^2}{\omega_n^2+\omega^2}.$$ As such, $$|H(200\pi)|=\frac{1}{\sqrt{2}}=\frac{\omega_n^2}{\omega_n^2+(200\pi)^2} \implies \omega_n = ?$$

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  • $\begingroup$ I tried it however for it to work, I needed to dived the final omega by 2 for it to work. Maybe because I need to remove the negative frequencies of the frequency response? $\endgroup$ – Dimitris Pantelis Nov 15 '17 at 19:13
  • $\begingroup$ one question on your solution, because I lack the mathematical knowledge I guess: in the second line, how did you end up with that solution? (just some tips so I can look it up) $\endgroup$ – Dimitris Pantelis Nov 15 '17 at 19:23
  • $\begingroup$ @DimitrisPantelis I used $(a+b)^2=(a-b)^2+4ab$. $\endgroup$ – Math Lover Nov 15 '17 at 20:12
  • $\begingroup$ yeah, I should have seen that already. Thank :) $\endgroup$ – Dimitris Pantelis Nov 15 '17 at 20:43

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