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We already know that we can derive the parameters \$\omega_{n}\$ and \$\zeta\$ from a second order system which adopts the canonical form:

\$ H(s) = K\frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\$

How can I find the damping ratio and the natural frequency of a second-order system with one or two zeros:

\$ H(s) = \frac{K_1s^2 + K_2s + K_3}{s^2 + K_4s + K_5}\$

\$ H(s) = \frac{K_1s + K_2}{s^2 + K_3s + K_4}\$

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  • \$\begingroup\$ The numerator s-terms differentiate the classical 2nd order transient responses, so ROT performance parameters are not readily extracted. \$\endgroup\$
    – Chu
    Commented Feb 26, 2017 at 23:45
  • \$\begingroup\$ Natural frequency and damping factor are related to the denominator, its expression doesn't change in the cases of low, band or high pass systems. \$\endgroup\$ Commented Nov 11, 2017 at 15:50

1 Answer 1

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For a second-order system featuring a double zero, the transfer function could look that way:

$$ H(s)= H_0\frac{1 + a_1s + a_2s^2 }{ 1 + b_1s + b_2s^2} = H_0 \frac {N(s)}{D(s)}$$

you can rearrange both \$ N(s) \$ and \$ D(s) \$ under the form

\$ N(s) = 1 + (s/\omega_{0n}*Q_n) + (s/\omega_{0n})^2 \$ and \$ D(s) = 1 + (s/\omega_{0}*Q) + (s/\omega_{0})^2 \$

with \$ \omega_{0n} = 1/\sqrt{a_2} \$ and \$ Q_n = \sqrt{a_2}/a_1 \$ with \$ \omega_{0} = 1/\sqrt{b_2} \$ and \$ Q = \sqrt{b_2}/b_1 \$

In your first expression, factor \$ K_3 \$ on top and \$ K_5 \$ in \$ D(s) \$, you have

$$ H(s) = \frac{K_3}{K_5}\frac{(1 + \frac{sK_2}{K_3} + \frac{s^2K_1}{K_3}) }{ 1 + \frac{sK_4}{K_5} + \frac{s^2}{K_5}} $$ then apply what I described above.

Same for the second expression:

$$ H(s)= \frac{K_2}{K_4} \frac{(1 + sK_1/K_2) }{1 + \frac{sK_3}{K_4} + \frac{s^2}{K_4}} $$ then apply what I described above.

If you want to learn more about these techniques, have a look at the APEC presentation available here:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

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  • \$\begingroup\$ Hello Laptop2d, thanks! This is much nicer indeed. I will try to use LaTeX next time. \$\endgroup\$ Commented Mar 9, 2017 at 20:43

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