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I want to know how to get damping and the natural response in a system with a zero and complex pole pair. I calculated the following transfer function:
(from an active filter composed of 2 capacitors and 3 resistors)

$$ \frac{As}{Bs^2+Cs+D} $$

Where the poles of s are complex (\$s\rightarrow \alpha\pm j\beta\$) and \$A \neq D\$, I've seen solutions where \$A =D\$ and the system takes the form of:

$$ \frac{\omega_n^2 s}{s^2+2\zeta\omega_n s+\omega_n^2} $$ But that's not the case in my transfer function.

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  • \$\begingroup\$ Did you mean to say \$A\neq D\$? \$\endgroup\$
    – AJN
    Sep 17 '21 at 17:00
  • \$\begingroup\$ Clearly A doesn’t equal C. \$\endgroup\$
    – Andy aka
    Sep 17 '21 at 17:10
  • \$\begingroup\$ Yes, I ment \$ A \neq D\$. I've updated the question. \$\endgroup\$ Sep 17 '21 at 19:16
  • \$\begingroup\$ For damping you already have what you need: make the denominator monic and solve for \$\zeta\$. As for the response just take the inverse Laplace. You might not even need to perform any calculations since the 2nd order transfer function is well known. \$\endgroup\$ Sep 17 '21 at 19:30
  • \$\begingroup\$ After some research and thinking, I've come to the following answer of how to transform the function into the standard form: \$ \dfrac{s/B}{s^2+C/B s+ D/B}\$, but I'm not sure if I can just ignore the numerator in the transfer function \$\endgroup\$ Sep 17 '21 at 19:51
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The first thing I recommend is to rewrite the equation in a low-entropy format in which the numerator and the denominator are unitless. In your case, as underlined by a concerned citizen, the presence of the zero at the origin does not change the exercise.

A second-order polynomial can be put under the following normalized form: \$D(s)=1+b_1s+b_2s^2\$. Then, you can equate this expression with the normalized form where the quality factor \$Q\$ appears and find the correspondence between the terms: \$D(s)=1+\frac{s}{\omega_oQ}+(\frac{s}{\omega_o})^2\$. From there, if you do the maths ok, the you find \$\omega_o=\frac{1}{\sqrt{b_2}}\$ and \$Q=\frac{\sqrt{b_2}}{b_1}\$.

So, in your case, you have \$H(s)=\frac{\omega_n^2 s}{s^2+2\zeta\omega_n s+\omega_n^2}\$. Factor \$\omega_n^2\$ in the numerator and the denominator and you will have the above form for the identification of your coefficients.

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