I have been trying to design a state variable filter based on a block diagram provided. The transfer function which I have is of the form:

$$\frac{K_1K_2}{s^2 + K_1s + K_1K_2}$$

I need to obtain and derive expressions for the natural frequency and quality factor of the state variable filter in terms of \$K_1\$ and \$K_2\$. In order to do so I need to compare my T.F with a 2nd order normalized T.F for state variable filters. Below is the T.F I am comparing with from this website.

$$\frac{V_{\text{out}}}{V_{\text{in}}} = \frac{A_o\left(\frac{f}{f_o}\right)}{\left[1 + 2\zeta\frac{f}{f_o} + \left(\frac{f}{f_o}\right)^2\right]}$$

I am getting confused since the 2nd order normalized T.F does not have an \$s^2\$ term.

For the natural frequency I have got:

$$\omega_n\text{ or }\frac{f}{f_o} = \sqrt{K_1K_2}$$

and for the quality factor:

$$2\zeta\omega_n = K_1s$$ $$\zeta = \frac{K_1s}{2\sqrt{K_1K_2}}$$

My question is: Am I comparing to the right transfer function since it does not have s squared terms? If so have I proceeded correctly? If not, are there any tips on what I could do to derive the above expressions correctly?

  • \$\begingroup\$ Your TF equation is correct, the website is wrong, all the 's' variables are missing. Also \$2\zeta \omega_n=K_1\$; 's' is the Laplace variable and doesn't have a numerical value in this context - it's the coefficient of s that you need. \$\endgroup\$
    – Chu
    Jan 7, 2018 at 13:36
  • \$\begingroup\$ @Chu thanks for your comment, so I guess the explanation given below is indeed correct. \$\endgroup\$
    – rrz0
    Jan 7, 2018 at 13:42
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    \$\begingroup\$ Yes - the "tutorial" of the referenced website is totally wrong!! \$\endgroup\$
    – LvW
    Jan 9, 2018 at 12:02
  • \$\begingroup\$ @LvW, thanks, that was the main reason I was confused! \$\endgroup\$
    – rrz0
    Jan 9, 2018 at 13:39
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    \$\begingroup\$ I left a written comment/correction at the site, but it was not considered. My recommendation: Do not rely on such tutorials. Instead, consult a good textbook (several good books are available online). \$\endgroup\$
    – LvW
    Jan 9, 2018 at 14:12

1 Answer 1


The transfer function of a second-order low pass filter is given by (reference):

$$H(s)=\frac{\omega_0^2}{s^2+2\zeta\omega_0 s+\omega_0^2}\tag{1}$$

where \$\omega_0\$ is the resonant frequency (in radians), and \$\zeta\$ is the damping ratio.

Comparing \$(1)\$ to your transfer function you get




  • \$\begingroup\$ Thanks for your answer. You mentioned that is the transfer function of a second-order low pass filter, however the transfer function in question here is that of a state variable filter. Does it still apply? \$\endgroup\$
    – rrz0
    Jan 7, 2018 at 13:17
  • \$\begingroup\$ @Rrz0: Yes, a state variable filter is just a certain structure to implement a filter characteristic. Your transfer function is a second-order low pass filter. \$\endgroup\$
    – Matt L.
    Jan 7, 2018 at 13:21
  • \$\begingroup\$ Can you please elaborate on how you got the value for the damping factor? I am getting (1/2) K1/sqrt(K1K2) \$\endgroup\$
    – rrz0
    Jan 8, 2018 at 10:36
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    \$\begingroup\$ @Rrz0: ... which is the same as the one I got, because \$K_1/\sqrt{K_1}=\sqrt{K_1}\$ :) \$\endgroup\$
    – Matt L.
    Jan 8, 2018 at 11:02
  • \$\begingroup\$ The frequency wo is not the "resonant" frequency but the so called "pole frequency". At this frequency the phase shift is -90deg. In contrast, "resonance" is defined for a frequency dependenet system at the point where the phase sift is zero. \$\endgroup\$
    – LvW
    Jan 9, 2018 at 8:30

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