I was given this transfer function and instructed to use a Sallen-key cascaded with a 1st order filter to achieve it.


The first order transfer function is


So if I choose \$c=10\mu F\$, then \$R=159\Omega\$.

That leave the Sallen-key transfer function to be


So \$A\$ must be equal to \$1\$ and \$\frac{1}{R^2C^2}=628^2\$. The same values for R and C as the first order filter work here.

The problem is that middle term, \$628s=\frac{2}{RC}\$. This contradicts the other values. It works fine if this was a typo and that term was meant to be \$\frac{1}{2}\$ that value. Did I make a mistake here? I've wasted hours deriving and rederiving the transfer function for the filter to no avail.

  • \$\begingroup\$ Although the answer given suits, note the gain of first order transfer function did not necessarily have to be 628. In other words: \$(628)^3\$ doesn't have to be factored as \$(628) \cdot (628)^2\$ but could e.g. have been factored as \$(157) \cdot (1256)^2\$ as well \$\endgroup\$
    – Huisman
    Dec 9, 2019 at 10:43
  • \$\begingroup\$ The transfer function must written in a low-entropy format with a leading term while the numerator over the denominator is unitless: factor \$628^3\$ and you end-up with \$H(s)=\frac{1}{1+\frac{s}{\omega_p}}\frac{1}{1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2}\$. This is a unity-gain low-pass filter with a cutoff frequency of \$\omega_p=\frac{1}{628}\$ followed by a second-order filter with a unity \$Q\$ also tuned at \$\omega_0 = \omega_p\$. Rewriting transfer functions the proper way is key in understanding their asymptotical responses. \$\endgroup\$ Dec 9, 2019 at 12:39

1 Answer 1


Start with a 2nd order low pass filter transfer function in its general form: -

$$H(s) = \dfrac{\omega_n^2}{s^2 + 2\zeta\omega_ns + \omega_n^2}$$

And then mulitply this with a single order low pass filter transfer function: -


Giving you this: -

$$\dfrac{\omega_n}{s+\omega_n}\cdot \dfrac{\omega_n^2}{s^2 + 2\zeta\omega_ns + \omega_n^2}$$

Can you see that this is the same form as your equation and therefore, \$2\zeta\omega_n\$ must equal \$\omega_n\$ or put differently, \$\zeta\$ = 0.5.

The problem you have in your analysis is that you haven't factored-in the effect of zeta (\$\zeta\$) and a sallen key filter can quite often have a zeta below 1 because this can optimize the transfer function. Look up butterworth sallen key filters for example.

  • \$\begingroup\$ Is ζ the same as Q, defined as 1/(3-A)? I thought that term was meant to be ω/Q \$\endgroup\$
    – griffin175
    Dec 9, 2019 at 10:40
  • \$\begingroup\$ zeta = 1/(2Q) or Q = 1/(2*zeta) \$\endgroup\$
    – Andy aka
    Dec 9, 2019 at 10:41
  • \$\begingroup\$ That makes perfect sense. I can't believe I got stuck on this for 5 hours. Thank you \$\endgroup\$
    – griffin175
    Dec 9, 2019 at 10:48
  • \$\begingroup\$ I'm going to have to retract the "that makes sense" part of my statement. After investigating further I found that I was at least partially correct. Q is 1/(3-A) so Q must be 1/2 and therefore zeta = 1. The contradiction remains. The solution was as @Huisman mentioned, to factor the numerator differently. This leaves the first order filter with a gain of its own and this works with an active filter. \$\endgroup\$
    – griffin175
    Dec 10, 2019 at 3:38
  • \$\begingroup\$ You said The same values for R and C as the first order filter work here. and I assumed this was up for grabs. Yes, if you insist that R1=R2 and C1=C2 (for the sallen key) then you are forced into the scenario where Q = 0.5 and \$\zeta\$=1. And, it has to follow that the numerator is factored differently. \$\endgroup\$
    – Andy aka
    Dec 10, 2019 at 8:05

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