# Is this transfer function impossible with a low-pass Sallen-key filter?

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

$$H(s)=\frac{628^3}{(s+628)(s^2+628s+628^2)}$$

The first order transfer function is

$$H(s)=\frac{628}{s+628}$$

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

That leave the Sallen-key transfer function to be

$$H(s)=\frac{628^2}{(s^2+628s+628^2)}=\frac{\frac{A}{R^2C^2}}{s^2+\frac{2}{RC}s+\frac{1}{R^2C^2}}$$

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.

• 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 – Huisman Dec 9 '19 at 10:43
• 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. – Verbal Kint Dec 9 '19 at 12:39

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: -

$$\dfrac{\omega_n}{s+\omega_n}$$

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.

• Is ζ the same as Q, defined as 1/(3-A)? I thought that term was meant to be ω/Q – griffin175 Dec 9 '19 at 10:40
• zeta = 1/(2Q) or Q = 1/(2*zeta) – Andy aka Dec 9 '19 at 10:41
• That makes perfect sense. I can't believe I got stuck on this for 5 hours. Thank you – griffin175 Dec 9 '19 at 10:48
• 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. – griffin175 Dec 10 '19 at 3:38
• 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. – Andy aka Dec 10 '19 at 8:05