For a general filter, what is the way to find the cut-off frequency? In this link Okawa Electronic Design shows a formula for the cut-off frequency of a Sallen-key filter that looks like:


But how did he find that?

My thoughts. I was in the understanding that the cut-off frequency of any filter can be found by solving:


Where \$\left|\underline{\hat{\mathscr{H}}}\left(\text{j}\omega\right)\right|\$ is the the absolute value of the transfer function such that \$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|\$ is at it's maximum.

If I use that on the Sallen-key filter I will get the following formula:


   C1 C2 R1 R2 R3))(\[Sqrt](C1^2 R1^2 R3^2 + C2^2 (R1 R3 - R2 R4)^2 + 
     2 C1 C2 R1 R3 (R1 R3 - 
        R2 (R3 + 
           R4)) - \[Sqrt](4 C1^2 C2^2 R1^2 R2^2 R3^4 + (C1^2 R1^2 \
R3^2 + C2^2 (R1 R3 - R2 R4)^2 + 
          2 C1 C2 R1 R3 (R1 R3 - R2 (R3 + R4)))^2)))

I found that using Mathematica, but it shows a very different formula than \$(1)\$. What am I doing wrong?

  • \$\begingroup\$ As the answers below say, -3dB is the standard definition, especially if anything is being measured. Sometimes, for analytical purposes, if you have a double pole or a complex conjugate pair of poles (i.e. anything resonant), then the frequency of a pole-pair's real component is convenient, because you can use it to mentally reckon the roll-off -- e.g. with 2nd order, approx -40dB at 10x that frequency, -80dB at 100x etc. This frequency isn't necessarily the same as the -3dB frequency. \$\endgroup\$ – Pete W Feb 11 at 17:47
  • \$\begingroup\$ @PeteW I think you mean the frequency of a pole-pair's imaginary component. \$\endgroup\$ – Andy aka Feb 11 at 17:48
  • \$\begingroup\$ hmm yes. "s" vs omega again? E.g. lets say it's a double pole (real in s plane), which would be Q=1/2 I think \$\endgroup\$ – Pete W Feb 11 at 17:49

For a unity-gain high-pass Sallen Key filter like this: -

enter image description here

The frequency \$\frac{1}{2\pi\sqrt{R_1R_2C_1C_2}}\$ isn't the 3 dB cut-off frequency but the formula for the natural resonant frequency of the filter. However, the 3 dB cut of frequency will be equal to that frequency when Q = \$\frac{1}{\sqrt2}\$. This is because, for a 2nd order low pass (or high pass) filter, the gain at the natural resonant frequency is numerically (and formulaically) equal to Q.

I use the term "natural resonant frequency" but some people (such as @LvW) call it the pole frequency. Anyway, for a 2nd order high-pass filter with unity gain, the \$j\omega\$ transfer function is: -

$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-\frac{\omega^2}{\omega_n^2}-j2\zeta\frac{\omega}{\omega_n}}$$

And if you let \$\omega = \omega_n\$ you get: -

$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{j}{2\zeta}$$

And this implies that at the natural resonant frequency (\$\omega_n\$) there is a phase shift of 90° and the gain is Q.


Usually the definition of cutoff frequency for a lowpass filter is that frequency at which the filter's response is 3dB down from its response at DC.

Some people choose 6dB down, or other numbers.

The way you calculate that frequency for a given filter is to either plot the response and look for the 3dB point, or express the filter's response mathematically and grind through the math to find the frequency analytically.

  • 1
    \$\begingroup\$ There is a common agreement that the 3dB-cutoff applies to 1st order filters as well as 2nd-order filters with BUTTERWORTH response (Qp=0.7071) only. For all other approximatins (Chebyshev, elliptica,..), we have different definitions (based on the ripple within the passband). \$\endgroup\$ – LvW Feb 11 at 18:23

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