According to articles and notes, the cutoff frequency of a 2nd order Butterworth lowpass filter: enter image description here

is expressed by the formula: $\omega _c = {1\over\sqrt{R_1R_2C_1C_2}}$

and since $\omega _c = 2 \pi f_c$, alternatively


Where does this formula come from? I'm asked to write the equations and formulas which eventually lead to the formula of the cutoff frequency, but I've been very confused and tutorials and notes don't help at all. Thanks in advance.

  • $\begingroup$ can you analyze the circuit (replace the op-amp, $R_3$, $R_4$ with a dependent voltage source) using, say, the node-voltage method? You will get two equations and two unknowns. do this in the $s$-domain so that the impedance of the caps are $\frac{1}{j \omega C}$ . Then get an expression of the the output voltage in terms of the input voltage: $\frac{V_O(s)}{V_I(s)}$. That is $H(s)$, your transfer function. Once you get there, then come back to us. $\endgroup$ Jan 31 '20 at 2:31
  • $\begingroup$ BTW, this is a better question for the Electrical Engineering StackExchange group. But, over there, for inline $\LaTeX$ math expressions, you need to use \$ instead of $ as a delimiter. That can be confusing. (this is because, i presume, hardware engineers discuss the price of parts more than DSPers do.) $\endgroup$ Jan 31 '20 at 2:35
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    $\begingroup$ ooops. i meant to say that the impedance of the caps are $\frac1{sC}$. $\endgroup$ Jan 31 '20 at 3:45

First you need to find (calculate or look up) the transfer function of that circuit. It has the form


where the constants $a$, $b$, and $c$ depend on the values of the resistors and capacitors.

The $3$ dB cut-off frequency $\omega_c$ can be found by solving


(because $H(0)=\frac{a}{c}$).

Solving $(2)$ involves solving a quadratic equation in $\omega_c^2$.


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