The Butterworth second-order low-pass filter has the following transfer function:


We know that at \$s=\omega_c\$, the filter's attenuation is -3 dB, but mathematically, the only way to achieve that using the transfer function is to have a -1 coefficient in front of the \$\omega^2\$ term like so:


Thus, the gain at the cut-off frequency is \$20\cdot \log(0.707)=-3\ \mathrm{dB}\$

But every text book I researched shows the last term is +1, making the gain at the cut-off frequency equal to \$20\cdot\log(\frac{1}{(2+\sqrt2)})=20\cdot \log(0.2928)=-10\ \mathrm{dB}.\$

What went wrong?

  • \$\begingroup\$ Take the simple case where \$\omega=1\$. (Why not? There's no impact on the shape.) Then you just have the absolute value of \$\frac1{-1+\sqrt{2}j+1}=\frac1{\sqrt{2}j}\$. Which has a fairly obvious resulting magnitude since it's just a upward-pointing vector on the imaginary axis of said length. \$\endgroup\$ Nov 24, 2023 at 21:02

2 Answers 2


Your mistake is setting \$s=\omega_c\$.

Recall that \$s=\sigma + j\omega\$. In the case of sinusoidal signals \$\sigma=0\$ and only \$j\omega\$ persists. In that case, the Butterworth lowpass filter transfer function becomes

$$H(j\omega) = \frac{\omega_c^2}{(j\omega)^2+\sqrt{2}j\omega\omega_c+\omega_c^2} $$

When \$\omega=\omega_c\$ we have

$$H(j\omega_c) = \frac{\omega_c^2}{-\omega_c^2+j\sqrt{2}\omega_c^2 + \omega_c^2} \Leftrightarrow$$

$$H(j\omega_c) = \frac{\omega_c^2}{j\sqrt{2}\omega_c^2}=- \frac{j}{\sqrt{2}} = \frac{1}{\sqrt{2}}e^{-j\frac{\pi}{2}} $$

which shows that at the cut-off frequency the gain is \$-3 \ \text{dB}\$ and the phase shift is \$-90^\circ\$.


All 2nd order low pass filters have a gain at the natural resonant frequency (\$\omega_c\$ or \$\omega_n\$ or \$\omega_0\$) equal to the Q-factor of the circuit. Given that a Butterworth filter has a Q-factor of 0.7071, we can expect the gain to be 0.7071 or -3 dB. Here's an image from my basic website that shows the spectrum of a 2nd order filter where Q-factor is varied: -

enter image description here

This is the relevant formula that shows that gain at the natural resonant frequency is Q: -

enter image description here

Note: \$Q = \dfrac{1}{2\zeta}\$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.