# Second-order Butterworth gain calculation at cut-off frequency

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

$$\frac{\omega_c^2}{s^2+\sqrt2s\omega_c+\omega_c^2}$$

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:

$$\frac{\omega_c^2}{\omega_c^2+\sqrt2\omega_c\omega_c-\omega_c^2}=\\\frac{\omega_c^2}{\omega_c^2(1+\sqrt2-1)}=\\\frac{1}{\sqrt2}=0.707$$

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?

• 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. Nov 24, 2023 at 21:02

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: -

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

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