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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?

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  • \$\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

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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\$.

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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}\$

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