I'm trying to design a second order low-pass filter using the following transfer function:

\$\ H(s)=\frac{f_c^2}{s^2 + 2 f_c s +f_c^2} \$

with cutoff frequency fc = 3400 Hz

Whenever I plot this, the cutoff frequency is at -6 dB instead of -3 dB. I'm not really sure what I'm doing wrong. Here's my Matlab code:

fc = 3400;

s = 1i*logspace(0,6,1000);

H_d = fc^2 ./ (s.^2 + 2*fc*s + fc^2);

semilogx(abs(s), 20 * log10(abs(H_d)))

1 Answer 1


I don't recognize your transfer function as one of the classic low-pass-filter implementations. A Butterworth filter, for instance, would be:

$$ H(s) = \frac{f_c^2}{s^2+1.414f_cs+f_c^2} $$

note: for me, it's unusual to see the use of \$f\$ instead of \$w\$ in these formulas, but mathematically it should be equivalent, as long as you use \$s=jf\$ instead of \$s=jw\$

  • \$\begingroup\$ This worked, thank you! What I did was I expanded k / (s + fc)^2 and set the constant k to fc^2 to balance it out. I overlooked the damping ratio and saw that ζ = 0.707 for Butterworth filters (I set ζ to 1). Some other observations I made was that 2 = 6 dB and 1.414 = 3 dB. Source of equations \$\endgroup\$
    – kane40
    Oct 16, 2019 at 5:49
  • \$\begingroup\$ I'm glad you were able to trace back to the mistake. \$\endgroup\$
    – joribama
    Oct 16, 2019 at 5:59
  • \$\begingroup\$ A 2nd order low pass filter has a gain value at the resonant frequency equal to the circuit's Q. So when Q = 0.5, the gain at the resonant frequency is 0.5 or -6 dB. If Q is ten (\$\zeta\$ = 0.05) then gain is ten at resonance. \$\endgroup\$
    – Andy aka
    Oct 16, 2019 at 7:54

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