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Band stop filter (2000 rad/s, 400 rad/s) from 3rd order

Short answer

Frequency transformation maps a low-pass prototype by s substitution. For band-stop: s→B s/(s²+ω₀²). Series L becomes parallel tank with L'=L B/ω₀², C'=1/(L B); shunt C becomes series LC with L'=1/(C B), C'=C B/ω₀². For ω₀=2000 rad/s, B=400 rad/s, third-order Butterworth yields series arms 0.1 mH∥2.5 mF, shunt 1.25 mH in series with 0.2 mF.

What is frequency transformation? Design a band stop filter having center frequency 2000 rad/s and bandwidth 400 rad/s from a third order Butterworth low pass filter. [Refer Table of doubly terminated Butterworth element values]

Frequency transformation

Frequency transformation is a change of the complex frequency variable that converts a normalized low-pass prototype filter (cutoff 1 rad/s) into a high-pass, band-pass or band-stop filter (or a low-pass with a different cutoff). The magnitude response of the prototype is mapped onto the new frequency axis, so the same attenuation values appear at the corresponding new frequencies. For a band-stop filter, s→Bss2+ω02s \to \dfrac{Bs}{s^2+\omega_0^2}.

Design of the band-stop filter

Prototype (table of doubly terminated Butterworth values, n=3n = 3, Rs=RL=1 ΩR_s = R_L = 1\ \Omega): series L1=1L_1 = 1 H, shunt C2=2C_2 = 2 F, series L3=1L_3 = 1 H.

Element rules for LP → BS:

  • series LL → parallel tank in the series arm: L′=LBω02L' = \dfrac{LB}{\omega_0^2}, C′=1LBC' = \dfrac{1}{LB};
  • shunt CC → series LC in the shunt arm: L′=1CBL' = \dfrac{1}{CB}, C′=CBω02C' = \dfrac{CB}{\omega_0^2}.
L1=1 H:L1′=L1Bω02=4004×106=0.1 mH,C1′=1L1B=1400=2.5 mFC2=2 F:L2′=1C2B=1800=1.25 mH,C2′=C2Bω02=8004×106=0.2 mFL3=1 H:L3′=0.1 mH,C3′=2.5 mF\begin{aligned} L_1 = 1\ \text{H}:&\quad L_1' = \frac{L_1B}{\omega_0^2} = \frac{400}{4\times10^6} = 0.1\ \text{mH},\quad C_1' = \frac{1}{L_1B} = \frac{1}{400} = 2.5\ \text{mF} \\ C_2 = 2\ \text{F}:&\quad L_2' = \frac{1}{C_2B} = \frac{1}{800} = 1.25\ \text{mH},\quad C_2' = \frac{C_2B}{\omega_0^2} = \frac{800}{4\times10^6} = 0.2\ \text{mF} \\ L_3 = 1\ \text{H}:&\quad L_3' = 0.1\ \text{mH},\quad C_3' = 2.5\ \text{mF} \end{aligned}
  Rs=1   [ Z1 ]          [ Z3 ]
 o-/\/\--[    ]---+------[    ]---+-----o
 |                |               |     +
 Vs            [ Y2 ]           RL=1    Vo
 |                |               |     -
 o----------------+---------------+-----o
ArmPrototypeBand-stop element
Z1 (series)L1=1L_1 = 1 H0.1 mH ∥ 2.5 mF (tank)
Y2 (shunt)C2=2C_2 = 2 F1.25 mH in series with 0.2 mF
Z3 (series)L3=1L_3 = 1 H0.1 mH ∥ 2.5 mF (tank)

Check: 1/0.1 mH×2.5 mF=1/1.25 mH×0.2 mF=20001/\sqrt{0.1\text{ mH}\times2.5\text{ mF}} = 1/\sqrt{1.25\text{ mH}\times0.2\text{ mF}} = 2000 rad/s. Band edges: ω1,2=ω02+(B/2)2∓B/2=1810.0\omega_{1,2} = \sqrt{\omega_0^2 + (B/2)^2} \mp B/2 = 1810.0 and 2210.02210.0 rad/s (3 dB points).

These values are for 1 Ω1\ \Omega terminations. For practical values, magnitude-scale by kmk_m (e.g. km=1000k_m = 1000 for 1 kΩ terminations): multiply every L by kmk_m and divide every C by kmk_m.

✓ Answer checked by Claude Sonnet 5.5

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