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Band pass (ω1 = 100, ω2 = 10000) from normalized ladder

Short answer

Given ω₁=100 rad/s, ω₂=10000 rad/s, ω₀=1000 rad/s, B=9900 rad/s. Transform the normalized ladder: series 0.7654 H → 77.313 µH series with 12.934 mF; shunt 1.8485 F → 186.72 µF parallel with 5.3557 mH; series 1.8485 H → 186.72 µH series with 5.3557 mF; shunt 0.7654 F → 77.313 µF parallel with 12.934 mH. All sections resonate at 1000 rad/s, passband 100–10000 rad/s.

What is frequency transformation in filter design? How can you obtain a bandpass filter from given lowpass filter at normalized frequency? Obtain a bandpass filter having ω1 = 100 rad/s and ω2 = 10000 rad/s from following lowpass filter at normalized frequency. [Figure: source V1, 1 Ω source resistor, series 0.7654 H, shunt 1.8485 F, series 1.8485 H, shunt 0.7654 F, 1 Ω load (output V2)]

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.

Obtaining a band-pass filter from the normalized low-pass filter

Replace ss by s2+ω02Bs\dfrac{s^2+\omega_0^2}{Bs}, where ω0=ω1ω2\omega_0 = \sqrt{\omega_1\omega_2} and B=ω2−ω1B = \omega_2 - \omega_1. Applied to each element:

  • series LL → series L′=L/BL' = L/B with C′=B/(Lω02)C' = B/(L\omega_0^2);
  • shunt CC → parallel C′=C/BC' = C/B with L′=B/(Cω02)L' = B/(C\omega_0^2);
  • resistors unchanged.

This maps ω0→0\omega_0 \to 0 and ω1,ω2→∓1\omega_1, \omega_2 \to \mp1 of the low-pass prototype, so the LP passband becomes the band ω1\omega_1 to ω2\omega_2.

Design for ω1=100\omega_1 = 100 rad/s, ω2=10000\omega_2 = 10000 rad/s

ω0=100×10000=1000 rad/s,ω02=106,B=10000−100=9900 rad/s\omega_0 = \sqrt{100\times10000} = 1000\ \text{rad/s}, \quad \omega_0^2 = 10^6, \quad B = 10000 - 100 = 9900\ \text{rad/s}

Given ladder (4th-order Butterworth, 1 Ω terminations): series 0.7654 H, shunt 1.8485 F, series 1.8485 H, shunt 0.7654 F.

0.7654 H:L′=0.76549900=77.313 μH,C′=99000.7654×106=12.934 mF1.8485 F:C′=1.84859900=186.72 μF,L′=99001.8485×106=5.3557 mH1.8485 H:L′=186.72 μH,C′=5.3557 mF0.7654 F:C′=77.313 μF,L′=12.934 mH\begin{aligned} 0.7654\ \text{H}:&\quad L' = \frac{0.7654}{9900} = 77.313\ \mu\text{H},\quad C' = \frac{9900}{0.7654\times10^6} = 12.934\ \text{mF} \\ 1.8485\ \text{F}:&\quad C' = \frac{1.8485}{9900} = 186.72\ \mu\text{F},\quad L' = \frac{9900}{1.8485\times10^6} = 5.3557\ \text{mH} \\ 1.8485\ \text{H}:&\quad L' = 186.72\ \mu\text{H},\quad C' = 5.3557\ \text{mF} \\ 0.7654\ \text{F}:&\quad C' = 77.313\ \mu\text{F},\quad L' = 12.934\ \text{mH} \end{aligned}
  Rs=1  [ Z1 ]        [ Z3 ]
 o-/\/\-[    ]--+-----[    ]--+------+---o
 |              |             |      |   +
 Vs          [ Y2 ]        [ Y4 ]  RL=1  Vo
 |              |             |      |   -
 o--------------+-------------+------+---o
ArmPrototypeBand-pass element
Z10.7654 H77.313 μH in series with 12.934 mF
Y21.8485 F186.72 μF ∥ 5.3557 mH
Z31.8485 H186.72 μH in series with 5.3557 mF
Y40.7654 F77.313 μF ∥ 12.934 mH

Check: 1/77.313 μH×12.934 mF=10001/\sqrt{77.313\ \mu\text{H}\times12.934\text{ mF}} = 1000 rad/s. 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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