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Band pass (2000 rad/s, 400 rad/s) from 4th order

Short answer

For ω₀ = 2000 rad/s, B = 400 rad/s and fourth‑order Butterworth prototype (0.7654 H, 1.848 F, 1.848 H, 0.7654 F): Z1: 1.9135 mH series with 130.65 µF; Y2: 4.62 mF parallel with 54.113 µH; Z3: 4.62 mH series with 54.113 µF; Y4: 1.9135 mF parallel with 130.65 µH. All resonate at 2000 rad/s; band edges are 1810 and 2210 rad/s.

What is the importance of frequency transformation? Obtain a bandpass filter having ω0 = 2000 rad/s and B = 400 rad/s from fourth order Butterworth lowpass filter. [Refer table 2]

Importance of frequency transformation

Frequency transformation converts a normalized low-pass prototype into HP, BP or BS filters by a change of variable, so:

  • Only one set of approximation tables (low-pass poles and ladder element values) is needed; HP, BP and BS filters are derived from it.
  • The difficult approximation step (finding T(s)T(s) for given αmax\alpha_{max}, αmin\alpha_{min}) is done once, for the simpler low-pass case; specifications of the other types are converted to an equivalent LP specification.
  • It can be applied either to the transfer function (substitute for ss) or directly to each element of an LC ladder, so the circuit is obtained without re-deriving it.
  • It reduces design time and errors, and keeps the properties of the prototype (equiripple, maximally flat, etc.).

Band-pass filter from the fourth-order Butterworth low-pass filter

LP → BP: s→s2+ω02Bss \to \dfrac{s^2+\omega_0^2}{Bs}, ω0=2000\omega_0 = 2000 rad/s (ω02=4×106\omega_0^2 = 4\times10^6), B=400B = 400 rad/s.

Prototype (Table 2, n=4n = 4, Rs=RL=1 ΩR_s = R_L = 1\ \Omega, starting with a series inductor): L1=0.7654L_1 = 0.7654 H, C2=1.848C_2 = 1.848 F, L3=1.848L_3 = 1.848 H, C4=0.7654C_4 = 0.7654 F.

L1:L1′=0.7654400=1.9135 mH,C1′=4000.7654×4×106=130.65 μFC2:C2′=1.848400=4.62 mF,L2′=4001.848×4×106=54.113 μHL3:L3′=4.62 mH,C3′=54.113 μFC4:C4′=1.9135 mF,L4′=130.65 μH\begin{aligned} L_1:&\quad L_1' = \frac{0.7654}{400} = 1.9135\ \text{mH},\quad C_1' = \frac{400}{0.7654\times4\times10^6} = 130.65\ \mu\text{F} \\ C_2:&\quad C_2' = \frac{1.848}{400} = 4.62\ \text{mF},\quad L_2' = \frac{400}{1.848\times4\times10^6} = 54.113\ \mu\text{H} \\ L_3:&\quad L_3' = 4.62\ \text{mH},\quad C_3' = 54.113\ \mu\text{F} \\ C_4:&\quad C_4' = 1.9135\ \text{mF},\quad L_4' = 130.65\ \mu\text{H} \end{aligned}
  Rs=1  [ Z1 ]        [ Z3 ]
 o-/\/\-[    ]--+-----[    ]--+------+---o
 |              |             |      |   +
 Vs          [ Y2 ]        [ Y4 ]  RL=1  Vo
 |              |             |      |   -
 o--------------+-------------+------+---o
ArmBand-pass element
Z1 (series)1.9135 mH in series with 130.65 μF
Y2 (shunt)4.62 mF ∥ 54.113 μH
Z3 (series)4.62 mH in series with 54.113 μF
Y4 (shunt)1.9135 mF ∥ 130.65 μH

All LC pairs resonate at 2000 rad/s; the 3 dB band edges are 1810.0 and 2210.0 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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