Skip to main content

Band pass (10000 rad/s, 1000 rad/s) from given ladder

Short answer

Frequency transformation maps a normalized LP to other filter types, enabling reuse of prototype tables and preserving selectivity. For BP, s→(s²+ω₀²)/(Bs) with ω₀=10⁴ rad/s, B=10³ rad/s: shunt 1 F becomes 1 mF ∥ 10 µH; series 2 H becomes 2 mH series 5 µF; both ends same, 1 Ω terminations.

What is frequency transformation? What are its importance. The low pass filter shown below has a cutoff frequency at 1 rad/sec. Transform it into a band pass filter having center frequency at 10000 rad/sec and bandwidth of 1000 rad/sec. [Figure: 1 Ω source resistor, shunt 1 F capacitor, series 2 H inductor, shunt 1 F capacitor, 1 Ω load]

Frequency transformation is the change of the complex-frequency variable ss in a normalized low-pass prototype (cut-off ω=1\omega = 1 rad/s) so that the same circuit or transfer function gives a high-pass, band-pass or band-stop response, or a low-pass response at another frequency. Typical substitutions are s→s/ωps \to s/\omega_p (LP→LP), s→ωp/ss \to \omega_p/s (LP→HP) and s→s2+ω02Bss \to \dfrac{s^2+\omega_0^2}{Bs} (LP→BP).

Importance

  • Only one set of design tables (Butterworth, Chebyshev, Bessel low-pass prototypes) is needed; every other filter type is derived from it.
  • The approximation is done once, on the simple low-pass problem; the transformed filter keeps the same ripple and selectivity. Note that a band-pass or band-stop filter has twice the order of the prototype (here a 3rd-order low-pass gives a 6th-order band-pass).
  • Element values can be converted directly in the circuit (each LL and CC is replaced by a simple combination), so no new synthesis is needed.
  • Normalized values (ω=1\omega = 1, 1 Ω1\ \Omega) are easy to handle and are later scaled to practical values.

Band-pass transformation of the given ladder

Prototype (ωc=1\omega_c = 1 rad/s): Rs=1 ΩR_s = 1\ \Omega, shunt C1=1C_1 = 1 F, series L2=2L_2 = 2 H, shunt C3=1C_3 = 1 F, RL=1 ΩR_L = 1\ \Omega (3rd-order Butterworth). Given ω0=104\omega_0 = 10^4 rad/s, B=103B = 10^3 rad/s, ω02=108\omega_0^2 = 10^8.

LP → BP rules (substitute s→s2+ω02Bss \to \dfrac{s^2+\omega_0^2}{Bs}):

Prototype elementBecomesValues
Series LLseries L′L'–C′C' (series resonant)L′=L/BL' = L/B, C′=B/(Lω02)C' = B/(L\omega_0^2)
Shunt CCshunt L′′∥C′′L'' \parallel C'' (parallel resonant)C′′=C/BC'' = C/B, L′′=B/(Cω02)L'' = B/(C\omega_0^2)
ResistorunchangedRR

Each new pair resonates at ω0\omega_0 since L′C′=L′′C′′=1/ω02L'C' = L''C'' = 1/\omega_0^2.

C1′′=11000=1 mF,L1′′=10001×108=10 μHL2′=21000=2 mH,C2′=10002×108=5 μFC3′′=1 mF,L3′′=10 μH\begin{aligned} C_1'' &= \frac{1}{1000} = 1\ \text{mF}, & L_1'' &= \frac{1000}{1\times10^8} = 10\ \mu\text{H} \\ L_2' &= \frac{2}{1000} = 2\ \text{mH}, & C_2' &= \frac{1000}{2\times10^8} = 5\ \mu\text{F} \\ C_3'' &= 1\ \text{mF}, & L_3'' &= 10\ \mu\text{H} \end{aligned} 
o-[1 Ω]-+--[2 mH]-[5 uF]--+------+
        |                 |      |
  [10uH || 1mF]     [10uH || 1mF] [1 Ω]
        |                 |      |
o-------+-----------------+------+

Check: 2×10−3×5×10−6=10−8=1/ω022\times10^{-3}\times5\times10^{-6} = 10^{-8} = 1/\omega_0^2 and 10×10−6×10−3=10−810\times10^{-6}\times10^{-3} = 10^{-8}, so every arm resonates at 10410^4 rad/s.

Answer: shunt arms: 10 µH ∥ 1 mF (both ends); series arm: 2 mH in series with 5 µF; source and load 1 Ω.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Filter Design chapter 3 questions →