--- title: "Butterworth LPF prototyping by insertion loss" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:butterworth-lpf-prototyping" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/butterworth-lpf-prototyping site: BCT Engineering --- # Butterworth LPF prototyping by insertion loss RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2077 Chaitra, 8 marks - Video: [Design of prototype of Low pass filter (LPF) for maximally flat/ Butterworth response (N=5)](https://www.youtube.com/watch?v=QvZJUBPvecQ) — RF Design Basics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/butterworth-lpf-prototyping - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Describe in detail the procedures for prototyping Butterworth LPF using insertion loss method. ## Short answer Butterworth LPF prototype design: choose PLR=1+(ω/ωc)^(2N), find N from stopband loss. Normalized g-values from table or gk=2sin((2k-1)π/(2N)) give ladder with alternating shunt C and series L. Scale to required impedance and fc, then implement with stepped-impedance lines. ## Answer In the **insertion loss method**, the filter response is defined by the **power loss ratio**: ```text PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²) IL (dB) = 10 log PLR ``` A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response. #### Procedure for a Butterworth LPF prototype 1. **Specify the filter:** cut-off frequency fc, source/load impedance R0, and the minimum attenuation A (dB) at some stopband frequency ω. 2. **Select the maximally flat response:** ```text PLR = 1 + k²(ω/ωc)^(2N) ``` k² = 10^(Lr/10) − 1 sets the loss at ωc; for the usual 3 dB cut-off k = 1. The response is flat at ω = 0 and rolls off at 20N dB/decade. 3. **Find the order N:** ```text N ≥ log10(10^(A/10) − 1) / (2·log10(ω/ωc)) Example: A = 20 dB at ω = 2ωc N ≥ log10(99)/(2×0.301) = 3.31 → N = 4 ``` 4. **Get the normalised element values** (R0 = 1 Ω, ωc = 1 rad/s) from the table, or ```text g0 = 1, gk = 2 sin[(2k − 1)π/(2N)], g(N+1) = 1 ``` | N | g1 | g2 | g3 | g4 | g5 | |---|---|---|---|---|---| | 1 | 2.000 | 1 | | | | | 2 | 1.414 | 1.414 | 1 | | | | 3 | 1.000 | 2.000 | 1.000 | 1 | | | 4 | 0.765 | 1.848 | 1.848 | 0.765 | 1 | 5. **Draw the ladder:** gk alternate between shunt C and series L. Two equivalent forms exist: starting with a shunt C (π type) or a series L (T type). ```text Rs L2 L4 (π-type, N=4) o-/\/-+--UUU--+--UUU--+--o | | | C1 C3 RL | | | o-----+-------+-------+--o ``` 6. **Impedance and frequency scaling** to R0 and ωc: ```text L = gk·R0/ωc, C = gk/(R0·ωc), RL = g(N+1)·R0 ``` Example (N = 3, R0 = 50 Ω, fc = 2 GHz): C1 = C3 = 1.59 pF, L2 = 7.96 nH. 7. **Transform** (if needed) to high-pass, band-pass or band-stop using the standard frequency transformations. 8. **Implement at microwave frequency:** Richards' transformation (L → short-circuited stub, C → open-circuited stub, λ/8 at fc) with Kuroda identities, or stepped-impedance lines. Simulate and tune. Butterworth is chosen when a flat passband is needed; Chebyshev gives a sharper cut-off for the same N at the cost of ripple. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Butterworth power loss ratio and its key property? A. 1 + k²(ω/ωc)^2N, maximally flat at ω = 0 B. 1 + k²T_N²(ω/ωc), equal ripple C. 1/(1+ω²), rising D. 1 + k²N Answer: A. The first 2N−1 derivatives vanish at ω = 0; roll-off is 20N dB/decade. (Key point: Butterworth response) 2. For A = 20 dB at ω = 2ωc, what N is needed? A. N = 3 B. N = 4 C. N = 2 D. N = 5 Answer: B. N ≥ log10(99)/(2 log10 2) = 3.31, so N = 4. (Key point: Order) 3. Which formula gives Butterworth element values? A. gk = 2 sin[(2k−1)π/(2N)] B. gk = sin(kπ/N) C. gk = 2N/k D. gk = ln coth(k) Answer: A. For N = 4 it gives 0.765, 1.848, 1.848, 0.765. (Key point: g-values) 4. Which scaling formulas convert the prototype to actual values? A. L = gR0/ωc, C = g/(R0ωc) B. L = g/(R0ωc), C = gR0/ωc C. L = gωc/R0 D. C = gR0ωc Answer: A. Impedance and frequency scaling: L = gR0/ωc, C = g/(R0ωc), RL = g(N+1)R0. 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