--- title: "Butterworth/Chebyshev LPF and double π-section" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:butterworth-chebyshev-double-pi" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/butterworth-chebyshev-double-pi site: BCT Engineering --- # Butterworth/Chebyshev LPF and double π-section RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2076 Bhadra, 8+2 marks - Video: [Low pass filter implementation using stub||Richard's transformation and Kuroda's identities||](https://www.youtube.com/watch?v=-l5nTRQxbtg) — 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-chebyshev-double-pi - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question How is a low pass filter prototype based on Butterworth and Chebyshev approximations designed using insertion loss method? Implement a low pass filter double π-sections using microstrips. ## Short answer Insertion loss method uses PLR = 1/(1−|Γ|²). Butterworth: PLR = 1+k²(ω/ωc)^2N, gk = 2 sin[(2k−1)π/2N]. Chebyshev: PLR = 1+k²TN², with ripple formulas. Double π (N=5) at 2 GHz, 50 Ω: Butterworth C1=C5=0.984 pF, L2=L4=6.438 nH, C3=3.183 pF; Chebyshev 0.5 dB: C1=C5=2.715 pF, L2=L4=4.893 nH, C3=4.044 pF. Microstrip: shunt C as low-Z lines, series L as high-Z 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. #### Prototype design steps (both responses) 1. Specify fc, R0, ripple Lr (Chebyshev) and stopband attenuation. 2. Choose PLR and find the order N. 3. Get prototype values g1 … gN+1 (R0 = 1 Ω, ωc = 1 rad/s). 4. Scale: L = g·R0/ωc, C = g/(R0·ωc). 5. Realise with microstrip lines and simulate. #### Butterworth (maximally flat) ```text PLR = 1 + k²(ω/ωc)^(2N) gk = 2 sin[(2k − 1)π/(2N)] ``` Flat passband, monotonic roll-off of 20N dB/decade. #### Chebyshev (equal ripple) ```text PLR = 1 + k²·TN²(ω/ωc), TN = Chebyshev polynomial k² = 10^(Lr/10) − 1 (ripple Lr dB) β = ln[coth(Lr/17.37)], γ = sinh[β/(2N)] g1 = 2a1/γ, gk = 4a(k−1)ak / [b(k−1)g(k−1)] ak = sin[(2k−1)π/2N], bk = γ² + sin²(kπ/N) ``` Ripple in the passband, but a sharper cut-off for the same N. For odd N, g(N+1) = 1. #### Double π-section (N = 5) values A double π-section is C–L–C–L–C (two π sections sharing the middle C). Assume fc = 2 GHz, R0 = 50 Ω, ripple 0.5 dB. | k | Butterworth gk | C or L | Chebyshev 0.5 dB gk | C or L | |---|---|---|---|---| | 1 | 0.618 | 0.984 pF | 1.7058 | 2.715 pF | | 2 | 1.618 | 6.438 nH | 1.2296 | 4.893 nH | | 3 | 2.000 | 3.183 pF | 2.5409 | 4.044 pF | | 4 | 1.618 | 6.438 nH | 1.2296 | 4.893 nH | | 5 | 0.618 | 0.984 pF | 1.7058 | 2.715 pF | #### Microstrip implementation (stepped impedance, Chebyshev) Shunt C → low-Z line (20 Ω), βl = g·Zℓ/R0; series L → high-Z line (120 Ω), βl = g·R0/Zh. Assumed εr = 4.2, h = 1.58 mm. | Element | Z | βl | Length | Width | |---|---|---|---|---| | C1, C5 | 20 Ω | 39.09° | 8.61 mm | 11.27 mm | | L2, L4 | 120 Ω | 29.35° | 7.26 mm | 0.43 mm | | C3 | 20 Ω | 58.23° | 12.83 mm | 11.27 mm | (For the Butterworth version: C1 = C5 → 14.16°, L2 = L4 → 38.63°, C3 → 45.84°.) ```text 50Ω ____ ______ ____ 50Ω ====| |------| |------| |==== | C1 | L2 | C3 | L4 | C5 | |____| thin |______| thin |____| ``` The middle capacitor line is long (βl > 45°), so in practice it is split or tuned in simulation, and step discontinuities are included. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Compared to Butterworth for the same N, Chebyshev gives: A. Flat passband, slower cut-off B. Passband ripple but sharper cut-off C. No ripple and sharper cut-off D. Slower cut-off and ripple Answer: B. Equal-ripple PLR 1 + k²T_N² trades ripple for sharper roll-off. (Key point: Chebyshev vs Butterworth) 2. How is k² related to the ripple Lr in dB? A. k² = 10^(Lr/10) − 1 B. k² = Lr/10 C. k² = 10 log Lr D. k² = 1 − Lr Answer: A. k² = 10^(Lr/10) − 1. (Key point: Ripple constant) 3. What is the structure of a double π-section (N = 5)? A. L–C–L–C–L B. C–L–C–L–C, two π sections sharing the middle C C. C–C–L–L–C D. Two separate L-sections Answer: B. Two π sections share the middle shunt capacitor. (Key point: Double π) 4. Why is the middle capacitor line (58.23°) a problem? A. βl > 45°, so the short-line approximation fails; split or tune it B. It is too narrow C. Its impedance is 50 Ω D. It is an inductor Answer: A. Lines longer than about 45° are not accurate lumped approximations, so they are split or tuned in simulation. 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