Butterworth LPF prototype and π-section microstrip

In the insertion loss method, the filter response is defined by the power loss ratio:

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.

Butterworth (maximally flat) LPF prototype design

1. Choose the response: PLR = 1 + k²(ω/ωc)^(2N). With k = 1, the cut-off ωc is the 3 dB point. The response is as flat as possible at ω = 0 (first 2N−1 derivatives are zero) and falls at 20N dB/decade beyond ωc.
2. Find the order N from the required stopband loss. Example spec: at least 15 dB at 2fc.

N ≥ log10(10^(A/10) − 1) / (2·log10(ω/ωc))
  = log10(10^1.5 − 1) / (2·log10 2) = 2.47  →  N = 3
Check: IL(2fc) = 10 log(1 + 2^6) = 18.13 dB ≥ 15 dB

3. Read the element values for source R = 1 Ω, ωc = 1 rad/s from the Butterworth table, or gk = 2 sin[(2k−1)π/2N]. For N = 3: g1 = 1, g2 = 2, g3 = 1, g4 = 1 (load).
4. Scale impedance and frequency (R0 = 50 Ω, fc = 2 GHz assumed): C = g/(ωc·R0), L = g·R0/ωc.
5. Convert to distributed form (Richards/Kuroda or stepped-impedance) and simulate.

π-section (shunt C – series L – shunt C)

C1 = C3 = 1/(2π·2×10⁹·50)  = 1.5915 pF
L2      = 2·50/(2π·2×10⁹)  = 7.9577 nH

Microstrip (stepped-impedance) implementation

High-Z lines act as series inductors, low-Z lines as shunt capacitors: βl = g·Zℓ/R0 for a capacitor, βl = g·R0/Zh for an inductor. Assumed substrate εr = 4.2, h = 1.58 mm, Zℓ = 20 Ω, Zh = 120 Ω (widths from microstrip design formulas, λg at 2 GHz):

 50Ω     ____         ____     50Ω
 ======|    |=======|    |======
       | C1 |  L2   | C3 |
       |____| thin  |____|
       wide   line   wide

Feed lines are 50 Ω (3.13 mm). Since βl of L2 is slightly above 45°, the response is checked by simulation and lengths are fine-tuned.