Butterworth/Chebyshev LPF and double π-section

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.

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)

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)

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.

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.

(For the Butterworth version: C1 = C5 → 14.16°, L2 = L4 → 38.63°, C3 → 45.84°.)

 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.