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Insertion loss design of given LC ladder filter

Short answer

Insertion loss method gives exact control of response via chosen power loss ratio. A 5th-order Butterworth prototype (g₁=g₅=0.618, g₂=g₄=1.618, g₃=2.000) scaled to 50 Ω and 2 GHz yields C₁=C₅≈0..98 pF, L₂=L₄≈6..44 nH, C₃≈3..18 pF. Microstrip implementation: wide low-Z pads for shunt C, narrow high-Z lines for series L.

Explain, why microwave filters are designed using the Insertion Loss Method? Explain the detail design steps of the following microwave filter, and implement the circuit using microstrip. [Figure: low pass LC ladder between source Zs and load ZL: shunt capacitor C1, series inductor L2, shunt capacitor C3, series inductor L4, shunt capacitor C5; no component values given]

Why the insertion loss method is used

The older image-parameter method gives a passband and stop-band but cannot set the exact response, so it needs many trial adjustments. The insertion loss method starts from a chosen power loss ratio:

PLR = 1 / (1 − |Γ(ω)|²) = 1 + M(ω²)/N(ω²)
Butterworth : PLR = 1 + k²(ω/ωc)^(2N)
Chebyshev   : PLR = 1 + k² T_N²(ω/ωc)

so it gives:

  • full control of passband flatness or ripple, cut-off sharpness and phase,
  • a systematic synthesis from tables (g-values), with a clear trade-off between order N and performance,
  • a design that is easy to scale and transform to HP, BP or BS, and to convert to microstrip.

Design steps for the given 5-element ladder

The figure is a fifth-order low-pass ladder that starts with a shunt capacitor: C₁ (shunt), L₂ (series), C₃ (shunt), L₄ (series), C₅ (shunt). No values are given, so this example assumes a Butterworth response, N = 5, fc = 2 GHz and Z_s = Z_L = R₀ = 50 Ω.

Step 1: Specifications. fc = 2 GHz, maximally flat, 50 Ω terminations.

Step 2: Prototype g-values (g_k = 2 sin[(2k − 1)π/2N], N = 5):

g₁g₂g₃g₄g₅g₆
0.6181.6182.0001.6180.6181.0

Step 3: Impedance and frequency scaling (ωc = 2π × 2×10⁹ = 1.2566×10¹⁰ rad/s):

Ck = gk / (R0 ωc),  Lk = R0 gk / ωc
C1 = C5 = 0.618 / (50 × 1.2566e10) = 0.984 pF
L2 = L4 = 50 × 1.618 / 1.2566e10   = 6.438 nH
C3      = 2.000 / (50 × 1.2566e10) = 3.183 pF

Step 4: Convert to microstrip (stepped impedance). Choose Z_h = 120 Ω (narrow line) and Z_ℓ = 20 Ω (wide line):

inductor : βl = gk·R0 / Zh
capacitor: βl = gk·Zℓ / R0
C1, C5: βl = 0.618 × 20/50  = 14.16°
L2, L4: βl = 1.618 × 50/120 = 38.63°
C3    : βl = 2.000 × 20/50  = 45.84°

Each length is ℓ = (βl/360°)·λg at 2 GHz, where λg depends on the substrate (εr, h).

(Alternative: Richards' transformation turns each C into a λ/8 open stub of Z₀ = R₀/g (C₁: 80.9 Ω, C₃: 25 Ω) and each L into a λ/8 series short stub of Z₀ = R₀·g (L₂: 80.9 Ω). Kuroda identities then convert the series stubs into shunt stubs.)

Step 5: Simulate and tune the layout, adding the effects of step discontinuities, and adjust the lengths.

Microstrip implementation

 Stepped-impedance microstrip LPF (top view)
       ___        _____        ___
 =====|   |______|     |______|   |=====
 50 Ω | C1| L2   | C3  |  L4  | C5| 50 Ω
      |___|      |_____|      |___|
 wide = low Z0 (shunt C)
 narrow = high Z0 (series L)

Wide low-impedance pads make the shunt capacitors C₁, C₃ and C₅. Narrow high-impedance lines make the series inductors L₂ and L₄. All are printed on one substrate over a ground plane.

✓ Answer checked by Claude Opus 5.5 (high effort)

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