- 2078 Chaitra · 10 marks
Double-section series-arm microstrip filter
Short answer
A double-section series-arm filter is a 5-element T-type Butterworth LPF designed by insertion loss method. With N=5 for stopband attenuation, scaled to 50 Ω, 2 GHz: L1=L5=2.46 nH, C2=C4=2.58 pF, L3=7.96 nH. Microstrip: wide low-Z (20 Ω) lines for shunt C (8.17 mm), narrow high-Z (120 Ω) lines for series L (3.65 mm, 11.81 mm) on εr=4.2.
Justify and describe how a double-section series-arm type micro strip filter is designed using insertion loss method.
A double-section series-arm filter is a 5-element T-type ladder low-pass filter (series L – shunt C – series L – shunt C – series L). It is designed from a low-pass prototype by the insertion loss method and then built in microstrip.
Justification for the insertion loss method
| Image parameter method | Insertion loss method |
|---|---|
| Cascades simple sections | Synthesises the whole filter from a chosen response |
| Response cannot be specified exactly | Passband and stopband specified exactly |
| Needs trial and error | Systematic, uses tables |
| Mismatch at terminations | Correct terminations included |
| No trade-off control | Trade-off of order, ripple, roll-off |
So the insertion loss method gives a predictable, optimum design, which is why it is preferred for microwave filters.
Design steps
- Specification: fc, passband ripple, stopband attenuation, R0. Assume fc = 2 GHz, R0 = 50 Ω, maximally flat response, ≥ 30 dB at 4 GHz.
- Response and order: Butterworth PLR = 1 + (ω/ωc)^(2N).
N ≥ log10(10³ − 1)/(2 log10 2) = 4.98 → N = 5
- Prototype values (gk = 2 sin[(2k−1)π/2N]): g1 = 0.618, g2 = 1.618, g3 = 2.000, g4 = 1.618, g5 = 0.618, g6 = 1. Starting with a series element gives the series-arm (T) form: g1, g3, g5 are series inductors; g2, g4 are shunt capacitors.
- Impedance and frequency scaling: L = g·R0/ωc, C = g/(R0·ωc).
L1 = L5 = 0.618×50/(2π×2×10⁹) = 2.459 nH
C2 = C4 = 1.618/(50×2π×2×10⁹) = 2.575 pF
L3 = 2×50/(2π×2×10⁹) = 7.958 nH
- Microstrip realisation (stepped impedance): series L → short high-impedance line (Zh = 120 Ω), βl = g·R0/Zh; shunt C → short low-impedance line (Zℓ = 20 Ω), βl = g·Zℓ/R0. Substrate assumed εr = 4.2, h = 1.58 mm.
| Element | Z | βl | Length | Width |
|---|---|---|---|---|
| L1 | 120 Ω | 14.75° | 3.65 mm | 0.43 mm |
| C2 | 20 Ω | 37.08° | 8.17 mm | 11.27 mm |
| L3 | 120 Ω | 47.75° | 11.81 mm | 0.43 mm |
| C4 | 20 Ω | 37.08° | 8.17 mm | 11.27 mm |
| L5 | 120 Ω | 14.75° | 3.65 mm | 0.43 mm |
50Ω ____ ____ 50Ω
===--L1--| |--L3--| |--L5--===
thin | C2 | thin | C4 | thin
|____| |____|
wide wide
- Check and tune: simulate (the short-line approximation needs βl < 45°, so L3 is fine-tuned), include step discontinuities, then fabricate and measure.
✓ Answer checked by Claude Opus 5.5 (high effort)
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