- 2073 Bhadra · 2+6 marks
Justify insertion loss method for filter design
Short answer
Insertion loss method synthesises from power loss ratio PLR=1/(1−|Γ|²), giving exact specified response and real terminations, unlike image method. Choose Butterworth/Chebyshev PLR, find order N, use g tables, scale/transform, implement with transmission lines.
Justify and describe how a microwave filter is designed using insertion loss method.
In the insertion loss method, the filter is synthesised directly from a chosen power loss ratio PLR(ω), so the complete frequency response is controlled from the start.
Justification
| Image parameter method | Insertion loss method |
|---|---|
| Cascades constant-k and m-derived sections | Synthesises the whole network |
| Response is not exactly specified | Response specified exactly |
| Needs iteration to meet specs | Systematic, table based |
| Termination mismatch ignored | Real terminations included |
| No direct trade-off | Trades order vs roll-off vs ripple |
Hence it gives optimum, repeatable designs, and it is the standard method for microwave filters.
Description of the method
- Specify fc (or band edges), R0, passband ripple and stopband attenuation.
- Choose the response:
PLR = 1/(1 − |Γ(ω)|²), IL = 10 log PLR
Butterworth: PLR = 1 + k²(ω/ωc)^(2N)
Chebyshev: PLR = 1 + k²·TN²(ω/ωc)
Linear phase: flat group delay
- Find the order N from the stopband specification. Example: Butterworth, 15 dB at 2fc: N ≥ log(10^1.5 − 1)/(2 log 2) = 2.47 → N = 3.
- Low-pass prototype values gk from tables (R0 = 1, ωc = 1): for N = 3 Butterworth, 1, 2, 1.
- Scaling and transformation: L = gR0/ωc, C = g/(R0ωc); LP → HP, BP, BS by frequency substitution. For fc = 2 GHz, R0 = 50 Ω: C = 1.59 pF, L = 7.96 nH.
- Implementation: lumped elements become transmission-line sections using Richards' transformation, Kuroda identities, stepped-impedance lines or coupled lines.
- Simulate and tune, including junction and step discontinuities, then fabricate.
✓ Answer checked by Claude Opus 5.5 (high effort)
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