---
title: "Butterworth LPF prototype and π-section microstrip"
subject: "RF and Microwave Engineering"
chapter: 6
key: "semester_7/rf_and_microwave_engineering:butterworth-lpf-pi-microstrip"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/butterworth-lpf-pi-microstrip
site: BCT Engineering
---

# Butterworth LPF prototype and π-section microstrip

RF and Microwave Engineering · Chapter 6: RF Design Practices

- Asked in: 2079 Bhadra, 6+2 marks
- Video: [Design of prototype of Low pass filter (LPF) for maximally flat/ Butterworth response (N=5)](https://www.youtube.com/watch?v=QvZJUBPvecQ) — RF Design Basics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/butterworth-lpf-pi-microstrip
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6

## Question

How is a low pass filter prototype based on Butterworth approximation designed using insertion loss method? Implement a low pass filter π section using microstrips.

## Short answer

Butterworth LPF using insertion loss method: PLR=1+(ω/ωc)^(2N), N determined from stopband attenuation. Impedance/frequency scaling yields C1=C3=1.59 pF, L2=7.96 nH for 50 Ω, 2 GHz. Microstrip π-section: high-Z (120 Ω) line for L2 (βl=47.75°, 11.81 mm), low-Z (20 Ω) for C1,C3 (βl=22.92°, 5.05 mm) on εr=4.2, h=1.58 mm.

## Answer

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

```text
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.

```text
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)

```text
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):

| Element | Line Z | βl | Length | Width |
|---|---|---|---|---|
| C1 (g = 1) | 20 Ω | 22.92° | 5.05 mm | 11.27 mm |
| L2 (g = 2) | 120 Ω | 47.75° | 11.81 mm | 0.43 mm |
| C3 (g = 1) | 20 Ω | 22.92° | 5.05 mm | 11.27 mm |

```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the Butterworth PLR 1 + k²(ω/ωc)^2N give, with k = 1?
   A. Equal ripple in the passband
   B. A maximally flat response with 3 dB loss at ωc
   C. Zero loss at ωc
   D. A response with transmission zeros
   Answer: B. At ω = ωc, PLR = 2 (3 dB). The response is flat at ω = 0 and falls at 20N dB/decade. (Key point: Butterworth PLR)

2. For at least 15 dB at 2fc, what order N is needed?
   A. N = 2
   B. N = 3
   C. N = 5
   D. N = 1
   Answer: B. N ≥ log10(10^1.5 − 1)/(2 log10 2) = 2.47, so N = 3 (IL = 18.13 dB). (Key point: Order)

3. For N = 3, R0 = 50 Ω, fc = 2 GHz, the π-section elements are:
   A. C1 = C3 = 1.59 pF, L2 = 7.96 nH
   B. L1 = L3 = 1.59 nH, C2 = 7.96 pF
   C. C1 = C3 = 7.96 pF, L2 = 1.59 nH
   D. C1 = C3 = 3.18 pF, L2 = 3.98 nH
   Answer: A. g = 1, 2, 1: C = g/(R0 ωc) = 1.59 pF, L = 2R0/ωc = 7.96 nH. (Key point: Element values)

4. Why is L2 (βl = 47.75°) fine-tuned by simulation?
   A. The short-line inductor approximation needs βl < 45°
   B. It is too wide
   C. It is a 50 Ω line
   D. It has zero length
   Answer: A. βl above about 45° breaks the short-line approximation, so lengths are tuned. (Key point: Stepped-impedance lengths)

## Related in RF and Microwave Engineering

- [Butterworth/Chebyshev LPF and double π-section](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Abutterworth-chebyshev-double-pi) — RF and Microwave Engineering, chapter 6
- [Insertion loss method and microstrip LPF](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Ainsertion-loss-lpf-microstrip) — RF and Microwave Engineering, chapter 6
- [Insertion loss design of given LC ladder filter](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Afilter-ladder-insertion-loss-microstrip) — RF and Microwave Engineering, chapter 6
- [Butterworth LPF prototyping by insertion loss](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Abutterworth-lpf-prototyping) — RF and Microwave Engineering, chapter 6
- [Why insertion loss; double-section shunt-arm LPF](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Ainsertion-loss-shunt-arm-lpf) — RF and Microwave Engineering, chapter 6
- [LPF prototype conversion and second order HPF](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Alpf-prototype-hpf-conversion) — RF and Microwave Engineering, chapter 6

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