---
title: "Parabolic reflector diameter: 75%, 5 GHz"
subject: "Propagation and Antenna"
chapter: 3
key: "semester_6/propagation_and_antenna:parabolic-diameter-75-5ghz"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/parabolic-diameter-75-5ghz
site: BCT Engineering
---

# Parabolic reflector diameter: 75%, 5 GHz

Propagation and Antenna · Chapter 3: Antenna Classification

- Asked in: 2074 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/parabolic-diameter-75-5ghz
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-3

## Question

A parabolic reflector antenna having antenna efficiency 75% is designed for 5 GHz resonance frequency with 3 dB waveguide loss. If Effective Isotropic Radiated Power (EIRP) is calculated 50 dBW and transmitting power is 600 W, calculate its diameter.

## Short answer

Wavelength is 0.06 m at 5 GHz. Antenna gain from EIRP is 25.22 dB (332.5). Diameter equals (λ/π)√(G/η) = 0.402 m (40.2 cm).

## Answer

EIRP = transmitter power − waveguide loss + antenna gain (all in dB). Find the antenna gain first, then the diameter from **G = η(πD/λ)²**.

**Given**: η = 0.75, f = 5 GHz, waveguide loss L = 3 dB, EIRP = 50 dBW, Pₜ = 600 W.

#### Step 1: Wavelength

```text
λ = c/f = (3 × 10⁸)/(5 × 10⁹) = 0.06 m
```

#### Step 2: Transmit power in dBW

```text
Pₜ = 10 log₁₀(600) = 27.78 dBW
```

#### Step 3: Antenna gain

```text
EIRP = Pₜ − L + G
G    = 50 − 27.78 + 3 = 25.22 dB
G    = 10^(2.522) = 332.54
```

#### Step 4: Diameter

```text
D = (λ/π) √(G/η)
  = (0.06/π) √(332.54/0.75)
  = 0.019099 × √443.39
  = 0.019099 × 21.057
  = 0.4022 m
```

Check: 0.75 × (π × 0.4022/0.06)² = 0.75 × 443.4 ≈ 332.5 → 25.22 dB. ✓

HPBW ≈ 70λ/D = 70 × 0.06/0.4022 ≈ 10.4°.

**Answer: G = 25.22 dB (332.5); diameter D ≈ 0.402 m (40.2 cm).**

## Check yourself

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

1. Using EIRP = Pt - L + G, what is G (Pt = 27.78 dBW, L = 3 dB, EIRP = 50 dBW)?
   A. 19.22 dB
   B. 50 dB
   C. 25.22 dB
   D. 30.78 dB
   Answer: C. G = 50 - 27.78 + 3 = 25.22 dB. (Key point: Gain from EIRP)

2. What is 600 W in dBW?
   A. 36.9 dBW
   B. 60 dBW
   C. 18 dBW
   D. 27.78 dBW
   Answer: D. 10 log10(600) = 27.78. (Key point: Power in dBW)

3. What is lambda at 5 GHz?
   A. 0.06 m
   B. 0.6 m
   C. 0.006 m
   D. 0.05 m
   Answer: A. 3x10^8/5x10^9 = 0.06 m. (Key point: Wavelength)

4. What diameter results?
   A. About 4.02 m
   B. About 0.402 m
   C. About 0.0402 m
   D. About 1.2 m
   Answer: B. D = 0.019099 x 21.057 = 0.402 m. (Key point: Diameter)

## Close variants set in other exams

- [Parabolic reflector diameter: 85%, 3 GHz](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aparabolic-diameter-85-3ghz) — Propagation and Antenna, chapter 3, asked 4×
- [Parabolic reflector diameter: 75%, 3 GHz](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aparabolic-diameter-75-3ghz) — Propagation and Antenna, chapter 3, asked 4×

## Related in Propagation and Antenna

- [Short note: Effective isotropic radiated power](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aeirp-short-note) — Propagation and Antenna, chapter 2, asked 4×

## Related in other subjects

- [Numerical: Okumura path loss, EIRP and received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-eirp-received-power) — Wireless Communication, chapter 3, asked 3×