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

# Parabolic reflector diameter: 75%, 3 GHz

Propagation and Antenna · Chapter 3: Antenna Classification

- Asked in: 2077 Chaitra, 7 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-75-3ghz
- 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 3 GHz resonant frequency with 2.5 dB waveguide loss. Find out the antenna diameter if effective isotropic radiated power (EIRP) is calculated 46 dBW and transmitting power is 500 W.

## Short answer

EIRP = P_t_dB - L + G. P_t = 500 W = 26.99 dBW, L = 2.5 dB, EIRP = 46 dBW → G_dB = 46 - 26.99 + 2.5 = 21.51 dB, G = 141.6. λ = 0.1 m, η = 0.75. Diameter D = (λ/π)√(G/η) = (0.1/π)√(141.6/0.75) ≈ 0.437 m (43.7 cm).

## Answer

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

**Given**: η = 0.75, f = 3 GHz, waveguide loss L = 2.5 dB, EIRP = 46 dBW, Pₜ = 500 W.

#### Step 1: Wavelength

```text
λ = c/f = (3 × 10⁸)/(3 × 10⁹) = 0.1 m
```

#### Step 2: Transmit power in dBW

```text
Pₜ = 10 log₁₀(500) = 26.99 dBW
```

#### Step 3: Antenna gain

```text
EIRP = Pₜ − L + G
G    = 46 − 26.99 + 2.5 = 21.51 dB
G    = 10^(2.151) = 141.59
```

#### Step 4: Diameter

```text
D = (λ/π) √(G/η)
  = (0.1/π) √(141.59/0.75)
  = 0.03183 × √188.79
  = 0.03183 × 13.740
  = 0.4374 m
```

Check: 0.75 × (π × 0.4374/0.1)² = 0.75 × 188.8 ≈ 141.6 → 21.51 dB. ✓

With lower efficiency (75 % instead of 85 %), a slightly larger dish is needed for the same gain.

**Answer: G = 21.51 dB (141.6); diameter D ≈ 0.437 m (43.7 cm).**

## Check yourself

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

1. Which relation links EIRP, transmit power, loss and gain (in dB)?
   A. EIRP = Pt - L + G
   B. EIRP = Pt + L + G
   C. EIRP = Pt - L - G
   D. EIRP = G - Pt
   Answer: A. Waveguide loss reduces the power reaching the antenna; gain is added. (Key point: EIRP relation)

2. What is the antenna gain here (Pt = 26.99 dBW, L = 2.5 dB, EIRP = 46 dBW)?
   A. 46.0 dB
   B. 21.51 dB
   C. 16.51 dB
   D. 26.99 dB
   Answer: B. G = 46 - 26.99 + 2.5 = 21.51 dB, or 141.6. (Key point: Gain calculation)

3. Which formula gives the diameter from gain?
   A. D = lambda sqrt(G eta)
   B. D = (pi/lambda) sqrt(G eta)
   C. D = (lambda/pi) sqrt(G/eta)
   D. D = G lambda / pi
   Answer: C. From G = eta(piD/lambda)^2. (Key point: Diameter formula)

4. What diameter results at 3 GHz with eta = 0.75?
   A. 1.0 m
   B. 4.37 m
   C. 0.0437 m
   D. 0.437 m
   Answer: D. lambda = 0.1 m; D = 0.03183 x 13.74 = 0.437 m. (Key point: Final result)

## 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%, 5 GHz](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aparabolic-diameter-75-5ghz) — 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×