---
title: "Parabolic reflector diameter: 85%, 3 GHz"
subject: "Propagation and Antenna"
chapter: 3
key: "semester_6/propagation_and_antenna:parabolic-diameter-85-3ghz"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/parabolic-diameter-85-3ghz
site: BCT Engineering
---

# Parabolic reflector diameter: 85%, 3 GHz

Propagation and Antenna · Chapter 3: Antenna Classification

- Asked in: 2073 Bhadra, 8 marks; 2071 Bhadra, 8 marks (asked 2 times)
- 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-85-3ghz
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-3

## Question

A parabolic reflector antenna having the antenna efficiency 85% 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

For f=3 GHz, λ=0.1 m. P_t=500 W gives 26.99 dBW; with 2.5 dB loss and EIRP 46 dBW, gain is 21.51 dB (ratio 141.6). Using G=η(πD/λ)² with η=0.85 gives D=0.411 m.

## Answer

The EIRP of a transmitting system is the transmitter power minus feeder (waveguide) loss plus antenna gain (all in dB). From EIRP we find the gain, then the diameter from the parabolic gain formula **G = η(πD/λ)²**.

**Given**: η = 0.85, 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(dBW) = 10 log₁₀(500) = 26.99 dBW
```

#### Step 3: Antenna gain from EIRP

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

#### Step 4: Diameter

```text
G = η (πD/λ)²
D = (λ/π) √(G/η)
  = (0.1/π) √(141.59/0.85)
  = 0.03183 × √166.58
  = 0.03183 × 12.906
  = 0.4108 m
```

Check: 0.85 × (π × 0.4108/0.1)² = 0.85 × 166.58 ≈ 141.6 → 21.51 dB. ✓

Beamwidth for interest: HPBW ≈ 70λ/D = 70 × 0.1/0.4108 ≈ 17°.

**Answer: Antenna gain = 21.51 dB (141.6); required dish diameter D ≈ 0.411 m (41.1 cm).**

## Check yourself

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

1. EIRP in dB equals
   A. Pt + loss + G
   B. Pt - G
   C. Pt(dBW) - loss + G
   D. G - loss only
   Answer: C. Waveguide loss subtracts from power. (Key point: EIRP relation)

2. 500 W in dBW is about
   A. 50 dBW
   B. 2.7 dBW
   C. 500 dBW
   D. 27 dBW
   Answer: D. 10 log10(500) = 26.99 dBW. (Key point: Transmit power)

3. The required gain is
   A. 21.51 dB
   B. 72.5 dB
   C. 16.5 dB
   D. 46 dB
   Answer: A. G = 46 - 26.99 + 2.5 = 21.51 dB, which is 141.6 as a ratio. (Key point: Gain)

4. The dish diameter is about
   A. 4.11 m
   B. 0.411 m
   C. 1.2 m
   D. 0.1 m
   Answer: B. D = (lambda/pi) sqrt(G/eta) = 0.03183 x 12.906 = 0.411 m. (Key point: Diameter)

## Close variants set in other exams

- [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×
- [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×