---
title: "Received power: 15.5 km, 3.4 GHz, dish receiver"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:max-received-power-15-5km"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/max-received-power-15-5km
site: BCT Engineering
---

# Received power: 15.5 km, 3.4 GHz, dish receiver

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2076 Baisakh, 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/max-received-power-15-5km
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

What is the maximum power that can be received over a distance of 15.5 km line-of-sight free space with a 3.4 GHz frequency consisting of the transmitting antenna gain of 25 dB and receiving antenna diameter 6.4 m with 75% antenna efficiency? Where transmitted power is 250 W.

## Short answer

Using λ = 0.08824 m, receiving dish effective aperture Ae = ηπD^2/4 = 24.13 m², so Gr = 4πAe/λ² = 45.9 dB. Applying the Friis equation, maximum received power is Pr = Pt Gt Gr (λ/(4πd))^2 = 6.32×10⁻⁴ W (0.632 mW, −1.99 dBm).

## Answer

Maximum power is received when the antennas are matched and aligned in free space, so the Friis equation applies directly. The receiving gain is found from the dish size and efficiency.

#### Given

- d = 15.5 km, f = 3.4 GHz, Pt = 250 W
- Gt = 25 dB = 10^2.5 = 316.23
- Receiving dish D = 6.4 m, η = 0.75

#### Step 1: Wavelength

```text
λ = c/f = 3×10⁸ / 3.4×10⁹ = 0.08824 m
```

#### Step 2: Effective aperture and gain of the receiving dish

```text
Ae = η·πD²/4 = 0.75 × π × 6.4² / 4 = 24.13 m²
Gr = 4π·Ae/λ² = 4π × 24.13 / 0.08824²
   = 3.894×10⁴  = 45.90 dB
```

#### Step 3: Received power (Friis, aperture form)

```text
Pr = Pt·Gt·Ae / (4πd²)
   = 250 × 316.23 × 24.13 / (4π × (15500)²)
   = 1.9075×10⁶ / 3.0191×10⁹
   = 6.32×10⁻⁴ W
```

#### Check in dB

```text
LFS = 20 log(4πd/λ) = 126.88 dB
Pt  = 10 log 250    = 23.98 dBW
Pr  = 23.98 + 25 + 45.90 − 126.88 = −31.99 dBW
    = −1.99 dBm ≈ 0.632 mW
```

**Answer: Maximum received power ≈ 6.32×10⁻⁴ W = 0.632 mW (−31.99 dBW, −1.99 dBm).**

## Check yourself

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

1. How is the receiving dish gain found?
   A. Gr = 4πAe/λ^2 with Ae = ηπD^2/4
   B. Gr = λ^2/4πAe
   C. Gr = 25 dB, same as Gt
   D. Gr = ηπD^2/4
   Answer: A. Effective aperture Ae = η·πD^2/4 and Gr = 4πAe/λ^2. (Key point: Gain of dish)

2. At 3.4 GHz the wavelength is about:
   A. 0.0882 m
   B. 0.882 m
   C. 0.0088 m
   D. 0.34 m
   Answer: A. 3×10^8/3.4×10^9 = 0.0882 m. (Key point: Wavelength)

3. Gt = 25 dB corresponds to a ratio of about:
   A. 1000
   B. 25
   C. 250
   D. 316
   Answer: D. 10^2.5 = 316.23. (Key point: Transmit gain)

4. The maximum received power is about:
   A. 6.32×10^-6 W
   B. 1.99 W
   C. 6.32×10^-4 W
   D. 6.32×10^-2 W
   Answer: C. Pr = Pt·Gt·Ae/(4πd^2) = 0.632 mW (−31.99 dBW). (Key point: Result)

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