---
title: "Received power: 100 MHz, 0.75 km"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:received-power-100mhz-0-75km"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/received-power-100mhz-0-75km
site: BCT Engineering
---

# Received power: 100 MHz, 0.75 km

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2072 Asoj, 6 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/received-power-100mhz-0-75km
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

A 100 MHz circuit consists of a transmitting and receiving antenna of 30 dB and 25 dB gains respectively. The power radiated by the transmitting antenna is 120 W. Using Friis transmission equation find the received power at a distance of 0.75 km over a free space.

## Short answer

λ = 3 m, free space loss = 69.94 dB. Transmit power 120 W = 20.79 dBW. Pr = 20.79 + 30 + 25 − 69.94 = 5.85 dBW (3.84 W).

## Answer

Using the Friis transmission equation Pr = Pt·Gt·Gr·(λ/4πd)².

#### Given

- f = 100 MHz, d = 0.75 km = 750 m, Pt = 120 W
- Gt = 30 dB = 1000, Gr = 25 dB = 316.23

#### Step 1: Wavelength

```text
λ = c/f = 3×10⁸ / 10⁸ = 3 m
```

#### Step 2: Free space loss

```text
4πd/λ = 4π × 750 / 3 = 3141.6
LFS   = 20 log(3141.6) = 69.94 dB
```

#### Step 3: Received power (ratio form)

```text
Pr = 120 × 1000 × 316.23 × (1/3141.6)²
   = 3.795×10⁷ / 9.870×10⁶
   = 3.84 W
```

#### Check in dB

```text
Pt = 10 log 120 = 20.79 dBW
Pr = 20.79 + 30 + 25 − 69.94 = 5.85 dBW = 35.85 dBm
10^0.585 = 3.84 W
```

**Answer: Received power ≈ 3.84 W (5.85 dBW).**

## Check yourself

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

1. At 100 MHz λ equals:
   A. 0.3 m
   B. 30 m
   C. 1 m
   D. 3 m
   Answer: D. 3×10^8/10^8 = 3 m. (Key point: Wavelength)

2. Gt = 30 dB and Gr = 25 dB correspond to ratios:
   A. 30 and 25
   B. 1000 and 316.23
   C. 100 and 31.6
   D. 3000 and 625
   Answer: B. 10^3 and 10^2.5. (Key point: Gain conversion)

3. LFS = 20 log(4πd/λ) here is:
   A. 39.94 dB
   B. 69.94 dB
   C. 49.94 dB
   D. 89.94 dB
   Answer: B. 4π×750/3 = 3141.6, whose 20 log is 69.94 dB. (Key point: Free space loss)

4. The received power is about:
   A. 38.4 W
   B. 3.84 mW
   C. 0.384 W
   D. 3.84 W
   Answer: D. 20.79 + 30 + 25 − 69.94 = 5.85 dBW = 3.84 W. (Key point: Result)

## Related in Propagation and Antenna

- [Received power: 0.5 km, 1 GHz circuit](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-0-5km-1ghz) — Propagation and Antenna, chapter 5, asked 2×
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