---
title: "Numerical: free-space received power (165 W, 325 MHz)"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:free-space-received-power-165w"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/free-space-received-power-165w
site: BCT Engineering
---

# Numerical: free-space received power (165 W, 325 MHz)

Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks

- Asked in: 2080 Bhadra, 6 marks
- Video: [Free Space Propagation Model - Wireless Communication](https://www.youtube.com/watch?v=vEbfiBaskVM) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/free-space-received-power-165w
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

A wireless communication transmitter has an output power of 165 watts at a carrier frequency of 325 MHz. It is connected to an antenna with a gain of 12 dBi. The receiving antenna is 15 km away and has a gain of 6 dBi. Calculate the power delivered to the receiver, considering free-space propagation. (Assume that there are no other losses or mismatches in the system.)

## Short answer

Convert 165 W to 52.17 dBm. Free-space path loss at 325 MHz over 15 km is 106.2 dB. With antenna gains 12 and 6 dBi, received power is 52.17+12+6-106.2=-36.03 dBm, about 0.25 microwatt or 2.5e-7 W.

## Answer

For free space propagation, the received power in dB form (Friis equation) is:

```text
Pr(dBm) = Pt(dBm) + Gt(dBi) + Gr(dBi) − LFS(dB)
LFS = 20 log(4πd/λ)
    = 32.44 + 20 log f(MHz) + 20 log d(km)
```

Given: Pt = 165 W, f = 325 MHz, Gt = 12 dBi, Gr = 6 dBi, d = 15 km, no other losses.

#### Step 1: Transmit power in dBm
```text
Pt = 10 log(165 / 0.001) = 10 log(165 000) = 52.17 dBm
```

#### Step 2: Free space path loss
```text
λ   = 3×10⁸ / 325×10⁶ = 0.923 m
LFS = 20 log(4π × 15 000 / 0.923)
    = 106.20 dB
(check: 32.44 + 50.24 + 23.52 = 106.20 dB)
```

#### Step 3: Received power
```text
Pr = 52.17 + 12 + 6 − 106.20
   = −36.03 dBm
Pr = 10^(−36.03/10) mW = 2.50 × 10⁻⁴ mW = 0.25 µW
```

**Answer: Power delivered to the receiver ≈ −36.0 dBm, i.e. about 2.5 × 10⁻⁷ W (0.25 µW).**

## Check yourself

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

1. Which expression gives the received power in dBm for free-space propagation?
   A. Pr = Pt + Gt + Gr + LFS
   B. Pr = Pt + Gt + Gr − LFS
   C. Pr = Pt − Gt − Gr − LFS
   D. Pr = Pt + Gt − Gr + LFS
   Answer: B. Gains add to the transmit power and the free-space path loss is subtracted: Pr = Pt + Gt + Gr − LFS. (Key point: Friis equation in dB)

2. What is the 165 W transmit power in dBm?
   A. 22.17 dBm
   B. 52.17 dBm
   C. 32.17 dBm
   D. 42.17 dBm
   Answer: B. Pt = 10 log(165/0.001) = 10 log(165000) = 52.17 dBm. (Key point: Converting power to dBm)

3. For 325 MHz and d = 15 km, what is the free-space path loss?
   A. 106.2 dB
   B. 96.2 dB
   C. 116.2 dB
   D. 86.2 dB
   Answer: A. LFS = 32.44 + 20 log 325 + 20 log 15 = 32.44 + 50.24 + 23.52 = 106.2 dB. (Key point: Free-space path loss)

4. What is the received power for the link?
   A. −36.0 dBm
   B. −88.0 dBm
   C. +36.0 dBm
   D. −48.0 dBm
   Answer: A. Pr = 52.17 + 12 + 6 − 106.20 = −36.03 dBm, about 0.25 µW. (Key point: Final received power)

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