--- title: "Numerical: 50 W in dBm/dBW and received power" subject: "Wireless Communication" chapter: 3 key: "semester_7/wireless_communication:transmit-power-dbm-free-space" source: https://www.bctengineer.com/q/semester_7/wireless_communication/transmit-power-dbm-free-space site: BCT Engineering --- # Numerical: 50 W in dBm/dBW and received power Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks - Asked in: 2081 Baisakh, 1+1+2+2 marks - Video: [Transmitter produces 50W of power If 50W ...| Calculating Received Power in dBm with 50W Transmitter](https://www.youtube.com/watch?v=O4Goxh9Vjwg) — ALTEQUAL - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/transmit-power-dbm-free-space - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3 ## Question If a transmitter produces 50 watts of power, express the transmit power in units of (i) dBm, and (ii) dBW. If 50 watts is applied to a unity gain antenna with a 900 MHz carrier frequency, find the received power in dBm at a free space distance of 100 m from the antenna. What is Pr(10 km)? Assume unity gain for the receiver antenna. ## Short answer 50 W equals 46.99 dBm and 16.99 dBW. With unity gains at 900 MHz, lambda=0.333 m, Friis gives 3.52 microwatts at 100 m, or -24.54 dBm. At 10 km free-space loss adds 40 dB, so received power is -64.54 dBm. ## Answer #### (i) and (ii) Transmit power ```text Pt(dBm) = 10 log(50 W / 1 mW) = 10 log(50 000) = 46.99 dBm Pt(dBW) = 10 log(50 W / 1 W) = 10 log(50) = 16.99 dBW ``` #### Received power at 100 m Friis equation with Gt = Gr = 1, f = 900 MHz: ```text λ = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m Pr = Pt·Gt·Gr·λ² / [(4π)²d²] = 50 × 1 × 1 × (0.3333)² / [(4π)² × 100²] = 3.52 × 10⁻⁶ W = 3.52 µW Pr(dBm) = 10 log(3.52×10⁻⁶ / 10⁻³) = −24.54 dBm ``` #### Received power at 10 km In free space power falls as 1/d², i.e. 20 dB per decade. Using d0 = 100 m as reference: ```text Pr(10 km) = Pr(100 m) + 20 log(100/10 000) = −24.54 − 40 = −64.54 dBm ``` **Answer: Pt = 46.99 dBm = 16.99 dBW; Pr(100 m) = −24.5 dBm; Pr(10 km) = −64.5 dBm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. 50 W expressed in dBm is: A. 46.99 B. 16.99 C. 50 D. 30 Answer: A. 10 log(50000) = 46.99 dBm. (Key point: dBm) 2. 50 W expressed in dBW is: A. 50 B. 46.99 C. 30 D. 16.99 Answer: D. 10 log 50 = 16.99 dBW. (Key point: dBW) 3. Received power at 100 m is about: A. -4.5 dBm B. -24.5 dBm C. -44.5 dBm D. -64.5 dBm Answer: B. 3.52 uW = -24.54 dBm. (Key point: Pr at 100 m) 4. Going from 100 m to 10 km in free space changes power by: A. -40 dB B. -80 dB C. -20 dB D. -60 dB Answer: A. Distance x100 gives 20 log(100) = 40 dB loss. (Key point: Distance scaling) ## Related in Wireless Communication - [Numerical: free-space received power (165 W, 325 MHz)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afree-space-received-power-165w) — Wireless Communication, chapter 3 - [Numerical: received power, E-field, flux density, voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afree-space-received-power-e-field) — Wireless Communication, chapter 3, asked 2× - [Numerical: Okumura median path loss (unity gain)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-median-loss-unity-gain) — Wireless Communication, chapter 3, asked 3× - [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× - [Numerical: λ/4 monopole and two-ray received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amonopole-two-ray-received-power) — Wireless Communication, chapter 3 - [Numerical: received power with cable and mismatch loss](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Areceived-power-coax-mismatch) — Wireless Communication, chapter 3 ## Related in other subjects - [Received power: 50 km, 10 GHz link](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-50km-10ghz) — Propagation and Antenna, chapter 5, asked 2× - [Received power: 100 MHz, 0.75 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-100mhz-0-75km) — Propagation and Antenna, chapter 5 - [FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-10khz-100mhz-75khz) — Communication Systems, chapter 3 - [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×