--- title: "Numerical: maximum cell radius using Hata model" subject: "Wireless Communication" chapter: 3 key: "semester_7/wireless_communication:hata-max-cell-radius" source: https://www.bctengineer.com/q/semester_7/wireless_communication/hata-max-cell-radius site: BCT Engineering --- # Numerical: maximum cell radius using Hata model Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks - Asked in: 2073 Magh, 6 marks - Video: [OKUMURA AND HATA MODELS (SOLVED PROBLEMS)](https://www.youtube.com/watch?v=serB4I59bC0) — Prof. Barapate's Tutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/hata-max-cell-radius - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3 ## Question Let us consider a medium sized city and assume the typical GSM downlink parameters. The Base Station (BS) is transmitting with power 50 W. The minimum acceptable received power at Mobile Station (MS) is -91 dBm. The carrier frequency is 900 MHz, the height of BS is 30 m and height of MS is 1 m. Estimate the maximum cell radius and corresponding cell area using Hata Model. Hata Model (given): L50(urban) (dB) = 69.55 + 26.16 log fc − 13.82 log hte − a(hre) + (44.9 − 6.55 log hte) log d For medium sized city: a(hre) = (1.1 log fc − 0.7) hre − (1.56 log fc − 0.8) dB For large city: a(hre) = 8.29 (log 1.54 hre)² − 1.1 dB for fc ≤ 300 MHz; a(hre) = 3.2 (log 11.75 hre)² − 4.97 dB for fc ≥ 300 MHz L50(suburban) (dB) = L50(urban) − 2[log(fc/28)]² − 5.4 L50(rural) (dB) = L50(urban) − 4.78 (log fc)² + 18.33 log fc − 40.94 ## Short answer Maximum allowed path loss is 137.99 dB. Mobile correction a(hre) = -1.26 dB, so Hata gives L = 127.68 + 35.22 log d; solving yields d = 1.96 km. Hexagonal cell area is about 10.0 km². ## Answer The maximum cell radius is the distance at which Hata path loss equals the largest loss the link can tolerate. Antenna gains and cable losses are not given, so they are taken as 0 dB. #### Step 1: Maximum allowed path loss ```text Pt = 50 W = 10 log(50 000 mW) = 46.99 dBm Pr,min = −91 dBm Lmax = Pt − Pr,min = 46.99 − (−91) = 137.99 dB ``` #### Step 2: Mobile antenna correction (medium city) fc = 900 MHz, log 900 = 2.9542, hre = 1 m: ```text a(hre) = (1.1 log fc − 0.7)hre − (1.56 log fc − 0.8) = (3.2497 − 0.7)×1 − (4.6086 − 0.8) = 2.5497 − 3.8086 = −1.26 dB ``` #### Step 3: Hata equation (urban) hte = 30 m, log 30 = 1.4771: ```text L50 = 69.55 + 26.16 log fc − 13.82 log hte − a(hre) + (44.9 − 6.55 log hte) log d = 69.55 + 77.28 − 20.41 + 1.26 + (44.9 − 9.68) log d = 127.68 + 35.22 log d ``` #### Step 4: Solve for d ```text 127.68 + 35.22 log d = 137.99 log d = 10.31 / 35.22 = 0.2927 d = 10^0.2927 = 1.96 km ``` #### Step 5: Cell area For a hexagonal cell of radius R: ```text A = (3√3/2)·R² = 2.598 × 1.962² = 10.0 km² ``` (If a circular cell is assumed, A = πR² = 12.1 km².) **Answer: Maximum cell radius ≈ 1.96 km; hexagonal cell area ≈ 10.0 km² (≈ 12.1 km² if circular).** All parameters are within Hata's range (hte 30–200 m, hre 1–10 m, d 1–20 km). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 50 W (46.99 dBm) and sensitivity −91 dBm and 0 dB gains/cables, Lmax is: A. 44.0 dB B. 137.99 dB C. 91.0 dB D. 227 dB Answer: B. Lmax = 46.99 − (−91) = 137.99 dB. (Key point: Maximum allowed path loss) 2. For medium city, 900 MHz, hre = 1 m, a(hre) is: A. +1.26 dB B. +2.55 dB C. −1.26 dB D. −3.81 dB Answer: C. 2.5497 − 3.8086 = −1.26 dB, so −a(hre) adds 1.26 dB to loss. (Key point: Mobile antenna correction) 3. Hata reduces to 127.68 + 35.22 log d. Setting it to 137.99 gives: A. d = 10 km B. d = 1.96 km C. d = 19.6 km D. d = 0.2 km Answer: B. log d = 10.31/35.22 = 0.2927, d = 1.96 km. (Key point: Solving for distance) 4. Hexagonal cell area for R = 1.96 km is: A. 12.1 km^2 B. 5.0 km^2 C. 3.8 km^2 D. 10.0 km^2 Answer: D. A = 2.598 R^2 = 10.0 km^2 (12.1 km^2 if circular). (Key point: Cell area) ## Related in Wireless Communication - [Numerical: Hata path loss for medium city](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Ahata-medium-city-pathloss) — Wireless Communication, chapter 3 - [Numerical: coverage range of a 150 W base station](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acoverage-range-path-loss-exponent) — Wireless Communication, chapter 3 - [Okumura vs Hata; delay spread numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-hata-delay-spread) — Wireless Communication, chapter 3 - [Numerical: 12-cell system capacity at 2% GOS](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-capacity-12-cell-system) — Wireless Communication, chapter 2 - [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: 50 W in dBm/dBW and received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atransmit-power-dbm-free-space) — Wireless Communication, chapter 3 ## Related in other subjects - [Layout of a medium head Francis station](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Afrancis-medium-head-layout) — Power Plant Design, chapter 3, asked 2× - [Rural village demand, survey 2070 BS, base year 2073](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-as70) — Water Supply Engineering, chapter 3 - [Time of critical DO deficit, 50 l/s city](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-time-city-50ls) — Sanitary Engineering, chapter 6, asked 2× - [Modelling 500 km as medium, 50 km as long](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Awrong-model-500km-50km) — Power System Analysis I, chapter 5