--- title: "Numerical: rms delay spread; flat or selective for AMPS/GSM" subject: "Wireless Communication" chapter: 3 key: "semester_7/wireless_communication:rms-delay-spread-amps-gsm" source: https://www.bctengineer.com/q/semester_7/wireless_communication/rms-delay-spread-amps-gsm site: BCT Engineering --- # Numerical: rms delay spread; flat or selective for AMPS/GSM Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks - Asked in: 2071 Magh, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/rms-delay-spread-amps-gsm - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3 ## Question A wireless channel is characterized by the following power-delay profile: | Power [dB] | 0 | -10 | -20 | -23 | |---|---|---|---|---| | Delays [ns] | 0 | 100 | 200 | 400 | Determine the root mean square (rms) delay spread and the 90% coherence bandwidth of the above channel. Is this channel flat fading or frequency selective fading for: i) An AMPS system with transmission bandwidth 30 kHz? ii) A GSM system with transmission bandwidth 200 kHz? ## Short answer Convert powers to linear, compute mean excess delay, second moment, RMS delay spread. Result: σ_τ = 42.63 ns, B_c(90%) ≈ 1/(50 σ_τ) = 469.2 kHz. Both AMPS (30 kHz) and GSM (200 kHz) are less than B_c, so flat fading. ## Answer **Step 1 – Convert powers to linear values** ```text P (dB): 0 −10 −20 −23 P (lin): 1 0.1 0.01 0.00501 τ (ns): 0 100 200 400 ΣP = 1 + 0.1 + 0.01 + 0.00501 = 1.11501 ``` **Step 2 – Mean excess delay** ```text τ̄ = Σ P(τ_k) τ_k / Σ P(τ_k) = (1×0 + 0.1×100 + 0.01×200 + 0.00501×400) / 1.11501 = (0 + 10 + 2 + 2.005) / 1.11501 = 14.005 / 1.11501 = 12.56 ns ``` **Step 3 – Second moment** ```text τ̄² (mean square) = Σ P τ² / Σ P = (0.1×100² + 0.01×200² + 0.00501×400²) / 1.11501 = (1000 + 400 + 801.9) / 1.11501 = 1974.8 ns² ``` **Step 4 – RMS delay spread** ```text σ_τ = √(1974.8 − 12.56²) = √(1974.8 − 157.8) = √1817.0 = 42.63 ns ``` **Step 5 – Coherence bandwidth (90% correlation)** ```text B_c(90%) ≈ 1 / (50 σ_τ) = 1 / (50 × 42.63 × 10⁻⁹) = 469.2 kHz ``` (For reference, the 50% coherence bandwidth is 1/(5σ_τ) ≈ 4.69 MHz.) **Step 6 – Decide the type of fading** A channel is flat for a system when its signal bandwidth B_s is smaller than B_c. | System | B_s | Compare with B_c = 469.2 kHz | Fading type | |---|---|---|---| | AMPS | 30 kHz | 30 kHz < 469.2 kHz | Flat fading | | GSM | 200 kHz | 200 kHz < 469.2 kHz | Flat fading | **Answer:** σ_τ = 42.63 ns, B_c(90%) ≈ 469.2 kHz; the channel is flat fading for both AMPS (30 kHz) and GSM (200 kHz). GSM's 200 kHz is about 43% of B_c, so it is closer to the limit; GSM receivers still include an equalizer to handle more dispersive channels. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The profile 0, -10, -20, -23 dB converts to linear powers of approximately: A. 1, 0.1, 0.01, 0.005 B. 1, 0.9, 0.8, 0.77 C. 1, 10, 100, 200 D. 0, 0.1, 0.2, 0.23 Answer: A. P = 10^(dB/10), giving 1, 0.1, 0.01, 0.00501. (Key point: Convert dB to linear) 2. How is rms delay spread computed? A. Square root of (mean square delay minus square of mean delay) B. Mean excess delay squared C. Maximum delay divided by 2 D. Sum of powers Answer: A. sigma_tau = sqrt(tau2bar - (taubar)^2), using power-weighted moments. (Key point: RMS delay spread formula) 3. With sigma_tau = 42.63 ns, the 90% coherence bandwidth 1/(50 sigma_tau) is about: A. 4.69 MHz B. 469 kHz C. 46.9 kHz D. 46.9 MHz Answer: B. 1/(50 x 42.63 ns) = 469.2 kHz. (Key point: Coherence bandwidth) 4. Why is the channel flat for both AMPS and GSM? A. Both bandwidths (30 kHz, 200 kHz) are below B_c = 469 kHz B. Both bandwidths exceed B_c C. Doppler is small D. Delay spread is zero Answer: A. Flat fading occurs when the signal bandwidth is less than the coherence bandwidth. 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