Rayleigh Fading

Rayleigh Fading: Statistical Model for Non-Line-of-Sight Propagation

Rayleigh fading channels arise when a signal reaches the receiver via multiple scattered paths with no dominant component. This guide covers the statistical model, performance analysis, and practical implications.

Rayleigh Fading Model

The received signal is a sum of many equal-probability scattered components:

r(t) = Σ_k a_k × cos(ω_t + φ_k)

When number of scatterers → ∞ with no LOS, envelope distribution becomes Rayleigh:

PDF: p(r) = (r/σ²) × exp(-r²/(2σ²))

• σ² = variance of underlying Gaussian I/Q components
• Mean amplitude: E[r] = σ × √(π/2) ≈ 1.25σ
• Mean power: E[r²] = 2σ²

Probability Functions

Cumulative Distribution Function: P(R ≤ r) = 1 - exp(-r²/(2σ²))

Outage Probability (signal below threshold): P_outage = 1 - exp(-R_th²/(2σ²))

Fade Depths and Probabilities

Typical Rayleigh channel (σ = 1):

• Signal > 0 dB: 69% of time
• Signal > -3 dB: 50% of time
• Signal > -10 dB: 11% of time
• Signal < -20 dB: 0.1% of time (deep fade)

Fade Duration (at specific SNR threshold): Average fade duration ≈ (1/(2×f_d)) × (1 - exp(-R_th²/(2σ²))) / (Rayleigh_PDF(R_th))

High Doppler: Short fades, many crossings Low Doppler: Long fades, few crossings

Diversity Techniques

Selection Diversity: Choose strongest of N branches Outage probability: P_out = (1 - exp(-R_th²/(2σ²)))^N

Maximal Ratio Combining: Weight each branch by SNR, sum coherently Best performance among linear combiners

N=2 branches: Outage probability improves from 50% to 75% at -3 dB threshold

Performance with BPSK

Bit error rate in Rayleigh fading: P_e = (1/2) × [1 - √(E_b/(2N_0 + E_b))]

Comparison AWGN vs. Rayleigh (for BER = 10^-6):

• AWGN: E_b/N_0 ≈ 10.5 dB required
• Rayleigh (no diversity): > 20 dB required (10 dB penalty!)
• Rayleigh (2-branch diversity): ≈ 14 dB required

Interleaving and Coding

Interleaving: Scrambles bit sequence: deep fades affect scattered symbols

• Block interleaver: Length = (delay spread / symbol period)
• Convolutional interleaver: Less buffering

Coding: Combined with interleaving for burst error correction Typical: Convolutional code + interleaver (e.g., GSM)

Interview Q&A

Q1: What causes Rayleigh fading and why is it "no LOS"? A: Rayleigh fading arises from many scattered components (trees, buildings) with random phases and similar amplitudes arriving at receiver. Central Limit Theorem shows sum of many independent random components has Gaussian envelope. "No LOS" means dominant direct path absent—if LOS present, envelope becomes Rician (LOS component shifts mean away from zero). In urban environments with buildings blocking line-of-sight, Rayleigh is accurate model.

Q2: Explain fade depth statistics: why is -20 dB signal only 0.1% of time? A: Rayleigh PDF: p(r) ∝ r×exp(-r²/2σ²). Tail of distribution decays exponentially: P(R < r) = 1-exp(-r²/2σ²). For r = 0.1σ (-20 dB): P ≈ 1-exp(-0.01/2) ≈ 0.005 ≈ 0.5%. This exponential decay explains why deep fades are rare but catastrophic when they occur—they create error bursts that damage user experience disproportionately.

Q3: Why is 10 dB power penalty required in Rayleigh vs. AWGN for same BER? A: Rayleigh fading causes random gain variations: sometimes +6 dB (constructive), often 0 dB (average), sometimes -20 dB (deep fade). Average power insufficient—deep fades create errors because instantaneous SNR drops below threshold even though average SNR adequate. Outage probability floor appears at 10-12% despite increasing transmit power. Diversity breaks this floor by providing independent fading paths—deep fade unlikely on both paths simultaneously.

Q4: Describe selection diversity and its improvement factor. A: Selection diversity uses N independent fading channels; receiver selects strongest branch for transmission. Outage probability: P_out = P_single^N = [1-exp(-R_th²/(2σ²))]^N. With N=2: if single-path gives 50% outage at threshold, dual-path gives 25% outage. Improvement ≈ 2× (3 dB). Higher N yields exponential improvement: N=4 gives ~6 dB, N=6 gives ~8 dB improvement (diminishing returns).

Q5: Why is interleaving critical in fading channels? A: Fading causes burst errors—many consecutive bit errors during fade. Convolutional codes assume random error distribution; burst errors overwhelm code's correction capability. Interleaver scrambles bit sequence: deep fade affects scattered symbols rather than consecutive block, converting burst to pseudo-random errors. Code then corrects effectively. Without interleaving, burst errors from 10 ms fade uncorrectable.

Q6: Explain the fade crossing rate and its significance for channel estimation. A: Fade crossing rate = average number of times signal crosses threshold per second ≈ 2×f_d×(Rayleigh_PDF at threshold). High Doppler (mobile): many crossings (fades short, brief). Low Doppler (quasi-stationary): few crossings (fades long). Crossing rate determines channel estimation update rate: high crossing rate needs frequent updates (RLS); low rate allows less frequent updates. Too-slow estimation tracks outdated channel, degrading performance.