- 2081 Bhadra · 4+6 marks
Okumura vs Hata; delay spread numerical
Short answer
Okumura uses measured curves, Hata gives closed-form formula with narrower range. For the profile, mean excess delay is 4.38 µs, rms delay spread 1.37 µs, maximum excess delay (10 dB) 5 µs, and 50% coherence bandwidth about 146 kHz.
Compare Okumura model with Hata model. Calculate mean excess delay, rms delay spread and maximum excess delay for the multipath profile Pr(τ) given as: Pr(τ) = (-20 dB, -10 dB, -10 dB, 0 dB) for τ = (0, 1, 2, 5) second. Also estimate 50% coherence bandwidth.
Okumura vs Hata model
| Point | Okumura model | Hata model |
|---|---|---|
| Basis | Measurements in Tokyo, given as curves | Empirical formula fitted to Okumura's curves |
| Form | LF + Amu − G(hte) − G(hre) − GAREA | Closed-form equation in fc, hte, hre, d |
| Frequency | 150–1920 MHz (to 3 GHz) | 150–1500 MHz |
| Distance | 1–100 km | 1–20 km |
| hte | 30–1000 m | 30–200 m |
| Use in software | Hard (needs graph reading) | Easy |
| Environments | Urban, suburban, open with corrections | Urban formula + suburban/rural corrections |
| Accuracy | About 10–14 dB std. deviation | Very close to Okumura for d > 1 km |
Delay spread calculation
The delays are taken as µs, the usual unit for such a profile. (If they really are in seconds, the numbers are the same but every time is in s and Bc is in Hz.)
| τ (µs) | Pr (dB) | Pr (linear) |
|---|---|---|
| 0 | −20 | 0.01 |
| 1 | −10 | 0.1 |
| 2 | −10 | 0.1 |
| 5 | 0 | 1 |
ΣP = 0.01 + 0.1 + 0.1 + 1 = 1.21
Mean excess delay
τ̄ = Σ P(τk)τk / ΣP
= (0.01×0 + 0.1×1 + 0.1×2 + 1×5) / 1.21
= 5.3/1.21 = 4.38 µs
Second moment
τ²̄ = (0.01×0 + 0.1×1 + 0.1×4 + 1×25) / 1.21
= 25.5/1.21 = 21.07 µs²
RMS delay spread
στ = √(τ²̄ − τ̄²) = √(21.07 − 4.38²)
= √(21.07 − 19.19) = 1.37 µs
Maximum excess delay (10 dB) By definition, τmax(X dB) = τX − τ0, where τ0 is the first arriving component and τX is the largest delay at which a component is within X dB of the strongest one. The strongest component is 0 dB at 5 µs, so τX = 5 µs; the first arrival is at τ0 = 0.
τmax(10 dB) = 5 − 0 = 5 µs
50% coherence bandwidth
Bc ≈ 1/(5στ) = 1/(5 × 1.374 µs) = 145.5 kHz ≈ 146 kHz
Answer: τ̄ = 4.38 µs, στ = 1.37 µs, τmax(10 dB) = 5 µs, Bc(50%) ≈ 146 kHz.
✓ Answer checked by Claude Opus 5.5 (high effort)
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