- 2075 Bhadra · 8 marks
Numerical: max symbol rate without equalizer
Short answer
From the power delay profile, mean excess delay is 27.73 µs and rms delay spread is 27.02 µs. For no equalizer, Ts must be at least 10στ = 270.2 µs, so the greatest symbol rate is about 3.70 ksymbols/s.
Determine the smallest symbol period Ts, and thus the greatest symbol rate that must be sent through RF channel with given power delay profile without using an equalizer.
| Power [dB] | 0 | 0 | -10 | -20 |
|---|---|---|---|---|
| Delay [µs] | 0 | 50 | 75 | 100 |
Modulation provides suitable BER performance whenever στ/Ts ≤ 0.1.
To avoid an equalizer, the rms delay spread must be small compared with the symbol period: στ/Ts ≤ 0.1, i.e. Ts ≥ 10στ.
Step 1: Convert powers to linear
| τ (µs) | P (dB) | P (linear) |
|---|---|---|
| 0 | 0 | 1 |
| 50 | 0 | 1 |
| 75 | −10 | 0.1 |
| 100 | −20 | 0.01 |
ΣP = 1 + 1 + 0.1 + 0.01 = 2.11
Step 2: Mean excess delay
τ̄ = Σ P·τ / ΣP
= (1×0 + 1×50 + 0.1×75 + 0.01×100) / 2.11
= 58.5 / 2.11 = 27.73 µs
Step 3: Second moment
τ²̄ = Σ P·τ² / ΣP
= (0 + 2500 + 0.1×5625 + 0.01×10 000) / 2.11
= 3162.5 / 2.11 = 1498.82 µs²
Step 4: RMS delay spread
στ = √(τ²̄ − τ̄²)
= √(1498.82 − 27.73²)
= √(1498.82 − 768.71)
= √730.11 = 27.02 µs
Step 5: Smallest symbol period and greatest symbol rate
Ts ≥ στ/0.1 = 10 × 27.02 = 270.2 µs
Rs = 1/Ts = 1/270.2 µs = 3.70 × 10³ symbols/s
Answer: Ts(min) ≈ 270.2 µs, so the greatest symbol rate without an equalizer is about 3.70 ksymbols/s. (Coherence bandwidth for reference: Bc ≈ 1/(5στ) ≈ 7.4 kHz.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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