---
title: "Numerical: max symbol rate without equalizer"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:symbol-period-power-delay-profile"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/symbol-period-power-delay-profile
site: BCT Engineering
---

# Numerical: max symbol rate without equalizer

Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks

- Asked in: 2075 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/symbol-period-power-delay-profile
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

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.

## 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.

## Answer

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 |

```text
ΣP = 1 + 1 + 0.1 + 0.01 = 2.11
```

#### Step 2: Mean excess delay
```text
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
```text
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
```text
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
```text
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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. To avoid an equalizer, what must hold?
   A. Ts ≤ 0.1 στ
   B. Ts = Tc
   C. στ/Ts ≤ 0.1, i.e. Ts ≥ 10 στ
   D. Ts equals mean excess delay
   Answer: C. Delay spread must be small compared with the symbol period: Ts ≥ 10 στ. (Key point: No-equalizer condition)

2. With linear powers 1, 1, 0.1, 0.01 at 0, 50, 75, 100 µs, the mean excess delay is:
   A. 27.73 µs
   B. 58.5 µs
   C. 50 µs
   D. 37.5 µs
   Answer: A. Sum of P·τ = 58.5, ΣP = 2.11, giving 27.73 µs. (Key point: Mean excess delay)

3. The RMS delay spread is computed as:
   A. The second moment minus the mean
   B. The square root of (second moment − mean squared)
   C. The square of the mean excess delay
   D. The sum of all delays
   Answer: B. στ = sqrt(1498.82 − 27.73^2) = 27.02 µs. (Key point: RMS delay spread)

4. A tap at −10 dB has linear power:
   A. 10
   B. 0.01
   C. 0.5
   D. 0.1
   Answer: D. 10^(−10/10) = 0.1; −20 dB gives 0.01. (Key point: Converting dB powers)

5. The greatest symbol rate without an equalizer is about:
   A. 3.70 ksymbols/s
   B. 37 ksymbols/s
   C. 370 symbols/s
   D. 27 ksymbols/s
   Answer: A. Ts = 10 x 27.02 = 270.2 µs, so Rs = 1/270.2 µs = 3.70 ksymbols/s. (Key point: Symbol rate)

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