---
title: "Maximum error-free data rate for 1 MHz, 10 dB SNR"
subject: "Communication Systems"
chapter: 6
key: "semester_6/communication_systems:max-data-rate-1mhz-10db"
source: https://www.bctengineer.com/q/semester_6/communication_systems/max-data-rate-1mhz-10db
site: BCT Engineering
---

# Maximum error-free data rate for 1 MHz, 10 dB SNR

Communication Systems · Chapter 6: Baseband Digital Data Transmission

- Asked in: 2068 Jestha (old course), 10 marks
- Video: [Nyquist Bit Rate & Shannon Capacity | Signal-to-Noise Ratio | Computer Networks Course | Lecture 18](https://www.youtube.com/watch?v=_LxyNd-pDdk) — Mind Your Exam Channel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/max-data-rate-1mhz-10db
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6

## Question

Evaluate the maximum data rate that can be transmitted error free through a channel with a bandwidth of 1 MHz and a minimum of 10 dB SNR at the input of the channel decoder.

## Short answer

Maximum error-free rate is channel capacity. Convert 10 dB SNR to ratio 10, then C = 1 MHz × log2(11) ≈ 3.46×10^6 b/s = 3.46 Mbps.

## Answer

The maximum error-free data rate is the **channel capacity**, given by the Shannon–Hartley theorem for an AWGN channel:

$$
C = B \log_2\!\left(1 + \frac{S}{N}\right)
$$

#### Given
- Bandwidth $B = 1$ MHz $= 10^6$ Hz
- $(S/N)_{dB} = 10$ dB

#### Step 1: Convert SNR to a ratio
$$
\frac{S}{N} = 10^{(S/N)_{dB}/10} = 10^{10/10} = 10
$$

#### Step 2: Capacity
$$
\begin{aligned}
C &= 10^6 \log_2 (1 + 10) = 10^6 \log_2 11 \\
&= 10^6 \times \frac{\ln 11}{\ln 2} = 10^6 \times \frac{2.3979}{0.6931} \\
&= 10^6 \times 3.4594 \\
&= 3.459 \times 10^6\ \text{b/s}
\end{aligned}
$$

#### Interpretation
- The spectral efficiency is $C/B = 3.46$ b/s/Hz.
- Any rate up to about 3.46 Mb/s can, in theory, be sent with an arbitrarily small error probability using suitable (possibly very long) channel codes. Rates above this cannot be sent error-free.
- At the Nyquist rate of $2B = 2$ Msymbols/s, this rate needs $3.46/2 \approx 1.73$ bits/symbol. That is more than binary signalling can carry, so a practical scheme needs multilevel signalling, e.g. M-ary with $M \ge 4$, together with strong coding.
- The "10 dB at the input of the channel decoder" means that the SNR is measured where decoding happens, so it is the SNR that sets the capacity.

**Answer: Maximum error-free data rate $C \approx 3.46$ Mb/s.**

## Check yourself

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

1. Which formula gives the maximum error-free data rate for a noisy 1 MHz channel?
   A. C = B / log2(1+S/N)
   B. C = 2B log2 M
   C. C = B log2(1 + S/N)
   D. C = S/N x B
   Answer: C. The Shannon-Hartley capacity applies because the channel has noise (SNR given). (Key point: Formula)

2. What is the SNR ratio for 10 dB?
   A. 1.0
   B. 10
   C. 100
   D. 1
   Answer: B. S/N = 10^(10/10) = 10, a power ratio, not 100. (Key point: dB conversion)

3. What is C for B = 1 MHz and S/N = 10?
   A. 2 Mb/s
   B. 3.46 Mb/s
   C. 1 Mb/s
   D. 10 Mb/s
   Answer: B. C = 10^6 log2(11) = 3.459 x 10^6 b/s, i.e. 3.46 b/s/Hz. (Key point: Capacity result)

4. What does the result mean physically?
   A. Any rate above C can be sent with strong coding
   B. Rates up to C can be sent with arbitrarily small error; above C cannot
   C. The data rate must equal C exactly
   D. C is the symbol rate
   Answer: B. Rates up to about 3.46 Mb/s are achievable in theory with suitable codes; higher rates cannot be error-free. (Key point: Interpretation)

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## Related in other subjects

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