---
title: "Output SNR of an amplified line"
subject: "Telecommunications"
chapter: 2
key: "semester_8/telecommunications:line-snr-amplifier-numerical"
source: https://www.bctengineer.com/q/semester_8/telecommunications/line-snr-amplifier-numerical
site: BCT Engineering
---

# Output SNR of an amplified line

Telecommunications · Chapter 2: Transmission Media

- Asked in: 2080 Baisakh, 3 marks
- Video: [Signal to Noise Ratio - SNR Explained: Basics and Solved Examples](https://www.youtube.com/watch?v=Y4j3172zxmY) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/line-snr-amplifier-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-2

## Question

A 10 mW audio signal is to be transmitted over 100 km of line having an attenuation of 2 dB/km to produce a signal at the receiving end also of 10 mW. The noise level on the line is −138 dBm. The signal is amplified at various points along the line. The available amplifier has a 50 dB gain and a noise factor of 7 dB. Determine the output SNR in dB.

## Short answer

Total loss 200 dB, four 50-dB amplifiers divide line into 25-km sections. Input signal -40 dBm, effective noise -131 dBm, per-section SNR 91 dB. Four identical sections reduce SNR by 10 log10 4 = 6 dB, giving output SNR ≈ 85 dB.

## Answer

**Given:** signal power 10 mW = 10 dBm; line 100 km at 2 dB/km; amplifier gain 50 dB, noise factor $F$ = 7 dB; line noise $-138$ dBm; received signal must also be 10 mW.

**Step 1 – Number of amplifiers**
$$
\begin{aligned}
\text{Total line loss} &= 100 \times 2 = 200\ \text{dB} \\
\text{Amplifiers needed} &= \frac{200}{50} = 4
\end{aligned}
$$
So the line is divided into 4 sections of 25 km (50 dB loss each), each followed by a 50 dB amplifier. Output = 10 dBm, as required.

**Step 2 – Signal at each amplifier input**
$$
S_{in} = 10\ \text{dBm} - 50\ \text{dB} = -40\ \text{dBm}
$$

**Step 3 – Effective noise at each amplifier input** (line noise raised by the noise factor)
$$
N_{in,eff} = -138 + 7 = -131\ \text{dBm}
$$

**Step 4 – SNR due to one section**
$$
\left(\frac{S}{N}\right)_1 = -40 - (-131) = 91\ \text{dB}
$$

**Step 5 – Four identical sections:** every section adds the same noise power, so total noise is 4 times, i.e. SNR falls by $10\log_{10}4 = 6.02$ dB.
$$
\left(\frac{S}{N}\right)_{out} = 91 - 6.02 = 84.98\ \text{dB}
$$

**Answer: output SNR ≈ 85 dB.**

## Check yourself

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

1. Line loss is 100 km x 2 dB/km = 200 dB and each amplifier gives 50 dB. How many amplifiers are needed to get 10 mW out?
   A. 2
   B. 3
   C. 5
   D. 4
   Answer: D. 200 dB / 50 dB = 4 amplifiers, each after a 25 km section. (Key point: Number of amplifiers)

2. With 10 dBm at the start of a 50 dB section, what is the signal at the amplifier input?
   A. -40 dBm
   B. -60 dBm
   C. +60 dBm
   D. -50 dBm
   Answer: A. 10 dBm - 50 dB = -40 dBm. (Key point: Signal at amplifier input)

3. How is the effective noise at each amplifier input found?
   A. Subtract the noise factor: -138 - 7 = -145 dBm
   B. Add the noise factor to the line noise: -138 + 7 = -131 dBm
   C. Add the gain: -138 + 50 = -88 dBm
   D. Use line noise alone: -138 dBm
   Answer: B. Noise factor of 7 dB raises the line noise to -131 dBm. (Key point: Effective noise)

4. One section gives 91 dB SNR. With 4 identical sections, what is the output SNR?
   A. About 85 dB, as total noise is 4 times (-6 dB)
   B. 364 dB
   C. 91 dB, unchanged
   D. About 97 dB
   Answer: A. Noise powers add: 4x noise reduces SNR by 10log10(4) = 6.02 dB, giving 84.98 dB. (Key point: Cascade SNR)

## Related in Telecommunications

- [SS7 signaling points: STP, SSP, SCP](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Ass7-stp-ssp-scp) — Telecommunications, chapter 5
- [Talker and listener echo in four-wire circuit](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Afour-wire-echo-numerical) — Telecommunications, chapter 2
- [Impairments to voice channel transmission](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Avoice-channel-impairments) — Telecommunications, chapter 2
- [Device holding time and calls per hour](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adevice-holding-time-calls) — Telecommunications, chapter 6
- [TST vs STS and three-stage design](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atst-sts-design-crosspoints) — Telecommunications, chapter 4

## Related in other subjects

- [Initial line design for 200 MW over 120 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aline-design-200mw-120km) — Transmission and Distribution System Design, chapter 2
- [Voltage and circuits for 160 MW over 100 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-voltage-160mw-100km) — Transmission and Distribution System Design, chapter 2
- [Nominal-T 100 km line, 20 MW at 110 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-20mw) — Power System Analysis I, chapter 5
- [Empirical formula; voltage for 150 MW over 100 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aempirical-formula-150mw-100km) — Transmission and Distribution System Design, chapter 2