Output SNR of an amplified line

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.