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Transmission impairments; attenuation of cable

Short answer

Transmission impairments include attenuation, distortion, and noise. With attenuation -0.3 dB/km over 5 km, total loss is -1.5 dB, so power reduces from 2 mW to 2 × 10^(-0.15) ≈ 1.42 mW.

Explain transmission impairments in detail. If the signal at the beginning of a cable with -0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?

Transmission impairments

Impairments are the imperfections that make the received signal differ from the one transmitted. There are three main kinds.

  1. Attenuation: loss of signal energy with distance in the medium. It grows with distance and frequency. It is measured in dB: dB = 10 log₁₀(P₂/P₁). Amplifiers or repeaters restore the strength.
  2. Distortion: change of the signal shape because different frequency components travel at different speeds (delay distortion) or are attenuated differently. It causes inter-symbol interference; equalisers correct it.
  3. Noise: unwanted signals added in the channel.
    • Thermal noise: random electron motion, N = kTB.
    • Intermodulation noise: sums/differences of frequencies in non-linear media.
    • Crosstalk: coupling between adjacent wires.
    • Impulse noise: sudden spikes (lightning, switching) causing burst errors.

SNR = signal power / noise power decides the error rate, and the capacity is C = B log₂(1 + SNR).

Numerical

Attenuation = -0.3 dB/km, length = 5 km, P₁ = 2 mW.

Total loss = -0.3 × 5 = -1.5 dB

dB = 10 log₁₀(P₂/P₁) = -1.5

P₂/P₁ = 10^(-0.15) = 0.7079

P₂ = 2 mW × 0.7079 = 1.416 mW

Answer: power at 5 km ≈ 1.42 mW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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