CAP7001 · Unit 2: Physical layer
Data rate and transmission impairments
14 practice questions · in the MTE and ETE syllabus
Key points
- Nyquist: 2 B log2(L) for noiseless channels
- Shannon: B log2(1 + SNR), SNR as a ratio
- Impairments: attenuation, distortion, noise
- Total delay = transmission + propagation
Data rate limits:
- Noiseless channel (Nyquist): BitRate = 2 x B x log2(L).
- Noisy channel (Shannon): C = B x log2(1 + SNR), with SNR as a plain ratio. SNR in dB = 10 log10(SNR).
Impairments: attenuation (loss of energy, dB = 10 log10(P2/P1)), distortion (components travel at different speeds, so the shape changes) and noise (thermal, crosstalk, impulse).
Delay: transmission time = message bits / bit rate; propagation time = distance / speed.
Worked example: B = 3000 Hz and SNR = 30 dB means a ratio of 1000, so C = 3000 x log2(1001), about 29.9 kbps. A 1500-byte frame on a 10 Mbps, 2000 km link (2 x 10^8 m/s) takes 1.2 ms + 10 ms = 11.2 ms.
Exam pitfall: never put a dB value straight into Shannon's formula; convert it to a ratio first. Using 30 instead of 1000 gives about half the correct capacity. Decibels add along a chain: -3, -3, +10 and -4 dB net 0 dB.
Try two questions
A signal's power falls from 10 mW to 5 mW. The change in decibels is approximately:
- -3 dB
- -5 dB
- -10 dB
- +3 dB
Show the answer
A. dB = 10 log10(P2/P1) = 10 log10(0.5) ~ -3 dB (negative = attenuation).
What is crosstalk?
- Loss of signal energy as it travels along the cable
- Unwanted coupling of one wire's signal onto a neighbouring wire
- Random electron motion that adds energy to the signal
- Change in signal shape from unequal frequency speeds
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B. Crosstalk is an effect of one circuit or wire pair on another, which twisting of the wires reduces.
Why not A: That describes attenuation.
Why not C: That describes thermal noise.
Why not D: That describes distortion.