Assam ASSEB HS Skill Telecom study materials
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3.8 km Copper Pair: Loss, SNR and a Request to Extend the Loop

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3.8 km Copper Pair: Loss, SNR and a Request to Extend the Loop

A bank branch in Nagaon is connected to its regional office in Guwahati over a 3.8 km dedicated copper pair rented from the local operator. The branch's 2 Mbit/s leased-line modem has been dropping sync since the monsoon began, so a technician runs a manual line test during a quiet period. With a 1 kHz test tone injected at −3.0 dBm at the exchange end, the far-end level measures −19.5 dBm. A wideband noise measurement across the 3.1 kHz voice band reads −68 dBm. The service level agreement specifies a maximum insertion loss of 5.0 dB per kilometre and a minimum signal-to-noise ratio of 25 dB at the receiver. The modem's transmitter can deliver a maximum of +6 dBm into the pair. The branch manager wants to shift his counter to a new room, which would lengthen the loop by 1.2 km of the same cable, and has asked whether the existing modem will still work.

1. Using the measured levels, calculate the insertion loss per kilometre and the receive signal-to-noise ratio, and state whether the pair meets the SLA. Show every step of the arithmetic.

Read the worked answer

Insertion loss

Loss (dB) = transmit level − receive level = (−3.0) − (−19.5) = 16.5 dB over 3.8 km.

Loss per km = 16.5 ÷ 3.8 = 4.34 dB/km.

Against the 5.0 dB/km limit this passes, but the spare allowance is only 0.66 dB/km, i.e. 5.0 × 3.8 = 19.0 dB allowed versus 16.5 dB measured, so only 2.5 dB of headroom over the whole loop.

The 16.5 dB also converts to a power ratio of 10^(16.5/10) = 44.7, i.e. about 1/45 of the launched power reaches the far end.

Signal-to-noise ratio

SNR = signal − noise = (−19.5) − (−68) = 48.5 dB.

Against the 25 dB floor this passes with 23.5 dB spare.

Verdict: the pair is compliant on both criteria at 1 kHz. Note what the verdict does not say - the pass is at a single frequency and against an average noise figure, so it does not yet explain the sync drops.

2. Calculate the transmit level that would be needed after the 1.2 km extension to keep the same receive level. Decide whether the existing modem can be used, and state one operational risk of simply raising the transmit power.

Read the worked answer

Extra loss = 1.2 km × 4.34 dB/km = 5.21 dB.

New total loss = 16.5 + 5.21 = 21.71 dB.

To keep the far end at −19.5 dBm, the new transmit level must be:

Tx = −19.5 + 21.71 = +2.21 dBm.

The modem's ceiling is +6 dBm, so there is 3.79 dB of headroom and the extension is theoretically feasible.

But the decision should be cautious:

  • 4.34 dB/km is a measured average, not a worst case. At the 5.0 dB/km contractual limit the 5.0 km loop would need 25.0 dB, i.e. Tx = +5.5 dBm - only 0.5 dB of headroom.
  • The extension adds joints, and joints are where monsoon water ingress appears.

Risk of simply turning up the power: in a multi-pair binder, a higher transmit level raises near-end and far-end crosstalk into neighbouring pairs and can degrade other customers' circuits, while doing nothing for noise picked up after the transmit point. It can also overdrive the far-end receiver. A regenerator, a remote DSLAM, or migrating the branch to fibre is the sounder answer if the post-move measurement shows little margin.

3. Three weeks later the line still passes the 1 kHz test, yet the modem logs CRC errors and retrains several times a day. Explain why a passing test does not clear the line, and list the measurements you would take next.

Read the worked answer

A 1 kHz tone measures one point on the frequency response plus the metallic continuity of the pair. It is blind to exactly the impairments that break data transmission:

  • Frequency response: loss rises with frequency on long pairs; the 1 kHz figure can look healthy while 100 kHz–1 MHz is badly rolled off.
  • Impulse noise: lightning, switching transients and faulty fluorescent ballasts arrive as short bursts. They barely move a long-term average noise reading (−68 dBm looked fine) but they destroy frames and produce CRC errors and retrains.
  • Impedance discontinuities: bridged taps, bad joints and water in a joint cause reflections that are invisible at 1 kHz.
  • Longitudinal imbalance: poor balance lets power-line hum and radio ingress convert into metallic noise.

Next measurements, in order:

  1. Insertion loss at 3 kHz and at 100 kHz, or a full loss/frequency sweep, to expose the slope.
  2. Impulse noise counter over a 15-minute window, with a defined threshold, during the hours when retrains occur.
  3. Noise metallic (dBmC or 3 kHz flat) and longitudinal balance.
  4. TDR trace or insulation-resistance test to locate bridged taps, wet joints and poor bonds; check bonding and earthing at both ends.
  5. Finally a 2 Mbit/s BER test for 30–60 minutes to confirm the repair.

Given the failure began with the monsoon, a water-logged joint or a corroded connector is the leading suspect; the 1 kHz number was never going to find it.