8.P — Worked Problems: Wireless and Telecom

Problem 1 — Cellular capacity and cell splitting

A city area of 100 km² is served by cells of 2 km radius with a reuse factor of 7, and 350 channels are available. Find the capacity, then the effect of halving the cell radius.

Solution

Cell area (hexagonal, with R the distance to a vertex):

A=\frac{3\sqrt3}{2}R^2=2.598\times4=10.4\ \text{km}^2

Number of cells:

\frac{100}{10.4}=9.6 \;\to\; 10\ \text{cells}

Channels per cell:

\frac{350}{7}=50

Total capacity:

10\times50=500\ \text{simultaneous calls}

Halve the radius to 1 km. Area becomes 2.6 km², so 38 cells:

38\times50=1900\ \text{calls}

Just under four times the capacity, from the same 350 channels.

Check the reuse distance still works. D=R\sqrt{21}=1\times4.58=4.58 km. Interference ratio at n=3.5:

\frac{(4.58)^{3.5}}{6}=\frac{205}{6}=34=15.3\ \text{dB}

Adequate — the usual requirement is 12 to 18 dB.

The costs of splitting, and they are real:

• Four times the sites, each with equipment, backhaul, power and site rental.
• More handovers per call. A vehicle at 50 km/h crosses a 2 km cell in 144 s and a 1 km cell in 72 s. Twice the handover signalling, and twice the opportunity for a dropped call.
• More interference from imperfect planning, since more cells means more boundaries.

The general conclusion. Capacity scales as 1/R^2 and cost scales roughly as 1/R^2 too, so cost per subscriber stays roughly flat while capacity grows. That is why densification has been the industry's default answer for forty years, and why it continues even now.

Problem 2 — Timing advance and cell radius

A GSM cell must cover a 40 km stretch of coastline from one site. Is it possible, and what does it cost?

Solution

Round trip time to 40 km:

t=\frac{2\times40{,}000}{3\times10^8}=267\ \mu\text{s}

GSM's timing advance field is 6 bits, in steps of one bit period:

t_{bit}=\frac{1}{270.833\ \text{kbit/s}}=3.69\ \mu\text{s}

t_{max}=63\times3.69=233\ \mu\text{s} \;\Rightarrow\; d_{max}=\frac{233\times10^{-6}\times3\times10^8}{2}=35\ \text{km}

40 km exceeds the limit. A phone at 40 km cannot be told to advance enough, so its burst arrives late and lands in the next timeslot, corrupting another call.

The standard workaround is extended range mode: allocate two consecutive timeslots per call, so the late arrival lands in the second slot which is deliberately left empty.

d_{max}=70\ \text{km}

The cost: capacity halves, since each call now occupies two of the eight slots.

Then check the link budget at 40 km. At 900 MHz over sea, with a path loss exponent near 2.5 because the sea surface is a good reflector:

L\approx 32.4+20\log_{10}(0.04\ \text{km at 1 km ref})\ldots

more directly, free-space loss at 40 km and 900 MHz:

L=32.4+32.0+59.1=123.5\ \text{dB}

With a 43 dBm base station, 17 dBi antenna, 0 dBi phone antenna:

P_{rx}=43+17-123.5+0=-63.5\ \text{dBm}

Comfortable on the downlink. The uplink is the problem: a 33 dBm phone gives 33+0-123.5+17=-73.5 dBm, against a typical GSM sensitivity of -104 dBm. 30 dB of margin — it works, because over water there are no obstructions and the path is close to free space.

Answer: possible with extended range mode, at half the capacity. And this is exactly why coastal and desert cells are the ones that hit the timing limit — everywhere else, terrain limits the range long before timing does.

Problem 3 — GPS position from pseudoranges

Four satellites give pseudoranges. Explain the solution method and quantify what a 1 µs receiver clock error would cost if it were not solved for.

Solution

The four equations:

\sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}+cb=\rho_i

These are nonlinear, so the standard method is iterative:

1. Guess a position — the earth's centre, or the last known fix.
2. Linearise by taking the first-order Taylor expansion about that guess, giving a matrix equation \mathbf{H}\Delta\mathbf{x}=\Delta\boldsymbol\rho where \mathbf H contains the unit vectors to each satellite.
3. Solve the linear system for the correction.
4. Update and repeat. It converges in three or four iterations to millimetre consistency.

With more than four satellites, the system is overdetermined and is solved by least squares:

\Delta\mathbf x=(\mathbf H^T\mathbf H)^{-1}\mathbf H^T\Delta\boldsymbol\rho

And (\mathbf H^T\mathbf H)^{-1} is exactly the DOP matrix — its diagonal entries give the dilution of precision in each coordinate. So the geometry's effect on accuracy falls straight out of the solution method.

Cost of an unsolved 1 µs clock error:

300\ \text{m, in every pseudorange simultaneously}

Because it is common to all of them, the three spheres from three satellites would fail to intersect at all — they would enclose a region 300 m across rather than meeting at a point. The receiver would report a fix hundreds of metres wrong, or no fix at all.

And a quartz crystal drifts by 1 µs in about 100 milliseconds, so the error would grow visibly during the fix itself.

Which is why the fourth satellite is not a refinement but a requirement, and why every GPS receiver is also a precision clock.

Problem 4 — Ionospheric correction

A dual-frequency receiver measures 20,186,543.2 m on L1 (1575.42 MHz) and 20,186,548.9 m on L2 (1227.60 MHz). Find the ionospheric delay and the corrected range.

Solution

The L2 range is longer because the ionospheric delay scales as 1/f^2 and L2 is the lower frequency.

\rho_2-\rho_1=5.7\ \text{m}

Correction formula:

\rho_{corr}=\frac{f_1^2\rho_1-f_2^2\rho_2}{f_1^2-f_2^2}

f_1^2=(1575.42)^2=2{,}481{,}948, \qquad f_2^2=(1227.60)^2=1{,}507{,}002

f_1^2-f_2^2=974{,}946

Rather than multiplying two huge numbers, rearrange:

\rho_{corr}=\rho_1-\frac{f_2^2}{f_1^2-f_2^2}(\rho_2-\rho_1)=\rho_1-\frac{1{,}507{,}002}{974{,}946}\times5.7

=\rho_1-1.546\times5.7=\rho_1-8.81\ \text{m}

\rho_{corr}=20{,}186{,}543.2-8.81=20{,}186{,}534.4\ \text{m}

The ionosphere added 8.8 m to the L1 measurement, and dual-frequency measurement removed it exactly.

A single-frequency receiver would apply the broadcast Klobuchar model, which typically removes 50 to 60% of the delay — leaving 3.5 to 4.4 m of error here.

The improvement, quantified. With the ionosphere removed, the remaining error budget is satellite clock (2 m), ephemeris (2.5 m), troposphere (0.5 m), multipath (1 m) and receiver noise (0.3 m):

\sigma=\sqrt{4+6.25+0.25+1+0.09}=3.4\ \text{m}

against about 6.5 m for single frequency. Roughly a factor of two, which is exactly the improvement seen when phones gained dual-frequency GNSS around 2018.

And the correction amplifies noise. The factor of 1.546 multiplies the L2 measurement's noise, so the corrected range is noisier than either raw measurement — about three times noisier. For a static receiver this is averaged away; for a fast-moving one it matters, which is why some receivers smooth the correction over many seconds.

Problem 5 — Wi-Fi throughput and overhead

An 802.11n link negotiates 65 Mbit/s. Estimate the actual TCP throughput, and the effect of frame aggregation.

Solution

Per-frame overhead for a 1500-byte packet:

Item	Time
DIFS	34 µs
Average backoff (CW=15)	67.5 µs
PLCP preamble and header	20 µs
Data, 1500 bytes at 65 Mbit/s	185 µs
SIFS	16 µs
ACK	24 µs
Total	346 µs

Efficiency:

\frac{185}{346}=53\%

Throughput:

65\times0.53=34.5\ \text{Mbit/s}

Subtract TCP and IP headers (about 3% for 1500-byte packets) and TCP acknowledgements travelling the other way (roughly another 5%):

\approx31\ \text{Mbit/s}

Under half the negotiated rate, which matches what people measure.

Now aggregate 32 frames into one transmission (A-MPDU, which 802.11n introduced):

\text{data time}=32\times185=5920\ \mu\text{s}

Fixed overhead stays at 34+67.5+20+16+24=161 µs.

\text{efficiency}=\frac{5920}{6081}=97.4\%

\text{throughput}=65\times0.974=63\ \text{Mbit/s}

Nearly double, and this is why aggregation was the single most valuable feature in 802.11n.

The limits of aggregation. A longer transmission means a longer wait for other stations, hurting latency. And one corrupted bit used to destroy the whole aggregate — which is why block acknowledgement was added, allowing selective retransmission of only the damaged sub-frames.

And aggregation only helps when you have enough data to fill it. For a stream of small packets — a voice call, a game — there is nothing to aggregate, and the efficiency stays at 53% or worse. That is exactly the case OFDMA in Wi-Fi 6 addresses, by letting several stations share one transmission rather than each taking a full turn.

Problem 6 — Bluetooth Low Energy battery life

A wearable sends a 20-byte notification every 200 ms while connected. Radio current is 12 mA, sleep is 2 µA, and each connection event takes 2 ms. Battery is 100 mAh. Find the life, and the effect of changing the interval.

Solution

I_{avg}=12\times\frac{0.002}{0.2}+0.002=0.12+0.002=0.122\ \text{mA}

\text{life}=\frac{100}{0.122}=820\ \text{hours}=34\ \text{days}

Now try a 1 second interval:

I_{avg}=12\times\frac{0.002}{1}+0.002=0.024+0.002=0.026\ \text{mA}

\text{life}=\frac{100}{0.026}=3846\ \text{hours}=160\ \text{days}

Nearly five times longer.

And 4 seconds:

I_{avg}=0.006+0.002=0.008\ \text{mA} \;\Rightarrow\; 12{,}500\ \text{hours}=1.4\ \text{years}

The trade is latency. At a 4 second interval, a button press is reported up to 4 seconds later — unacceptable for a heart rate monitor during exercise, and fine for a temperature sensor.

The standard solution is connection parameter update: the device requests a short interval when the user is interacting and a long one when idle. A fitness tracker uses 30 ms during a workout and 1 second otherwise, getting responsiveness when it matters and battery life the rest of the time.

BLE's slave latency parameter refines it further: the peripheral may skip up to N connection events if it has nothing to send, while the central still checks in at the short interval. So the central can send data promptly while the peripheral sleeps through most events — the best of both, and it is why the parameter exists.

Notice where the current goes. At 200 ms the radio dominates (0.12 of 0.122 mA). At 4 seconds the sleep current dominates (0.002 of 0.008 mA). Beyond that point, reducing the radio duty cycle buys nothing and the only remaining lever is the sleep current — which is why the last few nanoamps in a datasheet matter so much for long-life designs.

Problem 7 — NFC field strength and range

An NFC reader produces 5 A/m at 1 cm. Find the field at 3 cm and 6 cm, and determine the maximum operating distance if the card needs 1.5 A/m.

Solution

Near field, so H\propto1/r^3:

H(3\ \text{cm})=5\times\left(\frac{1}{3}\right)^3=5\times0.037=0.185\ \text{A/m}

H(6\ \text{cm})=5\times\left(\frac16\right)^3=5\times0.00463=0.023\ \text{A/m}

Maximum range:

1.5=5\left(\frac{1}{r}\right)^3 \;\Rightarrow\; r^3=\frac{5}{1.5}=3.33 \;\Rightarrow\; r=1.49\ \text{cm}

Under 1.5 cm with this reader.

To reach 4 cm, the field at 1 cm would need to be

H_1=1.5\times4^3=96\ \text{A/m}

Nineteen times stronger, requiring roughly 19² = 360 times the power in the coil, since power goes as field squared.

That is why the range is what it is. Regulatory emission limits and the reader's power budget cap the field, and the 1/r^3 law does the rest.

What actually extends the range in practice is not more power but better coupling: a larger reader coil, a better-matched card antenna, and higher Q in both. A large reader antenna in a transit gate achieves 8 to 10 cm where a small one in a phone achieves 2 to 3 cm.

And the security consequence is worth stating. For an attacker to power a card from 50 cm would need

\left(\frac{50}{1.5}\right)^3=37{,}000\ \text{times the field}

which is over a billion times the power. The physics, not the protocol, is what makes covert skimming from a distance impractical.

Eavesdropping on an active transaction is different, since the attacker only has to hear the reader's much stronger field, and that has been demonstrated at several metres. But without the card's cryptogram being valid for a future transaction, what is captured has little value.

Problem 8 — Fibre link budget

Design a 120 km single-mode link at 1550 nm running at 10 Gbit/s. Transmitter +3 dBm, receiver sensitivity -28 dBm, fibre loss 0.20 dB/km, splices every 5 km at 0.05 dB, two connectors at 0.5 dB.

Solution

Losses:

\text{fibre}=120\times0.20=24.0\ \text{dB}

\text{splices}=\frac{120}{5}\times0.05=24\times0.05=1.2\ \text{dB}

\text{connectors}=2\times0.5=1.0\ \text{dB}

\text{total}=26.2\ \text{dB}

Received power:

3-26.2=-23.2\ \text{dBm}

Margin:

-23.2-(-28)=4.8\ \text{dB}

Adequate but thin. The usual requirement is 3 dB for ageing plus 3 dB for repairs — a repair splice adds loss, and a cable may be repaired several times over its life.

Now check dispersion, which is often the real limit.

\Delta t=D\cdot L\cdot\Delta\lambda=17\times120\times\Delta\lambda

For a directly modulated laser with 0.3 nm of chirp:

\Delta t=17\times120\times0.3=612\ \text{ps}

The bit period at 10 Gbit/s is 100 ps. The pulse has spread over six bit periods. The link will not work at all, regardless of the power budget.

Fixes:

Use an externally modulated laser with 0.05 nm linewidth: \Delta t=102 ps. Still marginal — the usual criterion is that dispersion should stay under about 0.3 of a bit period.

Add dispersion-compensating fibre. A module compensating 2000 ps/nm cancels most of the 2040 ps/nm accumulated. Its insertion loss is typically 6 to 10 dB, which destroys the power budget — so an amplifier is needed too.

Or use coherent detection with digital dispersion compensation, which handles it computationally at no optical cost.

Revised design: externally modulated transmitter, one EDFA at 60 km giving 20 dB of gain, and dispersion compensation at the amplifier site.

3-13.1+20-8-13.1=-11.2\ \text{dBm}

Margin now 16.8 dB ✓ with dispersion handled.

The lesson worth extracting. For links under about 80 km at 10 Gbit/s, loss is the limit and the sums are simple. Beyond that, dispersion becomes the binding constraint, and the design changes character entirely.

Problem 9 — Satellite link budget

A GEO satellite at 12 GHz transmits 100 W through a 34 dBi antenna. A home dish is 60 cm with 60% efficiency, and the receiver noise temperature is 100 K. Find the C/N in a 30 MHz transponder.

Solution

EIRP:

P_{tx}=10\log_{10}(100\times1000)=50\ \text{dBm}=20\ \text{dBW}

EIRP=20+34=54\ \text{dBW}

Path loss at 35,786 km and 12 GHz:

L=32.4+20\log_{10}(35{,}786)+20\log_{10}(12{,}000)=32.4+91.1+81.6=205.1\ \text{dB}

Dish gain:

\lambda=\frac{3\times10^8}{12\times10^9}=0.025\ \text{m}

G=0.6\left(\frac{\pi\times0.6}{0.025}\right)^2=0.6\times(75.4)^2=3411=35.3\ \text{dBi}

Received power:

C=54-205.1+35.3=-115.8\ \text{dBW}

Noise power in 30 MHz at 100 K:

N=10\log_{10}(kTB)=10\log_{10}(1.38\times10^{-23}\times100\times3\times10^7)

=10\log_{10}(4.14\times10^{-14})=-133.8\ \text{dBW}

Carrier to noise:

\frac CN=-115.8-(-133.8)=18.0\ \text{dB}

Adequate for 8-PSK with rate-2/3 coding, which DVB-S2 says needs about 6.6 dB — so 11.4 dB of margin.

Now apply rain fade. At 12 GHz, heavy rain gives about 1.5 dB/km through a 4 km cell:

A=6\ \text{dB}

\frac CN=12.0\ \text{dB}

Still 5.4 dB of margin ✓ — which is why Ku-band satellite television survives most rain, and why the picture freezes only in genuinely extreme downpours.

At Ka band, 30 GHz, the same storm gives 4 dB/km over 4 km = 16 dB. The margin would be gone entirely, which is why Ka-band systems depend on adaptive coding and modulation rather than on a fixed margin.

And check the dish size sensitivity. Going from 60 cm to 45 cm loses

20\log_{10}\frac{60}{45}=2.5\ \text{dB}

which eats half the rain margin. That is why installers specify dish size by region — northern Europe uses 60 cm where southern Spain manages with 45 cm, because both the satellite's beam and the rainfall differ.

Problem 10 — Erlang B trunk sizing

A call centre receives 200 calls per hour with an average duration of 4 minutes. How many lines are needed for 1% blocking? For 0.1%?

Solution

Offered traffic in erlangs:

A=\frac{200\times4}{60}=13.3\ \text{erlangs}

One erlang is one circuit occupied continuously, so 13.3 erlangs means an average of 13.3 calls in progress.

Erlang B:

B=\frac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}

Evaluating for increasing N with A=13.3:

N	Blocking
16	8.9%
18	4.6%
20	2.1%
21	1.4%
22	0.9%
25	0.24%
27	0.09%

For 1% blocking: 22 lines. For 0.1%: 27 lines.

The interesting numbers are the ratios.

\frac{22}{13.3}=1.65, \qquad \frac{27}{13.3}=2.03

Sixty-five percent more lines than the average load, to hold blocking at 1%. That is the cost of randomness — calls do not arrive evenly.

And the trunking efficiency effect. Now take a system with ten times the traffic, 133 erlangs:

N	Blocking
150	1.2%
153	0.9%

153 lines for 133 erlangs — only 15% more than the average, against 65% for the small system.

Large trunk groups are dramatically more efficient, because random peaks in one direction are offset by troughs in another. This is the mathematical argument for centralising resources, and it appears again as statistical multiplexing in packet networks and as pooling in cloud computing — the same formula, different vocabulary.

And the counter-argument. A single large group is a single point of failure. Two groups of 80 lines each carry less traffic than one group of 160, and that lost efficiency is the price of redundancy.

Problem 11 — VoIP latency budget

A VoIP call uses the G.711 codec with 20 ms packets over a network with 40 ms one-way delay and 15 ms of jitter. Assess the quality and suggest improvements.

Solution

The budget:

Component	Delay
Codec (G.711 has no algorithmic delay)	0 ms
Packetisation (20 ms of audio per packet)	20 ms
Serialisation on the access link	1 ms
Network transit	40 ms
Jitter buffer (typically 2× the jitter)	30 ms
Decode and playout	5 ms
Total one way	96 ms

Under the 150 ms threshold ✓ — the call will feel natural.

Round trip: 192 ms, which is what matters for the conversational turn-taking that people actually notice.

Now make the network 120 ms — an intercontinental path:

\text{total}=20+1+120+30+5=176\ \text{ms}

Above 150 ms. Usable, with a noticeable pause before the other person responds.

Improvements, in order of how much they buy:

Reduce the jitter buffer. It is set at twice the measured jitter for safety. An adaptive buffer that shrinks when the network is stable can often run at 1.2× the jitter, saving 12 ms — at the cost of occasional dropouts when conditions change.

Reduce packetisation to 10 ms — saves 10 ms of delay, and doubles the header overhead, since each packet carries 40 bytes of RTP/UDP/IP headers regardless of payload.

\text{20 ms packets: } \frac{160+40}{160}=125\%\ \text{overhead ratio}

\text{10 ms packets: } \frac{80+40}{80}=150\%

So bandwidth rises from 80 to 96 kbit/s for a 10 ms saving. Usually not worth it.

Reduce network delay. The only genuinely large lever, and it means routing, peering and physical distance — largely outside the application's control.

The one that matters most in practice: prioritise the traffic. Voice packets marked with a high DSCP class jump ahead of bulk traffic in every queue. On a congested link this can remove 50 to 100 ms of queuing delay, dwarfing every other optimisation. It is why quality of service marking exists and why VoIP over an unmanaged congested link is so much worse than the raw distance suggests.

Problem 12 — DTMF detection

Design a Goertzel-based DTMF detector at an 8 kHz sample rate. Choose the block length and check the frequency resolution.

Solution

Eight frequencies: 697, 770, 852, 941, 1209, 1336, 1477 Hz, plus 1633 Hz for the rarely used fourth column.

The specification requires detecting a tone pair present for at least 40 ms, and rejecting tones more than 3.5% off frequency.

Block length. At 8 kHz, 40 ms is 320 samples. Use N=205, the standard choice, giving 25.6 ms — short enough that two blocks fit in the minimum tone duration, allowing a confirmation.

Bin spacing:

\Delta f=\frac{8000}{205}=39\ \text{Hz}

Check each frequency lands close to a bin centre:

Tone	k = f N/f_s	Nearest integer	Error
697	17.86	18	0.14 bins = 5.5 Hz
770	19.73	20	0.27 = 10.5 Hz
852	21.83	22	0.17 = 6.6 Hz
941	24.11	24	0.11 = 4.3 Hz
1209	30.98	31	0.02 = 0.8 Hz
1336	34.23	34	0.23 = 9.0 Hz
1477	37.84	38	0.16 = 6.2 Hz
1633	41.85	42	0.15 = 5.9 Hz

Worst error 10.5 Hz on 770 Hz, which is 1.4% — within the 3.5% acceptance band ✓.

This is exactly why N=205 is the standard value. It is the block length that minimises the worst-case bin error across all eight frequencies at 8 kHz.

Computation:

8\ \text{filters}\times205\ \text{samples}=1640\ \text{multiply-accumulates per block}

Compare with a 256-point FFT:

\frac{256}{2}\times8=1024\ \text{complex operations}\approx4096\ \text{real MACs}

Goertzel wins by 2.5 times, and it also avoids the windowing and bin-offset problems that an FFT would have. This is the case where the specialised algorithm is genuinely the right choice, and it is why every telephone exchange used it.

The additional checks a real detector must do, because tone detection alone produces false positives on speech:

• Exactly one tone from each group, and nothing significant in the other bins.
• The two tones within 6 dB of each other — the "twist" test.
• Second harmonic check: speech has strong harmonics; a DTMF tone does not. Reject if the second harmonic is within 20 dB of the fundamental.
• Duration: present for two consecutive blocks, then absent, before reporting a digit.

Without the harmonic test, ordinary speech triggers false digits regularly — a phenomenon known as talk-off, and it was a real operational problem in early detectors.

Problem 13 — Comparing coverage technologies

A rural community of 200 homes is 15 km from the nearest fibre. Compare fixed wireless, satellite and fibre extension.

Solution

Fibre extension. 15 km of buried fibre at typical costs of €30,000 to €80,000 per km in rural terrain:

15 km at 50,000 euro per km = 750,000 euro.

Plus distribution within the community, say €200,000.

950,000 euro over 200 homes = 4,750 euro per home.

Delivers gigabit symmetric with 30-year life and essentially no ongoing cost.

Fixed wireless. A tower with a line of sight to the fibre point, using 5 GHz or licensed microwave backhaul.

• Backhaul link: €30,000
• Tower and base station: €80,000
• Customer equipment at €300 each: €60,000

170,000 euro over 200 homes = 850 euro per home.

Delivers 50 to 200 Mbit/s shared. With 200 homes on one sector and a 500 Mbit/s backhaul, and a typical 20:1 contention ratio, each home sees 50 Mbit/s at peak.

Requires line of sight, which in hilly terrain may mean several towers.

Satellite (LEO). €500 terminal per home plus €50/month:

200 terminals at 500 euro each = 100,000 euro of capital.200 homes at 50 euro a month = 120,000 euro per year.

Delivers 50 to 200 Mbit/s with 40 ms latency, and no infrastructure at all.

Over ten years:

Option	Capital	10-year total	Per home
Fibre	€950k	€1.0M	€5000
Fixed wireless	€170k	€400k	€2000
Satellite	€100k	€1.3M	€6500

Fixed wireless is cheapest over ten years. Satellite has the lowest entry cost and the highest running cost. Fibre is most expensive initially and cheapest beyond about fifteen years, and it is the only option that does not need replacing as demand grows.

The decision depends on three questions:

1. Is there line of sight? If not, fixed wireless is out.
2. What is the time horizon? Beyond 20 years fibre wins outright.
3. Is there subsidy? Most countries subsidise rural fibre specifically because the private economics do not work but the social value does.

The honest recommendation for 200 homes: fixed wireless now, fibre when subsidy is available. And satellite as the immediate stopgap and the permanent answer for the outlying properties that no shared infrastructure will ever reach economically.

Problem 14 — Diagnose a dropped-call problem

Subscribers in one area report calls dropping about 30 seconds in, always while driving. Coverage measurements show good signal everywhere. Diagnose it.

Solution

The two clues are decisive: it happens while moving, and about 30 seconds in.

Thirty seconds while driving at, say, 50 km/h is about 420 m — plausibly a cell boundary crossing.

Good signal everywhere rules out coverage. The problem is handover, not radio.

Candidates, and how to distinguish them:

Missing neighbour relation. Each cell holds a list of neighbours the phone should measure. If a neighbour is missing from the list, the phone never reports it, and the network never commands the handover — so the call holds on to a fading cell until it drops.

This is the most common cause of exactly this symptom, and it happens after any new cell is added, or after a cell is re-parameterised.

Test: check the neighbour lists for reciprocity. A relation defined from A to B but not B to A causes drops in one direction of travel only — which would show up as a problem on one carriageway and not the other.

Handover parameters wrong. Hysteresis too large or time-to-trigger too long means the phone reports too late.

Test: the drop location should be consistently just past the ideal handover point.

Overshooting cell. A distant cell with an antenna tilted too high provides a strong signal in an area it is not planned to serve. The phone camps on it, then loses it abruptly behind an obstruction, and the intended serving cell is not in its neighbour list.

Test: look at which cell the dropped calls were on. A cell serving traffic kilometres outside its planned footprint is the signature, and the fix is mechanical or electrical downtilt.

Interference. Two cells using the same frequency with insufficient reuse distance. Signal strength is good; signal quality is poor.

Test — and this is the decisive one. Measure signal quality separately from strength. RSRP high with SINR low means interference; both low means coverage. The complaint said signal was good, which points here or at handover, and the driving correlation points at handover.

The diagnostic sequence:

1. Pull the drop statistics per cell and find which cell the calls were on when they dropped.
2. Check that cell's neighbour list against the cells that actually provide coverage where the drops occur.
3. Run a drive test logging serving cell, RSRP, SINR and handover events, along the route where the complaints originate.
4. Look for the pattern: did the phone report a strong neighbour and receive no handover command? That is a missing relation or a parameter fault. Did it report nothing at all? That is a missing measurement configuration.

The most probable answer, given the specifics: a missing or one-directional neighbour relation, most likely introduced when a nearby cell was added or modified. It is the single most common cause of location-specific, movement-correlated drops, and modern networks run automatic neighbour relation discovery precisely because manual lists were so error-prone.

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Part 9 leaves signals behind entirely and covers the electronics that moves kilowatts rather than milliwatts.