7.P — Worked Problems: Communication Systems

Problem 1 — Thermal noise and receiver sensitivity

A receiver has a 200 kHz bandwidth and a 4 dB noise figure. Find the noise floor and the minimum signal for a 12 dB SNR. Then find the range if the transmitter is 30 dBm with unity antennas at 450 MHz.

Solution

Noise floor:

P_n=-174+10\log_{10}(200{,}000)+4=-174+53+4=-117\ \text{dBm}

Minimum signal:

P_{min}=-117+12=-105\ \text{dBm}

Allowable path loss:

L=30-(-105)=135\ \text{dB}

Range from the free-space formula:

135=32.4+20\log_{10}d_{km}+20\log_{10}(450)

20\log_{10}(450)=53.1

20\log_{10}d_{km}=135-32.4-53.1=49.5 \;\Rightarrow\; d=10^{2.475}=299\ \text{km}

Free space says 299 km, which is absurd for a ground-level link.

Redo with a realistic path loss exponent of n=3.5 for an urban environment. Using a 1 km reference where the free-space loss is 32.4+53.1=85.5 dB:

135=85.5+35\log_{10}d_{km} \;\Rightarrow\; \log_{10}d=\frac{49.5}{35}=1.414 \;\Rightarrow\; d=26\ \text{km}

Twenty-six kilometres, which is a believable figure for a 1 W radio at 450 MHz.

The lesson. Free-space calculations are useful only for satellite and clear line-of-sight links. For anything terrestrial, the path loss exponent is the single number that decides the answer, and getting it from measurement rather than theory is what link planning actually consists of.

Reducing the bandwidth to 12.5 kHz (narrowband voice) drops the noise floor by 10\log(200/12.5)=12 dB, allowing 12 dB more path loss and extending the range to 10^{(49.5+12)/35}=57 km. Narrow bandwidth buys range, which is why long-range links are always narrowband.

Problem 2 — Cascade noise figure

A receiver chain is: 2 dB of cable, then a 15 dB gain LNA with 1.2 dB NF, then a mixer with 8 dB loss and 8 dB NF, then a 30 dB IF amplifier with 6 dB NF. Find the total noise figure, and the improvement from moving the LNA before the cable.

Solution

Convert to ratios. Cable: F_1=1.58 (L=2 dB), G_1=0.631. LNA: F_2=1.32, G_2=31.6. Mixer: F_3=6.31, G_3=0.158. IF: F_4=3.98.

F=1.58+\frac{1.32-1}{0.631}+\frac{6.31-1}{0.631\times31.6}+\frac{3.98-1}{0.631\times31.6\times0.158}

=1.58+0.507+0.266+0.946=3.30

NF=10\log_{10}(3.30)=5.2\ \text{dB}

Now put the LNA first, before the cable:

F=1.32+\frac{1.58-1}{31.6}+\frac{6.31-1}{31.6\times0.631}+\frac{3.98-1}{31.6\times0.631\times0.158}

=1.32+0.018+0.266+0.946=2.55

NF=4.1\ \text{dB}

1.1 dB better, which is a 30% increase in effective range at n=2.

Where the remaining noise comes from. Look at the terms: the fourth one, 0.946, is the IF amplifier, and it is the largest contributor after the LNA itself. The mixer's 8 dB loss is what lets it through — loss between stages removes the earlier gain's protection.

The proper fix: more gain before the mixer. Raising the LNA to 25 dB divides the third and fourth terms by ten:

F=1.32+0.002+0.027+0.095=1.44 \;\Rightarrow\; NF=1.6\ \text{dB}

From 5.2 dB to 1.6 dB. But more gain before the mixer means the mixer sees larger signals and may generate intermodulation, so the real design is a compromise between noise figure and linearity, and that compromise is the central problem of receiver design.

Problem 3 — AM power and efficiency

A 5 kW AM transmitter is modulated to 80% by a single tone. Find the total power, the sideband power, the efficiency, and the antenna current if the antenna is 50 Ω.

Solution

P_{total}=P_c\left(1+\frac{\mu^2}{2}\right)=5000\left(1+\frac{0.64}{2}\right)=5000\times1.32=6600\ \text{W}

Sideband power:

P_{sb}=P_{total}-P_c=1600\ \text{W}

Efficiency:

\eta=\frac{1600}{6600}=24.2\%

Antenna current, unmodulated:

I_c=\sqrt{\frac{5000}{50}}=10\ \text{A}

Modulated:

I=\sqrt{\frac{6600}{50}}=11.5\ \text{A}

Which gives the standard field test:

\frac{I_{mod}}{I_c}=\sqrt{1+\frac{\mu^2}{2}}=1.149

A broadcast engineer reads the antenna ammeter and infers the modulation depth from the rise. A 15% rise means about 80% modulation. That measurement was standard practice for decades and required no other instrument.

Now the comparison that explains why SSB exists. Transmitting only one sideband, the same 800 W of information-bearing power in one sideband gives the same received audio as this 6600 W AM signal.

\text{power saving}=10\log_{10}\frac{6600}{800}=9.2\ \text{dB}

And the bandwidth halves. SSB needs an eighth of the power and half the spectrum for the same result, which is why every long-distance voice service that could adopt it did.

Problem 4 — FM bandwidth and noise advantage

An FM system has a 5 kHz deviation and a 3 kHz maximum message frequency. Find \beta, the bandwidth, and the SNR improvement over AM. Then repeat for broadcast FM.

Solution

Narrowband case:

\beta=\frac{5}{3}=1.67

B=2(5+3)=16\ \text{kHz}

\text{improvement}=3\beta^2(\beta+1)=3\times2.79\times2.67=22.3 = 13.5\ \text{dB}

AM would use 2\times3 = 6 kHz. So this FM system uses 2.7 times the bandwidth for 13.5 dB.

Broadcast FM: \Delta f=75 kHz, f_m=15 kHz.

\beta=5, \qquad B=2(75+15)=180\ \text{kHz}

\text{improvement}=3\times25\times6=450=26.5\ \text{dB}

AM for the same 15 kHz audio would use 30 kHz. FM uses six times the bandwidth for 26.5 dB.

Compare the two exchange rates:

	Bandwidth ratio	Gain
Narrowband FM	2.7×	13.5 dB
Broadcast FM	6×	26.5 dB

Doubling the bandwidth ratio roughly doubled the decibels, which is what 3\beta^2(\beta+1) predicts — the gain grows faster than linearly with bandwidth.

Add pre-emphasis and broadcast FM gains a further 13 dB at the top of the audio band, so the effective advantage over AM approaches 40 dB. That is a factor of ten thousand in noise power, and it is why FM radio sounds like a different medium rather than a better version of the same one.

The threshold caveat. All of this holds only above about 10 dB of input SNR. Below threshold FM collapses, which is why FM coverage has a hard edge and AM fades gradually.

Problem 5 — Superheterodyne image rejection

An AM receiver tunes 530–1700 kHz with a 455 kHz IF and high-side local oscillator injection. Find the LO range and the image range. Then repeat with a 10.7 MHz IF and comment.

Solution

Local oscillator: f_{LO}=f_{sig}+f_{IF}, so 985 to 2155 kHz.

Image: f_{im}=f_{sig}+2f_{IF}=f_{sig}+910 kHz, so 1440 to 2610 kHz.

The problem is immediately visible. Tuning to 530 kHz, the image is at 1440 kHz — which is inside the broadcast band. A strong local station at 1440 kHz will be received while you are tuned to 530.

How much preselector rejection is needed? The preselector must reject something 910 kHz away while passing 530 kHz. As a frequency ratio that is 1440/530 = 2.7, which is 1.4 octaves. A two-pole tuned circuit gives about 12 dB per octave beyond its corner, so around 17 dB — usually enough, and it is why cheap AM radios sometimes do pick up images.

With a 10.7 MHz IF: the image is at f_{sig}+21.4 MHz, so 21.9 to 23.1 MHz — far outside the broadcast band, and trivially rejected by any input filter.

So why not always use a high IF? Because the selectivity filter at the IF must now separate stations 9 or 10 kHz apart at 10.7 MHz, which is a Q of

Q=\frac{10.7\times10^6}{10{,}000}=1070

No LC circuit reaches that. A crystal filter can, but it is expensive.

At 455 kHz the required Q is only 45.5, which two or three coupled tuned circuits deliver easily and cheaply.

Hence double conversion, which is what every good receiver does: a first IF around 45 MHz for image rejection, then a second at 455 kHz for selectivity. You get both, at the cost of a second oscillator and mixer — and a second image to worry about, which is why the first IF filter must also be reasonably good.

Problem 6 — Digital modulation choice

A link has 5 MHz of bandwidth and 22 dB SNR, and must carry 20 Mbit/s. Which modulation, and how much margin?

Solution

Required spectral efficiency:

\eta=\frac{20}{5}=4\ \text{bit/s/Hz}

Shannon capacity check first:

C=5\times10^6\log_2(1+158)=5\times10^6\times7.31=36.6\ \text{Mbit/s}

20 Mbit/s is 55% of capacity — demanding but achievable with good coding.

Raw modulation. 4 bit/s/Hz suggests 16-QAM, which carries exactly 4 bits per symbol. But pulse shaping costs bandwidth: with \alpha=0.25,

B=\frac{R_s(1.25)}{2}\ \ldots

for a symbol rate R_s, needing R_s = 5\times10^6\times2/1.25=8 MBaud. At 4 bits per symbol that is 32 Mbit/s raw — enough room for rate-5/8 coding, or 20 Mbit/s of payload with substantial redundancy.

SNR check. 16-QAM needs 20.5 dB for 10^{-6} uncoded. Available 22 dB, so 1.5 dB of margin — thin.

With rate-1/2 LDPC coding, the required SNR drops by around 8 dB to about 12.5 dB. Margin becomes 9.5 dB, which is comfortable. But rate-1/2 coding needs R_s\times4\times0.5 = 20 Mbit/s, requiring R_s = 10 MBaud, which needs 10\times1.25/2=6.25 MHz. More bandwidth than available.

The resolution: 64-QAM with rate-1/2 coding.

R_s=\frac{20}{6\times0.5}=6.67\ \text{MBaud} \;\Rightarrow\; B=\frac{6.67\times1.25}{2}=4.17\ \text{MHz} \ ✓

64-QAM needs 26.5 dB uncoded, minus roughly 8 dB of coding gain = 18.5 dB. Available 22 dB, margin 3.5 dB ✓.

Answer: 64-QAM with rate-1/2 LDPC, using 4.17 of the 5 MHz, with 3.5 dB of margin.

The general lesson. A higher constellation with stronger coding often beats a lower constellation with weaker coding, because coding gain is cheap and bandwidth is not. That trade is exactly why every modern standard uses high-order QAM with powerful codes rather than robust modulation with light coding.

Problem 7 — OFDM parameter design

Design an OFDM system for a 10 MHz channel with a 3 µs delay spread, aiming for at least 90% efficiency after the cyclic prefix.

Solution

Cyclic prefix must exceed the delay spread:

T_{cp}\ge3\ \mu\text{s}

For 90% efficiency:

\frac{T_{useful}}{T_{useful}+T_{cp}}\ge0.9 \;\Rightarrow\; T_{useful}\ge9T_{cp}=27\ \mu\text{s}

Subcarrier spacing:

\Delta f=\frac{1}{T_{useful}}=\frac{1}{27\times10^{-6}}=37\ \text{kHz}

Number of subcarriers:

N=\frac{10\times10^6}{37{,}000}=270 \;\to\; \textbf{256} \text{ (power of two for the FFT)}

Recompute with 256: \Delta f = 10\ \text{MHz}/256=39.1 kHz, T_{useful}=25.6 µs.

Take $T_{cp}=3.2\ \mu$s (an eighth of T_{useful}, a common choice):

\text{efficiency}=\frac{25.6}{28.8}=88.9\%

Close enough, or use T_{cp}=1/16 = 1.6 µs if the delay spread allows, giving 94%.

Guard subcarriers. Use 200 of the 256 for data and pilots, leaving 56 at the band edges as a guard, which gives a usable bandwidth of 200\times39.1 = 7.8 MHz.

Data rate with 64-QAM and rate-3/4 coding, allowing 8 pilot subcarriers:

\frac{192\times6\times0.75}{28.8\ \mu\text{s}}=\frac{864}{28.8\times10^{-6}}=30\ \text{Mbit/s}

Check the frequency offset tolerance:

1\%\times39.1\ \text{kHz}=391\ \text{Hz}

At a 2 GHz carrier that is 0.2 ppm — demanding, requiring a temperature-compensated crystal and a carrier recovery loop.

And check Doppler. At 120 km/h (33 m/s) and 2 GHz:

f_d=\frac{v}{c}f=\frac{33}{3\times10^8}\times2\times10^9=220\ \text{Hz}

Within the 391 Hz budget, but only just, and combined with oscillator error it would be marginal. This is exactly why 5G defines wider subcarrier spacings for higher bands and higher mobility — the tolerance scales with the spacing.

Problem 8 — Shannon limit comparison

A satellite link has E_b/N_0 = 3 dB available. What is the maximum spectral efficiency, and which practical schemes work?

Solution

Solve \dfrac{E_b}{N_0}=\dfrac{2^\eta-1}{\eta} for \eta with E_b/N_0 = 2 (3 dB as a ratio).

Try \eta=2: (4-1)/2=1.5. Too low. Try \eta=2.5: (5.66-1)/2.5=1.86. Still low. Try \eta=2.8: (6.96-1)/2.8=2.13. Slightly high. Try \eta=2.7: (6.50-1)/2.7=2.04. Close.

\eta_{max}\approx2.65\ \text{bit/s/Hz}

Practical schemes at 3 dB:

• Uncoded QPSK needs 9.6 dB for 10^{-5} — 6.6 dB short. Will not work.
• QPSK with rate-1/2 turbo coding needs about 0.7 dB and gives \eta = 2\times0.5=1 bit/s/Hz. Works, with 2.3 dB margin.
• QPSK with rate-3/4 LDPC needs about 2.0 dB, giving \eta = 1.5. Works, 1 dB margin.
• 8-PSK with rate-2/3 LDPC needs about 4.5 dB, \eta=2. Does not work — 1.5 dB short.

Best practical choice: QPSK with rate-3/4 LDPC, achieving 1.5 bit/s/Hz against the theoretical 2.65.

The gap is 57% of capacity, which sounds poor until you note that the theoretical figure assumes infinite block length and infinite decoding complexity. Real systems with a 64,800-bit LDPC block — the DVB-S2 standard — get much closer, to about 2.2 bit/s/Hz at this SNR.

And DVB-S2 is worth naming as the concrete case. It offers 28 combinations of modulation and code rate, from QPSK rate-1/4 up to 32-APSK rate-9/10, and the satellite selects one per receiver based on measured conditions. The adaptive scheme delivers 30% more throughput than the fixed system it replaced, entirely by matching the scheme to the conditions rather than designing for the worst case.

Problem 9 — Antenna gain and link budget

Design a 20 km point-to-point link at 5.8 GHz for 100 Mbit/s. Transmit power 20 dBm, receiver noise figure 5 dB, bandwidth 40 MHz, required SNR 25 dB. Find the antenna sizes needed.

Solution

Path loss:

L=32.4+20\log_{10}(20)+20\log_{10}(5800)=32.4+26.0+75.3=133.7\ \text{dB}

Noise floor:

P_n=-174+10\log_{10}(40\times10^6)+5=-174+76+5=-93\ \text{dBm}

Required received power:

P_{rx}=-93+25=-68\ \text{dBm}

Required total antenna gain:

G_{tx}+G_{rx}=P_{rx}-P_{tx}+L=-68-20+133.7=45.7\ \text{dB}

Add margin. Rain fade at 5.8 GHz over 20 km is modest, perhaps 3 dB in heavy rain; add 5 dB for ageing, misalignment and multipath. Total needed: 54 dB, or 27 dBi each end.

Dish diameter for 27 dBi at \lambda = 5.17 cm, \eta = 0.6:

10^{2.7}=501=0.6\left(\frac{\pi D}{0.0517}\right)^2

\frac{\pi D}{0.0517}=\sqrt{835}=28.9 \;\Rightarrow\; D=0.476\ \text{m}

A 50 cm dish at each end.

Beamwidth:

\theta=\frac{70\times0.0517}{0.5}=7.2°

Alignment tolerance of a few degrees — achievable with a rigid mount and a signal strength meter.

Now the Fresnel check. First zone radius at midpath:

r_1=17.3\sqrt{\frac{20}{4\times5.8}}=17.3\sqrt{0.862}=16.1\ \text{m}

Sixteen metres of clearance needed at the midpoint, and 60% of that is 9.7 m as the practical minimum.

Plus earth curvature over 20 km:

h=\frac{d^2}{8\times6371\times10^3/1000}\ldots \text{ more simply } h=\frac{d_{km}^2}{12.75}\times\frac{1}{k}

With k=4/3 for standard refraction: h = 400/(12.75\times1.33)=23.6 m.

Total clearance required at midpath: about 34 m above the terrain. With 20 m masts at each end you have 20 m of straight-line height at midpath, which is not enough — the masts need to be taller, or the path needs higher ground at the ends.

This calculation is why microwave links sit on hills and towers, and why the terrain profile, not the electronics, usually decides whether a link is feasible.

Problem 10 — Transmission line reflections

A 3.3 V driver with 20 Ω output impedance drives a 15 cm track of 60 Ω characteristic impedance into a CMOS input. Analyse the ringing and design a termination.

Solution

Propagation time. FR4 gives about 6.7 ps/mm:

t_{prop}=150\times6.7=1005\ \text{ps}\approx1\ \text{ns}

Is it a transmission line? For an edge rate of 1 ns, the knee frequency is 350 MHz and \lambda/10 at that frequency is

\frac{c/\sqrt{4.2}}{350\times10^6\times10}=\frac{1.46\times10^8}{3.5\times10^9}=4.2\ \text{cm}

15 cm exceeds 4.2 cm, so yes — this must be treated as a transmission line.

The initial step. Source impedance and line impedance form a divider:

V_{initial}=3.3\times\frac{60}{20+60}=2.48\ \text{V}

At the load (CMOS input, essentially open, \Gamma=+1): the wave reflects and the voltage doubles to 4.95 V.

Overshoot of 1.65 V above the 3.3 V rail — enough to forward-bias the input protection diode and, repeated millions of times, to damage the input.

Back at the source, \Gamma_s = (20-60)/(20+60) = -0.5. The returning 2.48 V wave reflects as -1.24 V, travels back, doubles at the load to -2.48 V, bringing the load to 4.95-2.48=2.47 V.

Undershoot below the 3.3 V level, and potentially below the input threshold — a false low that could be clocked as data.

The ringing continues, each round trip 2 ns, decaying by half each time.

Series termination. Add R_s so that R_{source}+R_s = Z_0:

R_s=60-20=40\ \Omega \;\to\; \text{use } 39\ \Omega

Now the initial step is 3.3\times60/120=1.65 V — half the supply, which travels down the line. At the open load it doubles to 3.3 V — exactly right. The reflection returns to the source where \Gamma_s=0, so it is absorbed and nothing further happens.

Cost: the receiver sees only half the voltage for the first 1 ns, so any receiver connected partway along the track sees an invalid mid-level for that time. Series termination therefore works only for a single receiver at the far end.

For multiple loads along the line, use parallel termination at the far end: 60 Ω to a mid-rail supply, or a Thevenin pair. It costs 3.3/120 = 27 mA of continuous current per line, which for a 32-bit bus is 880 mA — usually unaffordable, and the reason series termination and point-to-point routing dominate modern design.

Problem 11 — Quarter-wave matching

Match a 300 Ω folded dipole to 75 Ω coaxial cable at 100 MHz using a quarter-wave transformer. Give the required impedance and physical length, and assess the bandwidth.

Solution

Z_0=\sqrt{Z_{in}Z_L}=\sqrt{75\times300}=150\ \Omega

Physical length. With a velocity factor of 0.66:

\lambda=\frac{3\times10^8}{100\times10^6}=3\ \text{m}, \qquad \lambda_{cable}=3\times0.66=1.98\ \text{m}

\ell=\frac{1.98}{4}=0.495\ \text{m}

Verify at 100 MHz. \beta\ell = \pi/2, so

Z_{in}=\frac{150^2}{300}=75\ \Omega \ ✓

Bandwidth check at 90 MHz. The line is now 0.9\times\lambda/4, so \beta\ell = 0.9\times\pi/2 = 1.414 rad, and \tan(1.414)=6.31:

Z_{in}=150\frac{300+j150(6.31)}{150+j300(6.31)}=150\frac{300+j946}{150+j1893}

Magnitudes: numerator 992\angle72.4°, denominator 1899\angle85.5°.

Z_{in}=150\times0.522\angle(-13.1°)=78.4\angle-13.1° = 76.4-j17.8\ \Omega

\Gamma=\frac{76.4-j17.8-75}{76.4-j17.8+75}=\frac{1.4-j17.8}{151.4-j17.8}=\frac{17.9\angle-85.5°}{152.4\angle-6.7°}=0.117

\text{SWR}=\frac{1.117}{0.883}=1.27

Acceptable. A single quarter-wave transformer typically holds SWR below 1.5 over about ±15% of the design frequency, which for FM broadcast (88–108 MHz, ±10%) is adequate.

For wider bandwidth, cascade two or three quarter-wave sections with intermediate impedances — a Chebyshev transformer — which trades a small ripple in SWR for a much wider band, exactly the same trade as the filter families of Chapter 5.2.

Practical note. 150 Ω coaxial cable is not a standard product. Two 75 Ω cables in series give 150 Ω but with awkward mechanics; the usual real solution is a printed transmission line on a circuit board, where any impedance is available by choosing the track width.

Problem 12 — Fresnel zone and link planning

A 15 km link at 2.4 GHz has a hill 6 km from one end that rises to 45 m. The antennas are at 60 m and 50 m above sea level, and the terrain is at sea level elsewhere. Does the link work?

Solution

Line-of-sight height at the hill. Interpolating between 60 m and 50 m over 15 km, at 6 km:

h_{LOS}=60-\frac{6}{15}(60-50)=60-4=56\ \text{m}

Clearance above the hill: 56-45=11 m.

Earth curvature bulge at 6 km from one end of a 15 km path, with k=4/3:

h_{bulge}=\frac{d_1d_2}{12.75k}=\frac{6\times9}{12.75\times1.33}=\frac{54}{16.96}=3.2\ \text{m}

Effective clearance: 11-3.2=7.8 m.

First Fresnel zone radius at that point:

r_1=17.3\sqrt{\frac{d_1d_2}{f_{GHz}\times d_{total}}}=17.3\sqrt{\frac{6\times9}{2.4\times15}}=17.3\sqrt{1.5}=21.2\ \text{m}

Required 60% clearance: 0.6\times21.2 = 12.7 m.

Available: 7.8 m. The link fails the Fresnel criterion.

How much loss? At 37% of the first Fresnel radius (7.8/21.2), diffraction loss is roughly 6 to 8 dB compared with free space.

Options, in order of cost:

1. Raise one antenna. To gain 5 m of clearance at the 6 km point, raise the 60 m end by 5\times15/9=8.3 m — so a 68 m mast. Straightforward if the structure allows.
2. Move the site to where the terrain is more favourable.
3. Accept the 7 dB loss and add it to the link budget. If the margin can absorb it, this is the cheapest answer — but the loss varies with atmospheric conditions, since k changes, and on a night with strong temperature inversion the effective earth curvature changes and the loss can be much worse.
4. Use a lower frequency. At 900 MHz the Fresnel radius is 17.3\sqrt{54/(0.9\times15)}=34.6 m, which is worse — lower frequency means a fatter Fresnel zone.

That last point is the counter-intuitive one worth stating: lower frequencies need more clearance, not less. They diffract better around obstacles, which partly compensates, but the Fresnel requirement itself grows.

Recommendation: raise the mast. It is a one-off cost and it makes the link deterministic rather than weather-dependent.

Problem 13 — CDMA capacity

A CDMA system has a 1.25 MHz bandwidth, a 9.6 kbit/s data rate per user, and each user requires 7 dB of E_b/N_0. Estimate how many users share one cell.

Solution

Processing gain:

G_p=\frac{1.25\times10^6}{9600}=130 = 21.1\ \text{dB}

With N users, each user sees N-1 interferers. Treating them as noise:

\frac{E_b}{N_0}=\frac{G_p}{N-1}

Required E_b/N_0 = 7 dB = 5.01:

5.01=\frac{130}{N-1} \;\Rightarrow\; N-1=26 \;\Rightarrow\; N=27\ \text{users}

Now apply the real-world factors.

Voice activity. People speak about 40% of the time in a conversation, and a CDMA system stops transmitting during silence. Capacity multiplies by 1/0.4 = 2.5:

N=27\times2.5=67

Sectorisation. Three 120° sectors, each with its own antenna, roughly triples capacity, though with about 15% loss from overlap:

N=67\times3\times0.85=171

Interference from other cells typically adds 60% to the interference, so divide by 1.6:

N=\frac{171}{1.6}=107\ \text{users per cell}

About a hundred simultaneous calls per cell site.

The property that distinguishes CDMA from TDMA: there is no hard limit. Add a 108th user and everyone's quality degrades slightly rather than the call being blocked. This is soft capacity, and it is why CDMA networks handle sudden load surges better — a stadium emptying degrades quality rather than blocking calls entirely.

And the near-far problem's cost. All of the above assumes perfect power control. With 1 dB of power control error, capacity drops about 20%, which is why the control loop runs 1500 times per second and why it was one of the hardest parts of the system to build.

Problem 14 — Diagnose a failing link

A 5 GHz point-to-point link worked at 40 dB SNR when installed. Six months later it shows 22 dB, with occasional dropouts. Nothing has been changed. Diagnose it.

Solution

18 dB lost. That is a factor of 63 in power, so something substantial has changed.

Work through the candidates by what would produce that number.

Antenna misalignment. With a 3° beamwidth, being 3° off costs 3 dB, 6° off costs perhaps 15 dB. 18 dB means roughly 7° of error — plausible from a loosened mount, wind damage or a settling pole.

Test: re-aim and watch the level. This is the first thing to check, because it is free.

Water in the feeder. Water ingress into coaxial cable raises the loss enormously and progressively. 18 dB over 30 m of cable is entirely consistent with a cracked connector boot letting rain in.

Test: time-domain reflectometry (Chapter 7.7). Water produces a negative reflection at the ingress point, and the distance calculation locates it.

A new obstruction. A tree has grown into the Fresnel zone, or a building has gone up. Trees are the classic case, since foliage attenuates strongly at 5 GHz and gets worse in leaf and worse again when wet.

Test: does the level vary with wind and rain? A link that is worse when the trees are wet and moving is diagnostic.

New interference. Another 5 GHz system nearby. Note that this would reduce SNR without reducing signal level, which distinguishes it immediately.

Test: measure the received signal level separately from SNR. If the signal is unchanged and the SNR has dropped, it is interference, not loss.

Equipment degradation. A failing power amplifier or a drifting low-noise amplifier.

Test: measure the transmitter's output power directly at the connector.

The diagnostic order, cheapest first:

1. Read the transmit power and the receive signal level separately. This one measurement splits the problem in half: signal down means loss, signal unchanged means interference or receiver degradation.
2. Inspect connectors and boots for water, and check the cable for a positive weight change.
3. Run TDR on the feeder.
4. Look along the path for new obstructions, ideally at the time of day when the fade is worst.
5. Spectrum analyse the band for new signals.

The most probable answer, and the one that fits "worked for six months then degraded gradually with occasional dropouts": water ingress into the feeder, or a tree in leaf. Both are progressive rather than sudden, both worsen in rain, and both are extremely common.

The lesson worth carrying. The dropouts are the informative symptom. A steady 18 dB loss suggests a fixed cause; intermittent dropouts on top of it suggest something weather-dependent — which points at water or foliage rather than at alignment, since a loosened mount would give a stable, if wrong, level.

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Part 8 follows all of this into the systems you carry: how a phone actually reaches a network, how GPS resolves your position from four satellites, and what happens in the 100 milliseconds when you tap a card.