Exercise set

Telecommunications Link Design Exercises

Practice telecom link design problems for RF margin, EIRP, noise floor, C/N0, optical budgets, dispersion, SINR, rain fade and guard bands.

These exercises practise telecommunications link design calculations. The goal is not only to close a link on paper. The goal is to make power, noise, bandwidth, timing, radar range, uncertainty, and validation assumptions visible enough that an engineer can defend the design in review.

Assume simplified screening models unless an exercise states otherwise. Real telecommunications systems also require regulatory review, measured spectrum occupancy, calibrated instruments, connector inspection, antenna alignment records, weather and fading assumptions, traffic loading, firmware configuration, and acceptance-test evidence.

Release Evidence Notes

Use these exercises as screening evidence for link-release decisions, not as proof that a carrier or service is ready by budget closure alone. A credible release should connect each calculation to the physical reference plane, traffic requirement, channel condition, implementation allowance, measured impairment, acceptance threshold and operational response.

The minimum evidence set is:

  • budget evidence for transmit power reference, EIRP, antenna gain, feeder loss, connector state, path loss basis, Fresnel clearance, pointing, polarization, radome or enclosure loss and uncertainty allowance;
  • receiver evidence for sensitivity basis, noise bandwidth, noise figure, implementation loss, C/N0 or Eb/N0 target, SINR, adjacent-channel rejection, overload risk and modem threshold;
  • optical and physical-layer evidence for source power, receiver overload, connector loss, splice loss, dispersion, return loss, OTDR or OLTS trace, route length, patch state and spare margin;
  • service evidence for committed data rate, latency, jitter, packet loss, availability target, route diversity, protection switching behavior, rain-fade policy, degraded MCS capacity and failover authority;
  • validation evidence for instrument calibration, reference plane, spectrum occupancy, weather basis, field alignment record, traffic-load test, outage accounting, alarm thresholds, rollback rule and release signoff.

Treat a numerical pass as provisional when it depends on clear-sky received power, visual line of sight without Fresnel clearance, vendor sensitivity without bandwidth basis, optical margin without connector evidence, average SINR, unverified route diversity or availability counted only as carrier-up time. Link budgets, optical budgets, radar range, rain-fade and service-assurance decisions should support release only when the evidence covers the same installed path and service target.

How to Use These Exercises

For each problem, define:

  1. the physical boundary where power, bandwidth, or latency is measured;
  2. whether values are absolute units such as dBm and dBW or ratios such as dB;
  3. the required service condition: data rate, error rate, latency, jitter, availability, or margin;
  4. the environmental and implementation allowances;
  5. the field measurement that would confirm the calculation.

The most common mistake is to calculate one impressive number and treat it as the whole design. A credible telecommunications review connects received power, receiver noise, modulation, coding, bandwidth, interference, latency, and operational margin.

Engineering Boundary Notes

Telecommunications link design is a cross-boundary review. RF power, optical power, receiver noise, bandwidth, modulation, coding, latency, jitter, availability and legal compliance are connected, but each has a different reference plane and acceptance criterion. A defensible result states which boundary is being released.

Power-budget closure alone is not service release. A link may have received-power margin while failing EIRP compliance, Fresnel clearance, receiver overload, dispersion, adjacent-channel rejection, SINR, rain availability, protection-switching hit time or delay-asymmetry requirement. Use these exercises as a way to expose the governing limit, not to collapse every impairment into one margin.

The hub-level calculation should also defer to specialist evidence when needed. Fiber trace evidence, RF field measurements, MCS behavior, packet latency tests and route-diversity drills are stronger than a generic budget when the release decision depends on those domains. The link-design page should make the integration boundary visible.

Common Release Mistakes

  • mixing dB margins from RF, optical, modem and service layers without a common reference plane;
  • treating legal EIRP correction as harmless when it consumes receiver or fade margin;
  • accepting an optical or RF path from budget closure without connector, Fresnel, alignment or measurement evidence;
  • using vendor sensitivity without bandwidth, implementation loss, modulation and coding basis;
  • counting availability from carrier-up time while rain fade, protection switching or packet loss affects service;
  • using average SINR or latency when the release criterion is tail behavior, outage or degraded MCS capacity;
  • approving a link design without instrument calibration, uncertainty allowance and rollback rule.

Scenario Map

ScenarioMain calculationEngineering decision
Clear-path RF budgetfree-space path loss and received powerDecide whether the link meets fade-margin requirements.
Regulatory RF releaseEIRP limit, conducted power reduction and legal link marginDecide whether a link remains both legal and robust after compliance correction.
Receiver sensitivitynoise bandwidth, noise figure and required SNRCheck whether bandwidth and implementation loss are included.
Satellite or long pathC/N0 and Eb/N0Separate carrier quality from bit-rate demand.
Optical physical layersensitivity, overload and chromatic-dispersion marginRelease only if the receiver is neither starved, saturated nor timing-limited by dispersion.
Spectrum/capacitysymbol rate, occupied bandwidth and Shannon sanity checkConfirm that payload, coded and occupied rates are not confused.
Radar detectionmonostatic radar equation and range marginCheck whether received echo power clears the detection threshold with guarded margin.
Service assurancelatency, jitter, SINR, availability, protection switching, delay asymmetry and uncertaintyDecide whether the delivered service, not only the carrier, passes.
Weather and adaptive-rate servicerain fade, residual SNR and MCS capacityDecide whether the committed service remains available during degraded RF conditions.
Path clearancefirst Fresnel radius and clearance fractionDecide whether an obstruction leaves enough RF clearance.
Installation lossespointing and polarization lossesCheck whether field alignment uncertainty consumes release margin.

Validation Package Checklist

Before treating a link-design result as release evidence, collect:

  1. service requirement, path identifier, physical boundary and reference plane;
  2. transmit power, EIRP, antenna, feeder, connector, optical or RF loss basis;
  3. receiver sensitivity, bandwidth, noise figure, implementation loss and overload limit;
  4. modulation, coding, SINR, C/N0, Eb/N0, spectral and regulatory constraints;
  5. Fresnel, pointing, polarization, dispersion, rain, fading and route-diversity assumptions;
  6. latency, jitter, packet loss, protection switching and availability evidence;
  7. field measurements, calibration record, uncertainty allowance and acceptance threshold;
  8. release decision, degraded-mode rule, rollback action or further specialist validation.

A point-to-point radio link operates at 5.8\ \text{GHz} over a line-of-sight distance of 12\ \text{km}. The transmitter output power is 24\ \text{dBm}. Transmit feeder loss is 2.0\ \text{dB}, transmit antenna gain is 23\ \text{dBi}, receive antenna gain is 23\ \text{dBi}, receive feeder loss is 1.5\ \text{dB}, and miscellaneous loss is 2.5\ \text{dB}.

Receiver sensitivity for the selected modulation and coding mode is -77\ \text{dBm}. The design requires at least 15\ \text{dB} margin after the budget. Estimate received power and decide whether the link passes the margin requirement.

Solution

Use the free-space path loss formula with distance in kilometers and frequency in MHz:

L_{fs,dB}=32.44+20\log_{10}(d_{km})+20\log_{10}(f_{MHz})

Here:

d=12\ \text{km}
f=5800\ \text{MHz}

Therefore:

L_{fs}=32.44+20\log_{10}(12)+20\log_{10}(5800)
L_{fs}=32.44+21.58+75.27=129.29\ \text{dB}

The received power is:

P_r=P_t-L_{tx}+G_t+G_r-L_{fs}-L_{rx}-L_{misc}
P_r=24-2.0+23+23-129.29-1.5-2.5
P_r=-65.29\ \text{dBm}

Available margin over receiver sensitivity:

M=P_r-P_{sens}
M=-65.29-(-77)=11.71\ \text{dB}

The link has about 11.7\ \text{dB} margin, which is below the required 15\ \text{dB}. It misses the requirement by:

15-11.71=3.29\ \text{dB}

Engineering Comment

The link closes against receiver sensitivity, but it does not meet the design margin requirement. Corrective options include higher antenna gain, shorter feeder runs, lower-loss cable, reduced data-rate mode, improved antenna height or alignment, or a different frequency/path plan. Increasing transmitter power may be limited by regulation and by adjacent-channel coexistence.

Plausibility Check

At 5.8\ \text{GHz} and 12\ \text{km}, a path loss near 129\ \text{dB} is plausible. Because both antennas are directional, the budget can still close, but only if installation and alignment preserve the assumed gains and losses.

Exercise 2: Receiver Noise Floor and Required Sensitivity

A receiver uses a 5.0\ \text{MHz} noise bandwidth. Receiver noise figure is 6.0\ \text{dB}. The selected waveform requires 13\ \text{dB} signal-to-noise ratio at the detector, and the implementation allowance is 2.0\ \text{dB}. Estimate the required received signal level. If the measured received power is -82.5\ \text{dBm}, find the operating margin.

Solution

The thermal noise density at room temperature is approximately:

-174\ \text{dBm/Hz}

Noise power over bandwidth is:

N_{dBm}=-174+10\log_{10}(B_{Hz})+NF

With:

B=5.0\times 10^6\ \text{Hz}
10\log_{10}(5.0\times 10^6)=66.99\ \text{dB}

So:

N=-174+66.99+6.0=-101.01\ \text{dBm}

The required received signal level is:

P_{req}=N+SNR_{req}+M_{impl}
P_{req}=-101.01+13+2=-86.01\ \text{dBm}

Measured operating margin:

M=P_{measured}-P_{req}
M=-82.5-(-86.01)=3.51\ \text{dB}

The receiver has about 3.5\ \text{dB} operating margin under the measured condition.

Engineering Comment

The calculation shows why bandwidth is part of receiver sensitivity. If the noise bandwidth doubles, the noise floor rises by about 3\ \text{dB} before any change in noise figure or modulation. A receiver sensitivity number without bandwidth, waveform, target error rate, and implementation loss is incomplete.

Plausibility Check

A 5\ \text{MHz} room-temperature noise floor near -107\ \text{dBm} before noise figure and near -101\ \text{dBm} after 6\ \text{dB} noise figure is reasonable. The 3.5\ \text{dB} margin is useful but not generous for an outdoor or interference-prone link.

Exercise 3: Carrier-to-Noise Density and Bit-Energy Margin

A satellite terminal uplink has effective isotropic radiated power:

EIRP=43\ \text{dBW}

The total propagation and implementation loss to the receiver is 198\ \text{dB}. The receiver figure of merit is:

G/T=9\ \text{dB/K}

Use:

k=-228.6\ \text{dBW/K/Hz}

The transmitted bit rate is 6.0\ \text{Mbit/s}. The modem requires E_b/N_0=10.5\ \text{dB} for the target error rate. Find available C/N_0, available E_b/N_0, and link margin.

Solution

Carrier-to-noise density in dB-Hz is:

\left(C/N_0\right)_{dBHz}=EIRP-L+\left(G/T\right)-k

Substitute values:

\left(C/N_0\right)=43-198+9-(-228.6)
\left(C/N_0\right)=82.6\ \text{dB-Hz}

Convert bit rate to dB-Hz:

10\log_{10}(R_b)=10\log_{10}(6.0\times10^6)=67.78\ \text{dB-Hz}

Available bit-energy ratio:

\left(E_b/N_0\right)_{available}=\left(C/N_0\right)-10\log_{10}(R_b)
\left(E_b/N_0\right)_{available}=82.6-67.78=14.82\ \text{dB}

Link margin against the modem requirement:

M=14.82-10.5=4.32\ \text{dB}

The link has about 4.3\ \text{dB} bit-energy margin before any additional reserved weather or interference allowance.

Engineering Comment

C/N_0 is useful because it separates carrier quality from bit rate. A higher bit rate consumes bit-energy margin even if received carrier power is unchanged. If the service requires an extra 3\ \text{dB} rain or interference allowance, the remaining margin would fall to about 1.3\ \text{dB}.

Plausibility Check

The result is dimensionally consistent: dBW, dB loss, dB/K, and Boltzmann’s constant in dBW/K/Hz produce dB-Hz. Subtracting bit rate in dB-Hz produces E_b/N_0 in dB.

Exercise 4: Optical Fiber Power Budget and Receiver Overload

A fiber-optic link uses a transmitter with average launched power:

P_{tx}=+1.0\ \text{dBm}

The receiver sensitivity is -18\ \text{dBm} and receiver overload occurs above -3\ \text{dBm}. The route has:

  • fiber length 18\ \text{km};
  • attenuation 0.35\ \text{dB/km};
  • eight connectors at 0.35\ \text{dB} each;
  • four splices at 0.10\ \text{dB} each;
  • wavelength filter loss 1.2\ \text{dB};
  • required design margin 3.0\ \text{dB} for aging, repairs, cleaning uncertainty, and future patching.

Find received optical power, available margin after design allowance, and overload margin.

Solution

Fiber attenuation:

L_{fiber}=0.35(18)=6.30\ \text{dB}

Connector loss:

L_{conn}=8(0.35)=2.80\ \text{dB}

Splice loss:

L_{splice}=4(0.10)=0.40\ \text{dB}

Total path loss before design margin:

L_{path}=6.30+2.80+0.40+1.20=10.70\ \text{dB}

Received optical power:

P_{rx}=P_{tx}-L_{path}
P_{rx}=1.0-10.70=-9.70\ \text{dBm}

Available margin after design allowance:

M=P_{rx}-P_{sens}-M_{design}
M=-9.70-(-18)-3.0=5.30\ \text{dB}

Overload margin:

M_{overload}=P_{overload}-P_{rx}
M_{overload}=-3-(-9.70)=6.70\ \text{dB}

The link has 5.3\ \text{dB} margin after the design allowance and is 6.7\ \text{dB} below overload.

Engineering Comment

This is a credible optical budget because it checks both weak-signal sensitivity and strong-signal overload. A short link with a high-power transmitter can fail by saturating the receiver, while a long or dirty link can fail by insufficient received power. Acceptance testing should record wavelength, reference method, connector condition, measured insertion loss, received power, and margin.

Plausibility Check

The fiber loss is not the whole budget. Connectors and filters consume 4.0\ \text{dB} here, which is more than half of the fiber attenuation. That is realistic in patch-heavy plants and data-center interconnects.

Exercise 5: Occupied Bandwidth and Capacity Sanity Check

A digital radio service must deliver 80\ \text{Mbit/s} payload throughput. It uses a code rate of 3/4, protocol and pilot overhead of 15\%, and 16-QAM modulation. Assume each 16-QAM symbol carries 4 coded bits before overhead. A raised-cosine pulse shape uses roll-off factor \alpha=0.25.

  1. Estimate gross coded bit rate.
  2. Estimate symbol rate.
  3. Estimate occupied bandwidth.
  4. Check whether a 40\ \text{MHz} channel has enough bandwidth.
  5. Use Shannon capacity as an ideal sanity check for the 80\ \text{Mbit/s} payload in 40\ \text{MHz}.

Solution

Gross coded bit rate including code rate and overhead:

\displaystyle R_{gross}=\frac{R_{payload}}{R_c}(1+\alpha_{oh})
\displaystyle R_{gross}=\frac{80}{0.75}(1.15)=122.7\ \text{Mbit/s}

For 16-QAM:

k=\log_2(16)=4\ \text{bits/symbol}

Symbol rate:

\displaystyle R_s=\frac{R_{gross}}{k}=\frac{122.7}{4}=30.7\ \text{Msymbol/s}

Occupied bandwidth approximation:

B\approx R_s(1+\alpha)
B\approx30.7(1.25)=38.3\ \text{MHz}

A 40\ \text{MHz} channel leaves:

40-38.3=1.7\ \text{MHz}

This is about:

\displaystyle \frac{1.7}{40}=4.3\%

of the channel width.

For an ideal Shannon check using payload rate:

C=B\log_2(1+SNR)

Rearrange:

SNR=2^{C/B}-1

With C/B=80/40=2.0\ \text{bit/s/Hz}:

SNR=2^2-1=3

In dB:

SNR_{dB}=10\log_{10}(3)=4.8\ \text{dB}

Engineering Comment

The occupied-bandwidth calculation is more restrictive than the ideal payload capacity check. The service fits a 40\ \text{MHz} channel only narrowly, and the result does not include spectral mask guard bands, transmitter filtering, oscillator error, adjacent-channel leakage, implementation loss, or fading margin. Shannon capacity is a lower-bound sanity check on required SNR, not a modem acceptance rule.

Plausibility Check

The payload spectral efficiency is 2.0\ \text{bit/s/Hz}, but the gross coded rate consumes about 3.07\ \text{bit/s/Hz} before roll-off is considered. This explains why engineering bandwidth estimates must distinguish payload, coded, symbol, and occupied-bandwidth rates.

Exercise 6: One-Way Latency and Jitter Acceptance

A telecommunications service crosses a 40\ \text{km} fiber route and a packet radio modem. Use refractive index n=1.47 for the fiber, so propagation speed is approximately c/n. The path includes:

  • two switches with 8\ \mu\text{s} processing delay each;
  • radio modem interleaving delay 1.2\ \text{ms};
  • one 1500 byte packet serialized at 100\ \text{Mbit/s};
  • 95th percentile queueing delay 0.70\ \text{ms};
  • encryption processing delay 0.15\ \text{ms}.

The service requirement is one-way 95th percentile latency below 4.0\ \text{ms} and peak-to-peak packet-delay jitter below 1.0\ \text{ms}. During acceptance testing, measured one-way delay ranges from 2.0\ \text{ms} to 3.1\ \text{ms}. Check latency and jitter acceptance.

Solution

Fiber propagation speed:

\displaystyle v=\frac{c}{n}=\frac{3.0\times10^8}{1.47}=2.04\times10^8\ \text{m/s}

Propagation delay:

\displaystyle t_p=\frac{40,000}{2.04\times10^8}=1.96\times10^{-4}\ \text{s}
t_p=0.196\ \text{ms}

Switch processing delay:

t_{switch}=2(8\ \mu\text{s})=16\ \mu\text{s}=0.016\ \text{ms}

Serialization delay for one 1500 byte packet:

N=1500(8)=12,000\ \text{bits}
\displaystyle t_s=\frac{12,000}{100\times10^6}=1.20\times10^{-4}\ \text{s}=0.120\ \text{ms}

Total one-way 95th percentile estimate:

t_{oneway}=t_p+t_{switch}+t_{interleave}+t_s+t_{queue,95}+t_{encryption}
t_{oneway}=0.196+0.016+1.200+0.120+0.700+0.150
t_{oneway}=2.382\ \text{ms}

Latency margin:

M_t=4.0-2.382=1.618\ \text{ms}

The 95th percentile latency estimate passes with about 1.62\ \text{ms} margin.

Peak-to-peak jitter from the acceptance record:

J_{pp}=t_{max}-t_{min}
J_{pp}=3.1-2.0=1.1\ \text{ms}

The jitter requirement is 1.0\ \text{ms}, so the jitter test fails by:

1.1-1.0=0.1\ \text{ms}

Engineering Comment

The service passes one-way latency but fails jitter. This is a realistic outcome: average or percentile latency can be acceptable while variation is too high for voice, synchronization, control, or measurement services. Corrective actions may include queue shaping, lower interleaver depth, traffic isolation, QoS verification, smaller packets, clock review, or removing bursty background traffic from the same service class.

Plausibility Check

Fiber propagation over 40\ \text{km} contributes only about 0.2\ \text{ms} one way. The dominant terms are modem interleaving and queueing. That means a route-shortening project would not fix the actual timing problem unless the physical distance were much larger.

Exercise 7: SINR with Co-Channel Interference

A receiver measures desired carrier power of (-70\ \text{dBm}). The thermal noise floor over the channel is (-96\ \text{dBm}), and a co-channel interferer is measured at (-82\ \text{dBm}). The selected modulation requires (18\ \text{dB}) SINR. Calculate combined interference-plus-noise power, SINR, and SINR deficit.

Solution

Compare noise power with interference power:

\displaystyle \frac{N}{I}=10^{(-96-(-82))/10}=10^{-1.4}=0.0398

The combined interference-plus-noise power is:

I+N=I(1+0.0398)=1.0398I

In dB relative to the interferer:

10\log_{10}(1.0398)=0.17\ \text{dB}

Therefore:

P_{I+N}=-82+0.17=-81.83\ \text{dBm}

SINR is:

SINR=P_C-P_{I+N}
SINR=-70-(-81.83)=11.83\ \text{dB}

SINR deficit:

18-11.83=6.17\ \text{dB}

Engineering Comment

The link may look acceptable against thermal noise alone, but the interferer dominates the impairment. Corrective options include frequency coordination, antenna pattern changes, filtering, site shielding, transmit-power coordination, scheduling, or a lower-order modulation mode.

Plausibility Check

Because the interferer is (14\ \text{dB}) stronger than the noise floor, adding noise changes the interference level by only (0.17\ \text{dB}). The SINR is therefore controlled almost entirely by co-channel interference.

Exercise 8: Availability Outage Budget

A committed service requires (99.99%) monthly availability. Use a 30 day month. Predicted rain fade and capacity fallback make the service unavailable for (38\ \text{min}) in the month, even though the radio carrier remains synchronized. Calculate the outage budget, actual service availability, and failure margin.

Solution

Total minutes in the month:

T=30(24)(60)=43200\ \text{min}

Allowed unavailable time:

T_{\text{out,allow}}=(1-0.9999)(43200)=4.32\ \text{min}

Actual service availability:

\displaystyle A=1-\frac{38}{43200}=0.999120

or:

A=99.9120\%

Outage excess:

38-4.32=33.68\ \text{min}

Outage ratio:

\displaystyle \frac{38}{4.32}=8.80

Engineering Comment

The service fails the availability objective by a wide margin. Carrier-up status is not enough when adaptive modulation or fading drops capacity below the committed service boundary.

Plausibility Check

A (99.99%) monthly target allows only a few minutes of outage. Tens of minutes of fallback or rain impairment can therefore be a major SLA failure even if users describe the link as mostly up.

Exercise 9: Route Diversity Availability

Two independent backhaul routes are available. Route A has availability (99.90%), and route B has availability (99.80%). Assume failures are independent and the service is available if either route is up. Calculate combined availability and expected annual outage minutes.

Solution

Route unavailability values are:

U_A=1-0.9990=0.0010
U_B=1-0.9980=0.0020

For independent routes, combined unavailability is:

U_{combined}=U_AU_B
U_{combined}=0.0010(0.0020)=0.000002

Combined availability:

A_{combined}=1-0.000002=0.999998

or:

A_{combined}=99.9998\%

Annual outage minutes:

T_{\text{out,y}}=0.000002(365)(24)(60)=1.05\ \text{min/y}

Engineering Comment

The result is excellent only if the independence assumption is credible. Shared towers, power, ducts, weather exposure, routers, timing sources, maintenance windows or software configuration can create common-cause failures that destroy the calculated diversity benefit.

Plausibility Check

Multiplying unavailability values is appropriate only for independent failures. If both routes share a single aggregation switch or power feed, the combined availability is not represented by this calculation.

Exercise 10: Measurement Uncertainty Guard Band

A field acceptance test calculates (5.3\ \text{dB}) link margin after planned design allowances. The minimum release margin is (3.0\ \text{dB}). The expanded measurement uncertainty is (1.2\ \text{dB}). Later, an unverified patch-panel condition could add (1.5\ \text{dB}) extra loss. Calculate guarded margin before and after the unverified loss, and state the release decision.

Solution

Guarded margin before the unverified loss:

M_g=5.3-1.2=4.1\ \text{dB}

Release margin above the minimum:

4.1-3.0=1.1\ \text{dB}

Guarded margin after the possible extra patch loss:

M_{g,\text{patch}}=5.3-1.2-1.5=2.6\ \text{dB}

Deficit relative to release requirement:

3.0-2.6=0.4\ \text{dB}

Engineering Comment

The link passes before the unverified patch condition but fails the guarded release check if that extra loss is credible. Acceptance should either inspect and measure the patch path, reserve additional margin, or release the service with a documented restriction.

Plausibility Check

A (0.4\ \text{dB}) deficit is small numerically but important in acceptance work because it appears after design allowances and measurement uncertainty. Guard bands are meant to prevent marginal links from being accepted on optimistic measurements.

Exercise 11: Fresnel Clearance Margin

A microwave path operates at:

f=18\ \text{GHz}

over a total path length:

D=22\ \text{km}

The controlling obstruction is near the midpoint, so:

d_1=11\ \text{km},\quad d_2=11\ \text{km}

Use the first-Fresnel-zone approximation:

\displaystyle r_1=17.32\sqrt{\frac{d_1d_2}{fD}}

where distances are in kilometers, frequency is in GHz and r_1 is in meters. The planning rule requires 60\% first-Fresnel clearance. The surveyed clearance above the obstruction is:

c=4.5\ \text{m}

Compute the required clearance, clearance margin and the additional height needed if only one end of the path can be raised.

Solution

First-Fresnel radius:

\displaystyle r_1=17.32\sqrt{\frac{11(11)}{18(22)}}
r_1=17.32\sqrt{0.3056}=9.57\ \text{m}

Required clearance:

c_{req}=0.60r_1=0.60(9.57)=5.74\ \text{m}

Clearance margin:

M_c=c-c_{req}=4.5-5.74=-1.24\ \text{m}

The path does not meet the 60\% clearance rule.

If only one endpoint is raised, the line height at the midpoint increases by half the endpoint height change. Required one-end height increase is therefore:

\Delta h_{one\ end}=2(1.24)=2.48\ \text{m}

If both endpoints can be raised equally, each endpoint would need about:

1.24\ \text{m}

of additional height.

Engineering Comment

Visual line of sight is not enough for a microwave release decision. The Fresnel screen should be tied to terrain data, rooftop equipment, future vegetation, antenna-height survey tolerance and measured received level. If the path has little RF margin, a small clearance deficit can explain modulation fallback even when cable and radio tests look normal.

Plausibility Check

At high microwave frequency the Fresnel radius is smaller than at lower frequencies, but a 22\ \text{km} path still has a radius near 10\ \text{m} at the midpoint. A 60\% rule therefore requires several meters of clearance, so a measured 4.5\ \text{m} clearance is credibly marginal.

Exercise 12: Pointing and Polarization Margin

A directional RF link has nominal margin before installation-loss allowances:

M_0=18.0\ \text{dB}

The release requirement is:

M_{req}=15.0\ \text{dB}

The antenna half-power beamwidth is:

\theta_{3dB}=4.0^\circ

Use the small-angle pointing-loss approximation:

\displaystyle L_p=12\left(\frac{\theta}{\theta_{3dB}}\right)^2

For normal conditions, expected pointing error is:

\theta=1.2^\circ

Polarization mismatch angle is:

\phi=12^\circ

Use:

L_{pol}=-20\log_{10}(\cos\phi)

Additional feeder and installation allowance is:

L_{inst}=0.8\ \text{dB}

Compute remaining margin. Then check a wind case where pointing error increases to 2.0^\circ.

Solution

Normal pointing loss:

\displaystyle L_p=12\left(\frac{1.2}{4.0}\right)^2=1.08\ \text{dB}

Polarization mismatch loss:

L_{pol}=-20\log_{10}(\cos12^\circ)=0.19\ \text{dB}

Total normal installation loss:

L_{total}=1.08+0.19+0.8=2.07\ \text{dB}

Remaining normal margin:

M_{normal}=18.0-2.07=15.93\ \text{dB}

Release margin above requirement:

15.93-15.0=0.93\ \text{dB}

The normal condition passes, but with less than 1\ \text{dB} spare margin.

Wind pointing loss:

\displaystyle L_{p,wind}=12\left(\frac{2.0}{4.0}\right)^2=3.00\ \text{dB}

Wind-case total loss:

L_{total,wind}=3.00+0.19+0.8=3.99\ \text{dB}

Wind-case remaining margin:

M_{wind}=18.0-3.99=14.01\ \text{dB}

Wind-case deficit:

15.0-14.01=0.99\ \text{dB}

The wind pointing case fails the release margin.

Engineering Comment

The nominal link has just enough margin, but the installation is sensitive to antenna movement. A release package should not hide pointing, polarization and feeder allowances inside one unexplained implementation loss. The corrective action could be mast stiffening, improved alignment, lower-gain wider-beam antenna, lower-rate mode, added fade margin or a service restriction for severe wind.

Plausibility Check

Pointing loss grows with the square of pointing error in this approximation. Increasing error from 1.2^\circ to 2.0^\circ raises pointing loss from about 1.1\ \text{dB} to 3.0\ \text{dB}. Since the normal spare margin is only 0.93\ \text{dB}, the wind case should fail.

Exercise 13: Timing-Service Delay Asymmetry Guard Band

A packet timing service is carried over a protected fiber route. A field test estimates one-way propagation and equipment delay as:

t_f=4.850\ \text{ms}

in the forward direction and:

t_r=4.970\ \text{ms}

in the reverse direction. The timing protocol assumes path symmetry unless an asymmetry correction is configured. For this simplified screen, the time-offset error from uncorrected asymmetry is:

\displaystyle e_t=\frac{t_r-t_f}{2}

The service requirement is:

|e_t|\leq1.5\ \mu\text{s}

An initial calibration applies asymmetry correction:

C_1=118\ \mu\text{s}

with expanded uncertainty:

U_{C1}=4\ \mu\text{s}

A later calibrated measurement proposes:

C_2=119.5\ \mu\text{s}

with expanded uncertainty:

U_{C2}=1.0\ \mu\text{s}

Use a guarded timing error:

\displaystyle e_{guard}=\left|\frac{\Delta t-C}{2}\right|+\frac{U_C}{2}

where:

\Delta t=t_r-t_f

Check the uncorrected case and both correction cases.

Solution

Path asymmetry is:

\Delta t=4.970-4.850=0.120\ \text{ms}

Convert to microseconds:

\Delta t=120\ \mu\text{s}

Uncorrected timing error:

\displaystyle e_{uncorrected}=\frac{120}{2}=60\ \mu\text{s}

The uncorrected service fails the 1.5\ \mu\text{s} requirement by:

60-1.5=58.5\ \mu\text{s}

For the first correction:

\displaystyle e_{res,1}=\left|\frac{120-118}{2}\right|=1.0\ \mu\text{s}

Uncertainty contribution:

\displaystyle \frac{U_{C1}}{2}=\frac{4}{2}=2.0\ \mu\text{s}

Guarded timing error:

e_{guard,1}=1.0+2.0=3.0\ \mu\text{s}

The first correction still fails the guarded release rule.

For the later correction:

\displaystyle e_{res,2}=\left|\frac{120-119.5}{2}\right|=0.25\ \mu\text{s}

Uncertainty contribution:

\displaystyle \frac{U_{C2}}{2}=\frac{1.0}{2}=0.50\ \mu\text{s}

Guarded timing error:

e_{guard,2}=0.25+0.50=0.75\ \mu\text{s}

The later calibrated correction passes the 1.5\ \mu\text{s} service requirement.

Engineering Comment

Latency and jitter acceptance do not prove timing accuracy. A symmetric two-way timing protocol can be wrong by half the path asymmetry even when packet delay looks stable. Timing-service release should record route symmetry, protection state, fiber pair assignment, equipment delay calibration, asymmetry correction, holdover behavior, switchovers and uncertainty evidence. A route-diversity event can change timing error if the new path has different asymmetry.

Plausibility Check

The forward and reverse delays differ by only 0.120\ \text{ms}, which looks small on a packet-latency scale. For a timing service, it equals 120\ \mu\text{s} of asymmetry and therefore 60\ \mu\text{s} of uncorrected time error. That is far above a 1.5\ \mu\text{s} requirement, so calibration uncertainty becomes release-critical.

Exercise 14: Rain-Fade MCS Capacity Gate

A microwave backhaul link has clear-sky received power:

P_{r,clear}=-55.0\ \text{dBm}

The receiver noise-plus-implementation reference for the channel is:

N=-91.0\ \text{dBm}

A screened rain event adds path attenuation:

A_{rain}=9.5\ \text{dB}

The high-capacity modulation and coding mode carries 180\ \text{Mbit/s} and requires:

SNR_{high,req}=28.0\ \text{dB}

The robust fallback mode carries 80\ \text{Mbit/s} and requires:

SNR_{robust,req}=18.0\ \text{dB}

The committed service capacity is:

R_{committed}=120\ \text{Mbit/s}

Calculate clear-sky SNR, rain SNR, remaining margin for each MCS in rain, and whether the committed service remains available during the screened rain event.

Solution

Clear-sky SNR:

SNR_{clear}=P_{r,clear}-N
SNR_{clear}=-55.0-(-91.0)=36.0\ \text{dB}

Clear-sky margin for the high-capacity mode:

M_{clear,high}=36.0-28.0=8.0\ \text{dB}

Received power during rain:

P_{r,rain}=P_{r,clear}-A_{rain}
P_{r,rain}=-55.0-9.5=-64.5\ \text{dBm}

Rain SNR:

SNR_{rain}=P_{r,rain}-N
SNR_{rain}=-64.5-(-91.0)=26.5\ \text{dB}

Rain margin for the high-capacity mode:

M_{rain,high}=26.5-28.0=-1.5\ \text{dB}

Rain margin for the robust mode:

M_{rain,robust}=26.5-18.0=8.5\ \text{dB}

Additional clear-sky margin needed to hold the high-capacity mode through the screened rain event:

M_{add}=28.0+9.5-36.0=1.5\ \text{dB}

The radio carrier can remain available in robust mode, but the committed service cannot remain available because:

80\ \text{Mbit/s}<120\ \text{Mbit/s}

Engineering Comment

The link is not a hard RF outage during the screened rain event, but it is a service outage for the committed capacity. That distinction should drive alarms, traffic shaping, SLA reporting and mitigation. Options include more antenna gain, lower-frequency path, shorter hop, route diversity, reduced committed rate during rain, or explicit priority traffic policy during fallback.

Plausibility Check

The clear-sky high-mode margin is 8.0\ \text{dB}, while the screened rain fade is 9.5\ \text{dB}. The high-capacity mode should therefore fail by about 1.5\ \text{dB}. The robust mode still has large SNR margin, so the carrier can stay up even though committed capacity is not met.

Exercise 15: Monostatic Radar Received Power and Range Margin

An X-band monostatic radar must detect a small target at a screened range of:

R=18\ \text{km}

The transmitter peak power is:

P_t=1.0\ \text{kW}

The antenna gain is:

G=34\ \text{dBi}

The operating frequency is:

f=9.4\ \text{GHz}

The target radar cross section is:

\sigma=2.0\ \text{m}^2

Combined implementation, atmospheric, processing and miscellaneous losses are:

L=6.0\ \text{dB}

The receiver detection threshold is:

P_{min}=-112\ \text{dBm}

and the release rule requires at least:

M_{req}=6.0\ \text{dB}

margin above that threshold. Estimate received echo power, raw detection margin, guarded margin and the maximum range that would meet the guarded margin if all other assumptions remain unchanged.

Solution

Convert transmitter power to dBW:

P_{t,dBW}=10\log_{10}(1000)=30.0\ \text{dBW}

Wavelength is:

\displaystyle \lambda=\frac{c}{f}
\displaystyle \lambda=\frac{3.0\times10^8}{9.4\times10^9}=0.0319\ \text{m}

The wavelength term is:

20\log_{10}(\lambda)=20\log_{10}(0.0319)=-29.9\ \text{dB}

Radar cross section in dBsm is:

\sigma_{dBsm}=10\log_{10}(2.0)=3.0\ \text{dBsm}

The monostatic radar equation in dB form is:

P_{r,dBW}=P_{t,dBW}+2G_{dBi}+20\log_{10}(\lambda)+\sigma_{dBsm}-30\log_{10}(4\pi)-40\log_{10}(R_m)-L

For R_m=18{,}000\ \text{m}:

30\log_{10}(4\pi)=33.0\ \text{dB}
40\log_{10}(18{,}000)=170.2\ \text{dB}

Substitute:

P_{r,dBW}=30.0+2(34)-29.9+3.0-33.0-170.2-6.0
P_{r,dBW}=-138.1\ \text{dBW}

Convert to dBm:

P_{r,dBm}=-138.1+30=-108.1\ \text{dBm}

Raw detection margin is:

M_{raw}=P_{r,dBm}-P_{min}
M_{raw}=-108.1-(-112)=3.9\ \text{dB}

Guarded release margin is:

M_{guard}=M_{raw}-M_{req}
M_{guard}=3.9-6.0=-2.1\ \text{dB}

The screened range fails the guarded detection rule.

Because monostatic radar received power scales as R^{-4}, range can be adjusted from the margin shortfall:

R_{allow}=R\,10^{M_{guard}/40}
R_{allow}=18(10^{-2.1/40})=16.0\ \text{km}

The same radar assumptions would meet the guarded rule only to about 16.0\ \text{km}, not 18\ \text{km}.

Engineering Comment

Radar range checks are extremely sensitive to distance because the outbound and return paths both matter. A small negative margin can imply a meaningful range reduction. The release review should state whether power is peak or average, whether antenna gain includes scan loss, how processing gain and pulse integration are counted, what radar cross section distribution is assumed, and whether clutter, false-alarm probability, receiver bandwidth, polarization, rain attenuation and target aspect are included.

The R^{-4} law also means that improving detection can come from several places: more transmit power, higher antenna gain, lower losses, better processing, lower detection threshold, shorter range or a larger target cross section. Those are not interchangeable operationally, so the calculation should end with a release or redesign decision rather than only a received-power number.

Plausibility Check

The raw echo is only 3.9\ \text{dB} above the receiver threshold, so it is plausible that a 6.0\ \text{dB} guarded release rule fails. The range correction is modest because the shortfall is about 2.1\ \text{dB} and radar power changes by 40\ \text{dB} per decade of range. Reducing range from 18\ \text{km} to about 16\ \text{km} is therefore consistent with the fourth-power range dependence.

Exercise 16: Protection Switching Hit Time and Buffer Margin

A protected backhaul service carries committed traffic at:

R=150\ \text{Mbit/s}

The protection-switching requirement allows service interruption no longer than:

T_{hit,max}=50\ \text{ms}

The equipment has a short-term packet buffer of:

B=1.0\ \text{MB}

Use decimal megabytes for this screen:

1.0\ \text{MB}=1{,}000{,}000\ \text{bytes}

The normal preconfigured protection path has:

  • failure detection time: 18\ \text{ms};
  • protection switch time: 22\ \text{ms};
  • forwarding reconfirmation time: 8\ \text{ms}.

Average packet payload is:

S_p=1200\ \text{bytes/packet}

Check the protection-switching hit time, data accumulated during the hit, approximate packet count and buffer margin. Then check a degraded mode where a controller waits for routing reconvergence, adding:

T_{conv}=35\ \text{ms}

Solution

Normal protection-switching hit time:

T_{hit}=18+22+8=48\ \text{ms}

Time margin:

M_T=50-48=2\ \text{ms}

The normal protection path meets the hit-time requirement, but only narrowly.

Convert hit time to seconds:

T_{hit}=0.048\ \text{s}

Data arriving during the hit is:

\displaystyle D_{hit}=\frac{RT_{hit}}{8}

with R in bit/s:

\displaystyle D_{hit}=\frac{150\times10^6(0.048)}{8}=900{,}000\ \text{bytes}

So:

D_{hit}=0.90\ \text{MB}

Approximate packet count:

\displaystyle N_p=\frac{900{,}000}{1200}=750\ \text{packets}

Buffer margin:

M_B=1.00-0.90=0.10\ \text{MB}

The normal protection mode passes both the hit-time and buffer screens, but with small margin.

For the degraded mode:

T_{hit,deg}=48+35=83\ \text{ms}

Hit-time margin:

M_{T,deg}=50-83=-33\ \text{ms}

Data arriving during the degraded hit:

\displaystyle D_{hit,deg}=\frac{150\times10^6(0.083)}{8}=1{,}556{,}250\ \text{bytes}

So:

D_{hit,deg}=1.56\ \text{MB}

Approximate degraded packet count:

\displaystyle N_{p,deg}=\frac{1{,}556{,}250}{1200}=1297\ \text{packets}

Degraded buffer margin:

M_{B,deg}=1.00-1.56=-0.56\ \text{MB}

The degraded mode fails both the hit-time requirement and the buffer screen. It should not be counted as protected-service availability for the committed traffic unless traffic is shaped, a larger buffer is provided, hitless switching is validated, or the service definition allows that interruption.

Engineering Comment

Route diversity is not enough by itself. A second path can exist and still fail the service requirement if detection, switching, forwarding confirmation or control-plane convergence takes too long. Availability accounting should distinguish carrier restoration, committed-rate restoration and packet-loss impact.

Release evidence should include a failure-injection test, timestamped packet capture, traffic rate, packet-size distribution, buffer occupancy, protection-controller state, alarm timestamps, route selected, clock behavior, QoS class and whether repeated switchovers or flapping links drain the buffer. If the service claims hitless or near-hitless protection, the measurement must be made at the service boundary, not only at the transport equipment port.

Plausibility Check

At 150\ \text{Mbit/s}, one millisecond carries 18{,}750 bytes. A 48\ \text{ms} hit therefore carries about 900{,}000 bytes, which fits inside a 1.0\ \text{MB} buffer with little spare room. Adding 35\ \text{ms} increases the hit to 83\ \text{ms} and pushes the required buffer above 1.5\ \text{MB}, so the degraded failure is expected.

Exercise 17: Fiber Chromatic Dispersion and Eye-Closure Margin

An optical route passes its received-power budget, but the service team is reviewing whether a (10\ \text{Gbit/s}) intensity-modulated link has enough dispersion margin.

The fiber length is:

L=42\ \text{km}

The chromatic dispersion coefficient is:

D=17\ \text{ps/(nm km)}

The transmitter spectral width is:

\Delta\lambda=0.12\ \text{nm}

Use the first-pass dispersion broadening estimate:

\Delta t_D=D\,L\,\Delta\lambda

The bit period, or unit interval, is:

\displaystyle UI=\frac{1}{R_b}

The release rule allows dispersion broadening no greater than:

0.35UI

Calculate dispersion broadening, the (10\ \text{Gbit/s}) unit interval, allowed broadening, and release margin. Then check a narrower transmitter with:

\Delta\lambda_{new}=0.04\ \text{nm}

Finally, check whether the original transmitter would pass the same dispersion fraction at:

R_b=2.5\ \text{Gbit/s}

Solution

Initial dispersion broadening:

\Delta t_D=17(42)(0.12)
\Delta t_D=85.7\ \text{ps}

For (10\ \text{Gbit/s}):

\displaystyle UI_{10G}=\frac{1}{10\times10^9}=100\ \text{ps}

Allowed broadening:

\Delta t_{allow}=0.35(100)=35\ \text{ps}

Dispersion margin:

M_D=35-85.7=-50.7\ \text{ps}

The original (10\ \text{Gbit/s}) configuration fails the dispersion screen even if optical received power is acceptable.

For the narrower transmitter:

\Delta t_{D,new}=17(42)(0.04)
\Delta t_{D,new}=28.6\ \text{ps}

New dispersion margin:

M_{D,new}=35-28.6=6.4\ \text{ps}

The narrower transmitter passes the simplified (10\ \text{Gbit/s}) dispersion screen.

For (2.5\ \text{Gbit/s}), the unit interval is:

\displaystyle UI_{2.5G}=\frac{1}{2.5\times10^9}=400\ \text{ps}

Allowed broadening at the same fraction is:

\Delta t_{allow,2.5G}=0.35(400)=140\ \text{ps}

Margin with the original transmitter at (2.5\ \text{Gbit/s}):

M_{D,2.5G}=140-85.7=54.3\ \text{ps}

A lower-rate mode passes the dispersion fraction, but it may not meet a committed (10\ \text{Gbit/s}) service requirement. The release decision should therefore distinguish physical-layer survivability from committed capacity.

Engineering Comment

Optical link design is not only received power. A link can pass sensitivity and overload checks while failing timing closure because chromatic dispersion broadens pulses relative to the unit interval. Corrective actions include a narrower-linewidth transmitter, dispersion compensation, shorter span, lower symbol rate, coherent receiver options or a different wavelength plan. Release evidence should include wavelength, fiber type, route length, spectral width, dispersion coefficient, receiver tolerance and measured eye or BER evidence.

Plausibility Check

The (42\ \text{km}) span and (0.12\ \text{nm}) source width produce about (86\ \text{ps}) of broadening, which is nearly a full (10\ \text{Gbit/s}) unit interval. That should fail a (35\ \text{ps}) allowance. Reducing spectral width by a factor of three reduces broadening by the same factor, to about (29\ \text{ps}), which explains why the narrower transmitter passes.

A fixed wireless link budget was prepared with conducted transmitter power:

P_t=27.0\ \text{dBm}

The transmit feeder loss is:

L_t=1.8\ \text{dB}

and the transmit antenna gain is:

G_t=24.0\ \text{dBi}

The licensed EIRP limit for the channel is:

EIRP_{limit}=47.0\ \text{dBm}

The original RF budget, before checking the EIRP limit, reported link margin over the required receiver threshold:

M_{raw}=7.0\ \text{dB}

The release rule requires at least:

M_{req}=5.0\ \text{dB}

Calculate current EIRP, regulatory excess, maximum legal conducted power, legal link margin after power reduction, and the remaining margin shortfall. Then check whether adding (1.0\ \text{dB}) of receive-side gain or reducing the required receiver threshold by (1.5\ \text{dB}) would pass without increasing EIRP.

Solution

EIRP is conducted transmitter power minus transmit feeder loss plus transmit antenna gain:

EIRP=P_t-L_t+G_t
EIRP=27.0-1.8+24.0=49.2\ \text{dBm}

Regulatory excess is:

E_{excess}=EIRP-EIRP_{limit}=49.2-47.0=2.2\ \text{dB}

The maximum legal conducted transmitter power is:

P_{t,max}=EIRP_{limit}+L_t-G_t
P_{t,max}=47.0+1.8-24.0=24.8\ \text{dBm}

So the transmitter must be reduced by:

\Delta P_t=27.0-24.8=2.2\ \text{dB}

For the same receive antenna, path and receiver threshold, reducing conducted power by (2.2\ \text{dB}) reduces received power and link margin by the same amount:

M_{legal}=M_{raw}-\Delta P_t=7.0-2.2=4.8\ \text{dB}

The release margin after legal correction is:

M_{release}=M_{legal}-M_{req}=4.8-5.0=-0.2\ \text{dB}

Therefore the current design is not releasable as documented. It either violates EIRP or falls just below the required legal link margin.

Adding receive-side gain does not increase transmit EIRP. With an extra (1.0\ \text{dB}) on the receive side:

M_{rx}=M_{legal}+1.0=5.8\ \text{dB}

Release margin becomes:

M_{rx,release}=5.8-5.0=0.8\ \text{dB}

That option passes the screen if the added gain is real at the installed reference plane.

Alternatively, lowering the required receiver threshold by (1.5\ \text{dB}) through a lower-rate or stronger-coding mode increases margin by the same amount:

M_{mode}=M_{legal}+1.5=6.3\ \text{dB}

Release margin becomes:

M_{mode,release}=6.3-5.0=1.3\ \text{dB}

That also passes, but only if the lower-rate mode still satisfies the committed service requirement.

Engineering Comment

Regulatory EIRP compliance is part of the link budget, not a paperwork check after the design is accepted. A link that closes only by exceeding the EIRP limit is not an engineering pass. Under an EIRP cap, increasing conducted power or transmit antenna gain may require reducing another transmit-side term, so receiver-side gain, shorter path, lower receiver threshold, interference reduction, better alignment, route diversity or a revised service rate may be the real corrective options. The release record should preserve conducted power, feeder loss, antenna gain, EIRP limit, measurement uncertainty, receiver reference plane and service-rate consequence.

Plausibility Check

The current EIRP is (49.2\ \text{dBm}), so it exceeds a (47.0\ \text{dBm}) limit by (2.2\ \text{dB}). Reducing transmitter power by that amount should reduce received margin by the same (2.2\ \text{dB}), moving the apparent (7.0\ \text{dB}) margin down to (4.8\ \text{dB}). The design is therefore only slightly short of a (5.0\ \text{dB}) legal margin, so a real (1.0\ \text{dB}) receive-side improvement plausibly restores release headroom.

Review Checklist

Before accepting a telecommunications link design, check:

  • whether all dB quantities have compatible references;
  • whether conducted transmit power, feeder loss, antenna gain and EIRP limit are reconciled before claiming RF margin;
  • whether receiver sensitivity includes bandwidth, noise figure, waveform, coding, and target error rate;
  • whether fade, rain, interference, aging, installation, and implementation margins are separated rather than counted twice;
  • whether optical budgets check sensitivity, overload, dispersion and connector evidence;
  • whether payload rate, coded rate, symbol rate, and occupied bandwidth are distinct;
  • whether SINR is checked when interference is credible;
  • whether Fresnel clearance and installation losses are checked before blaming rain fade or receiver sensitivity;
  • whether availability is counted at the committed service boundary;
  • whether adaptive-rate fallback is counted as unavailable when capacity falls below the committed service;
  • whether route-diversity assumptions are independent rather than shared;
  • whether protection-switching hit time, packet loss and buffer occupancy are measured at the committed service boundary;
  • whether measurement uncertainty is guarded before release;
  • whether radar calculations state peak or average power, RCS assumption, bandwidth, processing gain and false-alarm basis;
  • whether latency and jitter are measured under realistic load and failure states;
  • whether timing services check delay asymmetry, route protection state and calibration uncertainty;
  • whether field validation records enough context for later service assurance.

The calculation is finished only when the engineering decision is clear: pass, fail, derate, test again, or redesign with a stated reason.

Common Mistakes

  • Adding dBm, dBW and dB values without checking the reference, sign convention and whether a term is absolute power or a ratio.
  • Closing a link budget with transmit power or antenna gain that violates the applicable EIRP limit.
  • Counting the same margin twice, especially installation loss, implementation loss, rain fade, aging or measurement uncertainty.
  • Using receiver sensitivity from a narrower bandwidth or different coding mode than the one used by the service design.
  • Passing an RF power budget while ignoring SINR, adjacent-channel interference, desensitization, antenna alignment and spectrum occupancy.
  • Treating adaptive-rate fallback as service availability when the committed payload rate is no longer delivered.
  • Declaring route diversity independent when routes share towers, ducts, power, timing source, maintenance access or protection controller logic.
  • Counting a backup route as service availability without measuring protection-switching hit time, packet loss and buffer behavior.
  • Accepting an optical budget that passes sensitivity but violates receiver overload, connector cleanliness, dispersion, eye-margin or inspection requirements.
  • Blaming rain fade before checking Fresnel clearance, antenna pointing, polarization mismatch, tower movement and installation records.
  • Reporting latency or jitter from a quiet path while ignoring queueing, failover, protection switching, traffic shaping and load state.
  • Releasing timing service without guarding delay asymmetry, calibration uncertainty and path changes after protection switching.
  • Using radar range equations without stating peak versus average power, RCS assumption, clutter, false-alarm basis, processing gain and bandwidth.
REF

See also