Exercise set

Applied Electromagnetics and Wave Propagation Exercises

Worked electromagnetics exercises for wavelength, fields, impedance, EIRP, skin depth, path loss, waveguide cutoff and uncertainty.

Branch: Engineering Physics
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.1.7 · reviewed

These exercises practise applied electromagnetics and wave propagation as engineering physics. They focus on deciding which physical regime applies, calculating first-pass electromagnetic quantities, and identifying the evidence needed before trusting a model in hardware, field installation, or compliance testing.

The goal is not to memorize isolated formulas. The goal is to make the field path visible: where energy is stored, guided, radiated, reflected, absorbed, coupled, measured, or lost. A circuit, link, shield, antenna, waveguide, sensor, or optical path is credible only when the electromagnetic assumptions match the geometry and the validation method.

Assume simplified screening models unless an exercise states otherwise. Real systems should also check material tolerances, boundary conditions, connector geometry, surface finish, dielectric loss, cable routing, installation effects, temperature, moisture, nearby metal, instrument loading, calibration, and uncertainty.

How to Use These Exercises

For each exercise, define:

the physical boundary: source, path, receiver, material, enclosure, cable, antenna, or instrument;
the regime: lumped circuit, transmission line, guided wave, radiating wave, optical wave, or mixed regime;
the quantity that supports the engineering decision: field strength, received power, impedance, loss, delay, bandwidth, margin, or uncertainty;
the units and reference plane;
the validation evidence needed before release.

The common mistake is applying a familiar low-frequency circuit model after geometry, wavelength, rise time, return path, or boundary conditions have become the dominant physics.

Release Evidence Notes

Electromagnetic calculations should be tied to a physical boundary and regime. Record source, path, receiver, reference plane, frequency band, wavelength, rise time, material properties, geometry, grounding, shielding, cable route, antenna orientation, measurement bandwidth and calibration state. A field, impedance or gain result without that boundary is only a screening estimate.

Wave and coupling evidence should separate lumped, transmission-line, guided-wave, near-field and far-field assumptions. Wavelength, trace delay, reflection coefficient, waveguide cutoff, antenna far-field distance, enclosure aperture behavior and optical dispersion each depend on a different model boundary. A formula that is correct in one regime can be misleading in another.

Measurement evidence should include instrument and setup effects. Probe factor, cable loss, bandwidth, detector mode, chamber reflections, reference antenna, polarization, pointing, calibration date, fixture loading, uncertainty and environmental condition should be recorded before using a result for release. A narrow numerical pass can disappear when the measurement boundary changes.

Installation evidence should be treated as part of the electromagnetic system. Seams, connectors, shields, return paths, radomes, nearby metal, moisture, surface finish, cable bends, polarization mismatch, multipath and mounting hardware can dominate the result after installation. Catalogue or bench data should not be accepted without transfer evidence.

The practical release question is whether geometry, regime, materials, installation, measurement setup, uncertainty and functional performance all support the same action. If one layer conflicts, the result should trigger retest, model revision, shielding change, antenna alignment, reference-plane correction, margin increase or refusal to release the configuration.

Engineering Boundary Notes

These exercises use first-order electromagnetic screening models. They do not replace full-wave simulation, installed cable and enclosure testing, antenna range validation, EMC compliance testing, RF exposure assessment, thermal testing or calibrated optical link characterization. A calculated result is valid only for the stated geometry, frequency band, material state, boundary condition, reference plane and measurement setup.

Treat low-frequency circuit, transmission-line, guided-wave, radiating-wave and optical-wave models as different engineering boundaries. The same conductor, slot, connector or dielectric can move between regimes as frequency, edge rate, dimensions, installation and nearby metal change. Release evidence should state which model owns the decision and which adjacent regimes were screened out.

Common Release Mistakes

applying lumped-circuit intuition after wavelength, edge rate or return path controls the result;
using free-space path loss while ignoring antenna orientation, polarization, near-field distance or installation metalwork;
accepting shielding from material thickness alone without seams, apertures, bonds and cable penetrations;
comparing a measured emission or field value to a limit without bandwidth, detector mode, probe factor and uncertainty;
treating catalogue dielectric, antenna or cable data as installed evidence without transfer checks;
releasing an RF, EMC or optical change without reference-plane control and repeatable setup records.

Scenario Map

Scenario	Exercises	Primary check	Engineering decision
Wave regime and field coupling	1, 2, 3, 4	Wavelength, electric field, plane-wave impedance and induced voltage	Decide whether lumped, near-field or radiating assumptions are valid.
Material, shielding and discontinuity effects	5, 6, 7, 16	Skin depth, reflection coefficient, trace delay versus edge rate and dielectric loss-tangent heating	Review shielding, impedance control, termination, material loss and measurement reference planes.
Guided waves, antennas and apertures	8, 9, 10, 14, 15, 17, 18	Waveguide cutoff, far-field distance, enclosure slot fraction of wavelength, polarization and pointing loss, aperture gain, free-space path loss and EIRP power density	Validate guided-wave margin, antenna test distance, installed link margin, field exposure boundary and EMC aperture risk.
Optical, noise and uncertainty validation	11, 12, 13	Fiber dispersion, integrated noise and measurement uncertainty guard band	Decide whether link, probe and compliance results have enough margin for release.

Validation Package Checklist

source, path, receiver, material boundary and reference plane are named;
frequency, wavelength, rise time, bandwidth and applicable field regime are recorded;
cable, enclosure, antenna, waveguide, shield, dielectric and installation conditions are configuration controlled;
measurement bandwidth, detector mode, probe factor, cable loss, calibration and uncertainty are documented;
near-field, far-field, transmission-line, guided-wave and optical assumptions are not mixed without justification;
thermal, moisture, surface finish, connector and mounting effects are screened where they can dominate;
final release decision states accept, retest, realign, add shielding, change geometry, increase margin or hold.

Exercise 1: Wavelength and Regime Check

A system has a signal component at:

f=2.4\ \text{GHz}

Estimate the free-space wavelength. Then estimate the wavelength on a PCB trace with effective relative permittivity:

\epsilon_{eff}=4.0

The trace length is:

l=40\ \text{mm}

Decide whether the trace can be treated as electrically short using the conservative check:

\displaystyle l<\frac{\lambda}{10}

Solution

Free-space wavelength:

\displaystyle \lambda_0=\frac{c}{f}=\frac{3.0\times10^8}{2.4\times10^9}=0.125\ \text{m}

So:

\lambda_0=125\ \text{mm}

The approximate propagation velocity on the trace is:

\displaystyle v=\frac{c}{\sqrt{\epsilon_{eff}}}=\frac{3.0\times10^8}{2}=1.5\times10^8\ \text{m/s}

Wavelength on the trace:

\displaystyle \lambda_{trace}=\frac{v}{f}=\frac{1.5\times10^8}{2.4\times10^9}=0.0625\ \text{m}=62.5\ \text{mm}

One-tenth wavelength is:

\displaystyle \frac{\lambda_{trace}}{10}=6.25\ \text{mm}

The trace length is:

40\ \text{mm}>6.25\ \text{mm}

Therefore the trace should not be treated as an electrically short lumped connection at this frequency component.

Engineering Comment

The result does not mean every 40 mm trace is a failed design. It means propagation, return path, impedance, termination, connector discontinuities, and measurement reference plane must be reviewed. Digital systems should also check spectral content from edge rate, not only clock frequency.

Plausibility Check

The dielectric trace wavelength is half the free-space value because $\sqrt{\epsilon_{eff}}=2$ . One tenth of $62.5\ \text{mm}$ is only $6.25\ \text{mm}$ , so a $40\ \text{mm}$ trace is clearly not electrically short by this screen.

Exercise 2: Electric Field Across an Insulation Gap

A test fixture applies:

V=2.5\ \text{kV}

across an air-insulated gap:

d=1.2\ \text{mm}

Estimate the average electric field. Compare it with a simplified design limit:

E_{limit}=3.0\ \text{kV/mm}

Solution

Average field estimate:

\displaystyle E=\frac{V}{d}=\frac{2.5\ \text{kV}}{1.2\ \text{mm}}=2.083\ \text{kV/mm}

Margin to the simplified limit:

M=E_{limit}-E=3.0-2.083=0.917\ \text{kV/mm}

Relative margin:

\displaystyle M_{rel}=\frac{0.917}{3.0}=0.306

So the simplified margin is:

30.6\%

Engineering Comment

This is an average-field screen, not an insulation qualification. Real breakdown risk depends on edge radius, contamination, humidity, altitude, creepage path, transient overvoltage, material aging, partial discharge, and test waveform. A high-voltage review should use the worst local field and the installation environment, not only the nominal gap.

Plausibility Check

Dividing $2.5\ \text{kV}$ by $1.2\ \text{mm}$ gives a field a little above $2\ \text{kV/mm}$ . That is below the $3.0\ \text{kV/mm}$ simplified limit, leaving about $0.917\ \text{kV/mm}$ or $30.6\%$ margin.

Exercise 3: Plane-Wave Impedance and Power Density

An immunity test exposes equipment to a far-field plane wave with RMS electric field:

E_{rms}=3.0\ \text{V/m}

Use the free-space wave impedance:

\eta_0=377\ \Omega

Estimate RMS magnetic field and average power density.

Solution

For a plane wave:

\displaystyle H_{rms}=\frac{E_{rms}}{\eta_0}

Substitute:

\displaystyle H_{rms}=\frac{3.0}{377}=0.00796\ \text{A/m}

So:

H_{rms}=7.96\ \text{mA/m}

Average power density using RMS field:

\displaystyle S=\frac{E_{rms}^2}{\eta_0}=\frac{3.0^2}{377}=0.0239\ \text{W/m}^2

Engineering Comment

The plane-wave relation applies in the far field of the source and away from strong reflections. Near-field coupling can have very different electric-to-magnetic-field ratio. Test evidence should state antenna distance, chamber condition, field uniformity, polarization, modulation, dwell time, cable configuration, and functional performance criteria.

Plausibility Check

The magnetic field is small because free-space wave impedance is large: $3.0/377=0.00796\ \text{A/m}$ . Power density uses $E^2/\eta_0$ , so $9/377=0.0239\ \text{W/m}^2$ is consistent.

Exercise 4: Induced Voltage From a Changing Magnetic Field

A rectangular loop has area:

A=25\ \text{cm}^2

It is exposed to a sinusoidal magnetic field with peak amplitude:

B_{pk}=20\ \mu\text{T}

at:

f=50\ \text{kHz}

Assume the field is normal to the loop and the loop has one turn. Estimate peak induced voltage.

Solution

Convert area:

A=25\ \text{cm}^2=25\times10^{-4}=0.0025\ \text{m}^2

For sinusoidal flux:

V_{pk}=A\omega B_{pk}

where:

\omega=2\pi f=2\pi(50000)=314159\ \text{rad/s}

Substitute:

V_{pk}=0.0025(314159)(20\times10^{-6})=0.0157\ \text{V}

So:

V_{pk}=15.7\ \text{mV}

Engineering Comment

Millivolt-level pickup can be important in low-level sensor wiring, bridge circuits, thermocouples, photodiode amplifiers, and high-impedance inputs. Reducing loop area, twisting conductors, controlling return paths, shielding, filtering, and differential measurement can be more effective than adding software averaging after the interference has entered the signal chain.

Plausibility Check

The loop area is $0.0025\ \text{m}^2$ and the field is only $20\ \mu\text{T}$ , so the induced voltage should be small. The $50\ \text{kHz}$ frequency raises the flux-change rate enough to produce a millivolt-scale result, $15.7\ \text{mV}$ .

Exercise 5: Skin Depth and Shield Thickness

Estimate copper skin depth at:

f=10\ \text{MHz}

Use:

\sigma=5.8\times10^7\ \text{S/m}

and:

\mu=\mu_0=4\pi\times10^{-7}\ \text{H/m}

Then compare with a copper shield thickness:

t=70\ \mu\text{m}

Solution

Skin depth:

\displaystyle \delta=\sqrt{\frac{2}{\omega\mu\sigma}}

where:

\omega=2\pi f=2\pi(10^7)=6.283\times10^7\ \text{rad/s}

Substitute:

\displaystyle \delta=\sqrt{\frac{2}{(6.283\times10^7)(4\pi\times10^{-7})(5.8\times10^7)}}

\delta=2.09\times10^{-5}\ \text{m}

So:

\delta=20.9\ \mu\text{m}

Thickness in skin depths:

\displaystyle \frac{t}{\delta}=\frac{70}{20.9}=3.35

Approximate field-amplitude attenuation through a good conductor is proportional to:

e^{-t/\delta}=e^{-3.35}=0.035

Equivalent amplitude attenuation:

\displaystyle A_{dB}=20\log_{10}\left(\frac{1}{0.035}\right)=29.1\ \text{dB}

Engineering Comment

Skin depth is only one shielding mechanism. Real shield performance also depends on seams, apertures, connectors, cable penetrations, bonding impedance, magnetic permeability, corrosion, surface finish, and whether the coupling is near-field electric, near-field magnetic, or far-field radiation.

Plausibility Check

The shield is $70/20.9=3.35$ skin depths thick. An amplitude factor of $e^{-3.35}$ is small, about $0.035$ , so an attenuation estimate near $29\ \text{dB}$ is plausible before seam and aperture effects.

Exercise 6: Reflection at an Impedance Discontinuity

A transmission line has characteristic impedance:

Z_0=50\ \Omega

It is terminated by:

Z_L=75\ \Omega

Estimate voltage reflection coefficient, reflected power fraction, and VSWR.

Solution

Voltage reflection coefficient:

\displaystyle \Gamma=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{75-50}{75+50}=\frac{25}{125}=0.20

Reflected power fraction:

|\Gamma|^2=0.20^2=0.04

So:

4\%

of incident power is reflected.

VSWR:

\displaystyle VSWR=\frac{1+|\Gamma|}{1-|\Gamma|}=\frac{1.20}{0.80}=1.50

Engineering Comment

This mismatch may be acceptable in some broadband measurement setups and unacceptable in precision RF, radar, clock, or high-speed serial systems. The practical decision depends on edge rate, line length, allowable ringing, delivered power, transmitter tolerance, calibration plane, and whether multiple discontinuities create standing waves.

Plausibility Check

The load is higher than the line impedance, so the voltage reflection coefficient is positive. A coefficient of $0.20$ gives reflected power $0.20^2=4\%$ and a moderate VSWR of $1.50$ .

Exercise 7: Trace Delay and Edge-Rate Screening

A PCB trace has length:

l=120\ \text{mm}

The propagation velocity on the trace is:

v=1.6\times10^8\ \text{m/s}

A driver has 10-to-90 percent rise time:

t_r=1.2\ \text{ns}

Use the screening rule that transmission-line behavior should be reviewed when one-way delay exceeds:

\displaystyle \frac{t_r}{6}

Solution

Trace delay:

\displaystyle t_d=\frac{l}{v}=\frac{0.120}{1.6\times10^8}=7.5\times10^{-10}\ \text{s}

So:

t_d=0.75\ \text{ns}

Rise-time screen:

\displaystyle \frac{t_r}{6}=\frac{1.2}{6}=0.20\ \text{ns}

Since:

0.75\ \text{ns}>0.20\ \text{ns}

the trace should be reviewed as a transmission line.

Engineering Comment

Clock frequency alone would miss this risk. A slow repetition rate can still have fast edges. The review should include stackup, reference plane continuity, return current path, termination, connector/via discontinuities, receiver threshold, and measurement probe bandwidth.

Plausibility Check

The one-way trace delay is $0.75\ \text{ns}$ , while the edge-rate screen is only $1.2/6=0.20\ \text{ns}$ . The trace delay is several times larger than the threshold, so transmission-line review is justified.

Exercise 8: Rectangular Waveguide Cutoff

A rectangular waveguide has broad-wall dimension:

a=22.86\ \text{mm}

Estimate the cutoff frequency of the dominant TE10 mode:

\displaystyle f_c=\frac{c}{2a}

Then check operation at:

f=10\ \text{GHz}

and estimate guided wavelength:

\displaystyle \lambda_g=\frac{\lambda_0}{\sqrt{1-(f_c/f)^2}}

Solution

Convert:

a=0.02286\ \text{m}

Cutoff frequency:

\displaystyle f_c=\frac{3.0\times10^8}{2(0.02286)}=6.56\times10^9\ \text{Hz}

So:

f_c=6.56\ \text{GHz}

At $10\ \text{GHz}$ :

f>f_c

so the dominant mode can propagate.

Free-space wavelength:

\displaystyle \lambda_0=\frac{3.0\times10^8}{10\times10^9}=0.030\ \text{m}=30\ \text{mm}

Guided wavelength:

\displaystyle \lambda_g=\frac{30}{\sqrt{1-(6.56/10)^2}}\ \text{mm}

\displaystyle \lambda_g=\frac{30}{\sqrt{1-0.430}}\ \text{mm}=\frac{30}{0.755}\ \text{mm}=39.7\ \text{mm}

Engineering Comment

Waveguide operation is not just “above cutoff.” Bends, flanges, moisture, surface finish, mode purity, power handling, mismatch, and transitions affect performance. Near cutoff, dispersion and loss sensitivity increase, so operating frequency should be selected with margin.

Plausibility Check

The operating frequency $10\ \text{GHz}$ is above the $6.56\ \text{GHz}$ cutoff, so propagation is possible. The guided wavelength is longer than the $30\ \text{mm}$ free-space wavelength, which is expected above but not far from cutoff.

Exercise 9: Antenna Far-Field Distance

A directional antenna has largest dimension:

D=0.60\ \text{m}

It operates at:

f=5.8\ \text{GHz}

Estimate the Fraunhofer far-field distance:

\displaystyle R_{ff}=\frac{2D^2}{\lambda}

Then decide whether a test at:

R=5\ \text{m}

is a far-field pattern test.

Solution

Wavelength:

\displaystyle \lambda=\frac{c}{f}=\frac{3.0\times10^8}{5.8\times10^9}=0.0517\ \text{m}

Far-field distance:

\displaystyle R_{ff}=\frac{2(0.60)^2}{0.0517}=\frac{0.72}{0.0517}=13.9\ \text{m}

The test distance is:

5\ \text{m}<13.9\ \text{m}

So the test is not a far-field pattern test by this screen.

Engineering Comment

A near-field measurement may still be useful, but it should not be interpreted as a far-field gain or pattern result without transformation or a justified method. Antenna validation should state distance, polarization, chamber reflections, mounting structure, cable route, reference antenna, calibration, and whether the result is gain, pattern, return loss, or service-level performance.

Plausibility Check

The wavelength at $5.8\ \text{GHz}$ is about $0.0517\ \text{m}$ . With a $0.60\ \text{m}$ antenna dimension, the Fraunhofer distance grows to $13.9\ \text{m}$ , so a $5\ \text{m}$ test is clearly short of the far-field screen.

Exercise 10: Enclosure Slot as an Unintended Radiator

An enclosure seam has an electrically continuous slot length:

l=30\ \text{mm}

Check its size relative to wavelength at:

f_1=1\ \text{GHz}

and:

f_2=5\ \text{GHz}

Use free-space wavelength for a first-pass screen.

Solution

At $1\ \text{GHz}$ :

\displaystyle \lambda_1=\frac{3.0\times10^8}{1.0\times10^9}=0.300\ \text{m}=300\ \text{mm}

Slot fraction:

\displaystyle \frac{l}{\lambda_1}=\frac{30}{300}=0.10

At $5\ \text{GHz}$ :

\displaystyle \lambda_2=\frac{3.0\times10^8}{5.0\times10^9}=0.060\ \text{m}=60\ \text{mm}

Slot fraction:

\displaystyle \frac{l}{\lambda_2}=\frac{30}{60}=0.50

Engineering Comment

The same mechanical slot is a small fraction of a wavelength at $1\ \text{GHz}$ but half a wavelength at $5\ \text{GHz}$ , where it can become a much more efficient aperture. EMC design should control seam length, bonding, gasket continuity, cable penetrations, internal source placement, and whether high-frequency currents can excite the slot.

Plausibility Check

The slot is $30\ \text{mm}$ long. It is only $0.10\lambda$ at $1\ \text{GHz}$ but $0.50\lambda$ at $5\ \text{GHz}$ , so the same seam becomes much more significant at the higher frequency.

Exercise 11: Optical Fiber Chromatic Dispersion

A fiber link has length:

L=18\ \text{km}

Chromatic dispersion coefficient:

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

and source spectral width:

\Delta\lambda=0.10\ \text{nm}

Estimate pulse spreading:

\Delta t=D L \Delta\lambda

Compare with a 10 Gbit/s bit period:

T_b=100\ \text{ps}

Solution

Pulse spreading:

\Delta t=17(18)(0.10)=30.6\ \text{ps}

Fraction of bit period:

\displaystyle \frac{\Delta t}{T_b}=\frac{30.6}{100}=0.306

So dispersion spreading is:

30.6\%

of one bit period.

Engineering Comment

This is large enough to require a link review. The calculation omits receiver bandwidth, transmitter chirp, fiber type, modal dispersion, connector reflections, optical power margin, equalization, coding, temperature, and target bit error rate. Optical validation should include measured power, eye opening or equivalent receiver metric, wavelength, launch condition, connector cleanliness, and dispersion margin.

Plausibility Check

The dispersion product is $17(18)(0.10)=30.6\ \text{ps}$ . Against a $100\ \text{ps}$ bit period, that is $30.6\%$ , large enough to be a link-design concern.

Exercise 12: Noise Bandwidth in a Field Measurement

A small RF field probe produces a measured signal:

V_s=12\ \text{mV RMS}

The measurement chain has input-referred voltage-noise density:

e_n=40\ \text{nV}/\sqrt{\text{Hz}}

over bandwidth:

B=20\ \text{MHz}

Estimate RMS noise and SNR in decibels.

Solution

Integrated RMS noise:

V_n=e_n\sqrt{B}

V_n=(40\times10^{-9})\sqrt{20\times10^6}

\sqrt{20\times10^6}=4472

V_n=40\times10^{-9}(4472)=1.789\times10^{-4}\ \text{V}

So:

V_n=0.179\ \text{mV RMS}

SNR:

\displaystyle SNR_{dB}=20\log_{10}\left(\frac{V_s}{V_n}\right)

\displaystyle SNR_{dB}=20\log_{10}\left(\frac{12}{0.179}\right)=36.5\ \text{dB}

Engineering Comment

Bandwidth is part of the measurement. A wider bandwidth admits more noise and may also include unwanted signals. A narrow bandwidth can improve SNR but may miss broadband emissions, fast transients, or modulation. A defensible field measurement states bandwidth, detector mode, averaging, probe factor, calibration date, cable loss, reference plane, and environmental condition.

Plausibility Check

The square root of $20\ \text{MHz}$ is about $4472$ , so $40\ \text{nV}/\sqrt{\text{Hz}}$ integrates to $0.179\ \text{mV RMS}$ . A $12\ \text{mV}$ signal is about $67$ times larger, giving roughly $36.5\ \text{dB}$ SNR.

Exercise 13: Measurement Uncertainty Against an EMC Limit

An emissions measurement is:

M=2.0\ \text{dB}

below the limit. Independent standard uncertainty contributors are:

Source	Standard uncertainty
Field probe calibration	$1.5\ \text{dB}$
Cable and adapter repeatability	$0.5\ \text{dB}$
Setup repeatability	$1.0\ \text{dB}$

Combine the uncertainty by root-sum-square and decide whether the $2.0\ \text{dB}$ margin is robust.

Solution

Root-sum-square uncertainty:

u_c=\sqrt{1.5^2+0.5^2+1.0^2}

u_c=\sqrt{2.25+0.25+1.00}=\sqrt{3.50}=1.87\ \text{dB}

The measured margin is:

M=2.0\ \text{dB}

This is only:

2.0-1.87=0.13\ \text{dB}

above one combined standard uncertainty.

Engineering Comment

The result is not a comfortable pass. A $2\ \text{dB}$ apparent margin can disappear with calibration, setup, cable routing, temperature, firmware mode, enclosure assembly, or production variation. A release decision should define guard band, repeat test conditions, worst-case operating mode, cable configuration, sample variation, and corrective action if a production unit measures closer to the limit.

Plausibility Check

The largest uncertainty term is $1.5\ \text{dB}$ , and the root-sum-square combined value is $1.87\ \text{dB}$ . A measured margin of $2.0\ \text{dB}$ therefore clears one combined standard uncertainty by only $0.13\ \text{dB}$ .

Exercise 14: Antenna Polarization and Pointing Loss Margin

A wireless instrumentation link has a pre-installation received-power estimate:

P_{R,base}=-78.0\ \text{dBm}

This estimate already includes transmitter power, path loss, nominal antenna gains, and receiver feeder loss. The receiver sensitivity for the required performance is:

P_{min}=-90.0\ \text{dBm}

The release rule requires at least:

M_{req}=8.0\ \text{dB}

of received-power margin. During installation review, the antenna polarization angle error is:

\theta=35^\circ

The measured pointing loss is:

L_{point}=3.0\ \text{dB}

and an extra feeder loss from the installed cable route is:

L_{feed}=1.5\ \text{dB}

Estimate polarization mismatch loss, installed received power, received-power margin, and whether the link can be released. Then check a corrected installation with $\theta=10^\circ$ and $L_{point}=1.0\ \text{dB}$ .

Solution

For linear polarization angle error, a first-pass mismatch loss is:

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

For the as-installed case:

L_{pol,35}=-20\log_{10}(\cos35^\circ)

L_{pol,35}=1.73\ \text{dB}

Total extra installed loss is:

L_{extra}=L_{pol}+L_{point}+L_{feed}

L_{extra}=1.73+3.0+1.5=6.23\ \text{dB}

Installed received power is:

P_{R,inst}=-78.0-6.23=-84.23\ \text{dBm}

Margin above receiver sensitivity is:

M_{inst}=P_{R,inst}-P_{min}

M_{inst}=-84.23-(-90.0)=5.77\ \text{dB}

Release shortfall is:

M_{req}-M_{inst}=8.0-5.77=2.23\ \text{dB}

The as-installed link should not be released under the $8.0\ \text{dB}$ margin rule.

For the corrected installation:

L_{pol,10}=-20\log_{10}(\cos10^\circ)=0.13\ \text{dB}

Corrected extra loss is:

L_{extra,c}=0.13+1.0+1.5=2.63\ \text{dB}

Corrected received power is:

P_{R,c}=-78.0-2.63=-80.63\ \text{dBm}

Corrected margin is:

M_c=-80.63-(-90.0)=9.37\ \text{dB}

Margin above the release rule is:

9.37-8.0=1.37\ \text{dB}

The corrected installation passes this simplified margin screen.

Engineering Comment

The calculation shows why an antenna that is acceptable on paper can fail after installation. Polarization, pointing, feeder routing, nearby metal, mounting height, weatherproofing, connector torque, radome condition, ground plane, cable bend radius, and local interference can all consume margin. Release evidence should include installed received power or SNR, antenna orientation, service performance, interference scan, photos, cable loss record, and a retest trigger if the mounting changes.

Plausibility Check

A $35^\circ$ polarization error is not catastrophic by itself, but it costs about $1.7\ \text{dB}$ . Adding $3.0\ \text{dB}$ pointing loss and $1.5\ \text{dB}$ feeder loss consumes more than $6\ \text{dB}$ , so a nominal $12\ \text{dB}$ sensitivity margin falls below the required $8\ \text{dB}$ guard. Correcting alignment restores enough margin to pass.

Exercise 15: Antenna Aperture Gain Plausibility Check

A microwave antenna uses a rectangular aperture with physical area:

A=0.120\ \text{m}^2

The aperture efficiency estimate is:

\eta_a=0.55

The operating frequency is:

f=10\ \text{GHz}

Use:

\displaystyle \lambda=\frac{c}{f}

and:

\displaystyle G=\frac{4\pi\eta_aA}{\lambda^2}

The release target requires at least:

G_{req}=30.0\ \text{dBi}

Calculate wavelength, linear gain, gain in dBi and effective aperture. Then check whether a procurement claim of:

G_{claim}=32.0\ \text{dBi}

is plausible for the same aperture area. Finally, check a corrected aperture area:

A_2=0.145\ \text{m}^2

with the same efficiency.

Solution

Wavelength:

\displaystyle \lambda=\frac{3.0\times10^8}{10\times10^9}=0.030\ \text{m}

Linear gain estimate:

\displaystyle G=\frac{4\pi(0.55)(0.120)}{(0.030)^2}=921.5

Gain in dBi:

G_{dBi}=10\log_{10}(921.5)=29.65\ \text{dBi}

Release margin:

M_G=29.65-30.0=-0.35\ \text{dB}

The antenna narrowly fails the simplified gain release target.

Effective aperture:

A_e=\eta_aA=0.55(0.120)=0.066\ \text{m}^2

For the claimed $32.0\ \text{dBi}$ gain:

G_{claim,lin}=10^{32/10}=1585

Required aperture efficiency would be:

\displaystyle \eta_{req}=\frac{G_{claim,lin}\lambda^2}{4\pi A}

\displaystyle \eta_{req}=\frac{1585(0.030)^2}{4\pi(0.120)}=0.946

That claim would require about $94.6\%$ aperture efficiency, so it should not be accepted without measurement evidence.

For the corrected aperture:

\displaystyle G_2=\frac{4\pi(0.55)(0.145)}{(0.030)^2}=1113

G_{2,dBi}=10\log_{10}(1113)=30.47\ \text{dBi}

Corrected release margin:

M_{G,2}=30.47-30.0=0.47\ \text{dB}

The larger aperture passes the simplified gain screen, but with modest margin.

Engineering Comment

Aperture gain is a geometry-and-efficiency claim. A catalogue gain, a simulated pattern and an installed antenna can differ because of illumination taper, spillover, blockage, feed mismatch, radome loss, polarization, surface error, mounting structure and nearby metal. Release evidence should include the reference plane, frequency, polarization, measured gain or calibrated pattern data, return loss, installation geometry and uncertainty.

Plausibility Check

At $10\ \text{GHz}$ , wavelength is $3\ \text{cm}$ , so a $0.120\ \text{m}^2$ aperture is electrically large and a gain near $30\ \text{dBi}$ is plausible. The $32\ \text{dBi}$ claim is suspicious because it would require nearly ideal aperture efficiency for the stated area. Increasing area from $0.120$ to $0.145\ \text{m}^2$ increases gain by the same area ratio, which explains the small pass above the $30.0\ \text{dBi}$ release target.

Exercise 16: Dielectric Loss-Tangent Heating Screen

An RF dielectric window is installed in a region where the estimated RMS electric field is:

E_{rms}=5.0\ \text{kV/m}

The operating frequency is:

f=2.45\ \text{GHz}

The candidate polymer has relative permittivity:

\epsilon_r=3.2

and loss tangent:

\tan\delta=0.015

The volume of material exposed to the high-field region is:

V_m=8.0\times10^{-6}\ \text{m}^3

The simplified thermal resistance from the exposed material region to ambient is:

\theta=12\ ^\circ\text{C/W}

The release rule allows a temperature rise of at most:

\Delta T_{max}=12\ ^\circ\text{C}

Use the first-pass dielectric loss density estimate:

p_d=\omega\epsilon_0\epsilon_r\tan\delta E_{rms}^2

with:

\epsilon_0=8.85\times10^{-12}\ \text{F/m}

Estimate dielectric loss density, total heat generation, temperature rise and release margin. Then check a lower-loss material with:

\tan\delta=0.006

under the same field and geometry.

Solution

Angular frequency is:

\omega=2\pi f=2\pi(2.45\times10^9)=1.54\times10^{10}\ \text{rad/s}

Convert the electric field:

E_{rms}=5.0\ \text{kV/m}=5000\ \text{V/m}

For the first material:

p_d=(1.54\times10^{10})(8.85\times10^{-12})(3.2)(0.015)(5000)^2

p_d=1.64\times10^5\ \text{W/m}^3

Total dielectric heat generation is:

P_d=p_dV_m

P_d=(1.64\times10^5)(8.0\times10^{-6})=1.31\ \text{W}

Estimated temperature rise is:

\Delta T=P_d\theta=1.31(12)=15.7\ ^\circ\text{C}

Release margin is:

M_T=\Delta T_{max}-\Delta T=12.0-15.7=-3.7\ ^\circ\text{C}

The first material does not meet the simplified thermal release rule.

For the lower-loss material, loss density scales directly with loss tangent:

\displaystyle p_{d,new}=1.64\times10^5\left(\frac{0.006}{0.015}\right)=6.54\times10^4\ \text{W/m}^3

Total heat generation becomes:

P_{d,new}=(6.54\times10^4)(8.0\times10^{-6})=0.523\ \text{W}

Temperature rise becomes:

\Delta T_{new}=0.523(12)=6.28\ ^\circ\text{C}

Thermal margin becomes:

M_{T,new}=12.0-6.28=5.72\ ^\circ\text{C}

The lower-loss material passes this screening rule, provided the field estimate, material loss tangent, thermal path and environmental condition are validated for the released configuration.

Engineering Comment

Permittivity is not only a wavelength or capacitance parameter. In a high-frequency electric field, loss tangent can turn a dielectric support, radome, window, spacer or PCB material into a heat source. Release evidence should include the frequency, RMS field basis, material data at temperature, moisture condition, field nonuniformity, thermal path, duty cycle and whether the measured hot spot matches the model.

Plausibility Check

The loss density scales with frequency, loss tangent and the square of electric field. A 5 kV/m RMS field at 2.45 GHz is strong enough that a moderate loss tangent creates watt-level heating in only a few cubic centimetres of material. Reducing loss tangent from 0.015 to 0.006 reduces heat by 60 percent, so the temperature rise falling from about 15.7 to 6.28 degrees Celsius is consistent.

Exercise 17: Friis Path Loss and Installed RF Link Margin

A low-power telemetry link operates at:

f=2.4\ \text{GHz}

over a clear line-of-sight distance:

d=120\ \text{m}

The transmitter conducted power is:

P_t=10\ \text{dBm}

The antennas have gains:

G_t=6\ \text{dBi},\quad G_r=4\ \text{dBi}

The installed feeder losses are:

L_{tx}=1.5\ \text{dB},\quad L_{rx}=2.0\ \text{dB}

The polarization angle mismatch is:

\theta=20^\circ

Receiver sensitivity for the required data mode is:

P_{sens}=-82\ \text{dBm}

The release rule requires at least:

M_{fade}=12\ \text{dB}

of margin above sensitivity after installed losses. Use the free-space path-loss screen:

FSPL=32.44+20\log_{10}(f_{MHz})+20\log_{10}(d_{km})

and polarization mismatch loss:

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

Calculate received power and release margin. Then check the same installation at $d=250\ \text{m}$ .

Solution

Convert distance to kilometres:

d=120\ \text{m}=0.120\ \text{km}

Frequency in MHz:

f=2400\ \text{MHz}

Free-space path loss:

FSPL=32.44+20\log_{10}(2400)+20\log_{10}(0.120)

FSPL=32.44+67.60-18.42=81.62\ \text{dB}

Polarization mismatch loss:

L_{pol}=-20\log_{10}(\cos20^\circ)=0.54\ \text{dB}

Received power:

P_r=P_t+G_t+G_r-L_{tx}-L_{rx}-L_{pol}-FSPL

P_r=10+6+4-1.5-2.0-0.54-81.62

P_r=-65.7\ \text{dBm}

Link margin above receiver sensitivity:

M=P_r-P_{sens}

M=-65.7-(-82)=16.3\ \text{dB}

Release margin after the required fade allowance:

M_{release}=16.3-12.0=4.3\ \text{dB}

The $120\ \text{m}$ installation passes this simplified screen.

For $d=250\ \text{m}=0.250\ \text{km}$ :

FSPL_2=32.44+20\log_{10}(2400)+20\log_{10}(0.250)

FSPL_2=32.44+67.60-12.04=88.00\ \text{dB}

Received power becomes:

P_{r,2}=10+6+4-1.5-2.0-0.54-88.00=-72.0\ \text{dBm}

Link margin:

M_2=-72.0-(-82)=10.0\ \text{dB}

Release margin after fade allowance:

M_{release,2}=10.0-12.0=-2.0\ \text{dB}

The $250\ \text{m}$ case should not be released without higher antenna gain, lower feeder loss, better alignment, lower data-rate sensitivity, additional fade evidence or a shorter path.

Engineering Comment

Friis path loss is a useful first screen only when the path is close to clear line of sight and the antenna gains, polarization and reference planes are credible. Installed RF links can lose margin through cable routing, connector loss, radome wetting, antenna tilt, nearby metal, Fresnel-zone obstruction, multipath, interference, receiver desensitization and production variation. Release evidence should include installed RSSI or received-power measurement, antenna orientation, cable loss record, channel survey, service data rate, interference scan and the exact receiver sensitivity criterion.

Plausibility Check

At $2.4\ \text{GHz}$ and $120\ \text{m}$ , free-space loss around $82\ \text{dB}$ is plausible. The antenna gains and $10\ \text{dBm}$ transmitter power leave a received signal far above $-82\ \text{dBm}$ sensitivity, but the required fade allowance consumes most of that apparent margin. Increasing distance to $250\ \text{m}$ adds about $6.4\ \text{dB}$ of path loss, which is enough to turn a $4.3\ \text{dB}$ release margin into a failure.

Exercise 18: EIRP Power-Density and Near-Field Exclusion Gate

A directional RF test antenna is used in a development area. The transmitter conducted power at the antenna feed reference plane is:

P_t=30\ \text{dBm}

The antenna gain in the main beam is:

G_t=18\ \text{dBi}

The feeder and connector loss before the antenna is:

L_t=2\ \text{dB}

The operating frequency is:

f=5.8\ \text{GHz}

The largest antenna aperture dimension is:

D=0.45\ \text{m}

A maintenance walkway can place personnel at:

R_1=2.0\ \text{m}

from the antenna. The local screening rule limits far-field power density to:

S_{limit}=0.50\ \text{W/m}^2

Use the far-field screening relation:

\displaystyle S_p=\frac{EIRP}{4\pi R^2}

and the RMS field relation:

E_{rms}=\sqrt{S_p\eta_0}

with:

\eta_0=377\ \Omega

Calculate EIRP, the Fraunhofer far-field distance, estimated power density and RMS electric field at the walkway, the far-field exclusion distance for the stated power-density limit, and the release decision. Then check a controlled boundary at:

R_2=10\ \text{m}

Solution

EIRP in dBm is:

EIRP_{dBm}=P_t+G_t-L_t

EIRP_{dBm}=30+18-2=46\ \text{dBm}

Convert EIRP to watts:

EIRP_W=10^{(EIRP_{dBm}-30)/10}

EIRP_W=10^{(46-30)/10}=39.8\ \text{W}

Wavelength is:

\displaystyle \lambda=\frac{c}{f}

\displaystyle \lambda=\frac{3.0\times10^8}{5.8\times10^9}=0.0517\ \text{m}

The Fraunhofer far-field distance is:

\displaystyle R_{FF}=\frac{2D^2}{\lambda}

\displaystyle R_{FF}=\frac{2(0.45)^2}{0.0517}=7.83\ \text{m}

At the walkway distance:

\displaystyle S_{p,1}=\frac{39.8}{4\pi(2.0)^2}=0.792\ \text{W/m}^2

RMS electric field estimate:

E_{rms,1}=\sqrt{0.792(377)}=17.3\ \text{V/m}

The power-density estimate exceeds the screening limit:

0.792>0.50

The far-field distance check also fails:

2.0<7.83

So the walkway cannot be released from this far-field calculation. It is both above the power-density screen and inside the near-field region where the simple inverse-square estimate is not sufficient release evidence.

The far-field exclusion distance for the power-density limit alone is:

\displaystyle R_{S}=\sqrt{\frac{EIRP_W}{4\pi S_{limit}}}

\displaystyle R_S=\sqrt{\frac{39.8}{4\pi(0.50)}}=2.52\ \text{m}

Because the antenna far-field boundary is larger than the power-density-only radius:

R_{FF}>R_S

a paper release based on this far-field model should place the controlled boundary beyond the far-field distance or require a measured near-field survey.

At the controlled 10 m boundary:

\displaystyle S_{p,2}=\frac{39.8}{4\pi(10)^2}=0.0317\ \text{W/m}^2

RMS electric field is:

E_{rms,2}=\sqrt{0.0317(377)}=3.46\ \text{V/m}

The 10 m boundary passes the far-field and power-density screens:

10>7.83

and:

0.0317<0.50

The release decision is to block the 2 m walkway during transmission, move the controlled boundary to at least a validated far-field-safe location such as 10 m, or perform a qualified near-field measurement and risk review if work must occur closer to the antenna.

Engineering Comment

EIRP is not only a regulatory or link-budget number. It can also become a field-strength and exposure-boundary input, but only when the reference plane, antenna gain, feeder loss, direction and distance are clear. Near an aperture antenna, reactive and radiating fields may not follow the simple far-field inverse-square model, so a close-in personnel, EMC or sensor-saturation decision should not be released from far-field arithmetic alone.

Release evidence should record conducted power, antenna model, gain direction, frequency, aperture dimension, feeder losses, duty cycle, modulation state, beam direction, controlled-area geometry, warning/interlock state, measurement method and whether the evaluated point is in the far field or near field.

Plausibility Check

The EIRP is about $40\ \text{W}$ because $46\ \text{dBm}$ is 16 dB above 1 W. Spreading that power over a 2 m sphere gives a value below $1\ \text{W/m}^2$ but still above the stated $0.50\ \text{W/m}^2$ screen. The far-field distance is almost 8 m because a 45 cm aperture is electrically large at 5.8 GHz. Moving the boundary to 10 m reduces power density by a factor of 25 compared with 2 m and puts the calculation in the intended far-field regime.

Review Checklist

When reviewing applied electromagnetics or wave-propagation evidence, ask:

Is the physical boundary stated: source, path, receiver, material, enclosure, cable, antenna, instrument or reference plane?
Is the regime identified before equations are selected: lumped circuit, transmission line, guided wave, near field, far field, optical path or mixed regime?
Are wavelength, rise time, geometry and material properties checked before accepting a low-frequency approximation?
Are return path, reference plane, grounding, shielding and connector discontinuities represented?
Are skin depth, seam leakage, apertures and bonding impedance separated when interpreting shielding performance?
Are dielectric constant, loss tangent, moisture, temperature and duty cycle checked before accepting material behavior?
Are antenna measurements tied to distance, polarization, pointing, chamber condition, reference antenna and calibration state?
Is EIRP converted to field strength or power density only at a valid reference plane and field region?
Are RF link budgets tied to path length, antenna gains, feeder losses, polarization, sensitivity and fade-margin rule?
Are waveguide and optical calculations checked for cutoff, dispersion, loss, mode purity, launch condition and connector condition?
Are noise bandwidth, detector mode, averaging, probe factor, cable loss and calibration date included in measurement evidence?
Are uncertainty guard bands applied before comparing emissions, field strength, link margin or gain against a release limit?
Does installed geometry match the bench or catalogue configuration used for the calculation?
Does the result imply retest, shielding change, termination, antenna alignment, reference-plane correction, margin increase or release?

Common Mistakes

Using a lumped circuit model after trace length, cable length, aperture size or edge rate makes propagation dominant.
Treating a near-field measurement as far-field gain or field-strength evidence without distance and transformation justification.
Releasing an EIRP-based field exposure or EMC decision without checking far-field distance, antenna direction, duty cycle and measured near-field behavior.
Reporting shield attenuation from skin depth while seams, connectors, bonding and cable penetrations dominate leakage.
Treating dielectric constant as a velocity factor while ignoring loss tangent, heating, moisture and temperature drift.
Ignoring return path and reference plane when interpreting trace delay, reflection or induced-voltage calculations.
Accepting waveguide operation because frequency is above cutoff while dispersion, mode purity, bends and transitions are unresolved.
Using catalogue antenna gain while installed polarization, pointing, radome, mounting and nearby metal consume the margin.
Releasing a wireless path from transmitter power alone while path loss, feeder loss, polarization mismatch and receiver sensitivity control the link margin.
Treating optical dispersion as a fiber-only number while launch condition, receiver bandwidth, chirp and connector reflection remain unverified.
Improving SNR by narrowing bandwidth without checking whether the wanted transient, modulation or emission is still captured.
Comparing an EMC result with a limit before applying calibration, setup, cable-routing and production-variation uncertainty.
Releasing an RF or EMC configuration after cable routing, enclosure seams, firmware clocking or mounting hardware changed.

Applied electromagnetics is strongest when the selected model, physical geometry, installation state, measurement method and uncertainty all describe the same field path.

REF