RF Journey

01 / 12

From circuit ports to fields

A source reads +10 dBm at a connector. Does that guarantee the same electric-field strength one metre away as an antenna specified to have +10 dBm EIRP in your direction?

No. A connector measurement belongs to a defined circuit reference plane. A distant field depends on what reaches and is accepted by the antenna, radiation efficiency, directional gain, polarization, distance, field region, and the surrounding environment. EIRP already combines radiated power and directional gain; connector power does not.

PORTV, I, Z, PDefined at a terminal reference plane
STRUCTUREcharge and currentDistributed over conductors and dielectrics
FIELDE, D, H, BFunctions of position and time
OBSERVERprobe or antennaOrientation and location select a response

Interactive · sources and geometry

Field origins explorer

Predict the electric-field direction, then change the source and geometry. The drawings show relationships, not calibrated field maps.

First prediction: which way does E point between charged plates?

Physical arrangement

Voltage 10 V

Plate gap 10 mm

Current 100 mA

Read the model

Voltage across distance

For wide plates with a small gap, E ≈ V/d. Fringing near the edges means the real field is not perfectly uniform.

E and DE acts on charge; D accounts conveniently for free charge and material polarization.
H and BH connects to free current; B is magnetic flux density and enters the force law and induction.

Plate estimate |E|: 1000 V/m
Wire estimate H at 10 mm: 1.592 A/m
E direction: upper plate to lower plate
Selected geometry: plates

Assumptions: wide parallel plates, an infinitely long isolated wire with an implied distant return, or an ideal TEM coaxial line. These are local teaching models, not full-wave solutions.

Static takeaway: charge separation creates electric fields, current creates circulating magnetic fields, and conductor geometry determines how those fields occupy space.

Common misconceptionFields begin only when energy leaves an antenna.

Fields also store and carry energy inside capacitors, inductors, transmission lines, packages, and connectors. Radiation is one possible field behavior, not the definition of a field.

02 / 12

Electric fields: force and voltage across space

What local quantity turns a voltage between two terminals into a direction and strength at every point between them?

Electric field intensity E is defined through the force on a positive test charge. Its units are volts per metre, equivalent to newtons per coulomb. Voltage is not the field itself: the voltage difference from a to b is the negative line integral of E along a path.

$\begin{aligned}\mathbf F&=q(\mathbf E+\mathbf v\times\mathbf B)\\V_b-V_a&=-\int_a^b\mathbf E\cdot\mathrm{d}\boldsymbol\ell\end{aligned}$ The first relation is the Lorentz force. The second connects local electric field to a terminal voltage.

Electric flux density D separates the bookkeeping of free charge from material polarization. In a simple linear, isotropic medium, D = εE. At a dielectric boundary, E and D need not have the same component continuity rules, which is why keeping both symbols matters.

Think about itIf the plate voltage stays fixed while the ideal plate spacing halves, what happens to the central E-field estimate?

Answer

It doubles: |E| ≈ |V|/d. This is a geometry-dependent approximation away from fringing edges, not a universal conversion from volts to volts per metre.

03 / 12

Magnetic fields: current and circulation

Why does a loop probe respond strongly beside a current path even when a nearby electric-field probe does not?

Magnetic field intensity H circulates around free current. Magnetic flux density B is the quantity in the magnetic part of the Lorentz force and in Faraday induction. For a simple linear, isotropic material, B = μH; in magnetically complex media the relationship can be nonlinear, anisotropic, or history-dependent.

$\oint\mathbf H\cdot\mathrm{d}\boldsymbol\ell=I_{\mathrm{free}}+\frac{\mathrm d}{\mathrm dt}\int\mathbf D\cdot\mathrm d\mathbf A$ $\text{ideal long wire}: |H(r)| = \frac{I}{2\pi r}$ The first equation includes displacement current. The second is a symmetry-specific result.

Near-field probes make the distinction tangible: small loops couple primarily to changing magnetic flux, while small electric probes respond primarily to electric potential and field. Probe size and orientation affect both sensitivity and spatial resolution.

Common misconceptionB and H are interchangeable names for the same number.

They describe related but distinct quantities, with different units and boundary/source roles. Writing B = μH is a material model, not a definition that erases the distinction.

04 / 12

Maxwell’s four connections

What compact set of rules connects free charge, current, electric field, magnetic field, and time variation?

$\nabla\cdot\mathbf D=\rho_{\mathrm f}$ $\nabla\cdot\mathbf B=0$ $\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}$ $\nabla\times\mathbf H=\mathbf J_{\mathrm f}+\frac{\partial\mathbf D}{\partial t}$ Macroscopic differential form; D = εE, B = μH, and J = σE only for the stated simple material model.

The divergence equations connect fields to sources and the absence of isolated magnetic charge in this model. The curl equations connect changing electric and magnetic fields. Maxwell’s displacement-current term, ∂D/∂t, completes the current continuity needed for a changing electric field to participate in wave propagation.

Go deeperFrom the curl equations to a wave equation

In a source-free, homogeneous, lossless medium, taking the curl again and using the companion curl equation gives ∇²E − με ∂²E/∂t² = 0, with the same form for H. The propagation speed is v = 1/√(με).

05 / 12

A wave that travels through space

At 2.45 GHz, what exactly repeats every 408.163 ps and every 122.364 mm?

At one position, the sinusoidal field repeats in time after one period T. At one instant, the spatial pattern repeats after one wavelength λ. Frequency and wavelength are related by the phase velocity of the medium, not automatically by c in every material.

$\omega = 2\pi f \qquad \beta = \frac{2\pi }{\lambda } \qquad v = \frac{\omega }{\beta } = f\lambda$ $\mathrm{at} 2.45 \mathrm{GHz} \text{in vacuum}: T = 408.163 \mathrm{ps} \qquad \lambda _{0} = 122.364 \mathrm{mm}$ One quarter wavelength is 30.591 mm. The exact vacuum light speed used is 299,792,458 m/s.

Interactive · space, time, and orientation

Wave anatomy explorer

Keep frequency fixed at 2.45 GHz. Scrub phase to compare a spatial snapshot with the time history at one sensor. Animation starts paused.

E-field spatial snapshot
sensor time history
propagation direction

Phase / time scrubber 0°

Sensor position 0.99 λ

Frequency: 2.450 GHz
Period T: 408.163 ps
Vacuum wavelength λ₀: 122.364 mm
Sensor value E/Epk: 0.998

E(z,t) = x̂ Epk cos(ωt − βz + φ)With e^(jωt), a +z traveling phasor carries e^(−jβz). A negative βz phase term is travel, not attenuation.

Assumptions: monochromatic, uniform, lossless plane wave in vacuum. E, H, and +z are mutually perpendicular; E × H points in the power-flow direction.

Static takeaway: a spatial plot freezes one instant across many positions; a time plot records one position across many instants. At 2.45 GHz, T = 408.163 ps and λ₀ = 122.364 mm.

Common misconceptionAn electromagnetic wave needs matter to carry it.

Unlike a mechanical wave, an electromagnetic wave propagates in vacuum. Matter changes its speed, wavelength, impedance, dispersion, and loss; it is not required for vacuum propagation.

06 / 12

Plane waves and electromagnetic power flow

How do volts per metre and amperes per metre become watts per square metre?

A plane wave is a local idealization whose field is uniform over every plane perpendicular to propagation. For a lossless +z wave with E along +x, H lies along +y. Their ratio is the wave impedance η, and the cross product E × H points along energy flow.

$E = \hat{x} E_{\mathrm{pk}} \cos (\omega t - \beta z + \phi)$ $H = \hat{y} (\frac{E_{\mathrm{pk}}}{\eta }) \cos (\omega t - \beta z + \phi)$ $\begin{aligned}\langle\mathbf S\rangle&=\operatorname{Re}\{\mathbf E_{\mathrm{rms}}\times\mathbf H_{\mathrm{rms}}^*\}\\|\langle\mathbf S\rangle|&=\frac{E_{\mathrm{rms}}^2}{\eta}=\eta H_{\mathrm{rms}}^2\end{aligned}$ RMS phasors are used for average power, so there is no extra factor of one half.

Interactive · fields to watts

Plane-wave power explorer

Connect RMS field strength, wave impedance, Poynting-vector magnitude, area, orientation, and the conditional EIRP far-field estimate.

Prediction: if far-field distance doubles, what happens to power density?

Calculation path

E field RMS 1.00 V/m

Surface area 0.10 m²

Surface-normal angle 0°

Electric field RMS: 1.0000 V/m
Magnetic field RMS: 2.6544 mA/m
Average power density: 2.6544 mW/m²
Power through surface: 0.2654 mW

Savg = Erms²/η₀; Pnormal = Savg A cos θRMS phasors are used, so Re{Erms × Hrms*} has no extra 1/2. η₀ = 376.7303 Ω—not 50 Ω.

Assumptions: uniform lossless plane wave for the field path; directional EIRP, free-space spreading, polarization match, and valid far-field conditions for the EIRP path. A connector reading alone does not supply those facts.

Static takeaway: in a vacuum plane wave, Hrms = Erms/376.730 Ω and average power density is Erms²/376.730. For +10 dBm EIRP at 1 m, the conditional free-space result is 0.7958 mW/m² and 0.5475 V/m RMS.

Wave impedance is not port impedance

η₀ ≈ 376.7303 Ω relates E and H in a vacuum plane wave. A 50 Ω port impedance relates voltage and current at a circuit reference plane. Equal units do not make them interchangeable.

07 / 12

Materials set speed, wavelength, and loss

If frequency stays at 2.45 GHz when a wave enters an ideal εr = 4 medium, which quantities change?

Frequency is fixed by the source and remains continuous across a stationary boundary. In a lossless, nondispersive medium, v = 1/√(με), λ = v/f, and η = √(μ/ε). With εr = 4 and μr = 1, speed and wavelength halve while wave impedance also halves: v = c/2, λ = 61.182 mm, and η = 188.365 Ω.

Loss makes propagation constant and wave impedance complex. With the e^(jωt) convention and a +z field proportional to e^(−γz), γ = α + jβ: α causes amplitude attenuation while β causes phase progression. A real material may also be dispersive, so ε, μ, and σ can vary with frequency.

Interactive · media, interfaces, and conductors

Material and boundary explorer

Use one model to compare lossless bulk media, normal-incidence reflection and transmission, and good-conductor skin depth.

Prediction: from vacuum into a lower-impedance dielectric, what sign has reflected E?

View

Frequency 2.45 GHz

Relative permittivity εr 4.0

Lossy-medium σ 10^-2 S/m

Vacuum wavelength: 122.364 mm
Dielectric wavelength: 61.182 mm
Dielectric wave impedance: 188.365 Ω
Power R / T: 11.11% / 88.89%
Copper skin depth: 1.335 µm

γ = √[jωμ(σ + jωε)] = α + jβ · η = √[jωμ/(σ + jωε)]For the selected lossy bulk model: α = 0.942 Np/m, β = 102.701 rad/m, |η| = 188.35 Ω, ∠η = 0.53°.

Assumptions: linear, isotropic, homogeneous bulk media with μr = 1; lossless media for the boundary coefficients; good-conductor approximation for copper skin depth. εr is an ideal bulk value, not an FR-4 microstrip effective permittivity.

Static takeaway: at 2.45 GHz, an ideal εr = 4, μr = 1 medium has v = c/2, λ = 61.182 mm, and η = 188.365 Ω. Vacuum to that medium gives ΓE = −1/3, R = 1/9, τE = 2/3, and T = 8/9. Copper skin depth is about 1.335 µm.

Go deeperThe lossy-medium convention used here

γ = √[jωμ(σ + jωε)] and η = √[jωμ/(σ + jωε)]. The square-root branch is selected with nonnegative α and β for a passive +z wave using e^(jωt)e^(−γz). Changing the time convention changes intermediate signs, not the physics.

08 / 12

Conductors, dielectrics, and skin depth

At 2.45 GHz, does current fill a copper conductor uniformly through its thickness?

Not in the good-conductor approximation. Fields entering a conductor drive current and decay with depth. Skin depth δ is the distance over which field amplitude falls to 1/e of its surface value. Power density, proportional to field amplitude squared in this setting, falls to e^(−2) at one skin depth—not to 1/e.

$\delta \approx \sqrt{\frac{2}{\omega \mu \sigma}} = \frac{1}{\sqrt{\pi f\mu \sigma}}$ $\text{copper at} 2.45 \mathrm{GHz}: \delta \approx 1.335 \mathrm{\mu m}$ Using σ = 5.8 × 10⁷ S/m and μr ≈ 1. Surface roughness, plating, geometry, and temperature can still matter.

An ideal dielectric stores electric energy without conduction loss. A real dielectric has polarization loss and frequency dependence. A good conductor is the opposite limiting case, where σ dominates ωε. Most RF materials live between perfect textbook limits.

Think about itIf frequency increases by a factor of four while μ and σ stay fixed, what happens to skin depth?

Answer

It halves because δ is proportional to 1/√f. Current becomes more concentrated near the surface.

09 / 12

What happens at a material boundary?

If transmitted electric-field amplitude is 2/3 of incident amplitude, did one third of the incident power disappear?

No. Amplitude and power coefficients are different, and the wave impedances on the two sides differ. At normal incidence between lossless media, tangential E and H continuity sets the electric-field coefficients below. The reflected H sign reverses relative to reflected E so that its Poynting vector points back toward the source.

$\begin{aligned}\Gamma_E&=\frac{\eta_2-\eta_1}{\eta_2+\eta_1}\\\tau_E&=\frac{2\eta_2}{\eta_2+\eta_1}=1+\Gamma_E\end{aligned}$ $R=|\Gamma_E|^2\qquad T=\frac{\eta_1}{\eta_2}|\tau_E|^2\qquad R+T=1$ These power relations assume lossless media and normal incidence.

For vacuum into the ideal εr = 4, μr = 1 medium, η₂ = η₀/2. Therefore ΓE = −1/3, R = 1/9, τE = 2/3, and T = 8/9. The transmitted E amplitude is smaller, but its associated H amplitude is larger than an E-only comparison suggests because η₂ is lower.

Common misconceptionA negative reflection coefficient means negative reflected power.

The sign is phase information for the reflected E amplitude. Reflected power fraction is |Γ|² and is nonnegative.

10 / 12

Polarization: how the electric field moves

At a fixed point, does the electric-field tip trace a line, circle, or ellipse—and which way does it rotate?

Polarization describes the trajectory and orientation of the electric-field vector at a fixed position. Two orthogonal components with equal phase produce a line; equal amplitudes in quadrature produce a circle; the general case is an ellipse. Handedness is meaningless unless propagation direction, viewing direction, time convention, and standard are stated.

Interactive · electric-field trajectory

Polarization explorer

Build a polarization state from orthogonal electric-field components, then rotate an ideal linear receiver to see mismatch loss.

Presets

x amplitude Ax 1.00

y amplitude Ay 0.00

Phase δ = φy − φx 0°

Instantaneous phase ψ 0°

Linear receiver angle 45°

Convention lock

Linear polarization

Time convention e^(jωt), propagation +z with e^(−jβz), and δ = φy − φx. Per IEEE, the polarization plane is viewed looking in the propagation direction: clockwise is right-hand, counterclockwise is left-hand.

The plotted dot is one instant. The complete curve is the path traced by the E-field tip at a fixed point in space.

Classification: linear
Axial ratio: ∞ : 1
Receiver power fraction: 0.5000
Mismatch loss: -3.010 dB

Ex = Ax cos ψ · Ey = Ay cos(ψ + δ)For two matched linear polarizations, PLF = cos²θ. A 45° mismatch gives PLF = 0.5, or −3.010 dB. The general control also handles elliptical states.

Assumptions: ideal monochromatic plane wave, orthogonal x/y components, an ideal polarization-matched system except for the displayed linear receive-axis projection, and no multipath or depolarization.

Static takeaway: equal in-phase x and y components make 45° linear polarization; equal quadrature components make circular polarization; unequal or non-quadrature components generally make an ellipse. A 45° linear mismatch loses 3.010 dB.

For ideal linear transmit and receive polarizations separated by angle θ, polarization loss factor is cos²θ. At 45°, half the available power couples: 0.5 or −3.0103 dB. At 90°, the ideal model predicts zero, though real antennas, scattering, and multipath usually prevent perfect isolation.

11 / 12

Reactive near field, Fresnel region, and far field

How far from an antenna must you stand before E/H = η₀ and EIRP/(4πr²) are justified?

There is no universal distance independent of source size, wavelength, geometry, and required accuracy. Reactive near fields emphasize stored energy and source coupling. The radiating near field, often called the Fresnel region for electrically large antennas, contains radiated energy but may retain distance-dependent angular structure. In the far field, transverse field ratios and angular pattern become approximately independent of distance while amplitudes fall roughly as 1/r.

Interactive · distance from a finite source

Field-region explorer

Change wavelength, source size, and observation distance. Common boundaries appear as engineering estimates—not universal physical walls.

Frequency 2.45 GHz

Largest source dimension D 0.10 m

Observation distance r 1.00 m

Criteria check

What is justified at this distance?

r ≥ 2D²/λ (0.163 m)
r ≥ 10D (1.000 m)
r ≥ 10λ (1.224 m)

The 10× checks are visible teaching proxies for “much greater than,” not standards. Antenna-specific phase-error and pattern requirements decide the real test distance.

Wavelength λ: 122.364 mm
r / λ: 8.172
D / λ: 0.817
Reactive estimate: 0.056 m
Aperture far-field estimate: 0.163 m

Illustrative distance-term balance: 1/r³ 0% · 1/r² 2% · 1/r 98%These normalized terms show why the transition is gradual; their exact coefficients and angular dependence are source-specific.

Assumptions: λ = c/f in vacuum. The 0.62√(D³/λ) and 2D²/λ guides are common antenna estimates, most meaningful for electrically large apertures. Near fields can contain radiating energy; far field does not mean zero stored energy everywhere.

Static takeaway: field regions transition gradually. A common reactive-near-field estimate is 0.62√(D³/λ); a common far-field estimate is 2D²/λ, supplemented by r ≫ D and r ≫ λ checks.

Common misconceptionNothing radiates in the near field.

Radiating 1/r terms coexist with stronger reactive terms close to many sources. “Near field” describes the mixture and spatial behavior; it does not prohibit outward energy flow.

12 / 12

Putting it together: one 2.45 GHz field journey

What extra information turns a +10 dBm reading into a defensible field estimate one metre away?

Begin with an explicit conditional case: directional EIRP is +10 dBm = 0.010 W; the point is one metre away in vacuum; the far-field and free-space assumptions are valid; and the stated direction and polarization apply. Only then may spherical spreading and η₀ connect power density to RMS fields.

$S_{\mathrm{avg}}=\frac{0.010}{4\pi\times1^2}=0.795775\,\mathrm{mW/m^2}$ $E_{\mathrm{rms}}=\sqrt{\eta_0S_{\mathrm{avg}}}=0.547533\,\mathrm{V/m}$ $H_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{\eta_0}=1.453382\,\mathrm{mA/m}$ At 2 m, power density is one quarter and both field amplitudes are one half.

What must be known from connector to receiving structure
Stage	Required information	Consequence
Port	+10 dBm available or delivered?	10 mW only after reference plane and power definition are fixed
Feed	Mismatch, feed loss	Accepted antenna power may be less than connector power
Antenna	Efficiency, gain, pattern	Directional EIRP requires accepted power × gain in that direction
Space	r = 1 m, far field assumed	Savg = 0.795775 mW/m² for +10 dBm EIRP
Fields	η₀ = 376.7303 Ω	Erms = 0.547533 V/m; Hrms = 1.453382 mA/m
Receiver	Polarization and effective aperture	Captured power depends on orientation, pattern, loss, and environment

Replace “+10 dBm EIRP” with “+10 dBm at the connector” and the numerical field result is no longer determined. You still need reflected versus accepted power, feed loss, radiation efficiency, gain pattern in the observation direction, polarization, field-region validity, and environmental effects such as scattering and multipath.

Four field quantities that should not be collapsed into two labels
Symbol	Quantity	SI unit	Primary role here
E	Electric field intensity	V/m	Force per charge; voltage gradient or line integral
D	Electric flux density	C/m²	Free-charge accounting; D = εE in a simple linear medium
H	Magnetic field intensity	A/m	Circulation tied to free current and displacement current
B	Magnetic flux density	T	Magnetic induction; B = μH in a simple linear medium

The connected mental model

Ports specify terminal relationships. Charge, current, and geometry establish fields. Maxwell’s equations connect their sources and time variation. Materials set propagation and loss. Boundaries redistribute amplitude and power. Polarization and distance determine what an observer can receive.

Ungraded review

Check your understanding

Answer each question in your own words, then reveal the model answer.

01Why can +10 dBm at an antenna connector not determine the electric field one metre away?
Model answer
Connector power does not specify accepted power, feed loss, radiation efficiency, directional gain, polarization, field region, or environmental effects. A field estimate needs those conditions or a stated directional EIRP.
02Assume +10 dBm EIRP, free-space far-field conditions, and a point one metre away. What are the average power density and RMS fields?
Model answer
About 0.7958 mW/m², 0.5475 V/m, and 1.453 mA/m. At two metres, power density is one quarter and both field amplitudes are one half.
03A wave enters an ideal lossless medium with εr = 4 and μr = 1 at normal incidence. What changes, and what does ΓE = −1/3 mean?
Model answer
Frequency stays fixed while speed, wavelength, and wave impedance halve. The negative coefficient means a 180° electric-field phase reversal; reflected power is positive at 1/9, while transmitted power is 8/9.

Sources and further study

MIT’s open electromagnetics texts provide the mathematical backbone; NIST fixes the exact vacuum light speed; IEEE fixes antenna terminology and polarization viewing convention. The other sources informed teaching structure, measurement context, and accessible interaction design.

Primary theory and constants

Teaching and measurement context

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