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13.2 Electromagnetic Waves & Transmission Line Fields

Key Takeaways

  • Uniform plane wave propagation in lossy media is governed by the complex propagation constant γ=α+jβ=jωμ(σ+jωϵ)\gamma = \alpha + j\beta = \sqrt{j\omega\mu (\sigma + j\omega\epsilon)}, where α\alpha represents the attenuation constant in Np/m\text{Np/m} and β\beta represents the phase constant in rad/m\text{rad/m}.

  • Intrinsic impedance η=jωμσ+jωϵ\eta = \sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}} defines the ratio of electric to magnetic field strength, simplifying to η0=μ0ϵ0≈377 Ω\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377\ \Omega (or 120π Ω120\pi\ \Omega) in free space, with phase velocity vp=1μϵ=3×108 m/sv_p = \frac{1}{\sqrt{\mu \epsilon}} = 3 \times 10^8\ \text{m/s}.

  • The Poynting vector S=E×H\mathbf{S} = \mathbf{E} \times \mathbf{H} represents instantaneous electromagnetic power flux density (W/m2\text{W/m}^2), with time-average power density Pavg=12Re{E×H∗}=∣E0∣22η az\mathbf{P}_{\text{avg}} = \frac{1}{2} \text{Re}\{\mathbf{E} \times \mathbf{H}^*\} = \frac{|E_0|^2}{2\eta}\ \mathbf{a}_z.

  • Transmission line field behavior is modeled by distributed primary parameters (R, L, G, C) yielding characteristic impedance Z0=R+jωLG+jωCZ_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}}, which reduces to Z0=LCZ_0 = \sqrt{\frac{L}{C}} for lossless lines.

  • Impedance mismatch at line terminations creates wave reflections governed by the voltage reflection coefficient ΓL=ZL−Z0ZL+Z0\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} and Voltage Standing Wave Ratio VSWR=1+∣ΓL∣1−∣ΓL∣VSWR = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|}, while transient switching surges propagate as traveling step waves.

Last updated: August 2026

13.2 Electromagnetic Waves & Transmission Line Fields

High-voltage power transmission lines, high-frequency communication cables, and industrial bus ducts behave not as lumped parameters, but as distributed field systems. In the PRC REE Licensure Examination, candidates are tested on uniform plane wave propagation, intrinsic wave impedance, Poynting power flow, transmission line equations, characteristic impedance Z0Z_0, reflection coefficients, standing wave ratios (VSWRVSWR), and surge wave behavior.


1. Wave Equations & Propagation Parameters

Applying the curl operator to Maxwell's equations in a linear, isotropic, homogeneous medium with conductivity σ\sigma, permeability μ\mu, and permittivity ϵ\epsilon yields the vector wave equation for time-harmonic electric fields:

∇2Es−γ2Es=0\nabla^2 \mathbf{E}_s - \gamma^2 \mathbf{E}_s = 0

where Es\mathbf{E}_s is the phasor electric field intensity, and γ\gamma is the Complex Propagation Constant:

γ=α+jβ=jωμ(σ+jωϵ)(m−1)\gamma = \alpha + j\beta = \sqrt{j\omega\mu (\sigma + j\omega\epsilon)} \quad (\text{m}^{-1})

Attenuation Constant (α\alpha) & Phase Constant (β\beta)

  • Attenuation Constant (α\alpha): Measures wave amplitude reduction per unit distance due to dielectric and conduction losses, expressed in Nepers per meter (Np/m) or decibels per meter (dB/m\text{dB/m}, where 1 Np=8.686 dB1\ \text{Np} = 8.686\ \text{dB}).
  • Phase Constant (β\beta): Measures phase shift per unit distance, expressed in radians per meter (rad/m): β=2πλ=ωvp\beta = \frac{2\pi}{\lambda} = \frac{\omega}{v_p}

Intrinsic Impedance (η\eta) & Phase Velocity (vpv_p)

The ratio of the transverse electric field to transverse magnetic field for a forward-propagating wave is the Intrinsic Impedance η\eta:

η=ExHy=jωμσ+jωϵ(Ω)\eta = \frac{E_x}{H_y} = \sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}} \quad (\Omega)

In Free Space (σ=0,μr=1,ϵr=1\sigma = 0, \mu_r = 1, \epsilon_r = 1):

η0=μ0ϵ0=4π×10−78.854×10−12≈120π≈377 Ω\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = \sqrt{\frac{4\pi \times 10^{-7}}{8.854 \times 10^{-12}}} \approx 120\pi \approx 377\ \Omega vp=c=1μ0ϵ0≈3×108 m/sv_p = c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3 \times 10^8\ \text{m/s}

Low-Loss Dielectrics vs. Good Conductors

The ratio σωϵ\frac{\sigma}{\omega \epsilon} is known as the loss tangent (tan⁡δ\tan\delta):

  1. Good Dielectrics (σωϵ≪1\frac{\sigma}{\omega \epsilon} \ll 1):

    α≈σ2μϵ(Np/m),β≈ωμϵ(rad/m),η≈μϵ\alpha \approx \frac{\sigma}{2} \sqrt{\frac{\mu}{\epsilon}} \quad (\text{Np/m}), \quad \beta \approx \omega \sqrt{\mu \epsilon} \quad (\text{rad/m}), \quad \eta \approx \sqrt{\frac{\mu}{\epsilon}}
  2. Good Conductors (σωϵ≫1\frac{\sigma}{\omega \epsilon} \gg 1):

    α=β=πfμσ(m−1),η=(1+j)πfμσ\alpha = \beta = \sqrt{\pi f \mu \sigma} \quad (\text{m}^{-1}), \quad \eta = (1 + j)\sqrt{\frac{\pi f \mu}{\sigma}}

    The depth at which the wave amplitude decays to 1/e1/e (36.8%36.8\%) of its surface value is the Skin Depth (δ\delta):

    δ=1α=1πfμσ(m)\delta = \frac{1}{\alpha} = \frac{1}{\sqrt{\pi f \mu \sigma}} \quad (\text{m})

2. Poynting Vector & Electromagnetic Power Flow

Electromagnetic fields transport energy through space. The instantaneous rate of energy flow per unit area is defined by the Poynting Vector S\mathbf{S}:

S=E×H(W/m2)\mathbf{S} = \mathbf{E} \times \mathbf{H} \quad (\text{W/m}^2)

For time-harmonic fields, the Time-Average Poynting Vector Pavg\mathbf{P}_{\text{avg}} represents true real power density:

Pavg=12Re{Es×Hs∗}=∣E0∣22∣η∣e−2αzcos⁡(θn)az(W/m2)\mathbf{P}_{\text{avg}} = \frac{1}{2} \text{Re} \left\{ \mathbf{E}_s \times \mathbf{H}_s^* \right\} = \frac{|E_0|^2}{2 |\eta|} e^{-2\alpha z} \cos(\theta_n) \mathbf{a}_z \quad (\text{W/m}^2)

where θn\theta_n is the phase angle of the intrinsic impedance η\eta. Total power crossing a surface SS is obtained by integrating over the area:

Ptotal=∬SPavg⋅dA(Watts)P_{\text{total}} = \iint_S \mathbf{P}_{\text{avg}} \cdot d\mathbf{A} \quad (\text{Watts})

3. Distributed Transmission Line Parameters & Telegrapher's Equations

A uniform transmission line is modeled using per-unit-length distributed parameters:

  • RR: Series resistance per unit length (Ω/m\Omega/\text{m})
  • LL: Series inductance per unit length (H/m\text{H/m})
  • GG: Shunt conductance per unit length (S/m\text{S/m})
  • CC: Shunt capacitance per unit length (F/m\text{F/m})

Telegrapher's Equations

Applying Kirchhoff's laws to an incremental line section dzdz yields the time-domain Telegrapher's Equations:

∂v(z,t)∂z=−Ri(z,t)−L∂i(z,t)∂t\frac{\partial v(z,t)}{\partial z} = -R i(z,t) - L \frac{\partial i(z,t)}{\partial t} ∂i(z,t)∂z=−Gv(z,t)−C∂v(z,t)∂t\frac{\partial i(z,t)}{\partial z} = -G v(z,t) - C \frac{\partial v(z,t)}{\partial t}

In the frequency domain (phasor form):

d2Vs(z)dz2=γ2Vs(z),d2Is(z)dz2=γ2Is(z)\frac{d^2 V_s(z)}{dz^2} = \gamma^2 V_s(z), \quad \frac{d^2 I_s(z)}{dz^2} = \gamma^2 I_s(z)

where the line propagation constant is γ=(R+jωL)(G+jωC)=α+jβ\gamma = \sqrt{(R + j\omega L)(G + j\omega C)} = \alpha + j\beta.

Characteristic Impedance (Z_0)

The ratio of forward voltage wave to forward current wave is the Characteristic Impedance Z0Z_0:

Z0=R+jωLG+jωC(Ω)Z_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}} \quad (\Omega)
  • Lossless Line (R=0,G=0R = 0, G = 0):

    Z0=LC(Purely real resistance),α=0,β=ωLCZ_0 = \sqrt{\frac{L}{C}} \quad (\text{Purely real resistance}), \quad \alpha = 0, \quad \beta = \omega \sqrt{LC}
  • Distortionless Line (Condition: RL=GC\frac{R}{L} = \frac{G}{C}):

    Z0=LC,α=RG,β=ωLCZ_0 = \sqrt{\frac{L}{C}}, \quad \alpha = \sqrt{RG}, \quad \beta = \omega \sqrt{LC}

    A distortionless line transmits signals without amplitude distortion across frequency components because α\alpha is independent of frequency ω\omega.


4. Line Reflections, VSWR & Impedance Matching

When a transmission line with characteristic impedance Z0Z_0 is terminated by a load impedance ZL≠Z0Z_L \neq Z_0, a portion of the incident wave reflects back toward the source.

Voltage Reflection Coefficient (ΓL\Gamma_L)

The ratio of the reflected voltage wave V0−V_0^- to the incident voltage wave V0+V_0^+ at the load is:

ΓL=ZL−Z0ZL+Z0=∣ΓL∣ejθr\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = |\Gamma_L| e^{j\theta_r}

Key Load Termination Cases:

  1. Matched Load (ZL=Z0Z_L = Z_0): ΓL=0\Gamma_L = 0 (No reflected wave; maximum power transfer).
  2. Open-Circuit Load (ZL=∞Z_L = \infty): ΓL=+1\Gamma_L = +1 (Total voltage reflection in-phase; Vload=2V0+V_{\text{load}} = 2 V_0^+).
  3. Short-Circuit Load (ZL=0Z_L = 0): ΓL=−1\Gamma_L = -1 (Total voltage reflection 180∘180^\circ out-of-phase; Vload=0V_{\text{load}} = 0).
  4. Purely Reactive Load (ZL=jXZ_L = jX): ∣ΓL∣=1|\Gamma_L| = 1 (Total power reflection).

Voltage Standing Wave Ratio (VSWR)

The superposition of incident and reflected waves creates a standing wave pattern. The ratio of peak voltage to minimum voltage along the line is:

VSWR=VmaxVmin=1+∣ΓL∣1−∣ΓL∣VSWR = \frac{V_{\text{max}}}{V_{\text{min}}} = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|}
  • Bounds: 1≤VSWR≤∞1 \le VSWR \le \infty.
  • VSWR=1.0VSWR = 1.0 indicates a perfectly matched line (∣ΓL∣=0|\Gamma_L| = 0).

Input Impedance at Distance l from Load

The input impedance Zin(l)Z_{\text{in}}(l) looking into a lossless transmission line of length ll terminated by load ZLZ_L is:

Zin(l)=Z0[ZL+jZ0tan⁡(βl)Z0+jZLtan⁡(βl)]Z_{\text{in}}(l) = Z_0 \left[ \frac{Z_L + j Z_0 \tan(\beta l)}{Z_0 + j Z_L \tan(\beta l)} \right]
  • Quarter-Wave Line (l=λ/4l = \lambda/4, βl=π/2\beta l = \pi/2): Zin=Z02ZL(Quarter-wave impedance transformer)Z_{\text{in}} = \frac{Z_0^2}{Z_L} \quad (\text{Quarter-wave impedance transformer})
  • Half-Wave Line (l=λ/2l = \lambda/2, βl=π\beta l = \pi): Zin=ZL(Impedance repeats every half-wavelength)Z_{\text{in}} = Z_L \quad (\text{Impedance repeats every half-wavelength})

5. Transient Surge Propagation on Power Lines (Bounce Diagrams)

When a circuit breaker closes or lightning strikes an overhead transmission line, a high-voltage step wave V1V_1 travels along the line at the surge velocity v=1LCv = \frac{1}{\sqrt{LC}}.

  • Surge Impedance: Zc=LCZ_c = \sqrt{\frac{L}{C}}
  • Sending-End Reflection Coefficient: ΓS=ZS−Z0ZS+Z0\Gamma_S = \frac{Z_S - Z_0}{Z_S + Z_0}
  • Receiving-End Reflection Coefficient: ΓR=ZR−Z0ZR+Z0\Gamma_R = \frac{Z_R - Z_0}{Z_R + Z_0}

A Bewley Lattice Diagram (Bounce Diagram) tracks successive position-time reflections (t1=l/v,2t1,3t1t_1 = l/v, 2t_1, 3t_1) to compute transient voltage profiles at any line position.


Solved Board Exam Numerical Problems

Problem 1: Intrinsic Impedance & Power Density of an EM Wave

Question: An electromagnetic wave operating at f=100 MHzf = 100\ \text{MHz} propagates through a lossless, non-magnetic medium (μr=1.0\mu_r = 1.0) with a relative permittivity ϵr=9.0\epsilon_r = 9.0. The electric field amplitude is E0=120 V/mE_0 = 120\ \text{V/m}. Calculate: (a) the phase velocity vpv_p, (b) the wavelength λ\lambda, (c) the intrinsic impedance η\eta, and (d) the average power density PavgP_{\text{avg}}.

Solution:

  1. Calculate phase velocity vpv_p:

    vp=cϵr=3.0×108 m/s9.0=1.0×108 m/sv_p = \frac{c}{\sqrt{\epsilon_r}} = \frac{3.0 \times 10^8\ \text{m/s}}{\sqrt{9.0}} = 1.0 \times 10^8\ \text{m/s}
  2. Calculate wavelength λ\lambda:

    λ=vpf=1.0×108 m/s100×106 Hz=1.0 meter\lambda = \frac{v_p}{f} = \frac{1.0 \times 10^8\ \text{m/s}}{100 \times 10^6\ \text{Hz}} = 1.0\ \text{meter}
  3. Calculate intrinsic impedance η\eta:

    η=η0ϵr=377 Ω9.0=3773=125.67 Ω\eta = \frac{\eta_0}{\sqrt{\epsilon_r}} = \frac{377\ \Omega}{\sqrt{9.0}} = \frac{377}{3} = 125.67\ \Omega
  4. Calculate average power density PavgP_{\text{avg}}:

    Pavg=E022η=(120)22×125.67=14,400251.34=57.28 W/m2P_{\text{avg}} = \frac{E_0^2}{2\eta} = \frac{(120)^2}{2 \times 125.67} = \frac{14,400}{251.34} = 57.28\ \text{W/m}^2

Problem 2: Transmission Line Reflection & VSWR Calculation

Question: A 50 Ω50\ \Omega lossless transmission line is terminated by an antenna load impedance ZL=75+j50 ΩZ_L = 75 + j50\ \Omega. Determine: (a) the load reflection coefficient ΓL\Gamma_L in polar form, (b) the reflection coefficient magnitude ∣ΓL∣|\Gamma_L|, and (c) the Voltage Standing Wave Ratio (VSWRVSWR).

Solution:

  1. Compute load reflection coefficient ΓL\Gamma_L:

    ΓL=ZL−Z0ZL+Z0=(75+j50)−50(75+j50)+50=25+j50125+j50\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{(75 + j50) - 50}{(75 + j50) + 50} = \frac{25 + j50}{125 + j50}
  2. Convert numerator and denominator to polar coordinates:

    Numerator: 25+j50=252+502∠arctan⁡(50/25)=55.90∠63.43∘\text{Numerator: } 25 + j50 = \sqrt{25^2 + 50^2} \angle \arctan(50/25) = 55.90 \angle 63.43^\circ Denominator: 125+j50=1252+502∠arctan⁡(50/125)=134.63∠21.80∘\text{Denominator: } 125 + j50 = \sqrt{125^2 + 50^2} \angle \arctan(50/125) = 134.63 \angle 21.80^\circ ΓL=55.90∠63.43∘134.63∠21.80∘=0.4152∠41.63∘\Gamma_L = \frac{55.90 \angle 63.43^\circ}{134.63 \angle 21.80^\circ} = 0.4152 \angle 41.63^\circ
  3. Extract magnitude ∣ΓL∣|\Gamma_L|:

    ∣ΓL∣=0.4152|\Gamma_L| = 0.4152
  4. Calculate VSWRVSWR:

    VSWR=1+∣ΓL∣1−∣ΓL∣=1+0.41521−0.4152=1.41520.5848=2.42VSWR = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} = \frac{1 + 0.4152}{1 - 0.4152} = \frac{1.4152}{0.5848} = 2.42
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Transmission Line Wave Reflection and Standing Wave Metrics
Voltage Standing Wave Ratio (VSWR) vs. Reflection Coefficient Magnitude
Test Your Knowledge

What is the intrinsic impedance of a non-magnetic dielectric medium having a relative permittivity εr = 4.0?

A

377.0 Ω

B

754.0 Ω

C

94.2 Ω

D

188.5 Ω

Test Your Knowledge

A 50 Ω lossless transmission line is terminated by an open circuit (ZL = ∞). What are the values of the voltage reflection coefficient ΓL and the Voltage Standing Wave Ratio (VSWR)?

A

ΓL = +1, VSWR = ∞

B

ΓL = 0, VSWR = 1.0

C

ΓL = -1, VSWR = ∞

D

ΓL = +0.5, VSWR = 3.0

Test Your Knowledge

Which specific condition must be satisfied for a lossy transmission line to operate as a distortionless line?

A

R / C = G / L

B

R / L = G / C

C

R · C = G · L

D

R + jωL = G + jωC

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