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13.2 Electromagnetic Waves & Transmission Line Fields
PRC Registered Electrical Engineer REE
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13.2 Electromagnetic Waves & Transmission Line Fields
Key Takeaways Explain This Topic AI
Uniform plane wave propagation in lossy media is governed by the complex propagation constant γ = α + j β = j ω μ ( σ + j ω ϵ ) \gamma = \alpha + j\beta = \sqrt{j\omega\mu (\sigma + j\omega\epsilon)} γ = α + j β = jω μ ( σ + jω ϵ ) , where α \alpha α represents the attenuation constant in Np/m \text{Np/m} Np/m and β \beta β represents the phase constant in rad/m \text{rad/m} rad/m .
Intrinsic impedance η = j ω μ σ + j ω ϵ \eta = \sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}} η = σ + jω ϵ jω μ defines the ratio of electric to magnetic field strength, simplifying to η 0 = μ 0 ϵ 0 ≈ 377 Ω \eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 377\ \Omega η 0 = ϵ 0 μ 0 ≈ 377 Ω (or 120 π Ω 120\pi\ \Omega 120 π Ω ) in free space, with phase velocity v p = 1 μ ϵ = 3 × 10 8 m/s v_p = \frac{1}{\sqrt{\mu \epsilon}} = 3 \times 10^8\ \text{m/s} v p = μ ϵ 1 = 3 × 1 0 8 m/s .
The Poynting vector S = E × H \mathbf{S} = \mathbf{E} \times \mathbf{H} S = E × H represents instantaneous electromagnetic power flux density ( W/m 2 \text{W/m}^2 W/m 2 ), with time-average power density P avg = 1 2 Re { E × H ∗ } = ∣ E 0 ∣ 2 2 η a z \mathbf{P}_{\text{avg}} = \frac{1}{2} \text{Re}\{\mathbf{E} \times \mathbf{H}^*\} = \frac{|E_0|^2}{2\eta}\ \mathbf{a}_z P avg = 2 1 Re { E × H ∗ } = 2 η ∣ E 0 ∣ 2 a z .
Transmission line field behavior is modeled by distributed primary parameters (R, L, G, C) yielding characteristic impedance Z 0 = R + j ω L G + j ω C Z_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}} Z 0 = G + jω C R + jω L , which reduces to Z 0 = L C Z_0 = \sqrt{\frac{L}{C}} Z 0 = C L for lossless lines.
Impedance mismatch at line terminations creates wave reflections governed by the voltage reflection coefficient Γ L = Z L − Z 0 Z L + Z 0 \Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} Γ L = Z L + Z 0 Z L − Z 0 and Voltage Standing Wave Ratio V S W R = 1 + ∣ Γ L ∣ 1 − ∣ Γ L ∣ VSWR = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} V S W R = 1 − ∣ Γ L ∣ 1 + ∣ Γ 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 Z 0 Z_0 Z 0 , reflection coefficients, standing wave ratios ( V S W R VSWR V S W R ), 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 σ , permeability μ \mu μ , and permittivity ϵ \epsilon ϵ yields the vector wave equation for time-harmonic electric fields: ∇ 2 E s − γ 2 E s = 0 \nabla^2 \mathbf{E}_s - \gamma^2 \mathbf{E}_s = 0 ∇ 2 E s − γ 2 E s = 0
where E s \mathbf{E}_s E s is the phasor electric field intensity, and γ \gamma γ is the Complex Propagation Constant : γ = α + j β = j ω μ ( σ + j ω ϵ ) ( m − 1 ) \gamma = \alpha + j\beta = \sqrt{j\omega\mu (\sigma + j\omega\epsilon)} \quad (\text{m}^{-1}) γ = α + j β = jω μ ( σ + jω ϵ ) ( m − 1 )
Attenuation Constant ( α \alpha α ) & Phase Constant ( β \beta β )
Attenuation Constant ( α \alpha α ): 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} dB/m , where 1 Np = 8.686 dB 1\ \text{Np} = 8.686\ \text{dB} 1 Np = 8.686 dB ).
Phase Constant ( β \beta β ): Measures phase shift per unit distance, expressed in radians per meter (rad/m) : β = 2 π λ = ω v p \beta = \frac{2\pi}{\lambda} = \frac{\omega}{v_p} β = λ 2 π = v p ω
Intrinsic Impedance ( η \eta η ) & Phase Velocity ( v p v_p v p )
The ratio of the transverse electric field to transverse magnetic field for a forward-propagating wave is the Intrinsic Impedance η \eta η : η = E x H y = j ω μ σ + j ω ϵ ( Ω ) \eta = \frac{E_x}{H_y} = \sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}} \quad (\Omega) η = H y E x = σ + jω ϵ jω μ ( Ω )
In Free Space ( σ = 0 , μ r = 1 , ϵ r = 1 \sigma = 0, \mu_r = 1, \epsilon_r = 1 σ = 0 , μ r = 1 , ϵ r = 1 ): η 0 = μ 0 ϵ 0 = 4 π × 10 − 7 8.854 × 10 − 12 ≈ 120 π ≈ 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 η 0 = ϵ 0 μ 0 = 8.854 × 1 0 − 12 4 π × 1 0 − 7 ≈ 120 π ≈ 377 Ω v p = c = 1 μ 0 ϵ 0 ≈ 3 × 10 8 m/s v_p = c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3 \times 10^8\ \text{m/s} v p = c = μ 0 ϵ 0 1 ≈ 3 × 1 0 8 m/s
Low-Loss Dielectrics vs. Good Conductors
The ratio σ ω ϵ \frac{\sigma}{\omega \epsilon} ω ϵ σ is known as the loss tangent ( tan δ \tan\delta tan δ ):
Good Dielectrics ( σ ω ϵ ≪ 1 \frac{\sigma}{\omega \epsilon} \ll 1 ω ϵ σ ≪ 1 ): α ≈ σ 2 μ ϵ ( Np/m ) , β ≈ ω μ ϵ ( rad/m ) , η ≈ μ ϵ \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}} α ≈ 2 σ ϵ μ ( Np/m ) , β ≈ ω μ ϵ ( rad/m ) , η ≈ ϵ μ
Good Conductors ( σ ω ϵ ≫ 1 \frac{\sigma}{\omega \epsilon} \gg 1 ω ϵ σ ≫ 1 ): α = β = π f μ σ ( m − 1 ) , η = ( 1 + j ) π f μ σ \alpha = \beta = \sqrt{\pi f \mu \sigma} \quad (\text{m}^{-1}), \quad \eta = (1 + j)\sqrt{\frac{\pi f \mu}{\sigma}} α = β = π f μ σ ( m − 1 ) , η = ( 1 + j ) σ π f μ
The depth at which the wave amplitude decays to 1 / e 1/e 1/ e ( 36.8 % 36.8\% 36.8% ) of its surface value is the Skin Depth ( δ \delta δ ) : δ = 1 α = 1 π f μ σ ( m ) \delta = \frac{1}{\alpha} = \frac{1}{\sqrt{\pi f \mu \sigma}} \quad (\text{m}) δ = α 1 = π f μ σ 1 ( 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 : S = E × H ( W/m 2 ) \mathbf{S} = \mathbf{E} \times \mathbf{H} \quad (\text{W/m}^2) S = E × H ( W/m 2 )
For time-harmonic fields, the Time-Average Poynting Vector P avg \mathbf{P}_{\text{avg}} P avg represents true real power density: P avg = 1 2 Re { E s × H s ∗ } = ∣ E 0 ∣ 2 2 ∣ η ∣ e − 2 α z cos ( θ n ) a z ( W/m 2 ) \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) P avg = 2 1 Re { E s × H s ∗ } = 2∣ η ∣ ∣ E 0 ∣ 2 e − 2 α z cos ( θ n ) a z ( W/m 2 )
where θ n \theta_n θ n is the phase angle of the intrinsic impedance η \eta η . Total power crossing a surface S S S is obtained by integrating over the area: P total = ∬ S P avg ⋅ d A ( Watts ) P_{\text{total}} = \iint_S \mathbf{P}_{\text{avg}} \cdot d\mathbf{A} \quad (\text{Watts}) P total = ∬ S P avg ⋅ d A ( Watts )
3. Distributed Transmission Line Parameters & Telegrapher's Equations
A uniform transmission line is modeled using per-unit-length distributed parameters:
- R R R : Series resistance per unit length ( Ω / m \Omega/\text{m} Ω/ m )
- L L L : Series inductance per unit length ( H/m \text{H/m} H/m )
- G G G : Shunt conductance per unit length ( S/m \text{S/m} S/m )
- C C C : Shunt capacitance per unit length ( F/m \text{F/m} F/m )
- Telegrapher's Equations
Applying Kirchhoff's laws to an incremental line section d z dz d z yields the time-domain Telegrapher's Equations : ∂ v ( z , t ) ∂ z = − R i ( 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} ∂ z ∂ v ( z , t ) = − R i ( z , t ) − L ∂ t ∂ i ( z , t ) ∂ i ( z , t ) ∂ z = − G v ( 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} ∂ z ∂ i ( z , t ) = − G v ( z , t ) − C ∂ t ∂ v ( z , t )
In the frequency domain (phasor form): d 2 V s ( z ) d z 2 = γ 2 V s ( z ) , d 2 I s ( z ) d z 2 = γ 2 I s ( 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) d z 2 d 2 V s ( z ) = γ 2 V s ( z ) , d z 2 d 2 I s ( z ) = γ 2 I s ( z )
where the line propagation constant is γ = ( R + j ω L ) ( G + j ω C ) = α + j β \gamma = \sqrt{(R + j\omega L)(G + j\omega C)} = \alpha + j\beta γ = ( R + jω L ) ( G + jω C ) = α + j β .
Characteristic Impedance (Z_0)
The ratio of forward voltage wave to forward current wave is the Characteristic Impedance Z 0 Z_0 Z 0 : Z 0 = R + j ω L G + j ω C ( Ω ) Z_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}} \quad (\Omega) Z 0 = G + jω C R + jω L ( Ω )
Lossless Line ( R = 0 , G = 0 R = 0, G = 0 R = 0 , G = 0 ): Z 0 = L C ( Purely real resistance ) , α = 0 , β = ω L C Z_0 = \sqrt{\frac{L}{C}} \quad (\text{Purely real resistance}), \quad \alpha = 0, \quad \beta = \omega \sqrt{LC} Z 0 = C L ( Purely real resistance ) , α = 0 , β = ω L C
Distortionless Line (Condition: R L = G C \frac{R}{L} = \frac{G}{C} L R = C G ): Z 0 = L C , α = R G , β = ω L C Z_0 = \sqrt{\frac{L}{C}}, \quad \alpha = \sqrt{RG}, \quad \beta = \omega \sqrt{LC} Z 0 = C L , α = RG , β = ω L C
A distortionless line transmits signals without amplitude distortion across frequency components because α \alpha α is independent of frequency ω \omega ω .
4. Line Reflections, VSWR & Impedance Matching
When a transmission line with characteristic impedance Z 0 Z_0 Z 0 is terminated by a load impedance Z L ≠ Z 0 Z_L \neq Z_0 Z L = Z 0 , a portion of the incident wave reflects back toward the source.
Voltage Reflection Coefficient ( Γ L \Gamma_L Γ L )
The ratio of the reflected voltage wave V 0 − V_0^- V 0 − to the incident voltage wave V 0 + V_0^+ V 0 + at the load is: Γ L = Z L − Z 0 Z L + Z 0 = ∣ Γ L ∣ e j θ r \Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = |\Gamma_L| e^{j\theta_r} Γ L = Z L + Z 0 Z L − Z 0 = ∣ Γ L ∣ e j θ r
Key Load Termination Cases:
Matched Load ( Z L = Z 0 Z_L = Z_0 Z L = Z 0 ): Γ L = 0 \Gamma_L = 0 Γ L = 0 (No reflected wave; maximum power transfer).
Open-Circuit Load ( Z L = ∞ Z_L = \infty Z L = ∞ ): Γ L = + 1 \Gamma_L = +1 Γ L = + 1 (Total voltage reflection in-phase; V load = 2 V 0 + V_{\text{load}} = 2 V_0^+ V load = 2 V 0 + ).
Short-Circuit Load ( Z L = 0 Z_L = 0 Z L = 0 ): Γ L = − 1 \Gamma_L = -1 Γ L = − 1 (Total voltage reflection 180 ∘ 180^\circ 18 0 ∘ out-of-phase; V load = 0 V_{\text{load}} = 0 V load = 0 ).
Purely Reactive Load ( Z L = j X Z_L = jX Z L = j X ): ∣ Γ L ∣ = 1 |\Gamma_L| = 1 ∣ Γ 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: V S W R = V max V min = 1 + ∣ Γ L ∣ 1 − ∣ Γ L ∣ VSWR = \frac{V_{\text{max}}}{V_{\text{min}}} = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} V S W R = V min V max = 1 − ∣ Γ L ∣ 1 + ∣ Γ L ∣
Bounds: 1 ≤ V S W R ≤ ∞ 1 \le VSWR \le \infty 1 ≤ V S W R ≤ ∞ .
V S W R = 1.0 VSWR = 1.0 V S W R = 1.0 indicates a perfectly matched line ( ∣ Γ L ∣ = 0 |\Gamma_L| = 0 ∣ Γ L ∣ = 0 ).
Input Impedance at Distance l from Load
The input impedance Z in ( l ) Z_{\text{in}}(l) Z in ( l ) looking into a lossless transmission line of length l l l terminated by load Z L Z_L Z L is: Z in ( l ) = Z 0 [ Z L + j Z 0 tan ( β l ) Z 0 + j Z L tan ( β l ) ] 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] Z in ( l ) = Z 0 [ Z 0 + j Z L tan ( βl ) Z L + j Z 0 tan ( βl ) ]
Quarter-Wave Line ( l = λ / 4 l = \lambda/4 l = λ /4 , β l = π / 2 \beta l = \pi/2 βl = π /2 ): Z in = Z 0 2 Z L ( Quarter-wave impedance transformer ) Z_{\text{in}} = \frac{Z_0^2}{Z_L} \quad (\text{Quarter-wave impedance transformer}) Z in = Z L Z 0 2 ( Quarter-wave impedance transformer )
Half-Wave Line ( l = λ / 2 l = \lambda/2 l = λ /2 , β l = π \beta l = \pi βl = π ): Z in = Z L ( Impedance repeats every half-wavelength ) Z_{\text{in}} = Z_L \quad (\text{Impedance repeats every half-wavelength}) Z in = Z L ( 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 V 1 V_1 V 1 travels along the line at the surge velocity v = 1 L C v = \frac{1}{\sqrt{LC}} v = L C 1 .
Surge Impedance: Z c = L C Z_c = \sqrt{\frac{L}{C}} Z c = C L
Sending-End Reflection Coefficient: Γ S = Z S − Z 0 Z S + Z 0 \Gamma_S = \frac{Z_S - Z_0}{Z_S + Z_0} Γ S = Z S + Z 0 Z S − Z 0
Receiving-End Reflection Coefficient: Γ R = Z R − Z 0 Z R + Z 0 \Gamma_R = \frac{Z_R - Z_0}{Z_R + Z_0} Γ R = Z R + Z 0 Z R − Z 0
A Bewley Lattice Diagram (Bounce Diagram) tracks successive position-time reflections ( t 1 = l / v , 2 t 1 , 3 t 1 t_1 = l/v, 2t_1, 3t_1 t 1 = l / v , 2 t 1 , 3 t 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 MHz f = 100\ \text{MHz} f = 100 MHz propagates through a lossless, non-magnetic medium ( μ r = 1.0 \mu_r = 1.0 μ r = 1.0 ) with a relative permittivity ϵ r = 9.0 \epsilon_r = 9.0 ϵ r = 9.0 . The electric field amplitude is E 0 = 120 V/m E_0 = 120\ \text{V/m} E 0 = 120 V/m . Calculate: (a) the phase velocity v p v_p v p , (b) the wavelength λ \lambda λ , (c) the intrinsic impedance η \eta η , and (d) the average power density P avg P_{\text{avg}} P avg .
Solution:
Calculate phase velocity v p v_p v p : v p = c ϵ r = 3.0 × 10 8 m/s 9.0 = 1.0 × 10 8 m/s v_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} v p = ϵ r c = 9.0 3.0 × 1 0 8 m/s = 1.0 × 1 0 8 m/s
Calculate wavelength λ \lambda λ : λ = v p f = 1.0 × 10 8 m/s 100 × 10 6 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} λ = f v p = 100 × 1 0 6 Hz 1.0 × 1 0 8 m/s = 1.0 meter
Calculate intrinsic impedance η \eta η : η = η 0 ϵ r = 377 Ω 9.0 = 377 3 = 125.67 Ω \eta = \frac{\eta_0}{\sqrt{\epsilon_r}} = \frac{377\ \Omega}{\sqrt{9.0}} = \frac{377}{3} = 125.67\ \Omega η = ϵ r η 0 = 9.0 377 Ω = 3 377 = 125.67 Ω
Calculate average power density P avg P_{\text{avg}} P avg : P avg = E 0 2 2 η = ( 120 ) 2 2 × 125.67 = 14 , 400 251.34 = 57.28 W/m 2 P_{\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 P avg = 2 η E 0 2 = 2 × 125.67 ( 120 ) 2 = 251.34 14 , 400 = 57.28 W/m 2
Problem 2: Transmission Line Reflection & VSWR Calculation
Question: A 50 Ω 50\ \Omega 50 Ω lossless transmission line is terminated by an antenna load impedance Z L = 75 + j 50 Ω Z_L = 75 + j50\ \Omega Z L = 75 + j 50 Ω . Determine: (a) the load reflection coefficient Γ L \Gamma_L Γ L in polar form, (b) the reflection coefficient magnitude ∣ Γ L ∣ |\Gamma_L| ∣ Γ L ∣ , and (c) the Voltage Standing Wave Ratio ( V S W R VSWR V S W R ).
Solution:
Compute load reflection coefficient Γ L \Gamma_L Γ L : Γ L = Z L − Z 0 Z L + Z 0 = ( 75 + j 50 ) − 50 ( 75 + j 50 ) + 50 = 25 + j 50 125 + j 50 \Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{(75 + j50) - 50}{(75 + j50) + 50} = \frac{25 + j50}{125 + j50} Γ L = Z L + Z 0 Z L − Z 0 = ( 75 + j 50 ) + 50 ( 75 + j 50 ) − 50 = 125 + j 50 25 + j 50
Convert numerator and denominator to polar coordinates: Numerator: 25 + j 50 = 25 2 + 50 2 ∠ 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 Numerator: 25 + j 50 = 2 5 2 + 5 0 2 ∠ arctan ( 50/25 ) = 55.90∠63.4 3 ∘ Denominator: 125 + j 50 = 125 2 + 50 2 ∠ 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 Denominator: 125 + j 50 = 12 5 2 + 5 0 2 ∠ arctan ( 50/125 ) = 134.63∠21.8 0 ∘ Γ L = 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 Γ L = 134.63∠21.8 0 ∘ 55.90∠63.4 3 ∘ = 0.4152∠41.6 3 ∘
Extract magnitude ∣ Γ L ∣ |\Gamma_L| ∣ Γ L ∣ : ∣ Γ L ∣ = 0.4152 |\Gamma_L| = 0.4152 ∣ Γ L ∣ = 0.4152
Calculate V S W R VSWR V S W R : V S W R = 1 + ∣ Γ L ∣ 1 − ∣ Γ L ∣ = 1 + 0.4152 1 − 0.4152 = 1.4152 0.5848 = 2.42 VSWR = \frac{1 + |\Gamma_L|}{1 - |\Gamma_L|} = \frac{1 + 0.4152}{1 - 0.4152} = \frac{1.4152}{0.5848} = 2.42 V S W R = 1 − ∣ Γ L ∣ 1 + ∣ Γ L ∣ = 1 − 0.4152 1 + 0.4152 = 0.5848 1.4152 = 2.42