Electromagnetic Waves, Antennas, and RF: Boundary Waves - Exercises and Complete Worked Solutions

Electromagnetic Waves, Antennas, and RF: Boundary Waves - Exercises and Complete Worked Solutions

EM21 derived electromagnetic boundary conditions and used them to obtain reflection, transmission, refraction, Fresnel coefficients, Brewster angle, critical angle, total internal reflection, and standing-wave ratio. This companion article turns those results into a sequence of worked problems. The emphasis is on preserving the distinction among field amplitudes, Power fractions, propagation directions, and polarization conventions [1, 2, 3, 4, 5].

The most frequently used normal-incidence relations are

(1)

with, for lossless media,

(2)

At oblique incidence the interface geometry matters, and TE and TM polarizations must be treated separately.

How to use this problem set

Attempt the exercises in Part I before reading Part II. For every interface problem, proceed in the same order:

1. identify the medium parameters and propagation directions;
2. determine whether the problem is Normal or oblique incidence;
3. determine whether TE/ polarization matters;
4. apply field-amplitude coefficients first;
5. convert to power coefficients only after accounting for impedance and, at oblique incidence, the normal Poynting-flux projection.

Part I: Exercises

Exercise 1: boundary conditions with surface charge and surface current

An interface lies in the xy plane at z = 0, with unit normal n = z pointing from medium 1 into medium 2. The permittivities are

(3)

The normal electric field immediately below the interface is

(4)

and the free surface-charge density is

(5)

Find E_2n. If in addition

(6)

find the tangential magnetic field H₂.

Exercise 2: derive the normal-incidence coefficients

For a normally incident plane wave traveling from a lossless medium of impedance η₁ into a lossless medium of impedance η₂, derive

(7)

from continuity of tangential E and H. Show that

(8)

and verify that τ = 1 + Γ at normal incidence.

Exercise 3: air to a dielectric with 𝜖_r = 4

A 12 V/m sinusoidal plane wave in air strikes normally a lossless, nonmagnetic dielectric with 𝜖_r = 4. Find

1. η₂;
2. Γ and τ;
3. reflected and transmitted electric-field amplitudes;
4. R and T;
5. incident, reflected, and transmitted average power densities.

Exercise 4: dielectric to air and a transmitted field larger than the incident field

Reverse Exercise 3: a wave now travels normally from the 𝜖_r = 4 dielectric into air. Compute Γ, τ, R, and T. Explain how |τ| > 1 can occur without violating conservation of energy.

Exercise 5: prove normal-incidence power conservation

Starting from

(9)

for real positive η₁ and η₂, prove algebraically that

(10)

where

(11)

Exercise 6: Snell’s law from tangential phase matching

A plane wave travels from air, n₁ = 1, into glass, n₂ = 1.5, at an incidence angle

(12)

Using tangential phase matching, calculate the refraction angle 𝜃_t. Also state the reflection angle.

Exercise 7: TE Fresnel coefficient at 45^∘

For air-to-glass incidence with n₁ = 1, n₂ = 1.5, μ_r1 = μ_r2 = 1, and

(13)

calculate

(14)

Verify numerically that R_TE + T_TE = 1.

Exercise 8: TM Fresnel coefficient at 45^∘

Repeat Exercise 7 for TM polarization. Compute

(15)

and compare the reflected power with the TE result.

Exercise 9: Brewster angle and zero TM reflection

For air-to-glass incidence with n₁ = 1 and n₂ = 1.5:

1. calculate the Brewster angle 𝜃_B;
2. use Snell’s law to calculate 𝜃_t at Brewster incidence;
3. show that 𝜃_B + 𝜃_t = 90^∘;
4. substitute the angles into the TM Fresnel coefficient and verify that Γ_TM = 0.

Exercise 10: critical angle from glass to air

A wave travels from glass with n₁ = 1.5 into air with n₂ = 1. Find the critical angle. Classify the transmitted field for incidence angles of 30^∘, 41.0^∘, 41.81^∘, and 50^∘.

Exercise 11: total internal reflection and evanescent penetration depth

A 10 GHz wave travels from a nonmagnetic dielectric with n₁ = 1.5 into air at

(16)

Show that the wave is above the critical angle. The tangential Wavenumber is conserved, and the normal wavenumber in air becomes imaginary:

(17)

with

(18)

Calculate the evanescent 1∕e penetration depth

(19)

Also verify that the magnitude of the TE reflection coefficient is unity.

Exercise 12: SWR from a known reflection coefficient

A lossless incident region has

(20)

Find the standing-wave ratio, the reflected power fraction, and the normalized field extrema |E|_max∕|E_i| and |E|_min∕|E_i|.

Exercise 13: infer reflection coefficient from SWR

A measured standing-wave ratio is

(21)

Find |Γ| and the reflected power fraction R. If the incident electric-field amplitude is 10 V/m, find the maximum and minimum field magnitudes in the standing-wave pattern.

Exercise 14: phase reversal and standing-wave location

At normal incidence, compare the two cases

(22)

At the interface z = 0, calculate the total electric-field amplitude relative to E_i0. Explain how the sign of Γ shifts a field maximum into a field minimum at the boundary.

Exercise 15: perfect electric conductor

A normally incident plane wave in free space strikes a perfect electric Conductor at z = 0. Derive

(23)

Show that the tangential electric field vanishes at the surface, the reflected power fraction is unity, and the magnetic field amplitude at the surface is twice the incident magnetic-field amplitude. What is the ideal SWR?

Exercise 16: Julia Fresnel and SWR sweep

Write a Julia program that, for a nonmagnetic air-to-glass interface with n₁ = 1 and n₂ = 1.5,

1. sweeps 𝜃_i from 0^∘ to 89^∘;
2. computes 𝜃_t from Snell’s law;
3. evaluates Γ_TE and Γ_TM;
4. computes R_TE and R_TM;
5. locates the numerical minimum of R_TM and compares it with the analytic Brewster angle;
6. computes the normal-incidence SWR from Γ(0).

Part II: Complete Worked Solutions

Solution 1: boundary conditions with surface charge and surface current

The normal electric-flux boundary condition is

(24)

Therefore

(25)

First,

D1n	= 𝜖1E1n	(26)
	= (2𝜖0)(100)	(27)
	= 1.7708 × 10−9 C/m2.	(28)

Thus

D2n	= 1.7708 × 10−9 + 5.0 × 10−9	(29)
	= 6.7708 × 10−9 C/m2.	(30)

Since D_2n = 𝜖₂E_2n,

E2n	=	(31)
	≈ 191.2 V/m.	(32)

Hence

(33)

For the magnetic field,

(34)

Write the tangential jump as

(35)

Then

(36)

Matching components with 3x gives

(37)

Thus

(38)

Since H₁ = 2y A/m,

(39)

The result demonstrates that surface charge controls the jump in normal D, while surface current controls the jump in tangential H.

Solution 2: derive the normal-incidence coefficients

At the interface, tangential electric-field continuity gives

(40)

The magnetic fields satisfy

(41)

The minus sign belongs to the reflected magnetic field because the reflected Poynting vector points opposite to the incident one. Continuity of tangential H gives

(42)

Divide both equations by E_i0:

(43)

(44)

Substitute τ = 1 + Γ:

(45)

Therefore

(46)

so

(47)

Then

τ	= 1 + Γ	(48)
	= 1 +	(49)
	= .	(50)

Thus

(51)

Solution 3: air to a dielectric with 𝜖_r = 4

For a nonmagnetic lossless dielectric,

(52)

Therefore

(53)

(54)

The reflection coefficient is

Γ	=	(55)
	= −.	(56)

Hence

(57)

The negative sign means a 180^∘ phase reversal of the reflected electric field.

The transmission coefficient is

(58)

so

(59)

For E_i0 = 12 V/m,

(60)

(61)

The reflected power fraction is

(62)

The transmitted power fraction is

T	= τ2	(63)
	= 22	(64)
	= ≈ 0.8889.	(65)

Thus

(66)

For a sinusoidal plane wave with peak electric-field amplitude E₀,

(67)

Hence

Si	= ≈ 0.1911 W/m2,	(68)
Sr	= ≈ 0.02124 W/m2,	(69)
St	= ≈ 0.1699 W/m2.	(70)

Therefore

(71)

within rounding.

Solution 4: dielectric to air and a transmitted field larger than the incident field

Now

(72)

Thus

(73)

so

(74)

The transmission coefficient is

(75)

so

(76)

The power coefficients are

(77)

T	= τ2	(78)
	= 2	(79)
	= = 0.8889.	(80)

Thus

(81)

There is no energy paradox. The electric-field amplitude is larger in the higher-impedance medium, but power density scales as E₀²∕(2η). The impedance increase offsets the larger transmitted electric-field amplitude.

Solution 5: prove normal-incidence power conservation

For real impedances,

(82)

Also,

T	= 2	(83)
	= .	(84)

Therefore

R + T	=	(85)
	=	(86)
	=	(87)
	= 1.	(88)

Hence

(89)

This is the boundary-wave statement of average power conservation for a lossless interface.

Solution 6: Snell’s law from tangential phase matching

Tangential phase continuity gives

(90)

Thus

sin 𝜃t	= sin 30∘	(91)
	= (0.5)	(92)
	= .	(93)

Therefore

(94)

The law of reflection gives

(95)

Solution 7: TE Fresnel coefficient at 45^∘

Snell’s law gives

(96)

so

(97)

For nonmagnetic media,

(98)

The TE reflection coefficient is

(99)

Substitution gives

(100)

The TE transmission coefficient is

(101)

so

(102)

The reflected power fraction is

(103)

At oblique incidence,

(104)

This gives

(105)

Therefore

(106)

Solution 8: TM Fresnel coefficient at 45^∘

The refraction angle is unchanged:

(107)

For TM polarization,

(108)

Hence

(109)

The electric-field transmission coefficient is

(110)

which gives

(111)

Then

(112)

The transmitted power fraction is

(113)

so

(114)

Again,

(115)

At 45^∘, TM reflection is already much weaker than TE reflection because the incidence angle is approaching the Brewster angle.

Solution 9: Brewster angle and zero TM reflection

For nonmagnetic lossless media,

(116)

Therefore

(117)

Snell’s law gives

(118)

which yields

(119)

Thus

(120)

For nonmagnetic media η ∝ 1∕n. The TM numerator is

(121)

At the Brewster geometry this vanishes, so

(122)

The absence of reflected TM power is therefore a direct boundary-condition result, not an independent empirical rule.

Solution 10: critical angle from glass to air

For incidence from the higher-index medium,

(123)

Hence

(124)

The cases are:

• 30^∘ < 𝜃 _c: ordinary propagating refraction occurs;
• 41.0^∘ < 𝜃 _c: propagating refraction still occurs, but the transmitted ray is very close to grazing;
• 41.81^∘ ≈ 𝜃 _c: the transmitted propagation direction is tangent to the interface, 𝜃_t = 90^∘;
• 50^∘ > 𝜃 _c: total internal reflection occurs and the field in medium 2 is evanescent in the normal direction.

Solution 11: total internal reflection and evanescent penetration depth

The critical angle is

(125)

so 50^∘ is above critical.

At 10 GHz,

(126)

The evanescent decay constant is

κ	= k0	(127)
	≈ 118.62 m−1.	(128)

Therefore

(129)

To verify total reflection, write

(130)

where

(131)

Insert this into

(132)

The numerator and denominator are complex conjugate in magnitude, giving

(133)

The phase is nonzero, so total internal reflection means unit reflected power, not necessarily Γ = +1 or −1.

Solution 12: SWR from a known reflection coefficient

The standing-wave ratio is

SWR	=	(134)
	=	(135)
	= 4.	(136)

Thus

(137)

The reflected power fraction is

(138)

The normalized extrema are

(139)

(140)

Their ratio is 1.60∕0.40 = 4, as required.

Solution 13: infer reflection coefficient from SWR

Invert

(141)

Solving gives

(142)

For SWR = 3,

(143)

Thus

(144)

If |E_i| = 10 V/m,

(145)

(146)

Solution 14: phase reversal and standing-wave location

At z = 0,

(147)

For Γ = +0.5,

(148)

This is the maximum envelope value because incident and reflected electric fields are in phase at the boundary.

For Γ = −0.5,

(149)

This is the minimum envelope value because the reflected field is 180^∘ out of phase with the incident field at the boundary.

The magnitude |Γ| determines the ratio of maxima to minima, while the phase of Γ determines where those maxima and minima occur in space.

Solution 15: perfect electric conductor

At a PEC surface, the tangential electric field must vanish:

(150)

Therefore

(151)

so

(152)

The reflected power fraction is

(153)

No time-average power enters the ideal conductor.

For the incident wave,

(154)

Because the reflected wave reverses propagation direction while also having E_r0 = −E_i0, its magnetic field at the boundary points in the same tangential direction as H_i and has equal magnitude. Hence

(155)

The ideal electric-field Standing Wave has a zero minimum, so

(156)

Solution 16: Julia Fresnel and SWR sweep

A compact Julia implementation is

using Printf

n1 = 1.0
n2 = 1.5
eta0 = 376.730313668
eta1 = eta0/n1
eta2 = eta0/n2

theta_deg = collect(range(0.0, 89.0, length=2001))
Rte = similar(theta_deg)
Rtm = similar(theta_deg)

for (i, deg) in enumerate(theta_deg)
    ti = deg2rad(deg)
    tt = asin((n1/n2)*sin(ti))

    rte = (eta2*cos(ti) - eta1*cos(tt)) /
          (eta2*cos(ti) + eta1*cos(tt))

    rtm = (eta2*cos(tt) - eta1*cos(ti)) /
          (eta2*cos(tt) + eta1*cos(ti))

    Rte[i] = abs2(rte)
    Rtm[i] = abs2(rtm)
end

imin = argmin(Rtm)
theta_B_num = theta_deg[imin]
theta_B_exact = rad2deg(atan(n2/n1))

gamma0 = (eta2 - eta1)/(eta2 + eta1)
SWR0 = (1 + abs(gamma0))/(1 - abs(gamma0))

@printf("Numerical Brewster angle = %.4f deg\n", theta_B_num)
@printf("Analytic Brewster angle  = %.4f deg\n", theta_B_exact)
@printf("Normal-incidence Gamma    = %.6f\n", gamma0)
@printf("Normal-incidence SWR      = %.6f\n", SWR0)

For air to glass,

(157)

The numerical minimum should converge toward that value as the angular grid is refined.

At normal incidence,

(158)

Therefore

(159)

Thus

(160)

The computational sweep is useful because it makes three landmarks visible at once: TE reflectance rises with angle, TM reflectance reaches zero at Brewster incidence, and both approach unity near grazing incidence.

Summary of the problem set

The calculations in EM21E1 can be organized around four layers:

(161)

The main practical cautions are:

• a field-amplitude transmission coefficient is not itself a transmitted-power fraction;
• TE and TM coefficients differ at oblique incidence;
• the sign or phase of Γ determines where standing-wave extrema occur;
• |Γ| = 1 in total internal reflection does not imply a trivial reflection phase;
• the SWR depends only on |Γ|, not on the phase of Γ.

These ideas form the direct mathematical bridge from free-space boundary waves to transmission lines, impedance matching, radomes, multilayer media, and RF hardware.

References

References