Electromagnetic Waves, Antennas, and RF: Wave Propagation in Materials - Exercises and Complete Worked Solutions

Electromagnetic Waves, Antennas, and RF: Wave Propagation in Materials - Exercises and Complete Worked Solutions

EM20 derived the propagation constant, attenuation constant, phase constant, intrinsic impedance, and skin depth for electromagnetic waves in homogeneous linear materials. This companion article turns those formulas into a systematic set of worked problems. The main objective is to learn how to decide which propagation model applies before inserting numbers.

Throughout this article the time convention is

(1)

and a wave traveling in the +z direction is written

(2)

The exact material relations used repeatedly are

(3)

(4)

(5)

and

(6)

For a good Conductor,

(7)

These formulas are consequences of Maxwell’s equations and the constitutive relations, not independent empirical rules [1, 2, 3, 4].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. For every problem, first compute or estimate

(8)

This single dimensionless ratio often tells you whether the lossless, low-loss, exact lossy, or good-conductor formulas are appropriate.

Part I: Exercises

Exercise 1: read a lossy plane wave

A plane wave is written

(9)

Identify α, β, the wavelength, the phase velocity, the amplitude reduction after 3.0 m, and the power-density reduction after the same distance.

Exercise 2: lossless dielectric at 2.4 GHz

A nonmagnetic lossless dielectric has

(10)

At f = 2.4 GHz, calculate

1. v_p,
2. β,
3. λ,
4. intrinsic impedance η.

Compare the wavelength with the vacuum wavelength.

Exercise 3: attenuation in nepers and decibels

A wave propagates through a lossy medium with

(11)

After 8.0 m, determine

1. the field-amplitude ratio E(z)∕E(0),
2. the power-density ratio S(z)∕S(0),
3. the attenuation in dB.

Explain why both the amplitude calculation using 20 log ₁₀ and the Power calculation using 10 log ₁₀ give the same dB loss.

Exercise 4: infer the attenuation constant from measurements

The electric-field amplitude falls to 25% of its initial value after traveling 5.0 m in a homogeneous medium. Determine α in Np/m and the corresponding attenuation over the 5.0 m path in dB.

Exercise 5: exact lossy-dielectric propagation constants

Take

(12)

Using the exact formulas, calculate

1. p = σ∕(ω𝜖),
2. α,
3. β,
4. λ,
5. v_p,
6. the field-amplitude ratio after 10 m.

Exercise 6: test the low-loss approximation

For the medium in Exercise 5, use

(13)

Compare each approximation with the exact result and compute the percentage error.

Exercise 7: complex intrinsic impedance in a lossy dielectric

For the material in Exercise 5, calculate the exact complex intrinsic impedance

(14)

Express the answer in both rectangular and polar form. If the electric-field phasor amplitude is

(15)

find the magnetic-field phasor amplitude H₀ and state the phase relation between E and H.

Exercise 8: impedance of a lossless high-permittivity dielectric

A nonmagnetic, lossless material has 𝜖_r = 9. Find its intrinsic impedance and compare it with free space. If the electric-field amplitude is 30 V/m, find the magnetic-field amplitude.

Exercise 9: copper at 1 MHz

Treat copper as a good conductor with

(16)

Calculate

1. α and β,
2. skin depth δ,
3. wavelength inside the conductor,
4. phase velocity,
5. approximate intrinsic impedance in rectangular and polar form.

Exercise 10: frequency scaling in a good conductor

For copper, compare 1 MHz and 100 MHz. Without re-deriving the full expressions, predict how α, δ, and |η| scale with frequency. Then calculate their numerical values at 100 MHz from the 1 MHz values.

Exercise 11: how many skin depths are enough?

In a good conductor the field amplitude obeys

(17)

Determine

1. the number of skin depths required for the field amplitude to fall to 1%,
2. the number required for power density to fall to 1%,
3. the corresponding physical depths in copper at 1 MHz.

Exercise 12: verify the good-conductor approximation for aluminum

Take aluminum with

(18)

Calculate p = σ∕(ω𝜖₀) and explain why the good-conductor approximation is overwhelmingly valid. Compare the approximate α with the exact α.

Exercise 13: phase accumulation and delay through a lossy dielectric

Use the exact propagation result from Exercise 5 and let a wave travel L = 2.0 m. Find

1. the accumulated phase βL in radians,
2. the equivalent number of cycles,
3. the phase modulo 360^∘,
4. the propagation delay L∕v_p.

Exercise 14: idealized conductor thickness for 60 dB field attenuation

At 10 MHz, estimate the copper thickness required for 60 dB of attenuation due only to propagation inside the conductor. Use the good-conductor approximation. Express the result in millimeters and in skin depths. State an important limitation of interpreting this number as a complete shielding calculation.

Exercise 15: derive α and β from γ²

Start from

(19)

Derive the exact formulas for α and β. Then recover the limiting forms for

1. a low-loss dielectric, σ ≪ ω𝜖,
2. a good conductor, σ ≫ ω𝜖.

Exercise 16: Julia comparison of exact and approximate propagation models

Write a Julia program that sweeps frequency from 10³ to 10¹⁰ Hz for a material with

(20)

At each frequency calculate the exact α, β, and |η|, along with the low-loss and good-conductor approximations. Plot or tabulate enough points to identify where each approximation becomes reasonable. Use the included file EM20E1_propagation_sweep.jl as a reference implementation after attempting the exercise.

Part II: Complete Worked Solutions

Solution 1: read a lossy plane wave

Compare

(21)

with

(22)

Therefore

(23)

The wavelength is

(24)

The frequency is f = 10⁸ Hz, so

(25)

Thus

(26)

At z = 3 m,

(27)

Power density is proportional to field amplitude squared, so

(28)

Hence

(29)

Solution 2: lossless dielectric at 2.4 GHz

For a lossless material,

(30)

With μ_r = 1 and 𝜖_r = 2.25,

(31)

so

(32)

The angular frequency is

(33)

Then

(34)

Therefore

(35)

The intrinsic impedance is

(36)

Thus

(37)

The vacuum wavelength is

(38)

The dielectric wavelength is shorter by the factor 1.5:

(39)

Solution 3: attenuation in nepers and decibels

The amplitude ratio is

(40)

Thus

(41)

The power-density ratio is

(42)

so

(43)

Using field amplitude,

LdB	= 20 log 10(0.3829)	(44)
	≈−8.34 dB.	(45)

Equivalently,

(46)

Using power,

(47)

The two answers agree because power is proportional to the square of field amplitude.

Solution 4: infer the attenuation constant from measurements

We are given

(48)

Taking natural logarithms,

(49)

Therefore

(50)

The dB attenuation is

LdB	= 20 log 10(0.25)	(51)
	≈−12.04 dB.	(52)

Thus a four-to-one reduction in field amplitude corresponds to approximately 12.04 dB of attenuation.

Solution 5: exact lossy-dielectric propagation constants

The material parameters are

(53)

First,

(54)

Since p ≪ 1, the medium is low loss, but we will use the exact formulas.

The attenuation constant is

(55)

which gives

(56)

Similarly,

(57)

Then

(58)

and

(59)

After 10 m,

(60)

Thus the phase behavior is close to that of a lossless 𝜖_r = 4 dielectric, while the field amplitude falls to about 39% over 10 m.

Solution 6: test the low-loss approximation

For 𝜖 = 4𝜖₀ and μ = μ₀,

(61)

The exact result from Exercise 5 is

(62)

Therefore

(63)

For the phase constant,

(64)

Compared with β_exact ≈ 4.19275 rad/m,

(65)

The small errors confirm what the ratio p ≈ 0.0449 already suggested: the low-loss approximation is excellent here.

Solution 7: complex intrinsic impedance in a lossy dielectric

Use

(66)

Substitution gives

(67)

The magnitude is

(68)

and the phase is

(69)

Hence

(70)

Since

(71)

we obtain

(72)

and

(73)

Therefore

(74)

With this convention, E leads H by about 1.287^∘.

Solution 8: impedance of a lossless high-permittivity dielectric

For a lossless nonmagnetic material,

(75)

Thus

(76)

Therefore

(77)

This is one third of the free-space impedance. If E₀ = 30 V/m,

(78)

Hence

(79)

Solution 9: copper at 1 MHz

For a good conductor,

(80)

With f = 1 MHz, μ = μ₀, and σ = 5.8 × 10⁷ S/m,

(81)

The skin depth is

(82)

so

(83)

The wavelength in the conductor is

(84)

therefore

(85)

The phase velocity is

(86)

This very small phase velocity should not be confused with a signal or energy velocity in free space; the field is attenuated extremely rapidly.

The good-conductor intrinsic impedance is

(87)

Numerically,

(88)

Its magnitude and phase are

(89)

Solution 10: frequency scaling in a good conductor

For fixed μ and σ,

(90)

Increasing frequency from 1 MHz to 100 MHz multiplies f by 100, so changes by a factor of 10.

Therefore

(91)

(92)

and

(93)

The higher-frequency field penetrates less deeply even though the magnitude of the field ratio E∕H increases.

Solution 11: how many skin depths are enough?

For amplitude,

(94)

Hence

(95)

Therefore

(96)

For power,

(97)

so

(98)

Thus

(99)

For copper at 1 MHz, δ ≈ 66.085 μm. Hence

(100)

and

(101)

Power reaches 1% in fewer skin depths because it depends on the square of field amplitude.

Solution 12: verify the good-conductor approximation for aluminum

At 10 MHz,

(102)

Then

(103)

Therefore

(104)

by more than ten orders of magnitude. The conduction current dominates the displacement current.

The good-conductor result is

(105)

Using the exact formula gives the same value to the displayed precision:

(106)

The relative error is below 10⁻⁸% for these parameters. Thus

(107)

Solution 13: phase accumulation and delay through a lossy dielectric

From Exercise 5,

(108)

For L = 2 m,

(109)

The number of cycles is

(110)

Hence

(111)

Modulo one full cycle, the phase is

(112)

Thus

(113)

The propagation delay is

(114)

so

(115)

Solution 14: idealized conductor thickness for 60 dB field attenuation

For copper at 10 MHz,

(116)

Therefore

(117)

For 60 dB field attenuation,

(118)

Hence

(119)

Thus

(120)

In skin depths,

(121)

Therefore

(122)

This result describes absorption associated with propagation inside an idealized bulk conductor. A complete shielding calculation must also account for reflection at interfaces, finite geometry, apertures, seams, polarization, incidence angle, and possible coupling through cables or penetrations.

Solution 15: derive α and β from γ²

Start from

(123)

Expanding the left side gives

(124)

Equating real and imaginary parts,

(125)

(126)

Square both equations and add:

(α2 − β2)2 + (2αβ)2

= ω4μ2𝜖2 + ω2μ2σ2.

(127)

The left side is

(128)

Therefore

(129)

Now add this equation to

(130)

The result is

(131)

where

(132)

Hence

(133)

Similarly,

(134)

For p ≪ 1,

(135)

Then

(136)

and

(137)

Therefore

(138)

For p ≫ 1,

(139)

The ±1 terms are negligible relative to p, so

α	≈ ω,	(140)
β	≈ ω.	(141)

Since p = σ∕(ω𝜖),

(142)

The low-loss and good-conductor formulas are therefore limiting forms of the same exact propagation constant.

Solution 16: Julia comparison of exact and approximate propagation models

A compact implementation should calculate

(143)

at every frequency. The exact complex quantities are

(144)

Then

(145)

For comparison, calculate

(146)

and

(147)

The included Julia script prints a logarithmically spaced table and, if Plots.jl is available, produces comparison plots.

For the specified material,

(148)

The transition p ≈ 1 occurs near

(149)

Thus frequencies far below roughly 45 MHz are conductor-like for this material, while frequencies far above that scale become progressively more dielectric-like. This illustrates an important point: the propagation regime is set by both material properties and frequency.

What EM20E1 adds to the series

EM20 introduced the propagation formulas. EM20E1 develops the calculation habits needed to use them reliably:

(150)

The next boundary-value step is to ask what happens when η changes discontinuously at an interface. That leads directly to reflection, transmission, and refraction.

References

References