Electromagnetic Waves, Antennas, and RF: Wave Propagation in Materials - Exercises and Complete Worked Solutions
ElectromagneticWavesAntennasAndRFWavePropagationInMaterialsExercisesAndCompleteWorkedSolutions
2026-10-09 22:53:46
2026-10-09 23:28:27
Topic
electromagnetic wave propagation
attenuation constant
phase constant
propagation constant
intrinsic impedance
skin depth
good conductor
low-loss dielectric
dielectric
conductor
phase velocity
wavelength
RF attenuation
shielding
worked solutions
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\section*{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
\begin{equation}
e^{i\omega t},
\end{equation}
and a wave traveling in the $+z$ direction is written
\begin{equation}
\widetilde{\mathbf E}(z)=\mathbf E_0e^{-\gamma z},
\qquad
\gamma=\alpha+i\beta.
\end{equation}
The exact material relations used repeatedly are
\begin{equation}
\boxed{
\gamma
=\sqrt{i\omega\mu(\sigma+i\omega\epsilon)},
}
\end{equation}
\begin{equation}
\boxed{
\alpha
=\omega\sqrt{\frac{\mu\epsilon}{2}}
\left[
\sqrt{1+\left(\frac{\sigma}{\omega\epsilon}\right)^2}-1
\right]^{1/2},
}
\end{equation}
\begin{equation}
\boxed{
\beta
=\omega\sqrt{\frac{\mu\epsilon}{2}}
\left[
\sqrt{1+\left(\frac{\sigma}{\omega\epsilon}\right)^2}+1
\right]^{1/2},
}
\end{equation}
and
\begin{equation}
\boxed{
\eta
=\sqrt{\frac{i\omega\mu}{\sigma+i\omega\epsilon}}.
}
\end{equation}
For a good conductor,
\begin{equation}
\boxed{
\alpha\approx\beta\approx\sqrt{\frac{\omega\mu\sigma}{2}},
\qquad
\delta\approx\sqrt{\frac{2}{\omega\mu\sigma}},
\qquad
\eta\approx(1+i)\sqrt{\frac{\omega\mu}{2\sigma}}.
}
\end{equation}
These formulas are consequences of Maxwell's equations and the constitutive relations, not independent empirical rules \cite{Griffiths2017,Cheng1989,Ulaby2015,Balanis2012}.
\section*{How to use this problem set}
Attempt all exercises in Part I before consulting Part II. For every problem, first compute or estimate
\begin{equation}
p=\frac{\sigma}{\omega\epsilon}.
\end{equation}
This single dimensionless ratio often tells you whether the lossless, low-loss, exact lossy, or good-conductor formulas are appropriate.
\begin{center}
\includegraphics{EM20E1_fig01_attenuation_rates.png}
\vspace{0.4em}
\textit{Figure 1. Field amplitude decays as $e^{-\alpha z}$ while power density decays as $e^{-2\alpha z}$.}
\end{center}
\section*{Part I: Exercises}
\section*{Exercise 1: read a lossy plane wave}
A plane wave is written
\begin{equation}
E(z,t)=E_0e^{-0.18z}\cos(2\pi\times10^8t-5.2z).
\end{equation}
Identify $\alpha$, $\beta$, the wavelength, the phase velocity, the amplitude reduction after $3.0\,\text{m}$, and the power-density reduction after the same distance.
\section*{Exercise 2: lossless dielectric at 2.4 GHz}
A nonmagnetic lossless dielectric has
\begin{equation}
\epsilon_r=2.25,
\qquad
\mu_r=1,
\qquad
\sigma=0.
\end{equation}
At $f=2.4\,\text{GHz}$, calculate
\begin{enumerate}
\item[(a)] $v_p$,
\item[(b)] $\beta$,
\item[(c)] $\lambda$,
\item[(d)] intrinsic impedance $\eta$.
\end{enumerate}
Compare the wavelength with the vacuum wavelength.
\section*{Exercise 3: attenuation in nepers and decibels}
A wave propagates through a lossy medium with
\begin{equation}
\alpha=0.12\,\text{Np/m}.
\end{equation}
After $8.0\,\text{m}$, determine
\begin{enumerate}
\item[(a)] the field-amplitude ratio $E(z)/E(0)$,
\item[(b)] the power-density ratio $S(z)/S(0)$,
\item[(c)] the attenuation in dB.
\end{enumerate}
Explain why both the amplitude calculation using $20\log_{10}$ and the power calculation using $10\log_{10}$ give the same dB loss.
\section*{Exercise 4: infer the attenuation constant from measurements}
The electric-field amplitude falls to $25\%$ of its initial value after traveling $5.0\,\text{m}$ in a homogeneous medium. Determine $\alpha$ in Np/m and the corresponding attenuation over the $5.0\,\text{m}$ path in dB.
\section*{Exercise 5: exact lossy-dielectric propagation constants}
Take
\begin{equation}
f=100\,\text{MHz},
\qquad
\epsilon_r=4,
\qquad
\mu_r=1,
\qquad
\sigma=1.0\times10^{-3}\,\text{S/m}.
\end{equation}
Using the exact formulas, calculate
\begin{enumerate}
\item[(a)] $p=\sigma/(\omega\epsilon)$,
\item[(b)] $\alpha$,
\item[(c)] $\beta$,
\item[(d)] $\lambda$,
\item[(e)] $v_p$,
\item[(f)] the field-amplitude ratio after $10\,\text{m}$.
\end{enumerate}
\section*{Exercise 6: test the low-loss approximation}
For the medium in Exercise 5, use
\begin{equation}
\alpha_{\text{LL}}\approx\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}},
\qquad
\beta_{\text{LL}}\approx\omega\sqrt{\mu\epsilon}.
\end{equation}
Compare each approximation with the exact result and compute the percentage error.
\section*{Exercise 7: complex intrinsic impedance in a lossy dielectric}
For the material in Exercise 5, calculate the exact complex intrinsic impedance
\begin{equation}
\eta=\sqrt{\frac{i\omega\mu}{\sigma+i\omega\epsilon}}.
\end{equation}
Express the answer in both rectangular and polar form. If the electric-field phasor amplitude is
\begin{equation}
\widetilde E_0=10\angle0^\circ\,\text{V/m},
\end{equation}
find the magnetic-field phasor amplitude $\widetilde H_0$ and state the phase relation between $E$ and $H$.
\begin{center}
\includegraphics{EM20E1_fig02_intrinsic_impedance_phasor.png}
\vspace{0.4em}
\textit{Figure 2. A complex intrinsic impedance means the electric and magnetic phasors are not exactly in phase.}
\end{center}
\section*{Exercise 8: impedance of a lossless high-permittivity dielectric}
A nonmagnetic, lossless material has $\epsilon_r=9$. Find its intrinsic impedance and compare it with free space. If the electric-field amplitude is $30\,\text{V/m}$, find the magnetic-field amplitude.
\section*{Exercise 9: copper at 1 MHz}
Treat copper as a good conductor with
\begin{equation}
\sigma=5.8\times10^7\,\text{S/m},
\qquad
\mu\approx\mu_0,
\qquad
f=1.0\,\text{MHz}.
\end{equation}
Calculate
\begin{enumerate}
\item[(a)] $\alpha$ and $\beta$,
\item[(b)] skin depth $\delta$,
\item[(c)] wavelength inside the conductor,
\item[(d)] phase velocity,
\item[(e)] approximate intrinsic impedance in rectangular and polar form.
\end{enumerate}
\section*{Exercise 10: frequency scaling in a good conductor}
For copper, compare $1\,\text{MHz}$ and $100\,\text{MHz}$. Without re-deriving the full expressions, predict how $\alpha$, $\delta$, and $|\eta|$ scale with frequency. Then calculate their numerical values at $100\,\text{MHz}$ from the $1\,\text{MHz}$ values.
\section*{Exercise 11: how many skin depths are enough?}
In a good conductor the field amplitude obeys
\begin{equation}
E(z)=E(0)e^{-z/\delta}.
\end{equation}
Determine
\begin{enumerate}
\item[(a)] the number of skin depths required for the field amplitude to fall to $1\%$,
\item[(b)] the number required for power density to fall to $1\%$,
\item[(c)] the corresponding physical depths in copper at $1\,\text{MHz}$.
\end{enumerate}
\begin{center}
\includegraphics{EM20E1_fig03_skin_depth_levels.png}
\vspace{0.4em}
\textit{Figure 3. Each additional skin depth multiplies field amplitude by $e^{-1}$ and power by $e^{-2}$.}
\end{center}
\section*{Exercise 12: verify the good-conductor approximation for aluminum}
Take aluminum with
\begin{equation}
\sigma=3.5\times10^7\,\text{S/m},
\qquad
\mu\approx\mu_0,
\qquad
f=10\,\text{MHz}.
\end{equation}
Calculate $p=\sigma/(\omega\epsilon_0)$ and explain why the good-conductor approximation is overwhelmingly valid. Compare the approximate $\alpha$ with the exact $\alpha$.
\section*{Exercise 13: phase accumulation and delay through a lossy dielectric}
Use the exact propagation result from Exercise 5 and let a wave travel $L=2.0\,\text{m}$. Find
\begin{enumerate}
\item[(a)] the accumulated phase $\beta L$ in radians,
\item[(b)] the equivalent number of cycles,
\item[(c)] the phase modulo $360^\circ$,
\item[(d)] the propagation delay $L/v_p$.
\end{enumerate}
\section*{Exercise 14: idealized conductor thickness for 60 dB field attenuation}
At $10\,\text{MHz}$, estimate the copper thickness required for $60\,\text{dB}$ 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.
\section*{Exercise 15: derive $\alpha$ and $\beta$ from $\gamma^2$}
Start from
\begin{equation}
(\alpha+i\beta)^2
=-\omega^2\mu\epsilon+i\omega\mu\sigma.
\end{equation}
Derive the exact formulas for $\alpha$ and $\beta$. Then recover the limiting forms for
\begin{enumerate}
\item[(a)] a low-loss dielectric, $\sigma\ll\omega\epsilon$,
\item[(b)] a good conductor, $\sigma\gg\omega\epsilon$.
\end{enumerate}
\section*{Exercise 16: Julia comparison of exact and approximate propagation models}
Write a Julia program that sweeps frequency from $10^3$ to $10^{10}\,\text{Hz}$ for a material with
\begin{equation}
\epsilon_r=4,
\qquad
\mu_r=1,
\qquad
\sigma=0.01\,\text{S/m}.
\end{equation}
At each frequency calculate the exact $\alpha$, $\beta$, and $|\eta|$, 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 \texttt{EM20E1\_propagation\_sweep.jl} as a reference implementation after attempting the exercise.
\begin{center}
\includegraphics{EM20E1_fig04_approximation_regimes.png}
\vspace{0.4em}
\textit{Figure 4. The dimensionless ratio $p=\sigma/(\omega\epsilon)$ organizes the approximation hierarchy.}
\end{center}
\section*{Part II: Complete Worked Solutions}
\section*{Solution 1: read a lossy plane wave}
Compare
\begin{equation}
E(z,t)=E_0e^{-0.18z}\cos(2\pi\times10^8t-5.2z)
\end{equation}
with
\begin{equation}
E(z,t)=E_0e^{-\alpha z}\cos(\omega t-\beta z).
\end{equation}
Therefore
\begin{equation}
\boxed{\alpha=0.18\,\text{Np/m},}
\qquad
\boxed{\beta=5.2\,\text{rad/m}.}
\end{equation}
The wavelength is
\begin{equation}
\lambda=\frac{2\pi}{\beta}
=\frac{2\pi}{5.2}
\approx1.208\,\text{m}.
\end{equation}
The frequency is $f=10^8\,\text{Hz}$, so
\begin{equation}
\omega=2\pi\times10^8\,\text{rad/s}.
\end{equation}
Thus
\begin{equation}
v_p=\frac{\omega}{\beta}
\approx1.208\times10^8\,\text{m/s}.
\end{equation}
At $z=3\,\text{m}$,
\begin{equation}
\frac{E(3)}{E(0)}=e^{-(0.18)(3)}=e^{-0.54}\approx0.583.
\end{equation}
Power density is proportional to field amplitude squared, so
\begin{equation}
\frac{S(3)}{S(0)}=e^{-2(0.18)(3)}=e^{-1.08}\approx0.340.
\end{equation}
Hence
\begin{equation}
\boxed{\lambda\approx1.208\,\text{m},\quad
v_p\approx1.208\times10^8\,\text{m/s},\quad
E(3)/E(0)\approx0.583,\quad
S(3)/S(0)\approx0.340.}
\end{equation}
\section*{Solution 2: lossless dielectric at 2.4 GHz}
For a lossless material,
\begin{equation}
v_p=\frac{c}{\sqrt{\mu_r\epsilon_r}}.
\end{equation}
With $\mu_r=1$ and $\epsilon_r=2.25$,
\begin{equation}
\sqrt{\mu_r\epsilon_r}=1.5,
\end{equation}
so
\begin{equation}
\boxed{v_p\approx1.999\times10^8\,\text{m/s}.}
\end{equation}
The angular frequency is
\begin{equation}
\omega=2\pi(2.4\times10^9)
\approx1.508\times10^{10}\,\text{rad/s}.
\end{equation}
Then
\begin{equation}
\beta=\frac{\omega}{v_p}
\approx75.45\,\text{rad/m}.
\end{equation}
Therefore
\begin{equation}
\boxed{\lambda=\frac{2\pi}{\beta}\approx8.328\times10^{-2}\,\text{m}.}
\end{equation}
The intrinsic impedance is
\begin{equation}
\eta=\eta_0\sqrt{\frac{\mu_r}{\epsilon_r}}
=\frac{376.73}{1.5}
\approx251.15\,\Omega.
\end{equation}
Thus
\begin{equation}
\boxed{\eta\approx251.2\,\Omega.}
\end{equation}
The vacuum wavelength is
\begin{equation}
\lambda_0=\frac{c}{f}\approx0.1249\,\text{m}.
\end{equation}
The dielectric wavelength is shorter by the factor $1.5$:
\begin{equation}
\lambda=\frac{\lambda_0}{1.5}.
\end{equation}
\section*{Solution 3: attenuation in nepers and decibels}
The amplitude ratio is
\begin{equation}
\frac{E(z)}{E(0)}=e^{-\alpha z}
=e^{-(0.12)(8)}
=e^{-0.96}
\approx0.3829.
\end{equation}
Thus
\begin{equation}
\boxed{E(8)/E(0)\approx0.383.}
\end{equation}
The power-density ratio is
\begin{equation}
\frac{S(z)}{S(0)}=e^{-2\alpha z}
=e^{-1.92}
\approx0.1466,
\end{equation}
so
\begin{equation}
\boxed{S(8)/S(0)\approx0.1466.}
\end{equation}
Using field amplitude,
\begin{align}
L_{\text{dB}}
&=20\log_{10}(0.3829)\\
&\approx-8.34\,\text{dB}.
\end{align}
Equivalently,
\begin{equation}
L_{\text{dB}}=-8.686\alpha z
=-8.686(0.12)(8)
\approx-8.34\,\text{dB}.
\end{equation}
Using power,
\begin{equation}
10\log_{10}(0.1466)\approx-8.34\,\text{dB}.
\end{equation}
The two answers agree because power is proportional to the square of field amplitude.
\section*{Solution 4: infer the attenuation constant from measurements}
We are given
\begin{equation}
0.25=e^{-\alpha(5)}.
\end{equation}
Taking natural logarithms,
\begin{equation}
\ln(0.25)=-5\alpha.
\end{equation}
Therefore
\begin{equation}
\boxed{
\alpha=-\frac{\ln(0.25)}{5}
\approx0.2773\,\text{Np/m}.
}
\end{equation}
The dB attenuation is
\begin{align}
L_{\text{dB}}
&=20\log_{10}(0.25)\\
&\approx-12.04\,\text{dB}.
\end{align}
Thus a four-to-one reduction in field amplitude corresponds to approximately $12.04\,\text{dB}$ of attenuation.
\section*{Solution 5: exact lossy-dielectric propagation constants}
The material parameters are
\begin{equation}
\epsilon=4\epsilon_0,
\qquad
\mu=\mu_0,
\qquad
\omega=2\pi(100\times10^6).
\end{equation}
First,
\begin{equation}
p=\frac{\sigma}{\omega\epsilon}
\approx0.04494.
\end{equation}
Since $p\ll1$, the medium is low loss, but we will use the exact formulas.
The attenuation constant is
\begin{equation}
\alpha
=\omega\sqrt{\frac{\mu\epsilon}{2}}
\left(\sqrt{1+p^2}-1\right)^{1/2},
\end{equation}
which gives
\begin{equation}
\boxed{\alpha\approx9.416\times10^{-2}\,\text{Np/m}.}
\end{equation}
Similarly,
\begin{equation}
\boxed{\beta\approx4.19275\,\text{rad/m}.}
\end{equation}
Then
\begin{equation}
\boxed{
\lambda=\frac{2\pi}{\beta}
\approx1.4986\,\text{m},
}
\end{equation}
and
\begin{equation}
\boxed{
v_p=\frac{\omega}{\beta}
\approx1.4986\times10^8\,\text{m/s}.
}
\end{equation}
After $10\,\text{m}$,
\begin{equation}
\frac{E(10)}{E(0)}=e^{-\alpha(10)}
\approx e^{-0.9416}
\approx0.390.
\end{equation}
Thus the phase behavior is close to that of a lossless $\epsilon_r=4$ dielectric, while the field amplitude falls to about $39\%$ over $10\,\text{m}$.
\section*{Solution 6: test the low-loss approximation}
For $\epsilon=4\epsilon_0$ and $\mu=\mu_0$,
\begin{equation}
\alpha_{\text{LL}}
\approx\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}}
\approx9.4183\times10^{-2}\,\text{Np/m}.
\end{equation}
The exact result from Exercise 5 is
\begin{equation}
\alpha_{\text{exact}}
\approx9.4159\times10^{-2}\,\text{Np/m}.
\end{equation}
Therefore
\begin{equation}
\boxed{
\frac{\alpha_{\text{LL}}-\alpha_{\text{exact}}}{\alpha_{\text{exact}}}\times100\%
\approx+0.025\%.
}
\end{equation}
For the phase constant,
\begin{equation}
\beta_{\text{LL}}
\approx\omega\sqrt{\mu\epsilon}
\approx4.19169\,\text{rad/m}.
\end{equation}
Compared with $\beta_{\text{exact}}\approx4.19275\,\text{rad/m}$,
\begin{equation}
\boxed{
\frac{\beta_{\text{LL}}-\beta_{\text{exact}}}{\beta_{\text{exact}}}\times100\%
\approx-0.025\%.
}
\end{equation}
The small errors confirm what the ratio $p\approx0.0449$ already suggested: the low-loss approximation is excellent here.
\section*{Solution 7: complex intrinsic impedance in a lossy dielectric}
Use
\begin{equation}
\eta=\sqrt{\frac{i\omega\mu}{\sigma+i\omega\epsilon}}.
\end{equation}
Substitution gives
\begin{equation}
\boxed{
\eta\approx(188.223+i\,4.227)\,\Omega.
}
\end{equation}
The magnitude is
\begin{equation}
|\eta|
=\sqrt{188.223^2+4.227^2}
\approx188.270\,\Omega,
\end{equation}
and the phase is
\begin{equation}
\angle\eta
=\tan^{-1}\left(\frac{4.227}{188.223}\right)
\approx1.287^\circ.
\end{equation}
Hence
\begin{equation}
\boxed{\eta\approx188.270\angle1.287^\circ\,\Omega.}
\end{equation}
Since
\begin{equation}
\widetilde H_0=\frac{\widetilde E_0}{\eta},
\end{equation}
we obtain
\begin{equation}
|\widetilde H_0|
=\frac{10}{188.270}
\approx5.312\times10^{-2}\,\text{A/m},
\end{equation}
and
\begin{equation}
\angle\widetilde H_0
=0^\circ-1.287^\circ
=-1.287^\circ.
\end{equation}
Therefore
\begin{equation}
\boxed{
\widetilde H_0
\approx0.0531\angle(-1.287^\circ)\,\text{A/m}.
}
\end{equation}
With this convention, $E$ leads $H$ by about $1.287^\circ$.
\section*{Solution 8: impedance of a lossless high-permittivity dielectric}
For a lossless nonmagnetic material,
\begin{equation}
\eta=\eta_0\sqrt{\frac{\mu_r}{\epsilon_r}}.
\end{equation}
Thus
\begin{equation}
\eta=\frac{376.73}{3}
\approx125.58\,\Omega.
\end{equation}
Therefore
\begin{equation}
\boxed{\eta\approx125.6\,\Omega.}
\end{equation}
This is one third of the free-space impedance. If $E_0=30\,\text{V/m}$,
\begin{equation}
H_0=\frac{E_0}{\eta}
\approx\frac{30}{125.58}
\approx0.2389\,\text{A/m}.
\end{equation}
Hence
\begin{equation}
\boxed{H_0\approx0.239\,\text{A/m}.}
\end{equation}
\section*{Solution 9: copper at 1 MHz}
For a good conductor,
\begin{equation}
\alpha\approx\beta\approx\sqrt{\frac{\omega\mu\sigma}{2}}.
\end{equation}
With $f=1\,\text{MHz}$, $\mu=\mu_0$, and $\sigma=5.8\times10^7\,\text{S/m}$,
\begin{equation}
\boxed{
\alpha\approx\beta\approx1.5132\times10^4\,\text{m}^{-1}.
}
\end{equation}
The skin depth is
\begin{equation}
\delta=\frac{1}{\alpha}
\approx6.6085\times10^{-5}\,\text{m},
\end{equation}
so
\begin{equation}
\boxed{\delta\approx66.1\,\mu\text{m}.}
\end{equation}
The wavelength in the conductor is
\begin{equation}
\lambda=\frac{2\pi}{\beta}
\approx4.1523\times10^{-4}\,\text{m},
\end{equation}
therefore
\begin{equation}
\boxed{\lambda\approx0.415\,\text{mm}.}
\end{equation}
The phase velocity is
\begin{equation}
v_p=\frac{\omega}{\beta}
\approx415.2\,\text{m/s}.
\end{equation}
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
\begin{equation}
\eta\approx(1+i)\sqrt{\frac{\omega\mu}{2\sigma}}.
\end{equation}
Numerically,
\begin{equation}
\boxed{
\eta\approx(2.609\times10^{-4})(1+i)\,\Omega.
}
\end{equation}
Its magnitude and phase are
\begin{equation}
\boxed{
|\eta|\approx3.690\times10^{-4}\,\Omega,
\qquad
\angle\eta\approx45^\circ.
}
\end{equation}
\section*{Solution 10: frequency scaling in a good conductor}
For fixed $\mu$ and $\sigma$,
\begin{equation}
\alpha\propto\sqrt{f},
\qquad
\delta\propto\frac{1}{\sqrt{f}},
\qquad
|\eta|\propto\sqrt{f}.
\end{equation}
Increasing frequency from $1\,\text{MHz}$ to $100\,\text{MHz}$ multiplies $f$ by $100$, so $\sqrt{f}$ changes by a factor of $10$.
Therefore
\begin{equation}
\boxed{
\alpha_{100\,\text{MHz}}
\approx10(1.5132\times10^4)
=1.5132\times10^5\,\text{m}^{-1},
}
\end{equation}
\begin{equation}
\boxed{
\delta_{100\,\text{MHz}}
\approx\frac{66.1\,\mu\text{m}}{10}
=6.61\,\mu\text{m},
}
\end{equation}
and
\begin{equation}
\boxed{
|\eta|_{100\,\text{MHz}}
\approx10(3.690\times10^{-4})
=3.690\times10^{-3}\,\Omega.
}
\end{equation}
The higher-frequency field penetrates less deeply even though the magnitude of the field ratio $E/H$ increases.
\section*{Solution 11: how many skin depths are enough?}
For amplitude,
\begin{equation}
0.01=e^{-z/\delta}.
\end{equation}
Hence
\begin{equation}
\frac{z}{\delta}=-\ln(0.01)=4.605.
\end{equation}
Therefore
\begin{equation}
\boxed{z_{1\%\,\text{amplitude}}=4.605\,\delta.}
\end{equation}
For power,
\begin{equation}
0.01=e^{-2z/\delta},
\end{equation}
so
\begin{equation}
\frac{z}{\delta}=\frac{-\ln(0.01)}{2}=2.303.
\end{equation}
Thus
\begin{equation}
\boxed{z_{1\%\,\text{power}}=2.303\,\delta.}
\end{equation}
For copper at $1\,\text{MHz}$, $\delta\approx66.085\,\mu\text{m}$. Hence
\begin{equation}
\boxed{
z_{1\%\,\text{amplitude}}
\approx304\,\mu\text{m}=0.304\,\text{mm},
}
\end{equation}
and
\begin{equation}
\boxed{
z_{1\%\,\text{power}}
\approx152\,\mu\text{m}=0.152\,\text{mm}.
}
\end{equation}
Power reaches $1\%$ in fewer skin depths because it depends on the square of field amplitude.
\section*{Solution 12: verify the good-conductor approximation for aluminum}
At $10\,\text{MHz}$,
\begin{equation}
\omega=2\pi\times10^7\,\text{rad/s}.
\end{equation}
Then
\begin{equation}
p=\frac{\sigma}{\omega\epsilon_0}
\approx6.29\times10^{10}.
\end{equation}
Therefore
\begin{equation}
\boxed{p\gg1,}
\end{equation}
by more than ten orders of magnitude. The conduction current dominates the displacement current.
The good-conductor result is
\begin{equation}
\alpha_{\text{GC}}
\approx\sqrt{\frac{\omega\mu_0\sigma}{2}}
\approx3.71718\times10^4\,\text{m}^{-1}.
\end{equation}
Using the exact formula gives the same value to the displayed precision:
\begin{equation}
\alpha_{\text{exact}}
\approx3.71718\times10^4\,\text{m}^{-1}.
\end{equation}
The relative error is below $10^{-8}\%$ for these parameters. Thus
\begin{equation}
\boxed{\text{the good-conductor approximation is effectively exact here.}}
\end{equation}
\section*{Solution 13: phase accumulation and delay through a lossy dielectric}
From Exercise 5,
\begin{equation}
\beta\approx4.19275\,\text{rad/m},
\qquad
v_p\approx1.49858\times10^8\,\text{m/s}.
\end{equation}
For $L=2\,\text{m}$,
\begin{equation}
\boxed{
\beta L
\approx(4.19275)(2)
=8.38549\,\text{rad}.
}
\end{equation}
The number of cycles is
\begin{equation}
N=\frac{\beta L}{2\pi}
\approx1.33459.
\end{equation}
Hence
\begin{equation}
\boxed{N\approx1.335\ \text{cycles}.}
\end{equation}
Modulo one full cycle, the phase is
\begin{equation}
0.33459(360^\circ)
\approx120.45^\circ.
\end{equation}
Thus
\begin{equation}
\boxed{\phi_{\text{mod }360}\approx120.45^\circ.}
\end{equation}
The propagation delay is
\begin{equation}
t_d=\frac{L}{v_p}
\approx\frac{2}{1.49858\times10^8}
\approx1.3346\times10^{-8}\,\text{s},
\end{equation}
so
\begin{equation}
\boxed{t_d\approx13.35\,\text{ns}.}
\end{equation}
\section*{Solution 14: idealized conductor thickness for 60 dB field attenuation}
For copper at $10\,\text{MHz}$,
\begin{equation}
\alpha
\approx\sqrt{\frac{\omega\mu_0\sigma}{2}}
\approx4.7851\times10^4\,\text{m}^{-1}.
\end{equation}
Therefore
\begin{equation}
\delta=\frac{1}{\alpha}
\approx2.0898\times10^{-5}\,\text{m}
=20.90\,\mu\text{m}.
\end{equation}
For $60\,\text{dB}$ field attenuation,
\begin{equation}
60=8.686\alpha t.
\end{equation}
Hence
\begin{equation}
t=\frac{60}{8.686\alpha}
\approx1.4436\times10^{-4}\,\text{m}.
\end{equation}
Thus
\begin{equation}
\boxed{t\approx0.144\,\text{mm}.}
\end{equation}
In skin depths,
\begin{equation}
\frac{t}{\delta}
\approx6.908.
\end{equation}
Therefore
\begin{equation}
\boxed{t\approx6.91\,\delta.}
\end{equation}
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.
\section*{Solution 15: derive $\alpha$ and $\beta$ from $\gamma^2$}
Start from
\begin{equation}
(\alpha+i\beta)^2
=-\omega^2\mu\epsilon+i\omega\mu\sigma.
\end{equation}
Expanding the left side gives
\begin{equation}
\alpha^2-\beta^2+i\,2\alpha\beta
=-\omega^2\mu\epsilon+i\omega\mu\sigma.
\end{equation}
Equating real and imaginary parts,
\begin{equation}
\alpha^2-\beta^2=-\omega^2\mu\epsilon,
\end{equation}
\begin{equation}
2\alpha\beta=\omega\mu\sigma.
\end{equation}
Square both equations and add:
\begin{align}
(\alpha^2-\beta^2)^2+(2\alpha\beta)^2
&=\omega^4\mu^2\epsilon^2+\omega^2\mu^2\sigma^2.
\end{align}
The left side is
\begin{equation}
(\alpha^2+\beta^2)^2.
\end{equation}
Therefore
\begin{equation}
\alpha^2+\beta^2
=\omega^2\mu\epsilon
\sqrt{1+\left(\frac{\sigma}{\omega\epsilon}\right)^2}.
\end{equation}
Now add this equation to
\begin{equation}
\alpha^2-\beta^2=-\omega^2\mu\epsilon.
\end{equation}
The result is
\begin{equation}
2\alpha^2
=\omega^2\mu\epsilon
\left[
\sqrt{1+p^2}-1
\right],
\end{equation}
where
\begin{equation}
p=\frac{\sigma}{\omega\epsilon}.
\end{equation}
Hence
\begin{equation}
\boxed{
\alpha
=\omega\sqrt{\frac{\mu\epsilon}{2}}
\left[
\sqrt{1+p^2}-1
\right]^{1/2}.
}
\end{equation}
Similarly,
\begin{equation}
\boxed{
\beta
=\omega\sqrt{\frac{\mu\epsilon}{2}}
\left[
\sqrt{1+p^2}+1
\right]^{1/2}.
}
\end{equation}
For $p\ll1$,
\begin{equation}
\sqrt{1+p^2}\approx1+\frac{p^2}{2}.
\end{equation}
Then
\begin{equation}
\alpha
\approx\omega\sqrt{\frac{\mu\epsilon}{2}}
\frac{p}{\sqrt2}
=\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}},
\end{equation}
and
\begin{equation}
\beta\approx\omega\sqrt{\mu\epsilon}.
\end{equation}
Therefore
\begin{equation}
\boxed{
\alpha_{\text{LL}}\approx\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}},
\qquad
\beta_{\text{LL}}\approx\omega\sqrt{\mu\epsilon}.
}
\end{equation}
For $p\gg1$,
\begin{equation}
\sqrt{1+p^2}\approx p.
\end{equation}
The $\pm1$ terms are negligible relative to $p$, so
\begin{align}
\alpha&\approx\omega\sqrt{\frac{\mu\epsilon}{2}}\sqrt{p},\\
\beta&\approx\omega\sqrt{\frac{\mu\epsilon}{2}}\sqrt{p}.
\end{align}
Since $p=\sigma/(\omega\epsilon)$,
\begin{equation}
\boxed{
\alpha\approx\beta\approx\sqrt{\frac{\omega\mu\sigma}{2}}.
}
\end{equation}
The low-loss and good-conductor formulas are therefore limiting forms of the same exact propagation constant.
\section*{Solution 16: Julia comparison of exact and approximate propagation models}
A compact implementation should calculate
\begin{equation}
\epsilon=\epsilon_r\epsilon_0,
\qquad
\mu=\mu_r\mu_0,
\qquad
p=\frac{\sigma}{\omega\epsilon}
\end{equation}
at every frequency. The exact complex quantities are
\begin{equation}
\gamma=\sqrt{i\omega\mu(\sigma+i\omega\epsilon)},
\qquad
\eta=\sqrt{\frac{i\omega\mu}{\sigma+i\omega\epsilon}}.
\end{equation}
Then
\begin{equation}
\alpha=\operatorname{Re}(\gamma),
\qquad
\beta=\operatorname{Im}(\gamma).
\end{equation}
For comparison, calculate
\begin{equation}
\alpha_{\text{LL}}=\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}},
\qquad
\beta_{\text{LL}}=\omega\sqrt{\mu\epsilon},
\end{equation}
and
\begin{equation}
\alpha_{\text{GC}}=\beta_{\text{GC}}
=\sqrt{\frac{\omega\mu\sigma}{2}}.
\end{equation}
The included Julia script prints a logarithmically spaced table and, if \texttt{Plots.jl} is available, produces comparison plots.
For the specified material,
\begin{equation}
p=\frac{0.01}{2\pi f(4\epsilon_0)}.
\end{equation}
The transition $p\approx1$ occurs near
\begin{equation}
f\approx\frac{\sigma}{2\pi\epsilon}
\approx4.49\times10^7\,\text{Hz}.
\end{equation}
Thus frequencies far below roughly $45\,\text{MHz}$ are conductor-like for this material, while frequencies far above that scale become progressively more dielectric-like. This illustrates an important point: \emph{the propagation regime is set by both material properties and frequency}.
\section*{What EM20E1 adds to the series}
EM20 introduced the propagation formulas. EM20E1 develops the calculation habits needed to use them reliably:
\begin{equation}
\boxed{
\text{material data}
\longrightarrow
p=\frac{\sigma}{\omega\epsilon}
\longrightarrow
\text{choose regime}
\longrightarrow
\alpha,\beta,\eta,\delta
\longrightarrow
\text{attenuation and phase}.
}
\end{equation}
The next boundary-value step is to ask what happens when $\eta$ changes discontinuously at an interface. That leads directly to reflection, transmission, and refraction.
\section*{References}
\begin{thebibliography}{9}
\bibitem{Griffiths2017}
D. J. Griffiths,
\emph{Introduction to Electrodynamics}, 4th ed.,
Cambridge University Press, 2017.
\bibitem{Cheng1989}
D. K. Cheng,
\emph{Field and Wave Electromagnetics}, 2nd ed.,
Addison-Wesley, 1989.
\bibitem{Ulaby2015}
F. T. Ulaby and U. Ravaioli,
\emph{Fundamentals of Applied Electromagnetics}, 7th ed.,
Pearson, 2015.
\bibitem{Balanis2012}
C. A. Balanis,
\emph{Advanced Engineering Electromagnetics}, 2nd ed.,
Wiley, 2012.
\bibitem{Jackson1999}
J. D. Jackson,
\emph{Classical Electrodynamics}, 3rd ed.,
Wiley, 1999.
\end{thebibliography}