05 electromagnetic_waves_master_notes

Topic 5: Electromagnetic Waves — Master Study Notes

Instructor: Naymur Rahman / Sanglap Sir Primary References:

• Field and Wave Electromagnetics (2nd Edition) — David K. Cheng (Chapter 8)
• Elements of Electromagnetics (7th Edition) — Matthew N.O. Sadiku (Chapter 10)
• Lecture Slides: L 11.pdf, L 12.pdf, L 13.pdf, L 14.pdf, L 15_updated.pdf
• Topper Notes: 03 solution to em eqns.pdf, Sanglap Sir-2309008.pdf

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1. Wave Equations & Helmholtz Equations

1.1 Source-Free Simple Medium Wave Derivation

To model how electromagnetic fields travel through space as wave disturbances, we start with Maxwell’s equations in a source-free ($\rho =0,\vec{J}=0$), linear, isotropic, and homogeneous medium characterized by permeability $\mu$ and permittivity $ϵ$:

$\nabla \cdot \vec{E}=0\text{(Gauss’s Law for Electric Field)}(1)$ $\nabla \cdot \vec{H}=0\text{(Gauss’s Law for Magnetic Field)}(2)$ $\nabla \times \vec{E}=-\mu \frac{\partial \vec{H}}{\partial t}\text{(Faraday’s Law)}(3)$ $\nabla \times \vec{H}=ϵ\frac{\partial \vec{E}}{\partial t}\text{(Ampere’s Circuital Law)}(4)$

To decouple these curl equations into a single differential equation of one variable, we take the curl of Faraday’s Law (Equation 3):

$\nabla \times (\nabla \times \vec{E})=-\mu \frac{\partial }{\partial t}(\nabla \times \vec{H})(5)$

Using the vector identity for the curl of a curl on the left-hand side: $\nabla \times (\nabla \times \vec{E})=\nabla (\nabla \cdot \vec{E})-{\nabla }^2\vec{E}(6)$

Substitute Equation (1) ($\nabla \cdot \vec{E}=0$) and Equation (4) ($\nabla \times \vec{H}=ϵ\frac{\partial \vec{E}}{\partial t}$) into Equation (6): $-{\nabla }^2\vec{E}=-\mu ϵ\frac{{\partial }^2\vec{E}}{\partial t^2}$

This simplifies to the 3D Homogeneous Vector Wave Equation for $\vec{E}$:

${\nabla }^2\vec{E}-\mu ϵ\frac{{\partial }^2\vec{E}}{\partial t^2}=0\text{[PYQ: 2018, 2019, 2024]}$

By taking the curl of Ampere’s Law (Equation 4) and repeating the same steps, we obtain the 3D Homogeneous Vector Wave Equation for $\vec{H}$:

${\nabla }^2\vec{H}-\mu ϵ\frac{{\partial }^2\vec{H}}{\partial t^2}=0(8)$

Comparing this to the classical wave equation ${\nabla }^2\psi -\frac{1}{u^2}\frac{{\partial }^2\psi }{\partial t^2}=0$, we find that electromagnetic fields propagate with a speed:

$u=\frac{1}{\sqrt{\mu ϵ}}\text{[m/s]}(9)$

In a vacuum (free space), this speed is exactly the speed of light: $c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}\approx 3\times 10^8\text{[m/s]}(10)$

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1.2 Time-Harmonic Fields & Helmholtz Equations

In engineering applications, we deal with fields that vary sinusoidally with time (time-harmonic fields represented by the phasor factor $e^{j\omega t}$). The time derivative transforms in the phasor domain: $\frac{\partial }{\partial t}\to j\omega \text{and}\frac{{\partial }^2}{\partial t^2}\to -{\omega }^2$

Substituting this into the wave equations (7) and (8) yields the Homogeneous Vector Helmholtz Equations:

${\nabla }^2{\vec{E}}_s+k^2{\vec{E}}_s=0\text{[PYQ: 2016, 2021, 2023, 2025]}(11)$ ${\nabla }^2{\vec{H}}_s+k^2{\vec{H}}_s=0(12)$

where $k$ is the wave number (phase constant) defined as: $k=\omega \sqrt{\mu ϵ}\text{[rad/m]}(13)$

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1.3 High-Yield Proof: Superposition of Alternate Solutions

Theorem: If $\vec{E}$ and $\vec{H}$ are solutions of source-free Maxwell’s equations in a simple medium ($ϵ,\mu$), then show that ${\vec{E}}^{\prime }=\eta \vec{H}$ and ${\vec{H}}^{\prime }=-\vec{E}/\eta$ are also solutions. [PYQ: 2024, 2025]

Proof: We must verify that ${\vec{E}}^{\prime }$ and ${\vec{H}}^{\prime }$ satisfy all four source-free Maxwell’s equations:

1. Divergence Equations: $\nabla \cdot {\vec{E}}^{\prime }=\nabla \cdot (\eta \vec{H})=\eta (\nabla \cdot \vec{H})$ Since $\nabla \cdot \vec{H}=0$, it follows that $\nabla \cdot {\vec{E}}^{\prime }=0$. (Verified) $\nabla \cdot {\vec{H}}^{\prime }=\nabla \cdot \left(-\frac{\vec{E}}{\eta }\right)=-\frac{1}{\eta }(\nabla \cdot \vec{E})$ Since $\nabla \cdot \vec{E}=0$, it follows that $\nabla \cdot {\vec{H}}^{\prime }=0$. (Verified)

2. Faraday’s Curl Law: $\nabla \times {\vec{E}}^{\prime }=\nabla \times (\eta \vec{H})=\eta (\nabla \times \vec{H})$ Using phasor Ampere’s Law ($\nabla \times \vec{H}=j\omega ϵ\vec{E}$): $\nabla \times {\vec{E}}^{\prime }=\eta (j\omega ϵ\vec{E})=j\omega (\eta ϵ)\vec{E}$ Since $\eta =\sqrt{\frac{\mu }{ϵ}}⟹\eta ϵ=\sqrt{\frac{\mu }{ϵ}}ϵ=\sqrt{\mu ϵ}$, and $\frac{\mu }{\eta }=\frac{\mu }{\sqrt{\mu /ϵ}}=\sqrt{\mu ϵ}$: $\eta ϵ=\frac{\mu }{\eta }$ Substituting this back: $\nabla \times {\vec{E}}^{\prime }=j\omega \left(\frac{\mu }{\eta }\right)\vec{E}=-j\omega \mu \left(-\frac{\vec{E}}{\eta }\right)=-j\omega \mu {\vec{H}}^{\prime }$ This is exactly $\nabla \times {\vec{E}}^{\prime }=-j\omega \mu {\vec{H}}^{\prime }$, verifying Faraday’s Law. (Verified)

3. Ampere’s Curl Law: $\nabla \times {\vec{H}}^{\prime }=\nabla \times \left(-\frac{\vec{E}}{\eta }\right)=-\frac{1}{\eta }(\nabla \times \vec{E})$ Using phasor Faraday’s Law ($\nabla \times \vec{E}=-j\omega \mu \vec{H}$): $\nabla \times {\vec{H}}^{\prime }=-\frac{1}{\eta }(-j\omega \mu \vec{H})=j\omega \left(\frac{\mu }{\eta }\right)\vec{H}$ Since $\frac{\mu }{\eta }=\eta ϵ$: $\nabla \times {\vec{H}}^{\prime }=j\omega ϵ(\eta \vec{H})=j\omega ϵ{\vec{E}}^{\prime }$ This is exactly $\nabla \times {\vec{H}}^{\prime }=j\omega ϵ{\vec{E}}^{\prime }$, verifying Ampere’s Law. (Verified)

Thus, the alternative fields ${\vec{E}}^{\prime }$ and ${\vec{H}}^{\prime }$ satisfy Maxwell’s equations and represent a valid physical wave propagation state. [Q.E.D.]

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1.4 Retarded Potential Wave Equation Derivation

Theorem: Starting from the homogeneous wave equation, show that the scalar potential at a distance $R$ at time $t$ depends on the charge density at an earlier time $(t-R/u)$. [PYQ: 2018]

Proof: The non-homogeneous wave equation for scalar potential $V$ is: ${\nabla }^{2}V-\mu ϵ\frac{{\partial }^{2}V}{\partial t^2}=-\frac{{\rho }_v}{ϵ}$

For a point charge at the origin in a homogeneous medium, we first analyze the solution in the region away from the source (${\rho }_v=0$), where the equation becomes homogeneous: ${\nabla }^{2}V-\mu ϵ\frac{{\partial }^{2}V}{\partial t^2}=0(14)$

Because of spherical symmetry, the potential $V$ depends only on the radial distance $R$ (and is independent of $\theta$ and $\varphi$). The Laplacian ${\nabla }^{2}V$ in spherical coordinates simplifies to: ${\nabla }^{2}V=\frac{1}{R^2}\frac{\partial }{\partial R}\left(R^2\frac{\partial V}{\partial R}\right)=\frac{{\partial }^{2}V}{\partial R^2}+\frac{2}{R}\frac{\partial V}{\partial R}(15)$

Substituting Equation (15) into Equation (14): $\frac{{\partial }^{2}V}{\partial R^2}+\frac{2}{R}\frac{\partial V}{\partial R}-\mu ϵ\frac{{\partial }^{2}V}{\partial t^2}=0(16)$

To solve this 2nd-order partial differential equation, we use a mathematical substitution: $V(R,t)=\frac{1}{R}u(R,t)(17)$

Taking partial derivatives: $\frac{\partial V}{\partial R}=-\frac{u}{R^2}+\frac{1}{R}\frac{\partial u}{\partial R}$ $\frac{{\partial }^{2}V}{\partial R^2}=\frac{2u}{R^3}-\frac{2}{R^2}\frac{\partial u}{\partial R}+\frac{1}{R}\frac{{\partial }^{2}u}{\partial R^2}$ $\frac{{\partial }^{2}V}{\partial t^2}=\frac{1}{R}\frac{{\partial }^{2}u}{\partial t^2}$

Substitute these derivatives back into Equation (16): $\left(\frac{2u}{R^3}-\frac{2}{R^2}\frac{\partial u}{\partial R}+\frac{1}{R}\frac{{\partial }^{2}u}{\partial R^2}\right)+\frac{2}{R}\left(-\frac{u}{R^2}+\frac{1}{R}\frac{\partial u}{\partial R}\right)-\frac{\mu ϵ}{R}\frac{{\partial }^{2}u}{\partial t^2}=0$

Simplifying this yields the classical 1D wave equation: $\frac{{\partial }^{2}u}{\partial R^2}-\mu ϵ\frac{{\partial }^{2}u}{\partial t^2}=0(18)$

The general solution of this wave equation is a superposition of outward-propagating and inward-propagating waves: $u(R,t)=f\left(t-R\sqrt{\mu ϵ}\right)+g\left(t+R\sqrt{\mu ϵ}\right)(19)$

Since $u=1/\sqrt{\mu ϵ}$, and physically waves must travel outward from the source charge at the origin (the inward wave $g(t+R/u)$ violates physical causality for an isolated source), we select only the outward wave: $u(R,t)=f\left(t-\frac{R}{u}\right)$

Thus, the scalar potential is: $V(R,t)=\frac{1}{R}f\left(t-\frac{R}{u}\right)(20)$

To find the function $f$, we compare this to the static case. As the frequency approaches zero ($u\to \infty$ or $t-R/u\to t$), the potential must reduce to the electrostatic potential for a point charge $Q(t)$ at the origin: $V(R)=\frac{Q(t)}{4\pi ϵR}$

Equating this to (20) under static conditions: $\frac{1}{R}f(t)=\frac{Q(t)}{4\pi ϵR}⟹f(t)=\frac{Q(t)}{4\pi ϵ}$

Therefore, for a time-varying charge, the functional form of $f(t-R/u)$ is: $f\left(t-\frac{R}{u}\right)=\frac{Q(t-R/u)}{4\pi ϵ}$

Substituting this back into (20) yields: $V(R,t)=\frac{Q(t-R/u)}{4\pi ϵR}(21)$

For a continuous volume charge distribution, we sum (integrate) the contributions of all charge elements: $V(R,t)=\frac{1}{4\pi ϵ}{\int }_{v^{\prime }}\frac{{\rho }_v(t-R/u)}{R}d{v}^{\prime }\text{[Proved]}$

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2. Plane Wave Propagation in Lossless & Lossy Media

2.1 Core Waveguide / Media Terminology

• Uniform Plane Wave: A wave whose electric ($\vec{E}$) and magnetic ($\vec{H}$) vectors lie entirely in planes perpendicular to the direction of wave propagation, having the same direction, magnitude, and phase in those planes.
• Complex Permittivity (${ϵ}_c$): Accounts for both capacitive storage and conduction losses: ${ϵ}_c={ϵ}^{\prime }-j{ϵ}^{\prime \prime }=ϵ-j\frac{\sigma }{\omega }(22)$
• Loss Tangent ($\tan {\delta }_c$): The ratio of conduction current density ($\vec{J}=\sigma \vec{E}$) to displacement current density (${\vec{J}}_d=j\omega ϵ\vec{E}$): $\tan {\delta }_c=\frac{{ϵ}^{\prime \prime }}{{ϵ}^{\prime }}=\frac{\sigma }{\omega ϵ}\text{[PYQ: 2015, 2019]}(23)$

[Figure: Vector diagram showing displacement current and conduction current phase relationships in complex planes. Refer to Sadiku 7th Edition, Fig 10.5]

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2.2 Mathematical Classification of Media

We classify media into Good Conductors and Lossy Dielectrics based on the value of their loss tangent:

Parameter	Lossy Dielectric (Good Insulator)	Good Conductor
Mathematical Condition	$\sigma \ll \omega ϵ⟹\tan {\delta }_c\ll 1$	$\sigma \gg \omega ϵ⟹\tan {\delta }_c\gg 1$
Dominant Current	Displacement Current dominates (${\vec{J}}_d\gg \vec{J}$)	Conduction Current dominates ($\vec{J}\gg {\vec{J}}_d$)
Wave Attenuation	Very low	Extremely high (confined to skin)

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2.3 The Complex Propagation Constant ($\gamma =\alpha +j\beta$)

In a source-free lossy medium, the phasor Helmholtz equation is solved using a complex propagation constant:

$\gamma =\alpha +j\beta =j\omega \sqrt{\mu {ϵ}_c}=j\omega \sqrt{\mu ϵ\left(1-j\frac{\sigma }{\omega ϵ}\right)}(24)$

By squaring both sides: ${\gamma }^2=(\alpha +j\beta {)}^2=({\alpha }^2-{\beta }^2)+j2\alpha \beta =-{\omega }^2\mu ϵ+j\omega \mu \sigma$

By equating real and imaginary parts, we solve for the general expressions of $\alpha$ (attenuation constant) and $\beta$ (phase constant):

$\alpha =\omega \sqrt{\frac{\mu ϵ}{2}\left[\sqrt{1+{\left(\frac{\sigma }{\omega ϵ}\right)}^2}-1\right]}\text{[Np/m]}(25)$ $\beta =\omega \sqrt{\frac{\mu ϵ}{2}\left[\sqrt{1+{\left(\frac{\sigma }{\omega ϵ}\right)}^2}+1\right]}\text{[rad/m]}(26)$

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2.4 Derivation of Approximations via Binomial Expansion

A. Low-Loss Dielectric ($\sigma \ll \omega ϵ$)

When the loss tangent is small, we simplify Equation (24) using the binomial expansion $(1-x{)}^{1/2}=1-\frac{1}{2}x+\frac{n(n-1)}{2!}{x}^2-\dots$ where $x=j\frac{\sigma }{\omega ϵ}$:

$\gamma =j\omega \sqrt{\mu ϵ}{\left(1-j\frac{\sigma }{\omega ϵ}\right)}^{1/2}$ $\gamma \approx j\omega \sqrt{\mu ϵ}\left[1-j\frac{\sigma }{2\omega ϵ}+\frac{1}{8}{\left(\frac{\sigma }{\omega ϵ}\right)}^2\right]$ $\gamma \approx j\omega \sqrt{\mu ϵ}\left[1+\frac{1}{8}{\left(\frac{\sigma }{\omega ϵ}\right)}^2\right]+\frac{\sigma }{2}\sqrt{\frac{\mu }{ϵ}}$

Since $\gamma =\alpha +j\beta$, equating real and imaginary parts yields:

$\alpha \approx \frac{\sigma }{2}\sqrt{\frac{\mu }{ϵ}}\text{[PYQ: 2019, 2025]}(27)$ $\beta \approx \omega \sqrt{\mu ϵ}\left[1+\frac{1}{8}{\left(\frac{\sigma }{\omega ϵ}\right)}^2\right]\text{[PYQ: 2019, 2025]}(28)$

B. Good Conductor ($\sigma \gg \omega ϵ$)

Since $\frac{\sigma }{\omega ϵ}\gg 1$, we can neglect the $1$ term inside Equation (24):

$\gamma =j\omega \sqrt{\mu \left(-j\frac{\sigma }{\omega }\right)}=\sqrt{j\omega \mu \sigma }$

Using the identity $\sqrt{j}=e^{j\pi /4}=\frac{1+j}{\sqrt{2}}$: $\gamma =\sqrt{\omega \mu \sigma }\left(\frac{1+j}{\sqrt{2}}\right)=(1+j)\sqrt{\frac{\omega \mu \sigma }{2}}=(1+j)\sqrt{\pi f\mu \sigma }$

Equating real and imaginary parts yields: $\alpha =\beta \approx \sqrt{\pi f\mu \sigma }\text{[PYQ: 2019, 2025]}(29)$

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2.5 Skin Depth ($\delta$)

• Definition: The skin depth ($\delta$) is defined as the distance an electromagnetic wave must propagate inside a good conductor for its field amplitude to decay to $1/e$ (approx. 36.8%) of its initial boundary value. [PYQ: 2015, 2016, 2021, 2025]

[Figure: Plot of Electric Field Amplitude decay as a function of depth z into a conducting wall, showing E decreasing exponentially to 0.368 Eo at z = delta. Refer to L 13.pdf, Slide 8]

Derivation: The electric field traveling in the $+z$ direction in a lossy medium decays as $e^{-\alpha z}$. Let the field at the boundary $z=0$ be $E_0$. At depth $z=\delta$: $E(\delta )=E_0{e}^{-\alpha \delta }\equiv E_0{e}^{-1}⟹\alpha \delta =1⟹\delta =\frac{1}{\alpha }$

For a good conductor, substituting Equation (29) for $\alpha$:

$\delta =\frac{1}{\sqrt{\pi f\mu \sigma }}\text{[m]}(30)$

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3. Phase Velocity, Group Velocity, & Dispersion

3.1 Terminology

• Phase Velocity ($u_p$): The speed at which a single constant-phase front of a single-frequency wave propagates through a medium: $u_p=\frac{\omega }{\beta }(31)$
• Group Velocity ($u_g$): The speed at which the envelope of a multi-frequency wave packet (containing information or signal energy) propagates: $u_g=\frac{d\omega }{d\beta }=\frac{1}{\frac{d\beta }{d\omega }}(32)$
• Dispersion: The phenomenon of signal distortion that occurs when different frequency components of a wave packet travel at different phase velocities (i.e., $u_p$ is a function of frequency $\omega$).

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3.2 Non-Dispersive vs. Dispersive Conditions

Taking the derivative of Equation (31) w.r.t $\omega$ yields: $\frac{d\beta }{d\omega }=\frac{d}{d\omega }\left(\frac{\omega }{u_p}\right)=\frac{1}{u_p}-\frac{\omega }{u_p^2}\frac{d{u}_p}{d\omega }$

Substitute this into Equation (32) to find the general relation between group and phase velocity:

$u_g=\frac{u_p}{1-\frac{\omega }{u_p}\frac{d{u}_p}{d\omega }}(33)$

From this expression, we classify three dispersion regimes:

1. No Dispersion: $\frac{d{u}_p}{d\omega }=0⟹u_g=u_p$ (Phase and group velocities are equal and constant; wave packet retains its shape perfectly). [PYQ: 2018]
2. Normal Dispersion: $\frac{d{u}_p}{d\omega }<0⟹u_g<u_p$ (Phase velocity decreases with frequency; higher frequencies travel slower).
3. Anomalous Dispersion: $\frac{d{u}_p}{d\omega }>0⟹u_g>u_p$ (Phase velocity increases with frequency; higher frequencies travel faster).

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3.3 High-Yield Waveguide/Plasma Relation: $u_p{u}_g=c^2$

In a plasma or waveguide, the phase constant is given by: $\beta =\frac{\omega }{c}\sqrt{1-{\left(\frac{{\omega }_p}{\omega }\right)}^2}=\frac{1}{c}\sqrt{{\omega }^2-{\omega }_p^2}$

Differentiating both sides w.r.t $\omega$: $2\beta \frac{d\beta }{d\omega }=\frac{1}{c^2}(2\omega )⟹\beta \frac{d\beta }{d\omega }=\frac{\omega }{c^2}$

Substitute $u_p=\frac{\omega }{\beta }$ and $u_g=\frac{1}{d\beta /d\omega }$: $\frac{\omega }{u_p}\left(\frac{1}{u_g}\right)=\frac{\omega }{c^2}⟹u_p{u}_g=c^2\text{[PYQ: 2018]}(34)$

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4. Wave Polarization

4.1 Definition

Polarization describes the time-varying path traced by the tip of the electric field vector ($\vec{E}(z,t)$) over a fixed plane perpendicular to the direction of wave propagation. [PYQ: 2015, 2016, 2017, 2021, 2023, 2024, 2025]

Let a wave propagate in the $+z$ direction. The most general expression for its transverse fields is: $\vec{E}(z,t)={\hat{a}}_x{E}_{x0}\cos (\omega t-\beta z)+{\hat{a}}_y{E}_{y0}\cos (\omega t-\beta z+\theta )(35)$ where $\theta$ is the phase difference by which the $y$-component leads the $x$-component.

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4.2 Categorization & Conditions

We classify wave polarization into three types:

A. Linear Polarization

The tip of the electric field vector traces out a straight line over time.

• Mathematical Conditions: $\theta =n\pi \text{for }n=0,\pm 1,\pm 2,\dots \text{[PYQ: 2016, 2025]}$
• Resultant Angle: $\varphi ={\tan }^{-1}\left(\frac{{\vec{E}}_y}{{\vec{E}}_x}\right)={\tan }^{-1}\left(\pm \frac{E_{y0}}{E_{x0}}\right)(\text{constant over time})$

B. Circular Polarization

The tip of the electric field vector traces out a circle over time.

• Mathematical Conditions: $E_{x0}=E_{y0}=E_0\text{and}\theta =\pm \left(n\pi +\frac{\pi }{2}\right)\text{[PYQ: 2015, 2016, 2024]}$
• Sub-classifications:
  1. Right-Hand Circular Polarization (RHCP / Negative Circular): The vector rotates clockwise when viewed in the direction of wave propagation ($\theta =-\pi /2$ or lagging by $90^{\circ }$). [PYQ: 2018, 2021, 2022]
  2. Left-Hand Circular Polarization (LHCP / Positive Circular): The vector rotates counter-clockwise when viewed in the direction of wave propagation ($\theta =+\pi /2$ or leading by $90^{\circ }$).

C. Elliptical Polarization

The tip of the electric field vector traces out an ellipse over time. This occurs under all other general phase differences and unequal amplitudes.

• Mathematical Conditions: $E_{x0}\ne E_{y0}\text{and}\theta \ne n\pi \text{[PYQ: 2015, 2016, 2024]}$

[Figure: Traced paths of Linear, Circular, and Elliptical electric field vectors over a full temporal cycle. Refer to Sadiku 7th Edition, Fig 10.11]

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4.3 High-Yield Proof: Resolution of Linear to Circular components

Theorem: Prove that a linearly polarized wave can be resolved into a right-hand circularly polarized wave and a left-hand circularly polarized wave of equal amplitude. [PYQ: 2017]

Proof: Let a linearly polarized wave be directed along the x-axis: $\vec{E}(z,t)={\hat{a}}_x{E}_0\cos (\omega t-\beta z)$

Let us define two equal-amplitude RHCP (${\vec{E}}_R$) and LHCP (${\vec{E}}_L$) wave components with amplitude $E_0/2$: ${\vec{E}}_L=\frac{E_0}{2}\left[{\hat{a}}_x\cos (\omega t-\beta z)-{\hat{a}}_y\sin (\omega t-\beta z)\right]$ ${\vec{E}}_R=\frac{E_0}{2}\left[{\hat{a}}_x\cos (\omega t-\beta z)+{\hat{a}}_y\sin (\omega t-\beta z)\right]$

Summing these two circularly polarized waves: ${\vec{E}}_L+{\vec{E}}_R=\frac{E_0}{2}\left[\cos (\omega t-\beta z)+\cos (\omega t-\beta z)\right]{\hat{a}}_x+\frac{E_0}{2}\left[-\sin (\omega t-\beta z)+\sin (\omega t-\beta z)\right]{\hat{a}}_y$ ${\vec{E}}_L+{\vec{E}}_R=E_0\cos (\omega t-\beta z){\hat{a}}_x=\vec{E}(z,t)$

Because the $y$-directed components are exactly equal and opposite, they cancel out, leaving only the $x$-directed linear field. Thus, any linearly polarized wave is a superposition of two counter-rotating circularly polarized waves of equal amplitude. [Q.E.D.]

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5. Reflection, Transmission, & Standing Waves (Normal Incidence)

5.1 Normal Incidence at a Plane Dielectric Boundary

[Figure: Uniform plane wave incident normally at z=0 interface separating Medium 1 (eta_1) and Medium 2 (eta_2). Arrows show vectors of Incident, Reflected, and Transmitted fields. Refer to L 15_updated.pdf, Slide 10]

Let an incident wave in Medium 1 travel along the $+z$ direction: ${\vec{E}}_i(z)={\hat{a}}_x{E}_{i0}{e}^{-{\gamma }_{1}z}{\vec{H}}_i(z)={\hat{a}}_y\frac{E_{i0}}{{\eta }_1}{e}^{-{\gamma }_{1}z}$

Due to the impedance mismatch, a reflected wave travels in the $-z$ direction in Medium 1, and a transmitted wave travels in the $+z$ direction in Medium 2: ${\vec{E}}_r(z)={\hat{a}}_x{E}_{r0}{e}^{+{\gamma }_{1}z}{\vec{H}}_r(z)=-{\hat{a}}_y\frac{E_{r0}}{{\eta }_1}{e}^{+{\gamma }_{1}z}$ ${\vec{E}}_t(z)={\hat{a}}_x{E}_{t0}{e}^{-{\gamma }_{2}z}{\vec{H}}_t(z)={\hat{a}}_y\frac{E_{t0}}{{\eta }_2}{e}^{-{\gamma }_{2}z}$

At the interface ($z=0$), applying electromagnetic boundary conditions (tangential $\vec{E}$ and $\vec{H}$ must be continuous): $E_i(0)+E_r(0)=E_t(0)⟹E_{i0}+E_{r0}=E_{t0}(36)$ $H_i(0)+H_r(0)=H_t(0)⟹\frac{E_{i0}}{{\eta }_1}-\frac{E_{r0}}{{\eta }_1}=\frac{E_{t0}}{{\eta }_2}(37)$

Dividing Equation (36) by $E_{i0}$: $1+\frac{E_{r0}}{E_{i0}}=\frac{E_{t0}}{E_{i0}}⟹1+\Gamma =\tau \text{[PYQ: 2015, 2016, 2017, 2019, 2021, 2022, 2023, 2024]}$

where:

• $\Gamma$ is the Reflection Coefficient: $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}(38)$
• $\tau$ is the Transmission Coefficient: $\tau =\frac{2{\eta }_2}{{\eta }_2+{\eta }_1}(39)$

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5.2 Creation of Standing Waves at a Conducting Interface

Question: Why is a standing wave created when a plane EM wave is incident normally on a plane conducting boundary? [PYQ: 2018, 2020]

Explanation: A perfect conductor has infinite conductivity ($\sigma \to \infty$), which reduces its intrinsic impedance to zero (${\eta }_2=0$). Substituting this into the reflection coefficient: $\Gamma =\frac{0-{\eta }_1}{0+{\eta }_1}=-1(40)$

This means the wave is completely reflected with a $180^{\circ }$ phase shift, and no wave is transmitted ($\tau =0$). The superposition of the incident wave traveling in the $+z$ direction and the reflected wave traveling in the $-z$ direction creates a Standing Wave because the two waves have equal amplitudes but opposite propagation directions. The fields do not travel; instead, they oscillate in place with fixed nodes (zeros) and antinodes (maxima).

Mathematical Proof: Let Medium 1 be a lossless dielectric ($\alpha =0⟹{\gamma }_1=j{\beta }_1$). The total electric field in Medium 1 is: ${\vec{E}}_1(z)={\vec{E}}_i(z)+{\vec{E}}_r(z)={\hat{a}}_x\left(E_{i0}{e}^{-j{\beta }_{1}z}+E_{r0}{e}^{j{\beta }_{1}z}\right)$

Substitute $E_{r0}=\Gamma E_{i0}=-E_{i0}$: ${\vec{E}}_1(z)={\hat{a}}_x{E}_{i0}\left(e^{-j{\beta }_{1}z}-e^{j{\beta }_{1}z}\right)$

Using Euler’s identity ($e^{j\theta }-e^{-j\theta }=2j\sin \theta$): ${\vec{E}}_{1s}(z)=-{\hat{a}}_{x}j2E_{i0}\sin ({\beta }_{1}z)(41)$

Converting back to the time domain: ${\vec{E}}_1(z,t)=\text{Re}\left\{{\vec{E}}_{1s}(z)e^{j\omega t}\right\}=\text{Re}\left\{-{\hat{a}}_{x}j2E_{i0}\sin ({\beta }_{1}z)e^{j\omega t}\right\}$ ${\vec{E}}_1(z,t)={\hat{a}}_{x}2E_{i0}\sin ({\beta }_{1}z)\sin (\omega t)\text{[PYQ: 2018, 2020]}(42)$

This is a classical standing wave equation. The spatial factor $\sin ({\beta }_{1}z)$ is completely decoupled from the temporal oscillation factor $\sin (\omega t)$.

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6. Oblique Incidence, Brewster’s Angle, & Doppler Effect

6.1 Brewster’s Angle Proof

Theorem: Prove that in the case of non-magnetic media (${\mu }_1={\mu }_2={\mu }_0$), Brewster’s angle (zero reflection) exists only for parallel polarization (P-polarization) and not for perpendicular polarization (S-polarization). [PYQ: 2022]

Proof:

1. For Perpendicular (S) Polarization: The reflection coefficient ${\Gamma }_{\perp }$ is given by: ${\Gamma }_{\perp }=\frac{{\eta }_2\cos {\theta }_i-{\eta }_1\cos {\theta }_t}{{\eta }_2\cos {\theta }_i+{\eta }_1\cos {\theta }_t}(43)$ For a non-magnetic medium, $\eta ={\eta }_0/\sqrt{{ϵ}_r}$. Setting ${\Gamma }_{\perp }=0⟹{\eta }_2\cos {\theta }_{iB}={\eta }_1\cos {\theta }_t$: $\frac{1}{\sqrt{{ϵ}_{r2}}}\cos {\theta }_{iB}=\frac{1}{\sqrt{{ϵ}_{r1}}}\cos {\theta }_t⟹{ϵ}_{r1}{\cos }^2{\theta }_{iB}={ϵ}_{r2}{\cos }^2{\theta }_t$ Using Snell’s Law ($\sin {\theta }_t=\sqrt{\frac{{ϵ}_{r1}}{{ϵ}_{r2}}}\sin {\theta }_{iB}⟹{\cos }^2{\theta }_t=1-\frac{{ϵ}_{r1}}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}$): ${ϵ}_{r1}(1-{\sin }^2{\theta }_{iB})={ϵ}_{r2}\left(1-\frac{{ϵ}_{r1}}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}\right)$ ${ϵ}_{r1}-{ϵ}_{r1}{\sin }^2{\theta }_{iB}={ϵ}_{r2}-{ϵ}_{r1}{\sin }^2{\theta }_{iB}⟹{ϵ}_{r1}={ϵ}_{r2}$ This requires both media to be identical, which means there is no boundary. Thus, no Brewster’s angle can exist for S-polarization.

2. For Parallel (P) Polarization: The reflection coefficient ${\Gamma }_{\parallel }$ is given by: ${\Gamma }_{\parallel }=\frac{{\eta }_2\cos {\theta }_t-{\eta }_1\cos {\theta }_i}{{\eta }_2\cos {\theta }_t+{\eta }_1\cos {\theta }_i}(44)$ Setting ${\Gamma }_{\parallel }=0⟹{\eta }_2\cos {\theta }_t={\eta }_1\cos {\theta }_{iB}$: $\sqrt{{ϵ}_{r1}}\cos {\theta }_t=\sqrt{{ϵ}_{r2}}\cos {\theta }_{iB}⟹{ϵ}_{r1}(1-{\sin }^2{\theta }_t)={ϵ}_{r2}(1-{\sin }^2{\theta }_{iB})$ Substituting Snell’s law: ${ϵ}_{r1}\left(1-\frac{{ϵ}_{r1}}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}\right)={ϵ}_{r2}(1-{\sin }^2{\theta }_{iB})$ ${ϵ}_{r1}-\frac{{ϵ}_{r1}^2}{{ϵ}_{r2}}{\sin }^2{\theta }_{iB}={ϵ}_{r2}-{ϵ}_{r2}{\sin }^2{\theta }_{iB}⟹{\sin }^2{\theta }_{iB}\left({ϵ}_{r2}-\frac{{ϵ}_{r1}^2}{{ϵ}_{r2}}\right)={ϵ}_{r2}-{ϵ}_{r1}$ Multiplying by ${ϵ}_{r2}$: ${\sin }^2{\theta }_{iB}\left({ϵ}_{r2}^2-{ϵ}_{r1}^2\right)={ϵ}_{r2}({ϵ}_{r2}-{ϵ}_{r1})$ ${\sin }^2{\theta }_{iB}=\frac{{ϵ}_{r2}({ϵ}_{r2}-{ϵ}_{r1})}{({ϵ}_{r2}-{ϵ}_{r1})({ϵ}_{r2}+{ϵ}_{r1})}=\frac{{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}$ Using ${\sin }^2\theta +{\cos }^2\theta =1⟹{\tan }^2\theta =\frac{{\sin }^2\theta }{1-{\sin }^2\theta }$: ${\tan }^2{\theta }_{iB}=\frac{\frac{{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}}{1-\frac{{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}}=\frac{{ϵ}_{r2}}{{ϵ}_{r1}}$ Taking the square root yields Brewster’s Angle for P-polarization: $\tan {\theta }_{iB}=\sqrt{\frac{{ϵ}_{r2}}{{ϵ}_{r1}}}\text{[PYQ: 2022]}(45)$

Since this has a valid real solution for any positive permittivities, Brewster’s angle exists strictly for parallel polarization. [Proved]

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6.2 Doppler Effect & Red-Shift

• Definition: The Doppler effect is the shift in the observed frequency of a wave due to the relative motion between the wave source and the observer. [PYQ: 2015, 2016, 2017, 2019, 2021, 2024]

[Figure: Concentric wavefront circles compressed in front of a moving transmitter and stretched behind it, illustrating frequency shift. Refer to Sadiku 7th Edition, Fig 10.19]

Mathematical Formula: If a source of frequency $f$ moves away from a stationary receiver with a relative velocity $v$ along the line of sight: $f^{\prime }=f\left(1-\frac{v}{c}\right)(46)$ If the source moves toward the receiver: $f^{\prime }=f\left(1+\frac{v}{c}\right)(47)$

Red-Shift and Cosmology: In astronomy, the light emitted by stars and galaxies shows a red-shift (a shift toward lower frequencies/longer wavelengths) if they are moving away from Earth. Since red is at the lower frequency end of the visible spectrum, this Doppler shift confirms that the universe is expanding. [PYQ: 2020, 2024]

Practical Examples:

1. Doppler Radar (Police speed traps): Measures the speed of moving vehicles by analyzing the frequency shift of the reflected radar signal.
2. Satellite Tracking: Determines the orbits of spacecraft based on the shift in their radio carrier frequencies as they pass over ground stations. [PYQ: 2020]

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7. Exhaustive PYQ Numerical Bank (Solved)

7.1 Type A: Lossless Wave Expression Propagation [PYQ: 2021]

Question: A uniform plane wave with $\vec{E}={\hat{a}}_x{E}_x$ propagates in a lossless simple medium (${ϵ}_r=4,{\mu }_r=1,\sigma =0$) in the $+z$ direction. Assume $E_x$ is sinusoidal with a frequency of $100\text{ MHz}$ and has a maximum magnitude of $10^{-4}\text{ V/m}$ at $t=0$ and $z=\frac{1}{8}\text{ m}$. (i) Write the instantaneous expression for $\vec{E}$. (ii) Write the instantaneous expression for $\vec{H}$.

Step-by-Step Solution:

1. Find the wave parameters: $\omega =2\pi f=2\pi \times 100\times 10^6=2\pi \times 10^8\text{[rad/s]}$ $\beta =\omega \sqrt{\mu ϵ}=\frac{\omega }{c}\sqrt{{\mu }_r{ϵ}_r}=\frac{2\pi \times 10^8}{3\times 10^8}\sqrt{1\times 4}=\frac{4\pi }{3}\text{[rad/m]}$

2. Determine the phase angle ($\varphi$): The general instantaneous expression for a wave propagating in the $+z$ direction is: $E_x(z,t)=E_0\cos (\omega t-\beta z+\varphi )$ Given $E_x=10^{-4}$ at $t=0,z=1/8$: $10^{-4}=10^{-4}\cos \left(0-\left(\frac{4\pi }{3}\right)\left(\frac{1}{8}\right)+\varphi \right)⟹\cos \left(-\frac{\pi }{6}+\varphi \right)=1$ $-\frac{\pi }{6}+\varphi =0⟹\varphi =\frac{\pi }{6}$

3. Write the instantaneous expression for $\vec{E}(z,t)$: $\vec{E}(z,t)={\hat{a}}_{x}10^{-4}\cos \left(2\pi \times 10^{8}t-\frac{4\pi }{3}z+\frac{\pi }{6}\right)\text{[V/m]}$

4. Find Intrinsic Impedance ($\eta$) and write $\vec{H}(z,t)$: $\eta =\sqrt{\frac{\mu }{ϵ}}=\frac{{\eta }_0}{\sqrt{{ϵ}_r}}=\frac{120\pi }{\sqrt{4}}=60\pi \approx 188.5[\Omega ]$ Since propagation is along $+z$ and $\vec{E}$ is along $+{\hat{a}}_x$, the magnetic field $\vec{H}$ must be along $+{\hat{a}}_y$ to satisfy the Poynting vector direction (${\hat{a}}_x\times {\hat{a}}_y={\hat{a}}_z$): $\vec{H}(z,t)={\hat{a}}_y\frac{E_0}{\eta }\cos \left(\omega t-\beta z+\varphi \right)$ $\vec{H}(z,t)={\hat{a}}_y\frac{10^{-4}}{60\pi }\cos \left(2\pi \times 10^{8}t-\frac{4\pi }{3}z+\frac{\pi }{6}\right)$ $\vec{H}(z,t)={\hat{a}}_{y}5.31\times 10^{-7}\cos \left(2\pi \times 10^{8}t-\frac{4\pi }{3}z+\frac{\pi }{6}\right)\text{[A/m]}$

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7.2 Type B: Seawater Wave Parameter Math [PYQ: 2018, 2024]

Question: The electric field intensity of a linearly polarized uniform plane wave propagating in the $+z$ direction in seawater is $E_x=100\cos (10^7\pi t)\text{ V/m}$ at $z=0$. For seawater, ${ϵ}_r=72,{\mu }_r=1$, and $\sigma =4\text{ S/m}$. (i) Determine: $\alpha$, $\beta$, $\eta$, $u_p$, $\lambda$, and skin depth $\delta$. (ii) Find the distance at which the wave amplitude decays to 1% of its value at $z=0$.

Step-by-Step Solution:

1. Check if Good Conductor or Lossy Dielectric: $\omega =10^7\pi \text{[rad/s]}$ $\omega ϵ=(10^7\pi )\cdot (72\cdot {ϵ}_0)=(10^7\pi )\cdot \left(72\cdot 8.854\times 10^{-12}\right)=0.02\text{[S/m]}$ $\frac{\sigma }{\omega ϵ}=\frac{4}{0.02}=200\gg 1⟹\text{Good Conductor}$

2. Calculate $\alpha$ and $\beta$: Since $\frac{\sigma }{\omega ϵ}\gg 1$, we use the good conductor approximations: $\alpha =\beta \approx \sqrt{\pi f\mu \sigma }=\sqrt{\frac{\omega \mu \sigma }{2}}=\sqrt{\frac{(10^7\pi )\cdot (4\pi \times 10^{-7})\cdot 4}{2}}=\sqrt{8{\pi }^2}\approx 8.89$ $\alpha =8.89\text{[Np/m]}\beta =8.89\text{[rad/m]}$

3. Calculate Intrinsic Impedance ($\eta$): ${\eta }_c\approx (1+j)\sqrt{\frac{\omega \mu }{2\sigma }}=(1+j)\sqrt{\frac{(10^7\pi )\cdot (4\pi \times 10^{-7})}{2\cdot 4}}=(1+j)\frac{\pi }{\sqrt{2}}\approx 2.22\angle 45^{\circ }[\Omega ]$

4. Calculate Phase Velocity ($u_p$), Wavelength ($\lambda$), and Skin Depth ($\delta$): $u_p=\frac{\omega }{\beta }=\frac{10^7\pi }{8.89}=3.53\times 10^6\text{[m/s]}$ $\lambda =\frac{2\pi }{\beta }=\frac{2\pi }{8.89}\approx 0.707\text{[m]}$ $\delta =\frac{1}{\alpha }=\frac{1}{8.89}\approx 0.112\text{[m]}=11.2\text{ cm}$

5. Calculate 1% decay distance ($z_1$): $E(z_1)=0.01E_0⟹E_0{e}^{-\alpha z_1}=0.01E_0$ $e^{-\alpha z_1}=0.01⟹-\alpha z_1=\ln (0.01)=-4.605$ $z_1=\frac{4.605}{8.89}\approx 0.518\text{[m]}=51.8\text{ cm}$

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7.3 Type C: Dispersive Lossy Medium [PYQ: 2017, 2019, 2021]

Question: A narrow-band signal propagates in a lossy dielectric medium which has a loss tangent of $0.3$ at $530\text{ kHz}$ (the carrier frequency of the signal). The dielectric constant of the medium is $2.5$. (i) Determine $\alpha$ and $\beta$. (ii) Determine $u_p$ and $u_g$. Is the medium dispersive?

Step-by-Step Solution:

1. Calculate $\alpha$ and $\beta$: The loss tangent is $\tan {\delta }_c=\frac{\sigma }{\omega ϵ}=0.3\ll 1⟹\text{Low-Loss Dielectric}$. $\omega =2\pi (530\times 10^3)=1.06\pi \times 10^6\text{[rad/s]}$ ${ϵ}^{\prime }={ϵ}_r{ϵ}_0=2.5{ϵ}_0$ Using the low-loss approximations: $\alpha \approx \frac{\sigma }{2}\sqrt{\frac{\mu }{ϵ}}=\frac{\omega ϵ\tan \delta }{2}\sqrt{\frac{\mu }{ϵ}}=\frac{\omega \tan \delta }{2}\sqrt{\mu ϵ}$ $\alpha \approx \frac{1.06\pi \times 10^6\cdot 0.3}{2}\left(\frac{\sqrt{2.5}}{3\times 10^8}\right)\approx 2.63\times 10^{-3}\text{[Np/m]}$ $\beta \approx \omega \sqrt{\mu ϵ}\left[1+\frac{1}{8}(\tan \delta {)}^2\right]=\left(\frac{1.06\pi \times 10^6\cdot \sqrt{2.5}}{3\times 10^8}\right)\left[1+\frac{1}{8}(0.3{)}^2\right]$ $\beta \approx (0.01755)\cdot [1.01125]\approx 0.0177\text{[rad/m]}$

2. Calculate velocities and check dispersion: $u_p=\frac{\omega }{\beta }=\frac{1.06\pi \times 10^6}{0.0177}\approx 1.88\times 10^8\text{[m/s]}$ For a low-loss dielectric, the group velocity is found by differentiating $\beta (\omega )$: $\beta \approx \omega \sqrt{\mu ϵ}\left(1+\frac{{\sigma }^2}{8{\omega }^2{ϵ}^2}\right)=\omega \sqrt{\mu ϵ}+\frac{{\sigma }^2}{8\omega ϵ}\sqrt{\frac{\mu }{ϵ}}$ $\frac{d\beta }{d\omega }\approx \sqrt{\mu ϵ}-\frac{{\sigma }^2}{8{\omega }^2ϵ}\sqrt{\frac{\mu }{ϵ}}=\sqrt{\mu ϵ}\left[1-\frac{1}{8}{\left(\frac{\sigma }{\omega ϵ}\right)}^2\right]$ $u_g=\frac{1}{d\beta /d\omega }\approx \frac{1}{\sqrt{\mu ϵ}\left[1-\frac{1}{8}(0.3{)}^2\right]}\approx \frac{3\times 10^8}{\sqrt{2.5}\cdot [0.98875]}\approx 1.91\times 10^8\text{[m/s]}$ Since $u_p\ne u_g$ ($u_g>u_p$), different frequencies in the signal travel at different speeds. Thus, the medium is dispersive (Anomalous Dispersion).

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7.4 Type D: Average Power Dissipated [PYQ: 2017, 2022, 2023, 2024]

Question: A sinusoidal electric intensity of amplitude $250\text{ V/m}$ and frequency $1\text{ GHz}$ exists in a lossy dielectric medium that has a relative permittivity of $2.5$ and a loss tangent of $0.001$. Find the average power dissipated in the medium per cubic meter.

Step-by-Step Solution:

1. Find the medium’s conductivity ($\sigma$): $\tan {\delta }_c=\frac{\sigma }{\omega ϵ}=0.001$ $\omega =2\pi f=2\pi \times 10^9\text{[rad/s]}$ $\sigma =0.001\cdot \omega {ϵ}_r{ϵ}_0=0.001\cdot (2\pi \times 10^9)\cdot 2.5\cdot \left(8.854\times 10^{-12}\right)$ $\sigma \approx 1.39\times 10^{-4}\text{[S/m]}$

2. Calculate the average power dissipated per unit volume ($P_d$): $P_{\text{avg}}=\frac{1}{2}\sigma E_0^2\text{[W/m³]}$ $P_{\text{avg}}=\frac{1}{2}\cdot \left(1.39\times 10^{-4}\right)\cdot (250{)}^2=4.34\text{[W/m³]}$

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7.5 Type E: Wave Reflections on a Perfect Conductor [PYQ: 2015, 2016]

Question: A $y$-polarized uniform plane wave propagating in the $+x$ direction in air ($x\le 0$) has a frequency of $100\text{ MHz}$ and impinges normally on a perfectly conducting plane at $x=0$. Assuming the electric field amplitude is $6\text{ mV/m}$, write the phasor and instantaneous expressions for: (i) Incident wave (ii) Reflected wave (iii) Total wave in air (iv) Determine the node location nearest to the conducting plane.

Step-by-Step Solution:

1. Find wave parameters in Medium 1 (air): $\omega =2\pi \times 100\times 10^6=2\pi \times 10^8\text{[rad/s]}$ ${\beta }_1=\frac{\omega }{c}=\frac{2\pi \times 10^8}{3\times 10^8}=\frac{2\pi }{3}\text{[rad/m]}$ ${\eta }_1={\eta }_0=120\pi \approx 377[\Omega ]$

2. Write expressions for Incident Wave (${\vec{E}}_i,{\vec{H}}_i$): Since propagation is along $+x$ and polarization is along $y$: ${\vec{E}}_{is}(x)={\hat{a}}_{y}6\times 10^{-3}{e}^{-j{\beta }_{1}x}={\hat{a}}_{y}6e^{-j\frac{2\pi }{3}x}\text{[mV/m]}$ ${\vec{E}}_i(x,t)={\hat{a}}_{y}6\cos \left(2\pi \times 10^{8}t-\frac{2\pi }{3}x\right)\text{[mV/m]}$ Using ${\vec{H}}_{is}=\frac{1}{{\eta }_1}{\hat{a}}_k\times {\vec{E}}_{is}=\frac{1}{{\eta }_0}{\hat{a}}_x\times ({\hat{a}}_y{E}_0)={\hat{a}}_z\frac{E_0}{{\eta }_0}$: ${\vec{H}}_{is}(x)={\hat{a}}_z\frac{6\times 10^{-3}}{120\pi }{e}^{-j\frac{2\pi }{3}x}={\hat{a}}_{z}1.59\times 10^{-5}{e}^{-j\frac{2\pi }{3}x}\text{[A/m]}$ ${\vec{H}}_i(x,t)={\hat{a}}_{z}1.59\times 10^{-5}\cos \left(2\pi \times 10^{8}t-\frac{2\pi }{3}x\right)\text{[A/m]}$

3. Write expressions for Reflected Wave (${\vec{E}}_r,{\vec{H}}_r$): For a perfect conductor, $\Gamma =-1$. The reflected wave travels along $-x$: ${\vec{E}}_{rs}(x)=\Gamma {\vec{E}}_{is}(-x)=-{\hat{a}}_{y}6e^{j\frac{2\pi }{3}x}\text{[mV/m]}$ ${\vec{E}}_r(x,t)=-{\hat{a}}_{y}6\cos \left(2\pi \times 10^{8}t+\frac{2\pi }{3}x\right)\text{[mV/m]}$ Using ${\vec{H}}_{rs}=\frac{1}{{\eta }_0}(-{\hat{a}}_x)\times {\vec{E}}_{rs}=\frac{1}{{\eta }_0}(-{\hat{a}}_x)\times (-{\hat{a}}_y{E}_0)={\hat{a}}_z\frac{E_0}{{\eta }_0}$: ${\vec{H}}_{rs}(x)={\hat{a}}_{z}1.59\times 10^{-5}{e}^{j\frac{2\pi }{3}x}\text{[A/m]}$ ${\vec{H}}_r(x,t)={\hat{a}}_{z}1.59\times 10^{-5}\cos \left(2\pi \times 10^{8}t+\frac{2\pi }{3}x\right)\text{[A/m]}$

4. Write expressions for Total Wave in Air (${\vec{E}}_1,{\vec{H}}_1$): ${\vec{E}}_{1s}(x)={\vec{E}}_{is}(x)+{\vec{E}}_{rs}(x)={\hat{a}}_{y}6\left(e^{-j{\beta }_{1}x}-e^{j{\beta }_{1}x}\right)=-j{\hat{a}}_{y}12\sin \left(\frac{2\pi }{3}x\right)\text{[mV/m]}$ ${\vec{E}}_1(x,t)={\hat{a}}_{y}12\sin \left(\frac{2\pi }{3}x\right)\sin \left(2\pi \times 10^{8}t\right)\text{[mV/m]}$

5. Determine the nearest node location ($x_{\text{node}}$): A node occurs where the total electric field is zero (${\vec{E}}_1=0$): $\sin \left(\frac{2\pi }{3}x\right)=0⟹\frac{2\pi }{3}x=-n\pi (\text{since }x\le 0)$ For the nearest node from the boundary ($n=1$): $\frac{2\pi }{3}x=-\pi ⟹x=-1.5\text{ m}$

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7.6 Type F: Decibel Decay and Conductivity [PYQ: 2020]

Question: Given the skin depth for graphite at $100\text{ MHz}$ is $0.20\text{ mm}$. (i) Determine the conductivity of graphite. (ii) Determine the distance that a $1\text{ GHz}$ wave travels in graphite such that its field intensity is reduced by $30\text{ dB}$.

Step-by-Step Solution:

1. Calculate Conductivity ($\sigma$): $\delta =0.20\text{ mm}=2\times 10^{-4}\text{ m}\text{at }f=10^8\text{ Hz}$ Using the skin depth formula for a good conductor: $\delta =\frac{1}{\sqrt{\pi f\mu \sigma }}⟹\sigma =\frac{1}{\pi f{\mu }_0{\delta }^2}$ $\sigma =\frac{1}{\pi \cdot 10^8\cdot (4\pi \times 10^{-7})\cdot (2\times 10^{-4}{)}^2}=\frac{1}{1.6{\pi }^2\times 10^{-5}}\approx 6.33\times 10^4\text{[S/m]}$

2. Calculate Attenuation constant (${\alpha }_2$) at $1\text{ GHz}$: Since $\sigma$ is constant, we find the new skin depth and attenuation constant at $f_2=10^9\text{ Hz}$: ${\alpha }_2=\sqrt{\pi f_2{\mu }_0\sigma }=\sqrt{\pi \cdot 10^9\cdot (4\pi \times 10^{-7})\cdot 6.33\times 10^4}=\sqrt{2.5\times 10^8}\approx 1.58\times 10^4\text{[Np/m]}$

3. Determine the distance ($z$) for a $30\text{ dB}$ reduction: In decibels, field reduction is defined as: $\text{dB attenuation}=20{\log }_{10}\left(\frac{E(z)}{E_0}\right)=-30\text{ dB}$ ${\log }_{10}\left(\frac{E(z)}{E_0}\right)=-1.5⟹\frac{E(z)}{E_0}=10^{-1.5}\approx 0.0316$ Since the amplitude decays exponentially ($e^{-{\alpha }_{2}z}$): $e^{-{\alpha }_{2}z}=10^{-1.5}⟹-{\alpha }_{2}z=\ln (10^{-1.5})=-3.454$ $z=\frac{3.454}{1.58\times 10^4}=2.18\times 10^{-4}\text{[m]}=0.218\text{ mm}$

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8. A+ Exam Hacks & Pitfalls

• The “Complex Permittivity” Trick: Before executing any plane wave math, always calculate $\frac{\sigma }{\omega ϵ}$. This immediately defines whether you are working with a good conductor or a low-loss dielectric, preventing you from using the wrong simplified formula set.
• Vector Notation on Field Parameters: If asked to write expressions for $\vec{E}$ and $\vec{H}$, always include their directional unit vectors (${\hat{a}}_x,{\hat{a}}_y,{\hat{a}}_z$) and hats. Leaving them as scalars will cost you significant marks.
• The Impedance Phase Lag: Remember that in a good conductor, the magnetic field lags the electric field by exactly $45^{\circ }$ ($\pi /4$ radians). When converting $\vec{H}$ back to the instantaneous time domain, you must subtract $45^{\circ }$ (or $\approx 0.785$ rad) from the cosine argument: $\cos (\omega t-\beta z-\pi /4)$.