Related Concepts: 5.02 Plane Waves in Lossless vs. Lossy Media & Skin Depth Calculations | 5.03 Poynting Vector, Power Flow & Normal Incidence Reflection (Gamma, Tau) | 5.04 Wave Polarization & Ionospheric Sky-Wave Radio Propagation | 5.06 Radio Wave Propagation Modes & Ionospheric Effects

5.07 Solved PYQ Numerical Bank - Waves & Propagation

Core Analytical Rule: Always compute $\frac{\sigma }{\omega ϵ}$ FIRST

Before applying any simplified propagation formulas, you must test the medium:

• If $\frac{\sigma }{\omega ϵ}\ll 1$, use Low-Loss Dielectric approximations.
• If $\frac{\sigma }{\omega ϵ}\gg 1$, use Good Conductor approximations.
• If $\sigma =0$, it is a Lossless Medium.

---

🌊 Plane Wave Decision Flowchart

graph TD
    A[Identify Medium Parameters: ε, μ, σ, ω] --> B{Calculate σ / ωε}
    B -->|σ = 0| C[Lossless Medium]
    B -->|σ / ωε < 0.1| D[Low-Loss Dielectric]
    B -->|σ / ωε > 10| E[Good Conductor]

    C --> C1["α = 0, β = ω√(με)"]
    D --> D1["α ≈ (σ/2)√(μ/ε), β ≈ ω√(με)"]
    E --> E1["α = β = √(πfμσ)"]

---

Type 1: Lossless Plane Wave — Instantaneous Fields

[PYQ: 2021] (08 marks)

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 that $E_x$ is sinusoidal with a frequency $100\text{ MHz}$ and has a maximum magnitude of $10^{-4}\text{ V/m}$ at $t=0$ and $z=1/8\text{ m}$.

1. Write the instantaneous expression for $\vec{E}$ at any $t$ and $z$.
2. Write the instantaneous expression for $\vec{H}$.

Solution:

1. Compute Wave Parameters:

   • Phase velocity: $v=\frac{c}{\sqrt{{\mu }_r{ϵ}_r}}=\frac{3\times 10^8}{\sqrt{1\times 4}}=1.5\times 10^8\text{ m/s}$
   • Angular frequency: $\omega =2\pi f=2\pi (100\times 10^6)=2\pi \times 10^8\text{ rad/s}$
   • Phase constant: $\beta =\frac{\omega }{v}=\frac{2\pi \times 10^8}{1.5\times 10^8}=\frac{4\pi }{3}\text{ rad/m}$
   • Intrinsic impedance: $\eta ={\eta }_0\sqrt{\frac{{\mu }_r}{{ϵ}_r}}=120\pi \sqrt{\frac{1}{4}}=60\pi \approx 188.5\Omega$

2. Formulate Electric Field $\vec{E}(z,t)$: The wave propagates in $+z$. The general form is $E_x(z,t)=E_{0x}\cos (\omega t-\beta z+\varphi )$. Given that at $t=0$ and $z=1/8\text{ m}$, $E_x$ is at its positive maximum ($E_{0x}=10^{-4}\text{ V/m}$): $\cos \left(0-\beta \left(\frac{1}{8}\right)+\varphi \right)=1⟹-\frac{4\pi }{3}\left(\frac{1}{8}\right)+\varphi =0⟹-\frac{\pi }{6}+\varphi =0⟹\varphi =\frac{\pi }{6}\text{ rad}$ $\boxed{\vec{E}(z,t)={\hat{a}}_{x}10^{-4}\cos \left(2\pi \times 10^{8}t-\frac{4\pi }{3}z+\frac{\pi }{6}\right)\text{ V/m}}$

3. Formulate Magnetic Field $\vec{H}(z,t)$: The direction of propagation is ${\hat{a}}_k={\hat{a}}_z$. Since ${\hat{a}}_H={\hat{a}}_k\times {\hat{a}}_E={\hat{a}}_z\times {\hat{a}}_x={\hat{a}}_y$: $\vec{H}(z,t)={\hat{a}}_y\frac{E_{0x}}{\eta }\cos (\omega t-\beta z+\varphi )$ $\boxed{\vec{H}(z,t)={\hat{a}}_y\frac{10^{-4}}{60\pi }\cos \left(2\pi \times 10^{8}t-\frac{4\pi }{3}z+\frac{\pi }{6}\right)\text{ A/m}}$

---

[PYQ: 2019] (15 marks)

A uniform plane wave with $\vec{E}={\hat{a}}_y{E}_y$ propagates in a lossless simple medium (${ϵ}_r=6,{\mu }_r=1.5,\sigma =0$) in the $+x$ direction. Assume that $E_y$ is sinusoidal with a frequency $150\text{ MHz}$ and has a maximum value of $10^{-4}\text{ V/m}$ at $t=0$ and $x=1/6\text{ m}$.

1. Write the instantaneous expression for $\vec{E}$ at any $t$ and $x$.
2. Write the instantaneous expression for $\vec{H}$.
3. Determine the location where $E_y$ is a positive maximum when $t=10^{-8}\text{ s}$.

Solution:

1. Compute Wave Parameters:

   • Phase velocity: $v=\frac{c}{\sqrt{{\mu }_r{ϵ}_r}}=\frac{3\times 10^8}{\sqrt{1.5\times 6}}=\frac{3\times 10^8}{3}=10^8\text{ m/s}$
   • Angular frequency: $\omega =2\pi f=2\pi (150\times 10^6)=3\pi \times 10^8\text{ rad/s}$
   • Phase constant: $\beta =\frac{\omega }{v}=\frac{3\pi \times 10^8}{10^8}=3\pi \text{ rad/m}$
   • Intrinsic impedance: $\eta ={\eta }_0\sqrt{\frac{{\mu }_r}{{ϵ}_r}}=120\pi \sqrt{\frac{1.5}{6}}=120\pi (0.5)=60\pi \approx 188.5\Omega$

2. Formulate Electric Field $\vec{E}(x,t)$: The wave propagates in $+x$. The general form is $E_y(x,t)=E_0\cos (\omega t-\beta x+\varphi )$. At $t=0,x=1/6\text{ m}$, $E_y$ reaches its maximum value ($E_0=10^{-4}\text{ V/m}$): $\cos \left(0-3\pi \left(\frac{1}{6}\right)+\varphi \right)=1⟹-\frac{\pi }{2}+\varphi =0⟹\varphi =\frac{\pi }{2}\text{ rad}$ $\boxed{\vec{E}(x,t)={\hat{a}}_{y}10^{-4}\cos \left(3\pi \times 10^{8}t-3\pi x+\frac{\pi }{2}\right)\text{ V/m}}$

3. Formulate Magnetic Field $\vec{H}(x,t)$: The direction of propagation is ${\hat{a}}_k={\hat{a}}_x$. Since ${\hat{a}}_H={\hat{a}}_k\times {\hat{a}}_E={\hat{a}}_x\times {\hat{a}}_y={\hat{a}}_z$: $\boxed{\vec{H}(x,t)={\hat{a}}_z\frac{10^{-4}}{60\pi }\cos \left(3\pi \times 10^{8}t-3\pi x+\frac{\pi }{2}\right)\text{ A/m}}$

4. Locate Positive Maximum: At $t=10^{-8}\text{ s}$, the cosine argument must equal $2m\pi$ (where $m=0,\pm 1,\pm 2,\dots$): $(3\pi \times 10^8)(10^{-8})-3\pi x+\frac{\pi }{2}=2m\pi ⟹3\pi -3\pi x+\frac{\pi }{2}=2m\pi$ Divide by $\pi$: $3.5-3x=2m⟹3x=3.5-2m⟹x=\frac{3.5-2m}{3}\text{ m}$ For the smallest positive location, choose $m=1$: $x=\frac{3.5-2(1)}{3}=\frac{1.5}{3}=0.5\text{ m}$ $\boxed{x=0.5\text{ m}}$

Trap: Phase Shift Offset

Forgetting the phase constant $\varphi$ is a common pitfall. Just because the wave is at a maximum at $t=0$, you cannot assume $\varphi =0$ if that maximum is located at a spatial coordinate other than the origin ($x\ne 0$).

---

Type 2: Good Conductor (Seawater) Attenuation

[PYQ: 2018, 2024] (14 marks)

The electric field intensity of a linearly polarized uniform plane wave propagating in the $+z$ direction in sea water is $E_x={\hat{a}}_{x}50\cos (10^5\pi t)\text{ V/m}$ at $z=0$. For sea water ${ϵ}_r=72,{\mu }_r=1,$ and $\sigma =4\text{ S/m}$. Determine:

1. The attenuation constant, phase constant, intrinsic impedance, phase velocity, wavelength, and skin depth.
2. The distance at which amplitude is $1\%$ of its value at $z=0$.

Solution:

1. Verify the Medium Classification: Given $\omega =10^5\pi \text{ rad/s}⟹f=50\text{ kHz}$. $\frac{\sigma }{\omega ϵ}=\frac{\sigma }{\omega {ϵ}_0{ϵ}_r}=\frac{4}{(10^5\pi )(72\times 8.854\times 10^{-12})}=\frac{4}{2.003\times 10^{-4}}\approx 19970\gg 1$ Since $\frac{\sigma }{\omega ϵ}\gg 1$, seawater at $50\text{ kHz}$ behaves as a Good Conductor.

2. Compute Conductor Parameters:

   • Attenuation ($\alpha$) & Phase ($\beta$) Constants: $\alpha =\beta =\sqrt{\pi f\mu \sigma }=\sqrt{\pi (50\times 10^3)(4\pi \times 10^{-7})(4)}=\sqrt{0.8{\pi }^2}=\pi \sqrt{0.8}\approx 2.8099$ $\boxed{\alpha =2.81\text{ Np/m},\beta =2.81\text{ rad/m}}$
   • Intrinsic Impedance (${\eta }_c$): In a good conductor, the phase angle of the impedance is fixed at $45^{\circ }$ (or $\pi /4\text{ rad}$): ${\eta }_c=\sqrt{\frac{\omega \mu }{\sigma }}\angle 45^{\circ }=\sqrt{\frac{10^5\pi \times 4\pi \times 10^{-7}}{4}}\angle 45^{\circ }=\sqrt{0.01{\pi }^2}\angle 45^{\circ }=0.1\pi \angle 45^{\circ }\Omega$ $\boxed{{\eta }_c=0.314\angle 45^{\circ }\Omega }$
   • Phase Velocity ($u_p$) and Wavelength ($\lambda$): $u_p=\frac{\omega }{\beta }=\frac{10^5\pi }{2.8099}\approx 1.118\times 10^5\text{ m/s}$ $\lambda =\frac{2\pi }{\beta }=\frac{2\pi }{2.8099}\approx 2.236\text{ m}$ $\boxed{u_p=1.118\times 10^5\text{ m/s},\lambda =2.24\text{ m}}$
   • Skin Depth ($\delta$): $\delta =\frac{1}{\alpha }=\frac{1}{2.8099}\approx 0.356\text{ m}$ $\boxed{\delta =0.356\text{ m}}$

3. Determine Distance for $1\%$ Amplitude: $E(z)=E_0{e}^{-\alpha z}⟹0.01E_0=E_0{e}^{-2.8099z}$ $\ln (0.01)=-2.8099z⟹-4.605=-2.8099z⟹z=\frac{4.605}{2.8099}\approx 1.638\text{ m}$ $\boxed{z=1.64\text{ m}}$

Trap: Angular Frequency Exponent

Pay close attention to the exponent in the wave definition. Some textbooks use older versions of this problem with $\omega =10^7\pi$, but KUET exam papers specifically written in 2018 and 2024 featured $\omega =10^5\pi$. Blindly using $10^7\pi$ will yield incorrect parameters.

---

Type 3: Low-Loss Dielectric & Dispersion

[PYQ: 2017] (12 marks)

A narrow-band signal propagates in a lossy dielectric medium which has a loss tangent of $0.2$ at $550\text{ kHz}$ (the carrier frequency of the signal). The dielectric constant of the medium is $2.5$.

1. Determine the attenuation constant $\alpha$ and phase constant $\beta$.
2. Determine the phase velocity $u_p$ and group velocity $u_g$. Is the medium dispersive?

[PYQ: 2021] (11 marks)

Same question as above, but with a loss tangent of $0.3$ at $530\text{ kHz}$.

Derivation of the Low-Loss Dispersion Relations:

In a low-loss dielectric where $\tan \delta =\frac{\sigma }{\omega ϵ}\ll 1$, we can approximate the propagation parameters using binomial expansions: $\alpha \approx \frac{\sigma }{2}\sqrt{\frac{\mu }{ϵ}}=\frac{\omega \tan \delta }{2v}$ $\beta \approx \omega \sqrt{\mu ϵ}\left[1+\frac{1}{8}(\tan \delta {)}^2\right]=\frac{\omega }{v}\left[1+\frac{1}{8}(\tan \delta {)}^2\right]$ Since $\tan \delta =\frac{\sigma }{\omega ϵ}$, the phase constant as a function of frequency is: $\beta (\omega )\approx \frac{\omega }{v}+\frac{{\sigma }^2}{8\omega v{ϵ}^2}$ Differentiating $\beta$ with respect to $\omega$ to find the group velocity: $\frac{d\beta }{d\omega }=\frac{1}{v}-\frac{{\sigma }^2}{8{\omega }^{2}v{ϵ}^2}=\frac{1}{v}\left[1-\frac{1}{8}(\tan \delta {)}^2\right]$ Since $u_g=\frac{1}{d\beta /d\omega }$, we obtain: $u_g\approx v{\left[1-\frac{1}{8}(\tan \delta {)}^2\right]}^{-1}$

Solution for 2017 ($\tan \delta =0.2,f=550\text{ kHz}$):

1. Compute Base Values: $\omega =2\pi (550\times 10^3)\approx 3.456\times 10^6\text{ rad/s}$ $v=\frac{c}{\sqrt{{ϵ}_r}}=\frac{3\times 10^8}{\sqrt{2.5}}\approx 1.897\times 10^8\text{ m/s}$ The ideal phase constant: ${\beta }_0=\frac{\omega }{v}=\frac{3.456\times 10^6}{1.897\times 10^8}\approx 0.01821\text{ rad/m}$

2. Calculate $\alpha$ and $\beta$ (1): $\alpha \approx {\beta }_0\left(\frac{\tan \delta }{2}\right)=(0.01821)\left(\frac{0.2}{2}\right)=0.001821\text{ Np/m}$ $\beta \approx {\beta }_0\left[1+\frac{1}{8}(\tan \delta {)}^2\right]=0.01821\left[1+\frac{0.04}{8}\right]=0.01821(1.005)\approx 0.01830\text{ rad/m}$ $\boxed{\alpha =1.82\times 10^{-3}\text{ Np/m},\beta =1.83\times 10^{-2}\text{ rad/m}}$

3. Calculate Velocities & Check Dispersion (2):

   • Phase velocity: $u_p=\frac{\omega }{\beta }=\frac{3.456\times 10^6}{0.01830}\approx 1.888\times 10^8\text{ m/s}$
   • Group velocity: $u_g\approx v{\left[1-\frac{1}{8}(\tan \delta {)}^2\right]}^{-1}\approx \frac{1.897\times 10^8}{1-0.005}=\frac{1.897\times 10^8}{0.995}\approx 1.907\times 10^8\text{ m/s}$ $\boxed{u_p=1.89\times 10^8\text{ m/s},u_g=1.91\times 10^8\text{ m/s}}$
   • Dispersion: Since $u_p$ varies with frequency and $u_p\ne u_g$, the medium is dispersive (specifically showing anomalous dispersion since $u_g>u_p$).

Solution for 2021 ($\tan \delta =0.3,f=530\text{ kHz}$):

• Base Values: $\omega =2\pi (530\times 10^3)\approx 3.330\times 10^6\text{ rad/s},v\approx 1.897\times 10^8\text{ m/s}$ ${\beta }_0=\frac{\omega }{v}\approx 0.01755\text{ rad/m}$
• Calculate constants: $\alpha \approx (0.01755)\left(\frac{0.3}{2}\right)=0.002632\text{ Np/m}$ $\beta \approx 0.01755\left[1+\frac{0.09}{8}\right]=0.01755(1.01125)\approx 0.01775\text{ rad/m}$ $\boxed{\alpha =2.63\times 10^{-3}\text{ Np/m},\beta =1.77\times 10^{-2}\text{ rad/m}}$
• Calculate velocities: $u_p=\frac{\omega }{\beta }=\frac{3.330\times 10^6}{0.01775}\approx 1.876\times 10^8\text{ m/s}$ $u_g=v{\left[1-\frac{1}{8}(0.09)\right]}^{-1}=\frac{1.897\times 10^8}{0.98875}\approx 1.919\times 10^8\text{ m/s}$ $\boxed{u_p=1.88\times 10^8\text{ m/s},u_g=1.92\times 10^8\text{ m/s}}(\text{Dispersive})$

---

Type 4: Average Power Dissipated in a Medium

[PYQ: 2017, 2022, 2023, 2024] (08 marks)

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.

Solution:

1. Extract conductivity ($\sigma$): From the definition of loss tangent: $\tan \delta =\frac{\sigma }{\omega ϵ}⟹\sigma =\omega {ϵ}_0{ϵ}_r\tan \delta$ $\sigma =(2\pi \times 10^9)\times (8.854\times 10^{-12}\times 2.5)\times 0.001$ $\sigma =(2\pi \times 10^9)\times (2.2135\times 10^{-11})\times 0.001\approx 1.391\times 10^{-4}\text{ S/m}$

2. Calculate Dissipated Power density: Because the electric field is given as a peak amplitude ($E_0=250\text{ V/m}$) rather than an RMS value, time-averaging introduces a factor of $1/2$: $P_{\text{av}}=\frac{1}{2}\sigma E_0^2=\frac{1}{2}(1.391\times 10^{-4})\times (250{)}^2$ $P_{\text{av}}=0.5\times (1.391\times 10^{-4})\times 62500\approx 4.346{\text{ W/m}}^3$ $\boxed{P_{\text{av}}=4.35{\text{ W/m}}^3}$

Trap: Peak Amplitude vs. RMS

Forgetting the $1/2$ factor is the most common error in this problem. Always check if the voltage/field intensity is given as “amplitude” (peak) or “RMS”. For peak value sinusoidal fields, the average power dissipated is $P_{av}=\frac{1}{2}\sigma E_0^2$, whereas for RMS it is simply $P_{av}=\sigma E_{\text{rms}}^2$.

---

Type 5: Normal Incidence on Perfect Conductor (Standing Waves)

[PYQ: 2015] (19 marks)

A y-polarized uniform plane wave (${\vec{E}}_i,{\vec{H}}_i$) with a frequency of $100\text{ MHz}$ propagates in air in the $+x$ direction and impinges normally on a perfectly conducting plane at $x=0$. Assuming the amplitude of ${\vec{E}}_i$ to be $6\text{ mV/m}$, write the phasor and instantaneous expression for:

1. ${\vec{E}}_i$ and ${\vec{H}}_i$ of the incident wave.
2. ${\vec{E}}_r$ and ${\vec{H}}_r$ of the reflected wave.
3. ${\vec{E}}_1$ and ${\vec{H}}_1$ of the total wave in air.
4. Determine the location nearest to the conducting plane where the total electric field ${\vec{E}}_1$ is zero. (Note: The 2016 paper asked an identical 8-mark subset covering only parts 1 & 2).

Solution:

1. Incident Wave Parameters (1):

   • Angular frequency: $\omega =2\pi f=2\pi (100\times 10^6)=2\pi \times 10^8\text{ rad/s}$
   • Phase constant (in air/free-space): $\beta =\frac{\omega }{c}=\frac{2\pi \times 10^8}{3\times 10^8}=\frac{2\pi }{3}\text{ rad/m}$
   • Intrinsic impedance (air): ${\eta }_0\approx 377\Omega \text{ (or }120\pi \Omega \text{)}$
   • Incident field amplitude: $E_0=6\text{ mV/m}=6\times 10^{-3}\text{ V/m}$ $H_0=\frac{E_0}{{\eta }_0}=\frac{6\times 10^{-3}}{377}\approx 15.9\times 10^{-6}\text{ A/m}=15.9\mu \text{A/m}$
   • Expressions: $\text{Phasors: }{\tilde{E}}_i={\hat{a}}_{y}6e^{-j\frac{2\pi }{3}x}\text{ mV/m},{\tilde{H}}_i={\hat{a}}_{z}15.9e^{-j\frac{2\pi }{3}x}\mu \text{A/m}$ $\boxed{{\vec{E}}_i(x,t)={\hat{a}}_{y}6\cos \left(2\pi \times 10^{8}t-\frac{2\pi }{3}x\right)\text{ mV/m}}$ $\boxed{{\vec{H}}_i(x,t)={\hat{a}}_{z}15.9\cos \left(2\pi \times 10^{8}t-\frac{2\pi }{3}x\right)\mu \text{A/m}}$

2. Reflected Wave (2): At a boundary with a perfect electrical conductor (PEC) at $x=0$, the total tangential electric field must be zero. This requires a reflection coefficient of $\Gamma =-1$.

   • Expressions: $\text{Phasors: }{\tilde{E}}_r=-{\hat{a}}_{y}6e^{+j\frac{2\pi }{3}x}\text{ mV/m},{\tilde{H}}_r={\hat{a}}_{z}15.9e^{+j\frac{2\pi }{3}x}\mu \text{A/m}$ \boxed{\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} \boxed{\vec{H}_r(x,t) = \hat{a}_z 15.9 \cos\left(2\pi \times 10^8 t + \frac{2\pi}{3}x\right)\text{ }\mu\text{A/m}

3. Total Wave in Air ($x<0$) (3):

   • Total Electric Field: ${\vec{E}}_1(x,t)={\vec{E}}_i(x,t)+{\vec{E}}_r(x,t)={\hat{a}}_{y}6\left[\cos \left(\omega t-\beta x\right)-\cos \left(\omega t+\beta x\right)\right]$ Using the trigonometric identity $\cos (A-B)-\cos (A+B)=2\sin A\sin B$: $\boxed{{\vec{E}}_1(x,t)={\hat{a}}_{y}12\sin \left(\frac{2\pi }{3}x\right)\sin \left(2\pi \times 10^{8}t\right)\text{ mV/m}}$
   • Total Magnetic Field: ${\vec{H}}_1(x,t)={\vec{H}}_i(x,t)+{\vec{H}}_r(x,t)={\hat{a}}_{z}15.9\left[\cos \left(\omega t-\beta x\right)+\cos \left(\omega t+\beta x\right)\right]$ Using $\cos (A-B)+\cos (A+B)=2\cos A\cos B$: $\boxed{{\vec{H}}_1(x,t)={\hat{a}}_{z}31.8\cos \left(\frac{2\pi }{3}x\right)\cos \left(2\pi \times 10^{8}t\right)\mu \text{A/m}}$

4. Find Nearest Node Location (4): A node (zero value) in the total electric field occurs when the spatial component vanishes: $\sin \left(\frac{2\pi }{3}x\right)=0⟹\frac{2\pi }{3}x=-m\pi (\text{since the wave is in air, }x<0)$ $x=-1.5m\text{ m}$ For the nearest location to the PEC plane excluding the boundary itself ($m=0$), choose $m=1$: $\boxed{x=-1.5\text{ m}}$

---

Type 6: Skin Depth & Decibel Attenuation

[PYQ: 2020] (10 marks)

Given the skin depth for graphite at $100\text{ MHz}$ is $0.20\text{ mm}$, determine:

1. The conductivity of graphite.
2. The distance that a $1\text{ GHz}$ wave travels in graphite such that its field intensity is reduced by $30\text{ dB}$.

Solution:

1. Calculate Conductivity (1): The skin depth for a good conductor is: $\delta =\frac{1}{\sqrt{\pi f\mu \sigma }}⟹\sigma =\frac{1}{\pi f\mu {\delta }^2}$ Given $\delta =0.20\text{ mm}=2\times 10^{-4}\text{ m}$, $f=10^8\text{ Hz}$, and assuming non-magnetic graphite ($\mu ={\mu }_0=4\pi \times 10^{-7}\text{ H/m}$): $\sigma =\frac{1}{\pi (10^8)(4\pi \times 10^{-7})(2\times 10^{-4}{)}^2}=\frac{1}{40{\pi }^2\times 4\times 10^{-8}}=\frac{10^6}{1.6{\pi }^2}\approx 63325\text{ S/m}$ $\boxed{\sigma =63.3\text{ kS/m}}$

2. Calculate Distance for $30\text{ dB}$ Reduction at $1\text{ GHz}$ (2):

   • First find the new attenuation constant $\alpha$ at $1\text{ GHz}$ ($f=10^9\text{ Hz}$).
   • For a good conductor, $\alpha =\frac{1}{\delta }=\sqrt{\pi f\mu \sigma }$.
   • Since $\alpha$ is proportional to $\sqrt{f}$, when frequency scales by a factor of 10 (from $100\text{ MHz}$ to $1\text{ GHz}$): ${\alpha }_{1\text{ GHz}}={\alpha }_{100\text{ MHz}}\times \sqrt{10}=\left(\frac{1}{0.2\times 10^{-3}}\right)\times \sqrt{10}=5000\sqrt{10}\approx 15811.4\text{ Np/m}$
   • A decibel reduction of $30\text{ dB}$ in field intensity (voltage/m) is governed by: $\text{Attenuation (dB)}=20{\log }_{10}\left(\frac{E_0}{E(z)}\right)=30⟹\frac{E_0}{E(z)}=10^{1.5}\approx 31.62$
   • Since $E(z)=E_0{e}^{-\alpha z}$: $e^{\alpha z}=31.62⟹\alpha z=\ln (31.62)\approx 3.454$ $z=\frac{3.454}{15811.4}\approx 2.184\times 10^{-4}\text{ m}=0.218\text{ mm}$ $\boxed{z=0.218\text{ mm}}$

---

Type 7: Ionospheric Plasma Frequency

[PYQ: 2019, 2023] (07/10 marks)

1. Derive the equation for the plasma frequency of an ionized medium.
2. If the total number of electrons in the ionosphere is around $3\times 10^9{\text{ cm}}^{-3}$, what is the minimum frequency above which radio communication can be established between a spacecraft and Earth?

Derivation of Plasma Frequency:

Let an electromagnetic wave with electric field $\vec{E}={\vec{E}}_0{e}^{j\omega t}$ act on a free electron of mass $m$ and charge $-e$. The equation of motion is: $m\frac{d\vec{v}}{dt}=-e\vec{E}⟹m(j\omega \vec{v})=-e\vec{E}⟹\vec{v}=-\frac{e\vec{E}}{j\omega m}$ The convection current density $\vec{J}$ from $N$ electrons per unit volume is: $\vec{J}=-Ne\vec{v}=\frac{N{e}^2}{j\omega m}\vec{E}=-j\frac{N{e}^2}{\omega m}\vec{E}$ Substituting this into Maxwell’s curl equation for $\vec{H}$: $\text{curl }\vec{H}=\nabla \times \vec{H}=\vec{J}+j\omega {ϵ}_0\vec{E}=j\omega {ϵ}_0\left(1-\frac{N{e}^2}{{\omega }^{2}m{ϵ}_0}\right)\vec{E}$ Thus, the equivalent relative permittivity of the ionized gas is: ${ϵ}_r=1-\frac{N{e}^2}{{\omega }^{2}m{ϵ}_0}=1-{\left(\frac{f_p}{f}\right)}^2$ Where the plasma cutoff frequency $f_p$ is defined as: $f_p=\frac{1}{2\pi }\sqrt{\frac{N{e}^2}{m{ϵ}_0}}$ Substituting electron constants ($e=1.6\times 10^{-19}\text{ C},m=9.1\times 10^{-31}\text{ kg},{ϵ}_0=8.85\times 10^{-12}\text{ F/m}$): $f_p\approx 9\sqrt{N}(\text{where }N{\text{ is in electrons/m}}^3)$

Solution (2):

1. Unit Conversion: The constant $9$ assumes $N$ is expressed in electrons per cubic meter. $N=3\times 10^9{\text{ electrons/cm}}^3\times {\left(10^2\text{ cm/m}\right)}^3=3\times 10^{15}{\text{ electrons/m}}^3$

2. Calculate Cutoff Frequency: To establish spacecraft communication, the signals must penetrate the ionosphere, requiring the operating frequency to be strictly greater than the plasma frequency: $f_{min}>f_p\approx 9\sqrt{N}$ $f_p=9\sqrt{3\times 10^{15}}=9\sqrt{30\times 10^{14}}=9\times 5.477\times 10^7\approx 4.93\times 10^8\text{ Hz}=493\text{ MHz}$ $\boxed{{\text{f}}_{min}=493\text{ MHz}}$

---

Type 8: Maximum Usable Frequency (MUF)

[PYQ: 2018] (10 marks)

A high-frequency radio link is to be established between two points at a distance of $250\text{ km}$ on the Earth’s surface. Considering the ionospheric height to be $200\text{ km}$ and its critical frequency to be $5\text{ MHz}$, calculate the maximum usable frequency (MUF) for the given path.

[PYQ: 2019] (10 marks)

Same path parameters, but with ground distance $D=1500\text{ km}$, ionospheric height $h=300\text{ km}$, and critical frequency $f_c=7\text{ MHz}$.

Solution for 2018 ($D=250\text{ km},h=200\text{ km},f_c=5\text{ MHz}$):

1. Apply flat-earth secant law: $\text{MUF}=f_c\sec {\theta }_i=f_c\sqrt{1+{\left(\frac{D}{2h}\right)}^2}$
2. Calculate: $\text{MUF}=5\sqrt{1+{\left(\frac{250}{2(200)}\right)}^2}=5\sqrt{1+(0.625{)}^2}=5\sqrt{1+0.3906}=5\sqrt{1.3906}\approx 5.89\text{ MHz}$ $\boxed{\text{MUF}=5.89\text{ MHz}}$

Solution for 2019 ($D=1500\text{ km},h=300\text{ km},f_c=7\text{ MHz}$):

1. Calculate: $\text{MUF}=7\sqrt{1+{\left(\frac{1500}{2(300)}\right)}^2}=7\sqrt{1+(2.5{)}^2}=7\sqrt{1+6.25}=7\sqrt{7.25}\approx 18.85\text{ MHz}$ $\boxed{\text{MUF}=18.85\text{ MHz}}$

---

Type 9: Line of Sight (LOS) Distance & Received Field Strength

[PYQ: 2017] (12 marks)

A VHF communication link is to be established with a $35\text{ W}$ transmitter at $90\text{ MHz}$. Find the maximum distance up to which line-of-sight communication is possible if the height of the transmitting and receiving antennas are $40\text{ m}$ and $25\text{ m}$, respectively. Also, determine the field strength at the receiving end.

Solution:

1. Calculate maximum LOS distance: Terrestrial radio links require the $4/3$ Earth curvature refraction model: $d=4.12\left(\sqrt{h_t}+\sqrt{h_r}\right)\text{ km}$ $d=4.12\left(\sqrt{40}+\sqrt{25}\right)=4.12(6.324+5)=4.12(11.324)\approx 46.65\text{ km}$ $\boxed{d\approx 46.7\text{ km}}$

2. Calculate Received Field Strength ($E$): For a space-wave path over flat terrain, the empirical field strength under the two-ray ground reflection model is: $E\approx \frac{88\sqrt{P_t}{h}_t{h}_r}{\lambda d^2}\text{ mV/m}$ Where:

   • $P_t$ = Transmitter power in kW $=35\text{ W}=0.035\text{ kW}$.
   • $h_t,h_r$ = Antenna heights in meters ($40\text{ m}$, $25\text{ m}$).
   • $d$ = Distance in km ($46.65\text{ km}$).
   • $\lambda$ = Wavelength in meters $=\frac{c}{f}=\frac{3\times 10^8}{90\times 10^6}=3.333\text{ m}$.

   Plugging in the parameters: $E=\frac{88\sqrt{0.035}\times 40\times 25}{3.333\times (46.65{)}^2}=\frac{88(0.187)\times 1000}{3.333\times 2176.2}\approx 2.27\text{ mV/m}$ $\boxed{E\approx 2.27\text{ mV/m}}$

---

Common Mistakes That Cost Marks

Avoid these numerical pitfalls:

1. Using incorrect units in the $9\sqrt{N}$ plasma frequency formula: Electron density is frequently given in electrons per ${\text{cm}}^3$. You must multiply by $10^6$ to convert to electrons per ${\text{m}}^3$ before calculating.
2. Peak Amplitude vs. RMS Power Dissipation: When calculating the average power dissipated per cubic meter ($P_{av}=\frac{1}{2}\sigma E_0^2$), verify if the electric field intensity $E_0$ is given as a peak amplitude or an RMS value. If it is RMS, do not include the $1/2$ factor.
3. Forgetting to verify the medium classification first: Do not blindly apply the good conductor or low-loss approximations. Always calculate the loss tangent $\frac{\sigma }{\omega ϵ}$ first to justify the formulas you use.
4. Ignoring the phase offset $\varphi$: In instantaneous plane wave equations, do not assume $\varphi =0$ if the spatial boundary conditions state that the maximum occurs at a coordinate away from the origin ($x\ne 0$ or $z\ne 0$).
5. Using Decibel formulas for field vs. power incorrectly: When calculating wave attenuation in dB, field intensity (voltage) reduction uses $20{\log }_{10}(E_0/E_z)=\text{dB}$, whereas power reduction uses $10{\log }_{10}(P_0/P_z)=\text{dB}$. Using the 10-log formula for electric fields will result in a 2x error in skin depth or distance.

---

PYQ Bank — Verbatim Questions & Answer Plans

Q1 — Lossless wave instantaneous fields [PYQ: 2021 — 08 Marks]

“A uniform plane wave with $\vec{E}={\hat{a}}_x{E}_x$ propagates in a lossless simple medium … Write instantaneous expressions for $\vec{E}$ and $\vec{H}$.”

Answer plan: See Type 1 for the complete step-by-step mathematical solution.

Q2 — Lossless wave propagation in +x direction [PYQ: 2019 — 15 Marks]

“A uniform plane wave with $\vec{E}={\hat{a}}_y{E}_y$ propagates in a lossless simple medium … determine instantaneous fields and positive maximum location.”

Answer plan: See Type 1 for the complete step-by-step mathematical solution.

Q3 — Seawater attenuation parameters [PYQ: 2018, 2024 — 14 Marks]

“The electric field intensity of a linearly polarized wave propagating in seawater is … Determine parameters and distance at which amplitude is 1%.”

Answer plan: See Type 2 for the complete step-by-step mathematical solution.

Q4 — Low-loss dielectric dispersion and velocity [PYQ: 2017, 2021 — 11/12 Marks]

“A narrow band signal propagates in a lossy dielectric medium … Determine $\alpha ,\beta ,u_p,u_g$. Is the medium dispersive?”

Answer plan: See Type 3 for the complete step-by-step mathematical solution.

Q5 — Average power dissipated in lossy dielectric [PYQ: 2017, 2022, 2023, 2024 — 08 Marks]

“A sinusoidal electric intensity of amplitude 250 V/m and frequency 1 GHz exists in a lossy dielectric … Find average power dissipated.”

Answer plan: See Type 4 for the complete step-by-step mathematical solution.

Q6 — Standing wave phasor and instantaneous fields [PYQ: 2015, 2016 — 08/19 Marks]

“A y-polarized wave propagates in air and impinges normally on a perfectly conducting plane at x=0. Write expressions for incident, reflected, and total waves.”

Answer plan: See Type 5 for the complete step-by-step mathematical solution.

Q7 — Graphite skin depth and 30 dB attenuation [PYQ: 2020 — 10 Marks]

“Given the skin depth for graphite at 100 MHz is 0.20 mm, determine conductivity and distance for 1 GHz wave to reduce by 30 dB.”

Answer plan: See Type 6 for the complete step-by-step mathematical solution.

Q8 — Spacecraft penetration frequency [PYQ: 2019, 2023 — 07/10 Marks]

“Derive the equation of plasma frequency of ionized medium … Find minimum frequency above which communication can be established.”

Answer plan: See Type 7 for the plasma frequency derivation and spacecraft communication math.

Q9 — MUF calculations [PYQ: 2018, 2019 — 08/10 Marks]

“A HF radio link has to be established … calculate the maximum usable frequency (MUF).”

Answer plan: See Type 8 for the step-by-step solutions to both the 250 km and 1500 km path variants.

Q10 — VHF link distance & field strength [PYQ: 2017 — 12 Marks]

“A VHF communication is to be established … Find LOS distance and field strength at receiving end.”

Answer plan: See Type 9 for the complete step-by-step mathematical solution.

---

Self-Check Before Moving On

• Can you determine if a medium is a good conductor or low-loss dielectric by calculating its loss tangent?
• Can you formulate the instantaneous and phasor expressions of electric and magnetic fields for a plane wave given boundary parameters?
• Can you calculate the attenuation constant, phase constant, intrinsic impedance, and skin depth for seawater at kHz frequencies?
• Can you derive the group velocity expression $u_g=v[1-\frac{1}{8}{\tan }^2\delta {]}^{-1}$ for a low-loss dielectric?
• Can you calculate the average power density dissipated per cubic meter, accounting for the peak-to-RMS factor?
• Can you construct standing wave equations for a normally incident wave hitting a perfect electrical conductor?
• Can you scale the attenuation constant with frequency ($\alpha \propto \sqrt{f}$) to solve skin depth problems?
• Can you derive the plasma cutoff frequency equation $f_p\approx 9\sqrt{N}$ from electron displacement physics?
• Can you apply the MUF Secant Law to solve single-hop ionospheric transmission paths?
• Can you compute the received field strength of a terrestrial VHF link using the two-ray ground reflection model?

---

Source: ECE 2105 field pyq.md, corrupted pyq.md, week 12 & 13 lecture dumps, Sadiku Ch 10 & 15, Masuk Sir PYQ solutions.