Electromagnetic Waves in Lossy Medium

Uniform Plane Wave in Lossy Medium

Power Dissipation

• A uniform plane wave propagates through a lossy medium.
• Objective: Find the power dissipated in a specific volume.
• Given parameters:
  • Power density: \,\sigma = 2 \,\text{S/m}
  • Electric field amplitude: E_o = 275 \,\text{V/m}
  • Frequency: f = 75 \,\text{GHz}
  • Relative permittivity: \varepsilon_r = 72
  • Length: l = 3\delta
  • Area: A = 1 \,\text{m}^2
  • Where \delta is skin depth

Lossy Medium Characteristics

• Distinction between lossy and low-loss mediums.
• Power density calculation:
  • \frac{\sigma}{\omega \varepsilon} = \frac{2}{2 \pi \times 75 \times 10^9 \times 72 \times 8.854 \times 10^{-12}} = 0.00667
  • \frac{\varepsilon''}{\varepsilon'} = 0.00667
• With f = 75 \,\text{GHz}, Eo = 275 \,\text{V/m}, \sigma = 2 \,\text{S/m}, \varepsilonr = 72, l = 3\delta, and A = 1 \,\text{m}^2

Further Calculations

• Attenuation constant:
  • \alpha = \omega \sqrt{\frac{\mu \varepsilon}{2}} \sqrt{\sqrt{1 + (\frac{\sigma}{\omega \varepsilon})^2} - 1} \approx \frac{\sigma}{2} \sqrt{\frac{\mu}{\varepsilon}}
• Intrinsic impedance:
  • \etac = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu0}{\varepsilon0 \varepsilonr}} = \frac{120\pi}{\sqrt{\varepsilon_r}} = \frac{120\pi}{\sqrt{72}} \approx 44.43 \Omega
• With f = 75 \,\text{GHz}, Eo = 275 \,\text{V/m}, \sigma = 2 \,\text{S/m}, \varepsilonr = 72, l = 3\delta, and A = 1 \,\text{m}^2

Power Dissipated

• Power dissipated is the difference between input and output power: P{diss} = P{in} - P_{out}
• Where P{in} is the input power and P{out} is the output power
• The Poynting vector: \vec{S}_{av} = \frac{1}{2} \text{Re}[\vec{E} \times \vec{H}^*]
• P = A S
• With l = 3\delta and A = 1 \,\text{m}^2

Power Density Calculation

• S{av} = \frac{|Eo|^2}{2 \eta} = \frac{275^2}{2 \times 44.43} = 851 \,\text{W/m}^2
• P{diss} = A (S{av,in} - S_{av,out}) = 851 - 2.11 \approx 848.9 \,\text{W}
• S{out} = S{in} e^{-2\alpha z} = 851 e^{-2\alpha z}, where \alpha = 0.00248
• The power decays exponentially with distance z.
• With l = 3\delta and A = 1 \,\text{m}^2

2018 Exam Q4

Problem Statement

• A linearly polarized uniform plane wave in a lossy medium (\mu \neq \mu0, \varepsilon \neq \varepsilon0) is propagating in the +z direction.
• At f = 1 \,\text{MHz}, the intrinsic impedance is \eta_c = 78 \angle 45^\circ \,\Omega and the skin depth is \delta = 0.25 \,\text{m}.
• The electric field magnitude is E1 = 100 \,\text{V/m}, and its phase at z = 0 and t = 0 is \phi1 = 60^\circ.
• The field vector at z = 0 and t = 0 points in the +y direction.

Objectives

1. Determine the wavelength in the medium (\lambda).
2. Determine the conductivity of the medium (\sigma).
3. Determine the intrinsic impedance of the medium (\eta_c).
4. Obtain expressions for the instantaneous electric field \vec{E}(z, t).
5. Obtain expressions for the instantaneous magnetic field \vec{H}(z, t).
6. Find the time-averaged power density vector \vec{S}_{av}.

Solution

Good Conductor Approximation

• Since it's a good conductor, \alpha = \beta = \frac{1}{\delta}.
• And \eta_c = (1 + j) \frac{\alpha}{\sigma} = \sqrt{\frac{j \omega \mu}{\sigma}}.

Wavelength Calculation

• \lambda = \frac{2\pi}{\beta} = 2 \pi \delta = 2 \pi (0.25) = 1.5708 \,\text{m}

Conductivity Calculation

• \delta = \sqrt{\frac{2}{\omega \mu \sigma}} \Rightarrow \sigma = \frac{2}{\omega \mu \delta^2} = \frac{2}{2 \pi f \mu \delta^2} = \frac{1}{\pi f \mu \delta^2}
• \sigma = \frac{1}{\pi \times 1 \times 10^6 \times 4\pi \times 10^{-7} \times 0.25^2} = 4.05 \,\text{S/m}
• Alternative:
  • \alpha = \sqrt{\pi f \mu \sigma} \Rightarrow \sigma = \frac{\alpha^2}{\pi f \mu} = \frac{4}{\pi f \mu}

Intrinsic Impedance Calculation

• \eta_c = (1 + j) \frac{\alpha}{\sigma} = (1 + j) \sqrt{\frac{\pi f \mu}{\sigma}}

Electric Field Expression

• \vec{E}(z, t) = \text{Re} [\hat{y} E0 e^{-\alpha z} e^{j(\omega t - \beta z + \phi0)}]
• \vec{E}(z, t) = \hat{y} 100 e^{-4z} \cos(2\pi \times 10^6 t - 4z + 60^\circ) \,\text{V/m}

Magnetic Field Expression

• \vec{H} = -\frac{1}{\eta} \hat{az} \times \vec{E} = -\hat{x} \frac{E0}{\eta} e^{-\alpha z} e^{j(\omega t - \beta z + \phi_0 - 45^\circ)}
• \vec{H}(z, t) = -\hat{x} 71.6 e^{-4z} \cos(2\pi \times 10^6 t - 4z + 15^\circ) \,\text{A/m}

Time-Averaged Power Density Vector

• \vec{S}{av} = \frac{1}{2} \text{Re} [\vec{E} \times \vec{H}^*] = \hat{z} \frac{1}{2} \frac{|E0|^2}{|\eta|} e^{-2\alpha z} \cos(\theta_{\eta})
• \vec{S}_{av} = \hat{z} \frac{1}{2} \frac{100^2}{78} e^{-8z} \cos(45^\circ) = \hat{z} 45.31 e^{-8z} \,\text{W/m}^2

2013 Test 1 Q2

Problem Statement

• A uniform plane, time-harmonic electromagnetic wave radiated by a short dipole antenna is given by:
  • E(R) = (\hat{\theta} + \hat{\varphi}) \frac{E_0}{R} e^{-\alpha R} e^{-j\beta R} \,\text{V/m}
• The antenna is used in a communication system with a submarine in seawater.
  • \varepsilon_r = 81
  • \mu_r = 1
  • \varepsilon'' = 8 \times 10^{-2}
• The system must operate to a depth of at least 1.2 km.
• Acceptable signal-to-noise ratio to a depth of 3 skin depths.

Solution Approach

1. Determine the skin depth in seawater.
2. Determine the attenuation constant.
3. Calculate the required frequency for the communication system.

Calculations

• Given: \varepsilon' = 81, \varepsilon'' = 8 \times 10^{-2}, \mu_r = 1
  • \frac{\varepsilon''}{\varepsilon'} = \frac{8 \times 10^{-2}}{81} = 0.000987 < < 1

Skin Depth

• Operating depth: d = 1.2 \,\text{km} = 1200 \,\text{m}

• d = 3\delta \quad \Rightarrow \quad \delta = \frac{d}{3} = \frac{1200}{3} = 400 \,\text{m} \quad \delta = 1/\alpha

• Frequency: f = 41.6 \,\text{MHz}

• Intrinsic impedance:

  • \etac = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu0 \mur}{\varepsilon0 \varepsilonr}} = \frac{120\pi}{\sqrt{\varepsilonr}} = \frac{120\pi}{\sqrt{81}} = \frac{40\pi}{3} \approx 41.89 \Omega

Magnetic Field Expression

• \vec{H}(R) = -\hat{\theta} \frac{E0}{\eta} \frac{e^{-\alpha R}}{R} + \hat{\varphi} \frac{E0}{\eta} \frac{e^{-\alpha R}}{R} = \frac{E(R)}{\eta}

Time-Averaged Power Density Vector

• \vec{S}{av} = \frac{1}{2} \text{Re} [\vec{E} \times \vec{H}^*] = \hat{R} \frac{|E0|^2}{2 \eta} \frac{e^{-2 \alpha R}}{R^2}
• Attenuation in dB:
  • \text{Attenuation} = 10 \log{10} \frac{S(R)}{S(0)} = 10 \log{10} e^{-2\alpha R} = 10 \log_{10} e^{-2 \times 3} = -26.01 \,\text{dB}

2019 Test 2 Q4

Problem Statement

• Given the following fields:
  • \vec{E} = [\hat{z}] E_0 e^{-jkx}
  • \vec{H} = [\hat{y}] H_0 e^{-jkx}

Calculations

• Need to apply these general formulas:
  • \eta = \sqrt{\frac{\mu}{\varepsilon}}
  • k = \omega \sqrt{\mu \varepsilon}

Numerical Values

• \frac{\omega}{\pi} = 10^8 \Rightarrow \omega = 2 \pi \times 10^8 \,\text{rad/s}
• f = \frac{\omega}{2 \pi} = 25 \,\text{MHz}
• k = \omega \sqrt{\mu \varepsilon} = 2 \pi f \sqrt{\mur \mu0 \varepsilonr \varepsilon0}

Rewrite magnetic Field

• Examine the rewritten form:

  • \vec{H} = [\hat{x}] H0 \cos(2 \pi 10^{8} t) [\hat{y}] H0 \cos(2 \pi 10^{8} t)

  Conclusion

• Wave polarization: Left-hand circular polarization (LHC).