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 $f = 75 \,\text{GHz}$, $E<em>o = 275 \,\text{V/m}$, $\sigma = 2 \,\text{S/m}$, $\varepsilon</em>r = 72$, $l = 3\delta$, 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:
  • $\eta<em>c = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu</em>0}{\varepsilon<em>0 \varepsilon</em>r}} = \frac{120\pi}{\sqrt{\varepsilon_r}} = \frac{120\pi}{\sqrt{72}} \approx 44.43 \Omega$
• With $f = 75 \,\text{GHz}$, $E<em>o = 275 \,\text{V/m}$, $\sigma = 2 \,\text{S/m}$, $\varepsilon</em>r = 72$, $l = 3\delta$, and $A = 1 \,\text{m}^2$

Power Dissipated

• Power dissipated is the difference between input and output power: $P<em>{diss} = P</em>{in} - P_{out}$
• Where $P<em>{in}$ is the input power and $P</em>{out}$ 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

• $S<em>{av} = \frac{|E</em>o|^2}{2 \eta} = \frac{275^2}{2 \times 44.43} = 851 \,\text{W/m}^2$
• $P<em>{diss} = A (S</em>{av,in} - S_{av,out}) = 851 - 2.11 \approx 848.9 \,\text{W}$
• $S<em>{out} = S</em>{in} e^{-2\alpha z} = 851 e^{-2\alpha z}$, 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 ($\mu \neq \mu<em>0, \varepsilon \neq \varepsilon</em>0$) 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 $E<em>1 = 100 \,\text{V/m}$, and its phase at $z = 0$ and $t = 0$ is $\phi</em>1 = 60^\circ$.
• 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

• $\vec{E}(z, t) = \text{Re} [\hat{y} E<em>0 e^{-\alpha z} e^{j(\omega t - \beta z + \phi</em>0)}]$
• $\vec{E}(z, t) = \hat{y} 100 e^{-4z} \cos(2\pi \times 10^6 t - 4z + 60^\circ) \,\text{V/m}$

Magnetic Field Expression

• $\vec{H} = -\frac{1}{\eta} \hat{a<em>z} \times \vec{E} = -\hat{x} \frac{E</em>0}{\eta} e^{-\alpha z} e^{j(\omega t - \beta z + \phi_0 - 45^\circ)}$
• $\vec{H}(z, t) = -\hat{x} 71.6 e^{-4z} \cos(2\pi \times 10^6 t - 4z + 15^\circ) \,\text{A/m}$

Time-Averaged Power Density Vector

• $\vec{S}<em>{av} = \frac{1}{2} \text{Re} [\vec{E} \times \vec{H}^*] = \hat{z} \frac{1}{2} \frac{|E</em>0|^2}{|\eta|} e^{-2\alpha z} \cos(\theta_{\eta})$
• $\vec{S}_{av} = \hat{z} \frac{1}{2} \frac{100^2}{78} e^{-8z} \cos(45^\circ) = \hat{z} 45.31 e^{-8z} \,\text{W/m}^2$

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:

  • $\eta<em>c = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu</em>0 \mu<em>r}{\varepsilon</em>0 \varepsilon<em>r}} = \frac{120\pi}{\sqrt{\varepsilon</em>r}} = \frac{120\pi}{\sqrt{81}} = \frac{40\pi}{3} \approx 41.89 \Omega$

Magnetic Field Expression

• $\vec{H}(R) = -\hat{\theta} \frac{E<em>0}{\eta} \frac{e^{-\alpha R}}{R} + \hat{\varphi} \frac{E</em>0}{\eta} \frac{e^{-\alpha R}}{R} = \frac{E(R)}{\eta}$

Time-Averaged Power Density Vector

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

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

• $\frac{\omega}{\pi} = 10^8 \Rightarrow \omega = 2 \pi \times 10^8 \,\text{rad/s}$
• $f = \frac{\omega}{2 \pi} = 25 \,\text{MHz}$
• $k = \omega \sqrt{\mu \varepsilon} = 2 \pi f \sqrt{\mu<em>r \mu</em>0 \varepsilon<em>r \varepsilon</em>0}$

Rewrite magnetic Field

• Examine the rewritten form:

  • $\vec{H} = [\hat{x}] H<em>0 \cos(2 \pi 10^{8} t)$ $[\hat{y}] H</em>0 \cos(2 \pi 10^{8} t)$

  Conclusion

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