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Exam Question Notes

2018 Exam Q5

• Medium 1: Permittivity \epsilon1, Permeability \mu1 = \mu0, Epsilon relative \epsilon{r1} = 1.
• Medium 2: Permittivity \epsilon2, Permeability \mu2 = \mu0, Epsilon relative \epsilon{r2} .
• Interface at y = 0.
• Incident wave: Propagates in the \hat{i} direction, Electric field \tilde{Ei}, Magnetic field \tilde{Hi}.
• Transmitted wave: Propagates in the \hat{t} direction, Electric field \tilde{Et}, Magnetic field \tilde{Ht}.
• Reflected wave: Propagates in the \hat{r} direction, Electric field \tilde{Er}, Magnetic field \tilde{Hr}.
• Incident electric field: \tilde{Ei} = \hat{x} E{io} \cos(2\pi \times 10^9 t - \pi y) V/m at y=0, where E_{io} = 40 V/m.
• The incident electric field is polarized in the x-direction and propagates along y-direction.
• The magnitude of the incident electric field is 40 V/m.

Calculations and Parameters

• Intrinsic impedance of medium 1: \eta1 = \sqrt{\frac{\mu1}{\epsilon1}} = \sqrt{\frac{\mu0}{\epsilon0 \epsilon{r1}}} = \frac{120\pi}{\sqrt{1}} = 120\pi = 377 \Omega.
• Wave number in medium 1: k1 = \omega \sqrt{\mu1 \epsilon1} = 2\pi \times 10^9 \sqrt{\mu0 \epsilon0 \epsilon{r1}} = \frac{2\pi \times 10^9}{3 \times 10^8} = \frac{20\pi}{3} rad/m.
• Incident Magnetic Field: \tilde{Hi} = \frac{1}{\etai} \hat{k} \times \tilde{E} = \frac{1}{\eta1} \hat{y} \times \tilde{Ei} = \frac{1}{120\pi} \hat{z} \cos(2\pi \times 10^9 t - \pi y) A/m.

Reflection Coefficient

• Reflection coefficient: \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1}.
• Given: \Gamma = \frac{60\pi - 120\pi}{60\pi + 120\pi} = \frac{-60\pi}{180\pi} = -\frac{1}{3} = -0.33.
• Reflected Electric Field: \tilde{Er}(y=0) = \Gamma \tilde{Ei}(y=0) = -0.2 \hat{x} \cos(2\pi \times 10^9 t + \pi y).
• Intrinsic impedance of medium 2: \eta2 = \sqrt{\frac{\mu2}{\epsilon2}} = \frac{120\pi}{\sqrt{\epsilon{r2}}} = 60\pi \Omega \implies \epsilon_{r2} = 4.

Transmission Coefficient

• Transmission coefficient: \tau = 1 + \Gamma = 1 + (-0.33) = 0.67.
• Transmitted Electric Field: \tilde{Et}(y=0) = \tau \tilde{Ei}(y=0) = 0.67 \hat{x} \cos(2\pi \times 10^9 t - 2\pi y).
• Wave number in medium 2: k2 = \omega \sqrt{\mu2 \epsilon2} = 2 \pi \times 10^9 \sqrt{\mu0 \epsilon0\epsilon{r2}} = \frac{2\pi \times 10^9 \times 2}{3 \times 10^8} = \frac{40 \pi}{3} rad/m.
• Transmitted Electric Field: \tilde{Et} = \hat{x} \tau E{io} \cos(2\pi \times 10^9 t - 2\pi y) = \hat{x} (1.2 \times 40) \cos(2\pi \times 10^9 t - 2\pi y) = 48 \hat{x} \cos(2\pi \times 10^9 t - 2\pi y).

Additional Problems

• Medium 1: \mu = 9\mu0, \epsilon = \epsilon0
• Medium 2: \mu = 4\mu0, \epsilon = \epsilon0
• Incident, reflected, and transmitted electric fields are denoted as \tilde{Ei}, \tilde{Er}, \tilde{E_t} respectively.
• Reflection coefficient: \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1}= -0.2.\tau = 1 + \Gamma = 0.8

2013 Test 1 Q1

• Incident electric field: \tilde{E_i} = (\hat{x} - j\hat{y}) 4 e^{-j \frac{2 \pi}{3} z}
• Medium 1: \mu1 = \mu0, \epsilon1 = \epsilon0
• Medium 2: \mu; \epsilon; \sigma = \infty (Perfect Electric Conductor, PEC)
• At the interface z = 0: \ ild{E_t} = 0
• Reflection coefficient: \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1} = \frac{0 - \eta1}{0 + \eta1} = -1
• Reflected electric field at z = 0: \tilde{Er}(z=0) = \Gamma \tilde{Ei}(z=0) = - (\hat{x} - j\hat{y}) 4
• Magnitude of Reflected Electric Field: |E_r| = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 5.66 V/m

Time Domain Expressions

• Reflected Electric Field: ild{Er}(z, t) = Re[\tilde{Er} e^{j\omega t}] = (\hat{x} - j\hat{y}) 4 e^{+j \frac{2 \pi}{3} z}
• \tilde{E_r}(z, t) = \hat{x} 4 \cos(\omega t + \frac{2 \pi}{3} z) - \hat{y} 4 \sin(\omega t + \frac{2 \pi}{3} z)

Circular Polarization

• The wave is Left Hand Circularly Polarized.
• Electric Field at z=0: \tilde{E}(0, t) = -\hat{x} 4 \cos(\omega t) - \hat{y} 4 \sin(\omega t)
• Amplitudes: ax = ay = 4
• Phase difference: \delta = 90^\circ
• Polarization angle: \psi = \tan^{-1} \left[ \frac{Ey(z, t)}{Ex(z, t)} \right] = \omega t; Varies with time.

Induced Current Density

• Boundary condition at PEC interface: \hat{n} \times (\tilde{H1} - \tilde{H2}) = \tilde{J_s}
• Since Medium 2 is PEC, \tilde{H_2} = 0
• Incident magnetic field: \tilde{Hi} = \frac{1}{\etao} \hat{z} \times \tilde{Ei} = \frac{1}{\etao} \hat{z} \times (\hat{x} - j\hat{y}) 4 e^{-j \frac{2 \pi}{3} z}
• Reflected magnetic field: \tilde{Hr} = - \frac{1}{\etao} \hat{z} \times \tilde{Er} = - \frac{1}{\etao} \hat{z} \times (\hat{x} - j\hat{y}) 4 e^{+j \frac{2 \pi}{3} z}
• Surface current density at z=0:\tilde{Js} = \hat{n} \times (\tilde{H1} - \tilde{H2}) = \hat{z} \times (\tilde{Hi} + \tilde{Hr}) = \hat{z} \times (\frac{2}{\etao} (\hat{x} - j\hat{y}) 4) = \frac{8}{\eta_o} (\hat{y} + j\hat{x})

Instantaneous Total Electric Field Intensity

• \tilde{E_1} = (\hat{x} - j\hat{y}) 4 e^{-j \frac{2 \pi}{3} z}
• \tilde{E2}(z, t) = Re[\tilde{Ei} + \tilde{E_r}] e^{j \omega t} = Re[(\hat{x} - j\hat{y}) e^{-j \frac{2 \pi}{3} z} + (-\hat{x} - j\hat{y}) e^{+j \frac{2 \pi}{3} z}] e^{j \omega t}
• \tilde{E_2}(z, t) = Re[\hat{x} (e^{-j \frac{2 \pi}{3} z} - e^{+j \frac{2 \pi}{3} z}) - j\hat{y} (e^{-j \frac{2 \pi}{3} z} + e^{+j \frac{2 \pi}{3} z})] e^{j \omega t}
• Instantaneous Total Electric Field: \tilde{E}(z,t) = \hat{x}2\sin(\frac{2 \pi z}{3})\sin(\omega t) - \hat{y}4\cos(\frac{2 \pi z}{3})

Normal Incidence on Lossy Dielectric Medium

• Uniform plane wave normally incident from free space on a lossy dielectric medium.
• Medium 1: Free space, \mu1 = \mu0, \epsilon1 = \epsilon0
• Medium 2: Lossy dielectric, \mu2 = \mu0, \epsilon2 = (2.25 - j0.675)\epsilon0
• Incident Electric Field: \tilde{Ei} = \hat{x} 10^{-6} e^{-j k1 z} V/m

Reflection Coefficient

• Relative Permittivity: \epsilon_r = 2.25
• Loss Tangent: tan \delta = \frac{\epsilon''}{\epsilon'} = \frac{0.675}{2.25} = 0.3
• Since tan \delta << 1, it is a low loss medium.
• Intrinsic Impedance of Medium 1: \eta_1 = 120 \pi = 377 \Omega
• Approximate Intrinsic Impedance of Medium 2: \eta2 = \frac{\etao}{\sqrt{\epsilon_r}} = \frac{377}{\sqrt{2.25}} = 251.3 \Omega
• Reflection Coefficient: \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1} = \frac{251.3 - 377}{251.3 + 377} = -0.2
• Exact calculation yields: \Gamma = 0.208 e^{j159.9^\circ}

Reflected and Transmitted Electric Fields

• Reflected Electric Field: \tilde{Er} = \hat{x} \Gamma Ei = \hat{x} 10^{-6} (0.208) e^{j159.9^\circ} e^{jkz} = \hat{x} 2.08 \times 10^{-7} e^{jkz} e^{j159.9^\circ}
• Transmission Coefficient: \tau = 1 + \Gamma = 1 + (-0.208e^{j159.9^\circ}) = 0.808 e^{j5.1^\circ}

Attenuation and Phase Constants

• Attenuation constant: \alpha = \omega \sqrt{\mu \epsilon} \left[ \sqrt{1 + (\frac{\epsilon ''}{\epsilon '})^2} - 1 \right]^{1/2}
• Approximate Attenuation Constant: \alpha_2 = \frac{\omega \mu \epsilon''}{2 \sqrt{\mu \epsilon'}} = 1.35 Np/m
• Exact value: \alpha_2 = 1.335 Np/m
• Phase constant: \beta = \omega \sqrt{\mu \epsilon} \left[ \sqrt{1 + (\frac{\epsilon ''}{\epsilon '})^2} + 1 \right]^{1/2}
• \omega = 1.8 \times 10^9 rad/s
• Phase Constant: \beta_2 = \omega \sqrt{\mu \epsilon'} = 9 rad/m
• Exact value: \beta_2 = 9.081 rad/m

Transmitted Electric Field

• \tilde{Et} = \hat{x} \tau Ei e^{-\alpha z} e^{-j \beta z} = \hat{x} (0.808 e^{j5.1^\circ} ) (10^{-6}) e^{-1.335 z} e^{-j 9.081 z}
• \tilde{E_t} = \hat{x} 8.08 \times 10^{-7} e^{-1.335 z} e^{-j (9.081 z - 5.1^\circ)}

Magnetic Fields and Power Density

• Reflected Magnetic Field: \tilde{Hr} = - \frac{1}{\eta1} \hat{z} \times \tilde{E_r}= \hat{y} \frac{2.08 \times 10^{-7}}{120 \pi}e^{j159.9^\circ}e^{jkz} = \hat{y} 5.778 \times 10^{-10} e^{j159.9^\circ}e^{jkz}
• Incident Magnetic Field: \tilde{Hi} = \frac{Ei}{\eta_0} =\hat{y} \frac{10^{-6}}{120*\pi}
• Transmitted Magnetic Field: \tilde{Ht} = \frac{1}{\eta2} \hat{z} \times \tilde{E_t}
• {\eta}_2 = 254 e^{j8.5^\circ}
• {\alpha}_2 = 1.335 \frac{Np}{m}
• {\beta}_2 = 9.081 \frac{rad}{m}
• {\tilde{H}}_t = \hat{y} 31.81*10^{-7} e^{j3.4^\circ} e^{-1.335z} e^{-j9.081z}

Time Averaged Power Density

• {S}_{av} = Re[\frac{1}{2}E \times H^*]
• Time averaged power density in region 1:
  {S}{av} ~\cong Re[\frac{1}{2} \frac{{\tilde{Ei}}^2}{\eta1} - \frac{{\tilde{Er}}^2}{\eta_1}]
• Time averaged power density in region 2:
  {S}{av} ~\cong Re[\frac{1}{2} \frac{{\tilde{Et}}^2}{\eta_2}]
• {S}_{av} ~\cong Re[\frac{1}{2} {(31.81*10^{-7})}^2 * 254 e^{j3.4^\circ} e^{-1.335z} e^{-j9.081z}]
• $${S}_{av} ~\cong 2.67 e^{-2*9.081z +j3.4}$