Wave Polarization and Electromagnetic Waves

Wave Polarization

Instantaneous Distribution

The instantaneous distribution equations are given as:

• Electric field: $E(z) = E<em>0 e^{-jkz} + E</em>0 e^{+jkz}$
• Magnetic field: $H = \frac{E}{\eta}$

Wave Polarization

• A plane wave propagates along the +z direction.
• A separate definition for H is not necessary.
• E is linearly polarized (vertical).

Polarization of a Uniform Plane Wave

• Describes the locus traced by the tip of the E vector.
• The vector is in the plane orthogonal to the direction of propagation.
• It is a function of time at a given point in space.
• Plane wave propagating along +z:
  • $\tilde{E}(z) = \hat{x}\tilde{E}<em>x(z) + \hat{y}\tilde{E}</em>y(z)$
  • $E<em>x(z) = E</em>{xo}e^{-jkz}$
  • $E<em>y(z) = E</em>{yo}e^{-jkz}$
  • $E<em>{xo} = a</em>x$
  • $E<em>{yo} = a</em>y e^{j\delta}$
  • $\tilde{E}(z) = (\hat{x}a<em>x + \hat{y}a</em>y e^{j\delta})e^{-jkz}$
  • $E(z, t) = Re[\tilde{E}(z)e^{j\omega t}]$
  • $= \hat{x}a<em>x \cos(\omega t - kz) + \hat{y}a</em>y \cos(\omega t - kz + \delta)$

Polarization State

• Describes the trace of E as a function of time at a fixed z.
• Includes magnitude of E and inclination angle.

Linear Polarization

• $\delta = 0$ or $\delta = \pi$
• A wave is linearly polarized if, for a fixed z, the tip of $E(z, t)$ traces a straight line segment as a function of time.
• This happens when $E<em>x(z, t)$ and $E</em>y(z, t)$ are in-phase ($\delta = 0$) or out-of-phase ($\delta = \pi$).
• Under these conditions, the equation simplifies to:
  • $E(z, t) = Re[\tilde{E}(z)e^{j\omega t}] = \hat{x}a<em>x \cos(\omega t - kz) + \hat{y}a</em>y \cos(\omega t - kz + \delta)$
• In-phase:
  • $\delta = 0$
  • $E(0, t) = (\hat{x}a<em>x + \hat{y}a</em>y)\cos(\omega t - kz)$
• Out-of-phase:
  • $\delta = \pi$
  • $E(0, t) = (\hat{x}a<em>x - \hat{y}a</em>y)\cos(\omega t - kz)$
• The field's magnitude is:
  • $|E(z, t)| = [a<em>x^2 + a</em>y^2]^{1/2} |\cos(\omega t - kz)|$
• The inclination angle is:
  • $\psi = \tan^{-1}(\frac{E<em>y}{E</em>x}) = \tan^{-1}(\mp \frac{a<em>y}{a</em>x})$ (out-of-phase).
• If $a_y = 0$, then $\psi = 0$° or 180°, and the wave is x-polarized.
• Conversely, if $a_x = 0$, then $\psi = 90$° or -90°, and the wave is y-polarized.

Linear Polarization cases

• $a<em>x = a</em>y$
• $\psi = -45$°

Circular Polarization

• $a<em>x = a</em>y = a$ and $\delta = \pi/2$
• $a<em>x = a</em>y = a$ and $\delta = -\pi/2$
• For $a<em>x = a</em>y = a$ and $\delta = \pi/2$, the equations become:
  • $\tilde{E}(z) = (\hat{x}a + \hat{y}ae^{j\pi/2})e^{-jkz} = a(\hat{x} + j\hat{y})e^{-jkz}$
  • $E(z, t) = Re[\tilde{E}(z)e^{j\omega t}] = \hat{x}a\cos(\omega t - kz) + \hat{y}a\cos(\omega t - kz + \pi/2) = \hat{x}a\cos(\omega t - kz) - \hat{y}a\sin(\omega t - kz)$

Circular Polarization cont.

• $a<em>x = a</em>y = a$ and $\delta = \pi/2$
• The corresponding field magnitude and inclination angle are:
  • $|E(z, t)| = [E<em>x^2(z, t) + E</em>y^2(z, t)]^{1/2} = [a^2 \cos^2(\omega t - kz) + a^2 \sin^2(\omega t - kz)]^{1/2} = a$
  • $\psi(z, t) = \tan^{-1}(\frac{E<em>y(z, t)}{E</em>x(z, t)}) = \tan^{-1}(\frac{-a\sin(\omega t - kz)}{a\cos(\omega t - kz)}) = -(\omega t - kz)$
• $a<em>x = a</em>y = a$ and $\delta = -\pi/2$
• $|E(z, t)| = a, \quad \psi = (\omega t - kz)$

Circular Polarization Visuals

• For a Left Hand Circular Polarized (LHCP) wave:
  • $E(z,t) = \hat{x}E<em>o \cos(\omega t - kz) - \hat{y}E</em>o \sin(\omega t - kz)$
• For a Right Hand Circular Polarized (RHCP) wave:
  • $E(z,t) = \hat{x}E<em>o \cos(\omega t - kz) + \hat{y}E</em>o \sin(\omega t - kz)$

LHCP and RHCP

• Illustrations of LHCP and RHCP waves at different times.

Elliptical Polarization: General Case

• When $a<em>x \neq a</em>y$, or the phase shift is not exactly $\pi/2$
• $\tan 2\psi = (\tan 2\alpha_0) \cos \delta$
• $\sin 2\chi = (\sin 2\alpha_0) \sin \delta$
• $(-\pi/2 \le \psi \le \pi/2)$
• $(-\pi/4 \le \chi \le \pi/4)$
• where $\alpha<em>0$ is an auxiliary angle defined by $\tan \alpha</em>0 = \frac{a<em>y}{a</em>x} \quad (0 \le \alpha_0 \le \frac{\pi}{2})$
• \psi > 0 if \cos \delta > 0
• \psi < 0 if \cos \delta < 0
• The polarization ellipse may be tilted.

Linear and Circular Polarization as Special Cases of Elliptical Polarization

• Axial Ratio (AR): $\text{AR} = \frac{\text{major axis}}{\text{minor axis}}$
• $1 \le \text{AR} \le \infty$

Example 7-2: RHC Polarized Wave

• An RHC polarized plane wave with an electric field magnitude of 3 (mV/m) is traveling in the +y-direction in a dielectric medium with $\varepsilon = 4\varepsilon<em>0$, $\mu = \mu</em>0$, and $\sigma = 0$.
• If the frequency is 100 MHz, obtain expressions for $E(y, t)$ and $H(y, t)$.
• Since the wave is traveling in the +y-direction, its field must have components along the x- and z-directions.
• By comparison with the RHC polarized wave, we assign the z-component of $\tilde{E}(y)$ a phase angle of zero and the x-component a phase shift of $\delta = -\pi/2$.
• $\tilde{E}(y) = \hat{x}\tilde{E}<em>x + \hat{z}\tilde{E}</em>z$

Example 7-2 cont.

• $\omega = 2\pi f = 2\pi \times 10^8 \text{ (rad/s)}$
• $k = \frac{\omega}{c} \sqrt{\varepsilon_r} = \frac{2\pi \times 10^8 \sqrt{4}}{3 \times 10^8} = \frac{4\pi}{3} \text{ (rad/m)}$
• $\tilde{E}(y) = \hat{x}ae^{-j\pi/2}e^{-jky} + \hat{z}ae^{-jky} = (-\hat{x}j + \hat{z})3e^{-jky} \text{ (mV/m)}$
• $\tilde{H}(y) = \frac{1}{\eta} \hat{y} \times \tilde{E}(y) = \frac{1}{\eta} \hat{y} \times (-\hat{x}j + \hat{z})3e^{-jky} = \frac{3}{\eta} (-\hat{z}j - \hat{x})e^{-jky} \text{ (mA/m)}$

Example 7-2 cont..

• \eta = \frac{\eta0}{\sqrt{\varepsilonr}} = \frac{120\pi}{\sqrt{4}} = 60\pi \text{ (\Omega)}
• The instantaneous fields $E(y, t)$ and $H(y, t)$ are:
  • $E(y, t) = Re[\tilde{E}(y)e^{j\omega t}] = Re[(-\hat{x}j + \hat{z})3e^{-jky}e^{j\omega t}] = 3[\hat{x}\sin(\omega t - ky) + \hat{z}\cos(\omega t - ky)] \text{ (mV/m)}$
  • $H(y, t) = Re[\tilde{H}(y)e^{j\omega t}] = Re[\frac{3}{\eta} (-\hat{z}j - \hat{x})e^{-jky}e^{j\omega t}] = \frac{1}{20\pi} [\hat{x}\cos(\omega t - ky) - \hat{z}\sin(\omega t - ky)] \text{ (mA/m)}$

2017 Exam Q5

• Given values:
  • $f = 10^8 \text{ Hz}$
  • $\epsilon_r = 2.25$
  • $\mu_r = 1$
  • $E_o = 0.1$
• Calculations:
  • $\omega = 2\pi f = 2\pi \times 10^8$
  • $k = \omega \sqrt{\mu \epsilon} = 2\pi \times 10^8 \sqrt{\mu<em>o \epsilon</em>o \epsilon_r} = \frac{2\pi \times 10^8}{c} \sqrt{2.25} = \frac{2\pi \times 10^8}{3 \times 10^8} 1.5 = \pi$
  • $\eta = \sqrt{\frac{\mu}{\epsilon}} = \sqrt{\frac{\mu<em>o}{\epsilon</em>o \epsilon_r}} = \frac{120\pi}{\sqrt{2.25}} = \frac{120\pi}{1.5} = 80\pi$
• $\tilde{E}(z, t) = \hat{x} E_o e^{-jkz} = \hat{x} 0.1 e^{-j\pi z}$
• $E(t, z) = \hat{x} \cos(2 \times 10^8 t - \pi z)$
• Linear polarization - Horizontal

2017 Exam Q5 - Linear Polarization

• Linear polarization condition: $a_y = 0$
• Given: $a<em>x = a; a</em>y = 0$
• $\delta = 0$
• $\psi = \tan^{-1}(\frac{E<em>y}{E</em>x}) = 0$

2013 Tut 1 Q2

• Given:
  • $a_x = 30$
  • $a_y = 30$
  • $\delta = -\frac{\pi}{6} = -30$°
• Calculations:
  • $\psi = \tan^{-1}(\frac{E<em>y}{E</em>x}) = \tan^{-1}(\frac{a<em>y}{a</em>x}) = \tan^{-1}(-\frac{30}{30}) = -\frac{\pi}{6}$
• $E(z, t) = \hat{x}30\cos(\omega t) + \hat{y}30\sin(\omega t - \frac{\pi}{6})$
• $\tilde{E} = \hat{x}30 - j\hat{y}6$

2013 Tut 1 Q2 - RHCP

• Given: $a<em>x = 30; a</em>y = 30$
• $E(0,t) = \hat{x}30\cos(\omega t) + \hat{y}30\sin(\omega t)$
• $\tilde{E} = \hat{x}30 - j\hat{y}6$

2013 Test 1 Q2 - Linear Polarization

• Linear polarization at 45°
• Given: $a<em>{\theta} = a</em>{\phi} = \frac{E_o}{\sqrt{2}}$
• $\delta = 0$
• $E(R, t) = \hat{\theta}E<em>{\theta} \cos(\omega t - kR + \phi</em>{\theta}) + \hat{\phi}E<em>{\phi} \cos(\omega t - kR + \phi</em>{\phi})$
• $\psi = \tan^{-1}(\frac{E<em>{\phi}}{E</em>{\theta}})$
• $\tilde{E}(z) = \hat{\theta} + \hat{\phi}$

2016 Exam Q5

• Given:
  • $a_x = 10$
  • $a_y = 10$
  • $\delta = \frac{\pi}{2}$
• Calculations:
  • $\psi = \tan^{-1}(\frac{E<em>y}{E</em>x})$
• $E(z,t) = -\hat{x}10\cos(\omega t + kz) - \hat{y}10\sin(\omega t + kz)$

2016 Exam Q5 - Left Hand Circular

• $E(z,t) = -\hat{x}10\cos(\omega t + kz) - \hat{y}10\sin(\omega t + kz)$
• $\hat{k} = -\hat{z}$

Lossy Media

• For a uniform plane wave with electric field $\tilde{E} = \hat{x}E_x(z)$ traveling along the z-direction, the wave equation reduces to:
  • $\frac{d^2 E<em>x(z)}{dz^2} - \gamma^2 E</em>x(z) = 0$
  • where $\gamma^2 = -\omega^2 \mu \epsilon_c = -\omega^2 \mu (\epsilon' - j\epsilon'')$
  • $\epsilon'' = \frac{\sigma}{\omega}$
• Since $\gamma$ is complex, we express it as $\gamma = \alpha + j\beta$, where $\alpha$ is the attenuation constant and $\beta$ is the phase constant.
• Replacing $\gamma$ with $(\alpha + j\beta)$, we obtain $\alpha^2 - \beta^2 + j2\alpha\beta = -\omega^2 \mu \epsilon' + j\omega^2 \mu \epsilon''$

Lossy Media cont.

• Complex algebra requires the real and imaginary parts on one side of an equation to equal the real and imaginary parts on the other side. Hence:
  • $\alpha^2 - \beta^2 = -\omega^2 \mu \epsilon'$
  • $2\alpha\beta = \omega^2 \mu \epsilon''$
• Solving these two equations for $\alpha$ and $\beta$ gives:
  • $\alpha = \omega \sqrt{\frac{\mu \epsilon'}{2}} \left[ \sqrt{1 + (\frac{\epsilon''}{\epsilon'})^2} - 1 \right]^{1/2} \text{ (Np/m)}$
  • $\beta = \omega \sqrt{\frac{\mu \epsilon'}{2}} \left[ \sqrt{1 + (\frac{\epsilon''}{\epsilon'})^2} + 1 \right]^{1/2} \text{ (rad/m)}$

Attenuation of E and H Fields

• $\tilde{E}(z) = \hat{x} \tilde{E}<em>x(z) = \hat{x} E</em>{x0} e^{-\gamma z} = \hat{x} E_{x0} e^{-\alpha z} e^{-j\beta z}$
• The associated magnetic field $\tilde{H}$ can be determined by:
  • $\nabla \times \tilde{E} = -j\omega \mu \tilde{H}$, or using $\tilde{H} = (\hat{k} \times \tilde{E})/\eta_c$
  • $\tilde{H}(z) = \hat{y} \tilde{H}<em>y(z) = \hat{y} \frac{\tilde{E}</em>x(z)}{\eta<em>c} = \hat{y} \frac{E</em>{x0}}{\eta_c} e^{-\alpha z} e^{-j\beta z}$
  • where \etac = \sqrt{\frac{\mu}{\epsilonc}} = \sqrt{\frac{\mu}{\epsilon'}} \left[ 1 - j\frac{\epsilon''}{\epsilon'} \right]^{-1/2} \text{ (\Omega)}

Attenuation and Skin Depth

• Magnitude of E field: $|\tilde{E}<em>x(z)| = |E</em>{x0} e^{-\alpha z} e^{-j\beta z}| = |E_{x0}| e^{-\alpha z}$
• Skin depth: $\delta_s = \frac{1}{\alpha} \text{ (m)}$
• The skin depth $\delta<em>s$ is the value of z at which $|\tilde{E}</em>x(z)|/|E_{x0}| = e^{-1} \approx 0.37$.
• At depth $z = 3\delta<em>s$, the field magnitude is less than 5% of its initial value, and at $z = 5\delta</em>s$, it is less than 1%.
• The skin depth characterizes how deep an electromagnetic wave can penetrate into a conducting medium.

Conductors and Dielectrics

• Materials behave as dielectric or conductor.
• Conduction current density vs. Displacement current density.
• \sigma >> \omega \epsilon: displacement current >> conduction current $\implies$ dielectric
• $\sigma \approx \omega \epsilon$: displacement current $\approx$ conduction current $\implies$ quasi conductor
• \sigma << \omega \epsilon: displacement current << conduction current $\implies$ conductor
• $\nabla \times \tilde{E} = \sigma + j\omega \epsilon$

Approximations for Good Conductors and Dielectrics

• Applying binomial expansion:
• Good dielectric: \sigma << \omega \epsilon \implies \frac{\epsilon''}{\epsilon'} << 1
• Good conductor: \sigma >> \omega \epsilon \implies \frac{\epsilon''}{\epsilon'} >> 100
  • $\alpha \approx \sqrt{\omega \mu \sigma / 2}$
  • $\beta \approx \sqrt{\omega \mu \sigma / 2}$
  • $\eta_c \approx (1 + j) \sqrt{\frac{\omega \mu}{\sigma}}$

Low and High Frequency Approximations

• Table summarizing expressions for $\alpha, \beta, \eta<em>c, v</em>p$, and $\lambda$ for various types of media.
• Any Medium:
  • $\alpha = \omega \sqrt{\frac{\mu \epsilon'}{2}} \left[ \sqrt{1 + (\frac{\epsilon''}{\epsilon'})^2} - 1 \right]^{1/2}$
  • $\beta = \omega \sqrt{\frac{\mu \epsilon'}{2}} \left[ \sqrt{1 + (\frac{\epsilon''}{\epsilon'})^2} + 1 \right]^{1/2}$
  • $\eta_c = \sqrt{\frac{\mu}{\epsilon'}} \left[ 1 - j\frac{\epsilon''}{\epsilon'} \right]^{-1/2}$
• Lossless Medium ($\sigma = 0$):
  • $\alpha = 0$
  • $\beta = \omega \sqrt{\mu \epsilon}$
  • $\eta_c = \sqrt{\frac{\mu}{\epsilon}}$
• Low-loss Medium (\frac{\epsilon''}{\epsilon'} << 1):
  • $\eta_c = \sqrt{\frac{\mu}{\epsilon}} \left( 1 + j \frac{\epsilon''}{2\epsilon'} \right)$
• Good Conductor (\frac{\epsilon''}{\epsilon'} > 1):
  • $\alpha = \sqrt{\pi f \mu \sigma}$
  • $\beta = \sqrt{\pi f \mu \sigma}$
  • $\eta_c = (1 + j) \sqrt{\frac{\pi f \mu}{\sigma}}$
• Notes: $\epsilon' = \epsilon; \epsilon'' = \frac{\sigma}{\omega}$