Wave Polarization and Electromagnetic Waves

Wave Polarization

Instantaneous Distribution

The instantaneous distribution equations are given as:

• Electric field: E(z) = E0 e^{-jkz} + E0 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}x(z) + \hat{y}\tilde{E}y(z)
  • Ex(z) = E{xo}e^{-jkz}
  • Ey(z) = E{yo}e^{-jkz}
  • E{xo} = ax
  • E{yo} = ay e^{j\delta}
  • \tilde{E}(z) = (\hat{x}ax + \hat{y}ay e^{j\delta})e^{-jkz}
  • E(z, t) = Re[\tilde{E}(z)e^{j\omega t}]
  • = \hat{x}ax \cos(\omega t - kz) + \hat{y}ay \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 Ex(z, t) and Ey(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}ax \cos(\omega t - kz) + \hat{y}ay \cos(\omega t - kz + \delta)
• In-phase:
  • \delta = 0
  • E(0, t) = (\hat{x}ax + \hat{y}ay)\cos(\omega t - kz)
• Out-of-phase:
  • \delta = \pi
  • E(0, t) = (\hat{x}ax - \hat{y}ay)\cos(\omega t - kz)
• The field's magnitude is:
  • |E(z, t)| = [ax^2 + ay^2]^{1/2} |\cos(\omega t - kz)|
• The inclination angle is:
  • \psi = \tan^{-1}(\frac{Ey}{Ex}) = \tan^{-1}(\mp \frac{ay}{ax}) (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

• ax = ay
• \psi = -45°

Circular Polarization

• ax = ay = a and \delta = \pi/2
• ax = ay = a and \delta = -\pi/2
• For ax = ay = 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.

• ax = ay = a and \delta = \pi/2
• The corresponding field magnitude and inclination angle are:
  • |E(z, t)| = [Ex^2(z, t) + Ey^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{Ey(z, t)}{Ex(z, t)}) = \tan^{-1}(\frac{-a\sin(\omega t - kz)}{a\cos(\omega t - kz)}) = -(\omega t - kz)
• ax = ay = 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}Eo \cos(\omega t - kz) - \hat{y}Eo \sin(\omega t - kz)
• For a Right Hand Circular Polarized (RHCP) wave:
  • E(z,t) = \hat{x}Eo \cos(\omega t - kz) + \hat{y}Eo \sin(\omega t - kz)

LHCP and RHCP

• Illustrations of LHCP and RHCP waves at different times.

Elliptical Polarization: General Case

• When ax \neq ay, 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 \alpha0 is an auxiliary angle defined by \tan \alpha0 = \frac{ay}{ax} \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\varepsilon0, \mu = \mu0, 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}x + \hat{z}\tilde{E}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{\muo \epsilono \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{\muo}{\epsilono \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: ax = a; ay = 0
• \delta = 0
• \psi = \tan^{-1}(\frac{Ey}{Ex}) = 0

2013 Tut 1 Q2

• Given:
  • a_x = 30
  • a_y = 30
  • \delta = -\frac{\pi}{6} = -30°
• Calculations:
  • \psi = \tan^{-1}(\frac{Ey}{Ex}) = \tan^{-1}(\frac{ay}{ax}) = \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: ax = 30; ay = 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{\theta} = a{\phi} = \frac{E_o}{\sqrt{2}}
• \delta = 0
• E(R, t) = \hat{\theta}E{\theta} \cos(\omega t - kR + \phi{\theta}) + \hat{\phi}E{\phi} \cos(\omega t - kR + \phi{\phi})
• \psi = \tan^{-1}(\frac{E{\phi}}{E{\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{Ey}{Ex})
• 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 Ex(z)}{dz^2} - \gamma^2 Ex(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}x(z) = \hat{x} E{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}y(z) = \hat{y} \frac{\tilde{E}x(z)}{\etac} = \hat{y} \frac{E{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}x(z)| = |E{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 \deltas is the value of z at which |\tilde{E}x(z)|/|E_{x0}| = e^{-1} \approx 0.37.
• At depth z = 3\deltas, the field magnitude is less than 5% of its initial value, and at z = 5\deltas, 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, \etac, vp, 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}