9/13/2017

Course Instructor Dr. Raymond C. Rumpf Office: A‐337 Phone: (915) 747‐6958 E‐Mail: [email protected]

EE 4347 Applied Electromagnetics

Topic 3a Electromagnetic Waves & Polarization

Electromagnetic Waves & Polarization These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited Slide 1

Lecture Outline • Maxwell’s Equations • Derivation of the Wave Equation • Solution to the Wave Equation • Intuitive Wave Parameters • Dispersion Relation • Electromagnetic Wave Polarization • Visualization of EM Waves

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Maxwell’s Equations

Electromagnetic Waves & Polarization Slide 3

Recall Maxwell’s Equations in Source Free Media  In source‐free media, we have and .J  0 v  0 Maxwell’s equations in the frequency‐domain become

Curl Equations  Divergence Equations Ej  B D 0   Hj  D B 0

Constitutive Relations  DE   BH 

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The Curl Equations Predict Waves After substituting the constitutive relations into the curl equations, we get   Ej  H Hj   E

A time‐harmonic magnetic field will A time‐harmonic electric field will induce a time‐harmonic electric induce a time‐harmonic magnetic field circulating about the magnetic field circulating about the electric field. field. A time‐harmonic circulating electric A time‐harmonic circulating field will induce a time‐harmonic magnetic field will induce a time‐ magnetic field along the axis of harmonic electric field along the circulation. axis of circulation.

An H induces an E. That E induces another H. That new H induces another E. That E induces yet another H. Ando so on.

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Visualization of Curl & Waves

Magnetic Field

Electric Field

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Derivation of the Wave Equation

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Wave Equation in Linear Media (1 of 2) Since the curl equations predict propagation, it makes sense that we derive the wave equation by combining the curl equations.   Ej   H Hj   E  Solve for H

1 1 H  E j

1 1    Ej   E j Electromagnetic Waves & Polarization Slide 8

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Wave Equation in Linear Media (2 of 2) The last equation is simplified to arrive at our final equation for waves in linear media.  1 E 2  E This equation is not very useful for performing derivations. It is typically used in numerical computations.

Note: We cannot simplify this further because, in general, the permeability is a function of position and cannot be brought outside of the curl operation.   1 EE  2 

Electromagnetic Waves & Polarization Slide 9

Wave Equation in LHI Media (1 of 2) In linear, homogeneous, and isotropic media two important simplifications can be made. First, in isotropic media the permeability and permittivity reduce to scalar quantities.   12 E E

Second, in homogeneous media  is a constant and can be brought to the outside of the curl operation and then brought to the right‐hand side of the equation.  E  2 E

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Wave Equation in LHI Media (2 of 2)  We now apply the vector identity A   AA 2  E 2 E   E 22EE   In LHI media, the divergence 22E EE  equation can be written in terms of E.   22EE 0 D 0   E 0   E 0  E 0

Electromagnetic Waves & Polarization Slide 11

Wave Number k and Propagation Constant  We can define the term as either  

k 22   or 22 k 2

This let’s us write the wave equation more simply as   22EkE 0 or 22EE 0

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Solution to the Wave Equation

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Components Decouple in LHI Media

We can expand our wave equation in Cartesian coordinates.  22EkE 0 22 2222 Eaxxˆˆˆˆˆˆ Ea yy Ea zz k Ea xx  k Ea yy  k Ea zz 0 22 22 22 EkEaxxxyyyzzzˆˆˆ  EkEa  EkEa 0  We see that the different field components have 22 decoupled from each other. EkExx 0 All three equations have the same numerical form so 22EkE 0 they all have the same solution. yy 22 Therefore, we only need the solution to one of them. EkEzz 0 22EkE 0

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General Solution to Scalar Wave Equation

Our final wave equation for LHI media is 22EkE 0 This could be handed off to a mathematician to obtain the following general solution.   Er Eejkr  Ee  jkr   00   forward wave backward wave

Electromagnetic Waves & Polarization Slide 15

General Solution to Vector Wave Equation Given the solution to scalar wave equation, we can write solutions for all three field components.   jkr   jkr  Erxx Ee Ee x   jkr   jkr  Eryy Ee Ee y   jkr   jkr  Erzz Ee Ee z We can assemble these three equations into a single vector equation.     E r Ex  raˆˆˆxy E  ra yz E  ra z  Eejk  r Ee  jk  r 00   forward wave backward wave Electromagnetic Waves & Polarization Slide 16

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General Expression for a Plane Wave

The solution to the wave equation gave us two plane waves. From the forward wave, we can extract a general expression for plane waves.    Er Pejk r Frequency‐domain  E rt,cos P t k r Time‐domain

We define the various parameters as  rxayazaˆˆˆposition  xyz k  wave vector  E  total electric field intensity 2f angular frequency  P  polarization vector t  time

Electromagnetic Waves & Polarization Slide 17

Magnetic Field Component Given that the electric field component of a plane wave is written as    Er Pejkr The magnetic field component is derived by substituting this solution into Faraday’s law.  E jH    1    PejHjk r    HkPe jkr   

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Solution in Terms of the Propagation Constant  The wave equation and it solution in terms of  is   22EkE 0  Er Eerr Ee   00 forward wave backward wave The general expressions for a plane wave are    Er Pe r Frequency‐domain  E rt,cos P  t r  Time‐domain

The magnetic field component is

  The wave vector and propagation 1   r constant are related through HPe   j   jk

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Visualization of an EM Wave (1 of 2)

We tend to draw and think of electromagnetic waves this way…

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Visualization of an EM Wave (2 of 2)

However, this is a more realistic visualization. It is important to remember that plane waves are also of infinite extent in all directions.

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Intuitive Wave Parameters

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Fundamental Vs. Intuitive Parameters

Fundamental Parameters Intuitive Parameters

These parameters are fundamental to These parameters collect specific solving Maxwell’s equations, but it is information about a wave from the difficult to specify how they affect a fundamental parameters. wave. This is because all of they all affect all properties of a wave.

Refractive index, n Magnetic Permeability,  Impedance,  Electric Permittivity,  Wavelength,  v Electrical Conductivity,  Velocity, Wave Number, k Propagation Constant,  Attenuation Coefficient,  Phase Constant, 

Electromagnetic Waves & Polarization Slide 23

Wave Velocity, v The scalar wave equation has been known since the 1700’s to be 2   wave disturbance 2   0   angular frequency v v  wave velocity If we compare our electromagnetic wave equation to the historical wave equation, we can derive an expression for wave velocity.  22EE 0  1 2    v 2  v   0 v

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Speed of Light in Vacuum, c0

In a vacuum,  = 0 and  = 0 and the velocity becomes the speed of light in a vacuum. 11 vc 0 299,792,458 m s  00 

When not in a vacuum,  = 0r and  = 0r and the velocity is reduced by a factor n called the refractive index. 11 11c v   0  00 rr    00   rr  n

n  rr

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Frequency is Constant, Wavelength Changes

Frequency is the most When a wave enters a different fundamental constant about a material, its speed and thus its wave. It never changes in linear wavelength change. materials. c  v  0   0 n n

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Speed, Frequency & Wavelength The speed of a wave, its frequency, and its wavelength are related through vf 

We are now in a position to derive an expression for wavelength. 112 f   f

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Wavelength  & Wave Number k Recall that we defined the wave number as 1 kn0r0r    00  rr  c0 The angular frequency  is related to wavelength  through the ordinary frequency f. cc 22f 00  2  0 n Substituting this into the first equation gives

112c0  2 kn 2   n k  cnc00  

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 Wave Vector, k  The wave vector conveys two pk ieces of information: (1) Magnitude conveys the wavelength  inside the medium, and (2) direction conveys the direction of the wave and is perpendicular to the wave fronts.

 kkakakax ˆˆˆxyyzz

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Magnitude Conveys Wavelength

Most fundamentally, the magnitude of the wave vector conveys the wavelength of the wave inside of the medium.

1 2  2  2 k1  k2  1 2

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Magnitude May Convey Refractive Index

When the frequency of a wave is known, the magnitude of the wave vector conveys refractive index.

1 2

 2 n  2 n 2 kkn1 kkn2 k  101 2020  0 0 0

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Material Impedance,  (1 of 3) Impedance is defined as the relationship between the amplitudes of E and H.

  EH00 Recall the relationship between E and H.   1   Er Pejkr and H k Pe  jkr   We can derive an expression for impedance by collecting all of the amplitude terms together in our expression for H.  1 Ek HkkEPekPeˆˆˆˆjkr 0  jkr 0  

This term is the amplitude of H

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Material Impedance,  (2 of 3) From the last slide, the amplitude of H is  Ek  Ek HkPeH00ˆ ˆ jk r   0

Dividing both sides of our expression by E0 gives an expression for impedance . E   0 Hk0 Since , the k    final expression for impedance is        k    

Electromagnetic Waves & Polarization Slide 33

Material Impedance,  (3 of 3) We can now revise our expression for the electric and magnetic field components of a wave as    kPˆ Er Pejkr and H e  jkr  where

Vacuum Impedance E0      0 376.73011346177  H  0 0 0

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Dispersion Relation

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Derivation in LHI Media

We started with the wave equation.  22EkE 0 We found the solution to be plane waves.    Er   Pejkr If we substitute our solution back into the wave equation, we get an equation called the dispersion relation.

2 2n 2 222 k knkkk0 x yz c0

The dispersion relation relates frequency to wave vector. For LHI media, it fixes the magnitude of the wave vector to be a constant.

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Index Ellipsoids

From the previous slide, the dispersion relation for a LHI material was: 222 22 kkkknxyz0 This defines a sphere called an “index ellipsoid.” The vector connecting the origin to a point on aˆ the surface of the sphere is the k‐vector for that z direction. Refractive index is calculated from this.  index ellipsoid kkn 0

For LHI materials, the refractive index is the same in all directions. Think of this as a map of the refractive aˆ index as a function of the wave’s direction y through the medium. aˆx

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What About Anisotropic Materials?

Isotropic Materials 222 22 kkkknabc0

Uniaxial Materials

222 22 2 kkkabc22 kk ab  k c 222kk000 nnnOEO

Biaxial Materials 222 kkkabc 2221 22 22 22 kknkknkkn000abc

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Electromagnetic Wave Polarization

What is Polarization?

Polarization is that property of a radiated electromagnetic wave which describes the time‐varying direction and relative magnitude of the electric field vector.

Linear Polarization (LP) Circular Polarization (CP)

Left‐Hand Circular Polarization (LCP)

To determine the handedness of CP, imagine watching the electric field in a plane while the wave is coming at you. Which way does it rotate?

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Orthogonality and Handedness We get from the curl equations that  EH From the divergence equations, we see that   Ek and Hk k    We conclude that , , and E H k  form an orthogonal triplet. E In fact, they follow the  right‐hand rule. H

Electromagnetic Waves & Polarization Slide 41

Possibilities for Wave Polarization    Recall that so the polE  k  arization vector must fP all within the plane perpendicular to .k We can decompose the polarization into two orthogonal directions, aˆ and .bˆ aˆ

 ˆ P paabˆ p b

 bˆ k

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Explicit Form to Convey Polarization Our electromagnetic wave can be now be written as   jk rˆ ˆ  jk r Er Pe paab pbe

pa and pb are in general complex numbers in order to convey the relative phase of each of these components.

jjab pEepEeaa bb

Substituting pa and pb into our wave expression gives    jjjkrj j   jkr Er Eeab aˆˆ Eebeˆˆ  Ea Eeba be a e  ab ab

We interpret b – a as the phase We interpret a as the phase difference between pa and pb. common to both pa and pb.

 ba   a     The final expression is: ˆ jˆ jjkr Er Eaab Eebee

Electromagnetic Waves & Polarization Slide 43

Determining Polarization of a Wave

To determine polarization, it is most convenient to write the expression for the wave that makes polarization explicity.

Ea  amplite along aˆ     E  amplite along bˆ Er Eaˆ Eebeejˆ jjkr b   ab   phase difference   common phase

We can now identify the polarization of the wave…

Polarization Designation Mathematical Definition Linear Polarization (LP)  = 0°

Circular Polarization (CP)  = ± 90°, Ea = Eb

Right‐Hand CP (RCP)  = + 90°, Ea = Eb

Left‐Hand CP (LCP)  = - 90°, Ea = Eb Elliptical Polarization Everything else

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Linear Polarization

A wave travelling in the +z direction is said to be linearly polarized if:   Exyz , ,  Pe jkz z P sin  xˆˆ cos  y P is called the polarization vector.

For an arbitrary wave, aˆ    Er Pejk r  Pabsinˆ cosˆ  abkˆ ˆ  k bˆ  k All components of P have equal phase.

Electromagnetic Waves & Polarization Slide 45

Linear Polarization

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Circular Polarization

A wave travelling in the +z direction is said to be circularly polarized if:   Exyz , ,  Pe jkz z P xˆˆ jy  P is called the polarization vector.

For an arbitrary wave,  j    Er Pejk r LCP  Pajbˆ ˆ  abkˆ ˆ  k  RCP  The two components of P have equal  j k amplitude and are 90 out of phase.

Electromagnetic Waves & Polarization Slide 47

LPx + LPy = LP45

A linearly polarized wave can always be decomposed as the sum of two linearly polarized waves that are in phase.

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LPx + jLPy = CP

A circularly polarized wave is the sum of two orthogonal linearly polarized waves that are 90° out of phase.

Electromagnetic Waves & Polarization Slide 49

RCP + LCP = LP

A LP wave can be expressed as the sum of a LCP wave and a RCP wave. The phase between the two CP waves determines the tilt of the LP wave polarization.

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Circular Polarization (1 of 2)

Engineering Right‐Hand Circular Polarization (RCP) x

y z

Physics/Optics Left‐Hand Circular Polarization (LCP)

Electromagnetic Waves & Polarization Slide 51

Circular Polarization (2 of 2)

Engineering Left‐Hand Circular Polarization (LCP) x

y z

Physics/Optics Right‐Hand Circular Polarization (RCP)

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Poincaré Sphere

The polarization of a wave can be mapped to a unique point on the Poincaré sphere. RCP Points on opposite sides of the sphere are orthogonal. ‐45° LP 90° LP See Balanis, Chap. 4.

0° LP +45° LP

LCP

Electromagnetic Waves & Polarization Slide 53

Why is Polarization Important? • Different polarizations can behave differently in a device • Orthogonal polarizations will not interfere with each other • Polarization becomes critical when analyzing devices on the scale of a wavelength • Focusing properties of lenses are different • Reflection/transmission can be different • Frequency of resonators • Cutoff conditions for filters, waveguides, etc.

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Example – Dissect a Wave (1 of 9)

The electric field component of a 5.6 GHz plane wave is given by:   jxjyjz573.0795 330.8676 240.8519 Ert, aˆx  0.4915 j 0.8550  e e e j573.0795xj 330.8676 yj 240.8519 z ajeeeˆy 1.4224  0.4702 jxjyjz573.0795 330.8676 240.8519 ajeeeˆz 0.7844 1.3885

1. Determine the wave vector. 2. Determine the wavelength inside of the medium. 3. Determine the free space wavelength. 4. Determine refractive index of the medium. 5. Determine the dielectric constant of the medium. 6. Determine the polarization of the wave. 7. Determine the magnitude of the wave.

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Example – Dissect a Wave (2 of 9)

Solution – Part 1 – Determine Wave Vector The standard form for a plane wave is    Er Pejk r Comparing this to the expression for the electric field shows that  Paˆˆxy0.4915 j 0.8550 a 1.4224 j 0.4702 a ˆ z 0.7844 j 1.3885   eejk r  j573.0795 x e  j 330.8676 y e  j 240.8519 z   The polarization vector will P be use again later. The wave vector is determined k from the second expression above to be

   jk y eeeeejk r jkx x y  jkz z  j573.0795 x e  j 330.8676 y e  j 240.8519 z  1 ka573.0795ˆˆˆxyz 330.8676 a 240.8519 a m

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Example – Dissect a Wave (3 of 9)

Solution – Part 2 – Wavelength inside the medium The wavelength inside the medium is related to the magnitude of the wave vector through  22  k     k

The magnitude of the wave vector is  222 kkkkxyz

222 573.0795 m111 330.8676 m 240.8519 m  704.239 m1

The wavelength is therefore 2  8.9224 cm 704.239 m1

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Example – Dissect a Wave (4 of 9)

Solution – Part 3 – Free space wavelength The free space wavelength is c 310 8 ms cf 0  53.5344 cm 00 0f 5.6 1091 s

Solution – Part 4 – Refractive index It follows that the refractive index of the medium is   53.5344 cm  00 n  6.0 n  8.9224 cm  Alternatively, we could determine the refractive index through k   81  kkck0 3 10 m s 704.239 m kkn n  6.0 0 91 kc002 f 2  5.6 10 s

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Example – Dissect a Wave (5 of 9)

Solution – Part 5 – Dielectric constant Assuming the medium has no magnetic response,

2 2 nnrr  6.0 36

Solution – Part 6 – Wave Polarization To determine the polarization, the electric field is written in the form that makes polarization explicit.     ˆ jjjkrˆ  Er Eaab Eebee  The choice for and is arbitrary, but they most both be aˆ bkˆ perpendicular to aˆ  ˆ P paabˆ p b

 bˆ k

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Example – Dissect a Wave (6 of 9) Solution – Part 6 – Wave polarization (cont’d) We determine a valid choice for by first picking any vecto aˆ r that is not in the same direction as k  va123ˆˆˆx ayz  a  The cross product will give us a vector perpendicular to k   kv aaaaˆˆˆˆ  0.2896x  0.8381yz  0.4622 kv ˆ We determine a valid choice for  using the cross product sob that it is perpendicular to both and aˆ k  ˆ ka ˆ baaa 0.5038ˆˆˆx  0.2771yz  0.8182 ka ˆ

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Example – Dissect a Wave (7 of 9) Solution – Part 6 – Wave polarization (cont’d)  To determine the component of the polarization vector in tP he and directions aˆ bˆ using the dot product.  pPaa ˆ 1.6971 V m  ˆ pPbjb 1.6971 V m

We can now write Ea and Eb from pa and pb by incorporating the phase difference into the parameter .

Ea 1.6971 V m

Eb 1.6971 V m  90 

The common phase between pa and pb is simply 0°.  0

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Example – Dissect a Wave (8 of 9) Solution – Part 6 – Wave polarization (cont’d) Finally, we have     ˆ jjjkrˆ  Er Eaab Eebee

Ea 1.6971 V m

Eb 1.6971 V m  90   0  1 ka573.0795ˆˆˆxyz 330.8676 a 240.8519 a m

From this, we determine that we have circular polarization (CP) because Ea = Eb and  = ±90°.

More specifically, this is left‐hand circular polarization (LCP) because  = -90°.

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Example – Dissect a Wave (9 of 9) Solution – Part 7 – Magnitude of electric field The magnitude of the wave is simply the magnitude of the polarization vector  Er P

22 EEab 1.6971 V m22 1.6971 V m  2.4 V m

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Visualization of EM Waves

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Waves in Materials (1 of 3)

Waves in Vacuum • H is 377× smaller than E.

E0 0 376.73  H0 • E and H are in phase

Im  0

• E  H  HkP

• Amplitude does not decay

  0

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Waves in Materials (2 of 3)

Waves in Dielectric • H is larger now, but still smaller than E. 1   

• E and H are still in phase

Im  0

• E  H  HkP

• Amplitude still does not decay   0

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More Realistic Wave (E Only)

It is important to remember that plane waves are of infinite extent in the x and y directions.

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More Realistic Wave (E & H)

It is important to remember that plane waves are of infinite extent in the x and y directions.

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