Plane Electromagnetic Waves

EE142 Dr. Ray Kwok

•reference: Fundamentals of Engineering Electromagnetics , David K. Cheng (Addison-Wesley) Electromagnetics for Engineers, Fawwaz T. Ulaby (Prentice Hall) Plane EM Wave - Dr. Ray Kwok

Source-free Maxwell Equations r r ε∇ ⋅E = ρf r in source-free medium ∇ ⋅ E = 0 r r ∂H (homogeneous, r ∂H ∇× E = −µ linear, isotropic) ∇× E = −µ r ∂t ∂t ρ = 0, J = 0 r ∇ ⋅H = 0 r ∇ ⋅ H = 0 r r r ∂E r ∇× H = J + ε ∂E f ∇× H = ε ∂t ∂t Plane EM Wave - Dr. Ray Kwok

Wave Equationr r ∂H ∇ × E = −µ r∂t r ∂E ∇ × H = ε ∂t r r r   r 2  ∂H  ∂ ∂ E ∇ × ()∇ × E = ∇×− µ  = −µ ()∇× H = −µε 2  ∂t  ∂t ∂t r r r r ∇ × ()()∇ × E = ∇ ∇ ⋅ E − ∇ 2E = −∇ 2E r r 2 2 ∂ E ∇ E = µε 2 ∂t 2 plane wave equation with v = 1/ µε Plane EM Wave - Dr. Ray Kwok

Traveling Wave x(f ± vt ) ≡ )u(f reverse / forward traveling wave ∂f ∂u = )u('f = )u('f ∂x ∂x ∂ 2f ∂u note: = )u("f = )u("f 1  2π 2π  ∂x 2 ∂x x ± vt =  x ± vt  k  λ λ  ∂f ∂u 1 = )u('f = ±vf )u(' = ()kx ± 2πft ∂t ∂t k ∂ 2f ∂u 1 = ±vf )u(" = v2 )u("f = ± ()ωt ± kx ∂t 2 ∂t k 2 2 x(f ± vt ) = f ()ωt ± kx ∂ f 1 ∂ f r r = wave equation f ()ωt − k ⋅ r ∂x 2 v2 ∂t 2 (3D) Plane EM Wave - Dr. Ray Kwok

Plane Wave r r 2 2 ∂ E ∇ E = µε Similarly for B or H field ∂t 2 r r r r r r (j ωt−k⋅ )r r 2 )t,r(E = Eoe 2 ∂ H r r ∇ H = µε 2 2 2 ∂t ∇ E = −k E r r r r r r H )t,r( = H e (j ωt−k⋅ )r ∂ 2E r o = −ω2E ∂t 2 − k 2 = −ω2µε k = ω µε ω 1 = v = k µε Plane EM Wave - Dr. Ray Kwok

Source-free EM wave r r (j ωt−kz ) )t,r(E = xˆE e V/m in vacuum with no current source. o r What is H ? (j ωt−kz ) ∇× E = yˆ{− jkE oe } r r kE xˆ yˆ zˆ H )t,r( = yˆ o e (j ωt−kz ) r ∂ ∂ ∂ ωµ ∇ × E = k 1 1 µε ε 1 ∂x ∂y ∂z = = = = ≡ (j ωt−kz ) ωµ fλµ vµ µ µ η Eoe 0 0 r r r r ηηη = wave Eo (j ωt−kz ) ∇ × E = −jωµ H H )t,r( = yˆ e Impedance r η (j ωt −kz ) ∇ × E = yˆ{}− jkE oe r r µE B = µH = yˆ o e (j ωt−kz ) r r kE vµ H )t,r( = yˆ o e (j ωt−kz ) ωµ r r E )t,r(B = yˆ o e (j ωt−kz ) v Plane EM Wave - Dr. Ray Kwok

EM wave properties

r (1) E & H are in phase. r (j ωt−kz ) )t,r(E = xˆEoe r r E r r r H )t,r( = yˆ o e (j ωt−kz ) (2) E ⊥ H ⊥ k η kˆ = Eˆ × Hˆ r r E B )t,r( = yˆ o e (j ωt−kz ) v Index of refraction (3) E c c o = v = = Bo n εrµr E µ o = η = Ho ε Plane EM Wave - Dr. Ray Kwok

Electromagnetic wave Plane EM Wave - Dr. Ray Kwok

Earlier exercise r r H )t,r( = xˆ 01.0 cos( 900 t + β )z A/m in vacuum with no current source. What is f, λ, direction of propagation, E ?

ω = 2 πf = 900, f = 143 Hz (1) E & H are in phase. λ = c/f = (3 x 10 8)/(143) = 2094 km r r r propagate to the –z direction (2) E ⊥ H ⊥ k r r kˆ = Eˆ × Hˆ ∇× H = jωε E E E(r) = a 3.77 cos[900t + 3(10 -6)z] V/m o = v y (3) Bo E o = η Ho

ηo = 377 Ω Plane EM Wave - Dr. Ray Kwok

Example r r H )t,r( = (xˆ 01.0 − ˆ 02.0y )sin( 10 9 t +βz − )1.0 A/m. What is E ?

β = ω/c = 10 9/(3 x 10 8) = 3.33 rad/m r r r (1) E & H are in phase. 9 V/m )t,r(E = Eo sin(10 t + 33.3 z− )1.0 r r r y (2) E ⊥ H ⊥ k E ˆ ˆ ˆ 2xˆ + yˆ k = E× H x Eˆ = = .0 894 xˆ + .0 447 yˆ 5 H

Eo (3) = v 2 2 2 → Ho = 0.01 + 0.02 Ho = 0.022 = Eo/ηo = E o/377 Bo Eo = 8.3 V/m E o = η r r 9 Ho )t,r(E = (3.8 xˆ .0 894 + yˆ .0 447 )sin( 10 t + 33.3 z − )1.0 η = 377 Ω r r o )t,r(E = (xˆ 4.7 + yˆ )7.3 sin( 10 9 t + 33.3 z − )1.0 V/m Plane EM Wave - Dr. Ray Kwok

Exercise r r 8 )t,r(E = xˆ 20 cos( 10 t + z3 − )y V/m in non-magnetic material. What is H ? r r r (1) E & H are in phase. 8 H )t,r( = Ho cos( 10 t + z3 − )y A/m r r r y r (2) E ⊥ H ⊥ k k k = −3zˆ + yˆ ˆ ˆ ˆ E − 3yˆ − zˆ k = E× H z Hˆ = = − .0 949 yˆ − .0 316 zˆ 10 H v = ω/k = 10 8/√10 = 3.16 x 10 7 m/s Eo √ → 2 (3) = v v = c/n = c/ εr εr = (c/v) = 90 √ √ √ Bo η = (µ/ε ) = ηo/ εr = 377/ 90 = 39.7 E H = E /η = 20/39.7 =0.5 o = η o o Ho r r 8 η = 377 Ω H )t,r( = (5.0 −yˆ .0 949 − ˆ .0z 316 ) cos( 10 t + z3 − )y o r r H )t,r( = −(yˆ .0 475 + ˆ .0z 158 )cos( 10 8 t + z3 − )y A/m Plane EM Wave - Dr. Ray Kwok

EM wave properties - generalized

(1) E & H are in phase. r 1 r r r r H = kˆ × E (2) E ⊥ H ⊥ k η kˆ = Eˆ × Hˆ

Index of refraction r r (3) E c c E = ηH × kˆ o = v = = Bo n εrµr E µ o = η = Ho ε Plane EM Wave - Dr. Ray Kwok

Earlier exercise (free space) r r H )t,r( = (xˆ 01.0 − yˆ 02.0 )sin( 10 9 t + βz − )1.0 A/m. What is E ?

β = ω/c = 10 9/(3 x 10 8) = 3.33 rad/m

r r E = ηH× kˆ r E = 377 (xˆ 01.0 − yˆ 02.0 )sin( ωt + βz − )1.0 ×(−ˆ)z r 9 E = (yˆ 77.3 + xˆ 54.7 )sin( 10 t + 33.3 z − )1.0 V/m. Plane EM Wave - Dr. Ray Kwok

Earlier exercise r r 8 )t,r(E = xˆ 20 cos( 10 t + z3 − )y V/m in non-magnetic material. r What is H ? k = −3zˆ + yˆ v = ω/k = 10 8/√10 = 3.16 x 10 7 m/s √ → 2 v = c/n = c/ εr εr = (c/v) = 90 √ √ √ η = (µ/ε ) = ηo/ εr = 377/ 90 = 39.7 r 1 r H = kˆ × E η r 1  − 3zˆ + yˆ  H =  ×[]xˆ 20 cos ()10 8 t + z3 − y 39 7.  10  r H = ()− yˆ .0 478 − ˆ .0z 159 cos ()10 8 t + z3 − y A/m Plane EM Wave - Dr. Ray Kwok

the work-energy theorem Poynting Theorem for electrodynamics

Work done on a charge δδδq by EM force: r r r r r ∂H r r r ∂E E ⋅ J = −µH ⋅ − ∇ ⋅ ()E × H − εE ⋅ r r r r r r ∂t ∂t dW = F ⋅ dl = δq(E + v × B)⋅ dtv r r r r r µ ∂H2 ε ∂E2 r r r E ⋅ J = − − − ∇ ⋅ ()E × H dW = δqE ⋅ vdt 2 ∂t 2 ∂t r r r r r r dW = ()ρδ V E ⋅ dtv = δVE ⋅ dtJ  2 2  r r dW 1  ∂H ∂E  = −∫ µ + ε dV − ∫∇ ⋅ ()E × H dV dW r r dt 2  ∂t ∂t  = ∫ E ⋅ dVJ r r dt  2 2  r r r r dW 1  ∂H ∂E  r r = −∫ µ + εo dV − ∫()E × H ⋅ ad ∂E dt 2  ∂t ∂t  ∇ × H = Jf + ε ∂t r dW ∂U r r r r r r r ≡ − EM − S⋅ ad ∂E dt ∂t ∫ E ⋅ J = E ⋅∇ × H − εE ⋅ Poynting Vector ∂t r r r r r r (math) Power flow out of the enclosed surface ∇ ⋅ ()E × H = H ⋅∇ × E − E ⋅∇ × H EM power stored in the volume r r r r  ∂H  r r E ⋅ ∇ × H = H ⋅− µ  − ∇ ⋅ ()E × H The decrease of the EM energy stored is to do work on  ∂t  a charge or due to the net energy escape through the   surface. Plane EM Wave - Dr. Ray Kwok

Poynting Vector r r r S ≡ E × H

Magnitude gives power intensity = power/area Direction gives direction of propagation r 1 r r S ≡ ℜ E[e × H*] Time average ave 2 2 E η 2 S = o = H ave 2η 2 o

This is similar to P = ½ (V 2/Z) = ½ ( I2Z) Plane EM Wave - Dr. Ray Kwok Example If solar illumination is characterized by a power density of 1 kW/m 2 at Earth’s surface, find (a) the total power radiated by the sun, (b) the total power intercepted by Earth, and (c) the electric field of the power density incident upon Earth’s surface, assuming that all the solar illumination is at a single frequency. The radius of Earth’s orbit around the 8 sun, Rs, is approximately 1.5 x 10 km, and Earth’s mean radius R E is 6380 km.

2 3 11 2 26 (a) Average power radiated Psun = S ave (4 πRs ) = (10 )(4 π)(1.5 x 10 ) = 2.8 x 10 W

2 3 6 2 17 (b) Average power intercepted P E = S ave (πRE ) = (10 )( π)(6.38 x 10 ) = 1.28 x 10 W

3 Eo = 2ηoSave = (2 377 )( 10 ) = 870 V/m Plane EM Wave - Dr. Ray Kwok

Linear Polarization

y Defined by the direction of E-field. For example: E 2 r r E1 )t,r( = xˆE 1o cos( ωt − β )z Horizontal polarization r r E1 x E2 )t,r( = yˆEo2 cos( ωt − β )z Vertical polarization

A linear combination of these…. y r r r

ETOTAL ETOTAL = E1 + E2 is still linearly polarized

x If E o1 = E o2 , then E TOTAL is at 45 degrees.

Any EM wave can be expressed as a linear combination of horizontal & vertical polarized field. Plane EM Wave - Dr. Ray Kwok

Circular Polarization

If E1 & E2 are 90 degrees out-of-phase, & E o1 = E o2 : r r E )t,r( = xˆE cos( ωt − β )z Horizontal polarization r 1 r 1o E )t,r( = yˆE sin( ωt − β )z Vertical polarization r 2 r r o2

ET = E1 + E2

RHCP E (t>0) T Right-Hand Circularly Polarized x r r )t,r(E = xÊo cos( ωt − β )z + yÊo sin( ωt − β )z E (t=0) r r T )t,r(E = xÊ e (j ωt −β )z + yÊ e (j ωt −βz−π )2/ r o o r (j ωt −β )z )t,r(E = (xˆ − jˆ E)y oe Plane EM Wave - Dr. Ray Kwok Plane EM Wave - Dr. Ray Kwok

LHCP r r )t,r(E = xÊo cos( ωt − β )z − yÊo sin( ωt − β )z r r )t,r(E = xÊ e (j ωt −β )z + yÊ e (j ωt −βz+π )2/ r o o r (j ωt−β )z )t,r(E = (xˆ + jˆ E)y oe y

ET(t=0) x

ET(t>0)

LHCP Plane EM Wave - Dr. Ray Kwok

Circular polarization in nature

Rose Chafer (a beetle) & others…

Reflection from its surface is LHCP., due to the helical molecular structure of their shells. Plane EM Wave - Dr. Ray Kwok Mantis Shrimp (Stomatopod Crustacean)

Able to sense LHCP & RHCP, over a wide spectrum of color, better than man-made instrument.

Most complex eye structure in animal kingdom.

Fastest animal (striking) & produces over 200 lbs of striking force !!!

Nature Photonics 3, 641-644 (2009) Roberts, Chiou, Marshall & Cronin,

Sheila Patek at TED.com

a popular dish known as "pissing shrimp" ( 攋尿蝦) Plane EM Wave - Dr. Ray Kwok

Circular polarization in starlight

Most starlight are slightly circularly polarized.

The interstellar medium ( low density dust ) can produce circularly polarized (CP) light from unpolarized light by sequential scattering from elongated interstellar grains (with complex index of refraction ) aligned in different directions. One possibility is twisted grain alignment along the line of sight due to variation in the galactic magnetic field; another is the line of sight passes through multiple clouds. Plane EM Wave - Dr. Ray Kwok

Polarizer

polarized θ polarized unpolarized

Io = ½ I1 Io = 2 = ½ 2 intensity I2 I1cos θ Iocos θ

Linearly polarized

complex n Plane EM Wave - Dr. Ray Kwok

Elliptical Polarization r r If E ≠ E o1 o2 )t,r(E = xÊ 1o cos( ωt − β )z ± yÊo2 sin( ωt − β )z r r )t,r(E = (xÊ m jyÊ e) (j ωt−β )z RHEP 1o o2 LHEP Or, if the phase difference is not 90 degrees r r (j ωt−β )z (j ωt −βz+δ) y )t,r(E = xÊoe + yÊoe r r jδ (j ωt−β )z E )t,r(E = (xˆ + êy E) oe 02 RHEP ( δ < 0) ET(t=0) x LHEP ( δ > 0) E01 ET(t>0) Special cases: if δ = −π/2, → RHCP if δ = π/2, → LHCP LHEP Plane EM Wave - Dr. Ray Kwok

Polarizationr Generalization r (j ωt −β )z (j ωt−βz+δ) )t,r(E = ˆax xe + ˆay ye axial ratio ≥ 1 a ξ R ≡ = 1 circular a η = ∞ linear 1 ellipticity angle tan χ ≡ ± χ > 0 when sin δ > 0 → LH R χ < 0 when sin δ < 0 → RH a y auxiliary angle tan ψo ≡ a x 0 ≤ ψο ≤ π/2

sin 2χ = sin( 2ψo )sin δ −π/4 ≤ χ ≤ π/4

tan 2γ = tan( 2ψo )cos δ rotation angle −π/2 ≤ γ ≤ π/2

γ > 0 if cos δ > 0 γ < 0 if cos δ < 0 Plane EM Wave - Dr. Ray Kwok

Polarization States Plane EM Wave - Dr. Ray Kwok e.g. what’s the polarization? r )t,z(E = xˆ3cos( ωt − kz + π )6/ − yˆ 4sin( ωt − kz + π )4/ r )t,z(E = xˆ e3 (j ωt−kz +π )6/ − yˆ e4 (j ωt−kz +π / 4−π )2/ r )t,z(E = xˆ e3 (j ωt−kz +π )6/ + yˆ e4 (j ωt −kz +π 4/ −π / 2+π) r )t,z(E = xˆ e3 (j ωt−kz +π )6/ + yˆ e4 (j ωt −kz +3π / )4 r )t,z(E = ()xˆ3 + yˆ e4 7j π 12/ e (j ωt−kz +π )6/ r δ = 105 o ˆ ˆ j105 o (j ωt−kz +π )6/ )t,z(E = (x3 + y e4 )e sin δ = sin( 105 o ) = .0 966 > 0 ⇒ χ > 0 ⇒ LH o a y 4 cos δ = cos( 105 ) = − .0 256 < 0 ⇒ γ < 0 tan ψo ≡ = y a x 3 o ay=4 ψo = 53 o o sin 2χ = sin( 2ψo )sin δ = sin( 106 )sin( 105 ) ψψψo ax=3 o χ = 34 χχχ x o o tan 2γ = tan( 2ψo )cos δ = tan( 106 )cos( 105 ) γ = −−− 69 o γ = −69 o LHEP Plane EM Wave - Dr. Ray Kwok

Linear combination Any LP EM wave can be written as linear combination of RHCP + LHCP r r e.g. )t,r(E = xˆEo cos( ωt − β )z r r E )t,r( = (xˆ − jˆ E)y e (j ωt −β )z r RHCP 1 r (j ωt −β )z E LHCP )t,r( = (xˆ + jˆ E)y 2e r r r r r r E )t,r( = E )t,r( + E )t,r( r T RHCP LHCP r (j ωt −β )z ET )t,r( = []xˆ ()()E1 + E 2 + jyˆ E2 − E1 e E E = E = o 2 1 2

More generally, any EM wave can be written as linear combination of RHEP + LHEP Plane EM Wave - Dr. Ray Kwok Exercise The amplitudes of an elliptically polarized plane wave traveling in a lossless,

nonmagnetic medium with εr = 4 and Hxo = 8 mA/m, and Hyo = 6 mA/m. Determine the average power flowing through an aperture in the y-z plane if its area is 20 m 2.

η = 60 π = 188.5 Ω 2 Save = 9.43 mW/m along the x-axis Pave = 0.19 W