Plane EM waves - San Jose State UniversityElectromagnetic Waves EE142 Dr. Ray Kwok •reference:...

PlaneElectromagnetic Waves

EE142Dr. 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

t

EH

0Ht

HE

0E

∂

∂ε=×∇

=⋅∇∂

∂µ−=×∇

=⋅∇

rr

r

rr

r

t

EJH

0Ht

HE

E

f

f

∂

∂ε+=×∇

=⋅∇∂

∂µ−=×∇

ρ=⋅∇ε

rrr

r

rr

r

in source-free medium

(homogeneous,

linear, isotropic)

ρ = 0, J = 0


Wave Equation

( ) ( )

( ) ( )

2

22

22

2

2

t

EE

EEEE

t

EH

tt

HE

t

EH

t

HE

∂

∂µε=∇

−∇=∇−⋅∇∇=×∇×∇

∂

∂µε−=×∇

∂

∂µ−=

∂

∂µ−×∇=×∇×∇

∂

∂ε=×∇

∂

∂µ−=×∇

rr

rrrr

rr

rr

rr

rr

plane wave equation with v2 = 1/µε


Traveling Wave

2

2

22

2

2

2

2

2

2

t

f

v

1

x

f

)u("fvt

u)u("vf

t

f

)u('vft

u)u('f

t

f

)u("fx

u)u("f

x

f

)u('fx

u)u('f

x

f

)u(f)vtx(f

∂

∂=

∂

∂

=∂

∂±=

∂

∂

±=∂

∂=

∂

∂

=∂

∂=

∂

∂

=∂

∂=

∂

∂

≡± reverse / forward traveling wave

wave equation

( )

( )

( )

( )rktf

kxtf)vtx(f

kxtk

1

ft2kxk

1

vt2

x2

k

1vtx

rr⋅−ω

±ω=±

±ω±=

π±=

λ

π±

λ

π=±

note:

(3D)


Plane Wave

µε==

ω

µεω=

µεω−=−

ω−=∂

∂

−=∇

=

∂

∂µε=∇

⋅−ω

1v

k

k

k

Et

E

EkE

eE)t,r(E

t

EE

22

2

2

2

22

)rkt(j

o

2

22

rr

rr

rrr

rr

rr

)rkt(j

o

2

22

eH)t,r(H

t

HH

rrrrr

rr

⋅−ω=

∂

∂µε=∇

Similarly for B or H field


Source-free EM wave)kzt(j

oeEx)t,r(E −ω=rr

)kzt(jo

)kzt(j

o

)kzt(j

o

ekE

y)t,r(H

ejkEyE

HjE

00eE

zyx

zyx

E

−ω

−ω

−ω

ωµ=

−=×∇

ωµ−=×∇

∂

∂

∂

∂

∂

∂=×∇

rr

r

rr

r

V/m in vacuum with no current source.

What is H ?

)kzt(jo

)kzt(jo

)kzt(jo

)kzt(jo

)kzt(j

o

ev

Ey)t,r(B

ev

EyHB

eE

y)t,r(H

1

v

1

f

1k

ekE

y)t,r(H

ejkEyE

−ω

−ω

−ω

−ω

−ω

=

µ

µ=µ=

η=

η≡

µ

ε=

µ

µε=

µ=

λµ=

ωµ

ωµ=

−=×∇

rr

rr

rr

rr

r

ηηηη = wave

Impedance


EM wave properties

)kzt(jo

)kzt(jo

)kzt(j

o

ev

Ey)t,r(B

eE

y)t,r(H

eEx)t,r(E

−ω

−ω

−ω

=

η=

=

rr

rr

rr

HEk

kHE

×=

⊥⊥rrr

(1) E & H are in phase.

ε

µ=η=

µε===

o

o

rro

o

H

E

c

n

cv

B

E

(2)

(3)

Index of refraction


Electromagnetic wave


Earlier exercise

)zt900cos(01.0x)t,r(H β+=rr

A/m in vacuum with no current source.

What is f, λ, direction of propagation, E ?

ω = 2πf = 900, f = 143 Hz

λ = c/f = (3 x 108)/(143) = 2094 km

propagate to the –z direction

EjHrr

ωε=×∇

E(r) = ay 3.77 cos[900t + 3(10-6)z] V/m

HEk

kHE

×=

⊥⊥rrr


η=

=

o

o

o

o

H

E

vB

E

(2)

(3)

ηo = 377 Ω


Example)1.0zt10sin()02.0y01.0x()t,r(H 9 −β+−=

rrA/m. What is E ?

β = ω/c = 109/(3 x 108) = 3.33 rad/m


HEk

kHE

×=

⊥⊥rrr

(2)

η=

=

o

o

o

o

H

E

vB

E(3)

ηo = 377 Ω

)1.0z33.3t10sin(E)t,r(E9

o −+=rrr

V/m

y447.0x894.05

yx2E +=

+=x

y

H

E

Ho2 = 0.012 + 0.022 → Ho = 0.022 = Eo/ηo = Eo/377

Eo = 8.3 V/m

)1.0z33.3t10sin()7.3y4.7x()t,r(E

)1.0z33.3t10sin()447.0y894.0x(3.8)t,r(E

9

9

−++=

−++=rr

rr

V/m


Exercise)yz3t10cos(20x)t,r(E 8 −+=

rrV/m in non-magnetic material.

What is H ?

v = ω/k = 108/√10 = 3.16 x 107 m/s

v = c/n = c/√εr → εr = (c/v)2 = 90

η = √(µ/ε) = ηo/√εr = 377/√90 = 39.7


HEk

kHE

×=

⊥⊥rrr

(2)

)yz3t10cos(H)t,r(H 8

o −+=rrr

A/m

Ho = Eo/η = 20/39.7 =0.5

)yz3t10cos()158.0z475.0y()t,r(H

)yz3t10cos()316.0z949.0y(5.0)t,r(H

8

8

−++−=

−+−−=rr

rr

A/m

z316.0y949.010

zy3H −−=

−−=

y

z

k

E

H

η=

=

o

o

o

o

H

E

vB

E(3)

ηo = 377 Ω

yz3k +−=r


EM wave properties - generalized

HEk

kHE

×=

⊥⊥rrr


ε

µ=η=

µε===

o

o

rro

o

H

E

c

n

cv

B

E

(2)

(3)

Index of refraction

Ek1

Hrr

×η

=

kHE ×η=rr


Earlier exercise (free space)

)1.0zt10sin()02.0y01.0x()t,r(H 9 −β+−=rr

A/m. What is E ?

β = ω/c = 109/(3 x 108) = 3.33 rad/m

)1.0z33.3t10sin()54.7x77.3y(E

)z()1.0ztsin()02.0y01.0x(377E

kHE

9 −++=

−×−β+ω−=

×η=

r

r

rr

V/m.


Earlier exercise)yz3t10cos(20x)t,r(E 8 −+=

rrV/m in non-magnetic material.

What is H ?

A/m

yz3k +−=r

( )[ ]

( ) ( )yz3t10cos159.0z478.0yH

yz3t10cos20x10

yz3

7.39

1H

Ek1

H

8

8

−+−−=

−+×

+−=

×η

=

r

r

rr

v = ω/k = 108/√10 = 3.16 x 107 m/s

v = c/n = c/√εr → εr = (c/v)2 = 90

η = √(µ/ε) = ηo/√εr = 377/√90 = 39.7


Poynting Theorem

( )

( )

( )

( )HEt

HHHE

HEEHHE

t

EEHEJE

t

EJH

dVJEdt

dW

dtJEVdtvEVdW

dtvEqdW

dtvBvEqdFdW

f

rrr

rrr

rrrrrr

rrrrrr

rrr

rr

rrrr

rr

rrrrlrr

×⋅∇−

∂

∂µ−⋅=×∇⋅

×∇⋅−×∇⋅=×⋅∇

∂

∂⋅ε−×∇⋅=⋅

∂

∂ε+=×∇

⋅=

⋅δ=⋅ρδ=

⋅δ=

⋅×+δ=⋅=

∫

Work done on a charge δδδδq by EM force:

( )

( )

( )

( )

adSt

U

dt

dW

adHEdVt

E

t

H

2

1

dt

dW

dVHEdVt

E

t

H

2

1

dt

dW

HEt

E

2t

H

2JE

t

EEHE

t

HHJE

EM

2

o

2

22

22

rr

rrrrr

rrrr

rrrr

rr

rrrr

rrrr

∫

∫∫

∫∫

⋅−∂

∂−≡

⋅×−

∂

∂ε+

∂

∂µ−=

×⋅∇−

∂

∂ε+

∂

∂µ−=

×⋅∇−∂

∂ε−

∂

∂µ−=⋅

∂

∂⋅ε−×⋅∇−

∂

∂⋅µ−=⋅

(math)

Poynting Vector

Power flow out of the enclosed surface

EM power stored in the volume

The decrease of the EM energy stored is to do work on

a charge or due to the net energy escape through the

surface.

the work-energy theorem

for electrodynamics


Poynting Vector

HESrrr

×≡

Magnitude gives power intensity = power/area

Direction gives direction of propagation

2

o

2

o

ave

ave

H22

ES

*]HE[e2

1S

η=

η=

×ℜ≡rrr

This is similar to P = ½ (V2/Z) = ½ (I2Z)

Time average


ExampleIf solar illumination is characterized by a power density of 1 kW/m2 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

sun, Rs, is approximately 1.5 x 108 km, and Earth’s mean radius RE is 6380 km.

(a) Average power radiated Psun = Save(4πRs2) = (103)(4π)(1.5 x 1011)2 = 2.8 x 1026 W

(b) Average power intercepted PE = Save(πRE2) = (103)(π)(6.38 x 106)2 = 1.28 x 1017 W

(c) Electric field intensity …

870)10)(377(2S2E

2

ES

3

aveoo

2

o

ave

==η=

η=

V/m


Linear Polarization

)ztcos(Ex)t,r(E 1o1 β−ω=rr

x

y Defined by the direction of E-field. For example:

Horizontal polarization

)ztcos(Ey)t,r(E 2o2 β−ω=rr

Vertical polarization

x

y

21TOTAL EEErrr

+=ETOTAL is still linearly polarized

If Eo1 = Eo2, then ETOTAL is at 45 degrees.

A linear combination of these….

E1

E2

Any EM wave can be expressed as a linear combination ofhorizontal & vertical polarized field.


Circular Polarization

)ztcos(Ex)t,r(E 1o1 β−ω=rr

21T EEErrr

+=

x

y

If E1 & E2 are 90 degrees out-of-phase, & Eo1 = Eo2:

Horizontal polarization

)ztsin(Ey)t,r(E 2o2 β−ω=rr

Vertical polarization

ET(t=0)

ET(t>0)RHCP

Right-Hand Circularly Polarized

)zt(j

o

)2/zt(j

o

)zt(j

o

oo

eE)yjx()t,r(E

eEyeEx)t,r(E

)ztsin(Ey)ztcos(Ex)t,r(E

β−ω

π−β−ωβ−ω

−=

+=

β−ω+β−ω=

rr

rr

rr


LHCP

x

y

ET(t=0)

ET(t>0)

)zt(j

o

)2/zt(j

o

)zt(j

o

oo

eE)yjx()t,r(E

eEyeEx)t,r(E


β−ω

π+β−ωβ−ω

+=

+=

β−ω−β−ω=

rr

rr

rr

LHCP


Circular polarization in nature

Rose Chafer (a beetle)& others…

Reflection from its surface is LHCP.,due to the helical molecular structureof their shells.


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)


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.


Polarizer

θunpolarized

Io

polarized

I1 = ½ Io

polarized

I2 = I1cos2θ = ½ Iocos2θ

complex n

Linearly polarized

intensity


Elliptical Polarization

)zt(j

2o1o

2o1o

e)EyjEx()t,r(E


β−ω=

β−ω±β−ω=

mrr

rrIf Eo1 ≠ Eo2

Or, if the phase difference is not 90 degrees

RHEP

LHEP

)zt(j

o

j

)zt(j

o

)zt(j

o

eE)eyx()t,r(E

eEyeEx)t,r(E

β−ωδ

δ+β−ωβ−ω

+=

+=rr

rr

RHEP (δ < 0)

LHEP (δ > 0)

Special cases:

if δ = −π/2, → RHCP

if δ = π/2, → LHCP

x

y

ET(t=0)

LHEP

ET(t>0)

E02

E01


Polarization Generalization)zt(j

y

)zt(j

x eayeax)t,r(E δ+β−ωβ−ω +=rr

0

0

cos)2tan(2tan

sin)2sin(2sin

a

atan

R

1tan

a

aR

o

o

x

y

o

<γ

>γ

δψ=γ

δψ=χ

≡ψ

±≡χ

≡η

ξaxial ratio ≥ 1= 1 circular

= ∞ linear

ellipticity angle

χ > 0 when sinδ > 0 → LHχ < 0 when sinδ < 0 → RH

auxiliary angle

0 ≤ ψο ≤ π/2

−π/4 ≤ χ ≤ π/4

rotation angle −π/2 ≤ γ ≤ π/2

if cosδ > 0

if cosδ < 0


Polarization States


e.g. what’s the polarization?

( )( ) )6/kzt(j105j

)6/kzt(j12/7j

)4/3kzt(j)6/kzt(j

)2/4/kzt(j)6/kzt(j

)2/4/kzt(j)6/kzt(j

ee4y3x)t,z(E

ee4y3x)t,z(E

e4ye3x)t,z(E

e4ye3x)t,z(E

e4ye3x)t,z(E

o π+−ω

π+−ωπ

π+−ωπ+−ω

π+π−π+−ωπ+−ω

π−π+−ωπ+−ω

+=

+=

+=

+=

−=

r

r

r

r

r)4/kztsin(4y)6/kztcos(3x)t,z(E π+−ω−π+−ω=

r

o

oo

o

o

oo

o

o

o

x

y

o

69

)105cos()106tan(cos)2tan(2tan

34

)105sin()106sin(sin)2sin(2sin

53

3

4

a

atan

−=γ

=δψ=γ

=χ

=δψ=χ

=ψ

=≡ψ00256.0)105cos(cos

LH00966.0)105sin(sin

105

o

o

o

<γ⇒<−==δ

⇒>χ⇒>==δ

=δ

x

y

ay=4

ax=3

χχχχ

γ γ γ γ = −−−− 69o

ψψψψo

LHEP


Linear combination

)ztcos(Ex)t,r(E o β−ω=rr

Any LP EM wave can be written as linear combination of RHCP + LHCP

( ) ( )[ ]

2

EEE

eEEyjEEx)t,r(E

)t,r(E)t,r(E)t,r(E

eE)yjx()t,r(E

eE)yjx()t,r(E

o12

)zt(j

1221T

LHCPRHCPT

)zt(j

2LHCP

)zt(j

1RHCP

==

−++=

+=

+=

−=

β−ω

β−ω

β−ω

rr

rrrrrr

rr

rr

e.g.

More generally, any EM wave can be written as linear combination of RHEP + LHEP


ExerciseThe 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 m2.

η = 60π = 188.5 ΩSave = 9.43 mW/m2 along the x-axis

Pave = 0.19 W

Plane EM waves - San Jose State UniversityElectromagnetic Waves EE142 Dr. Ray Kwok •reference:...

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Transcript of Plane EM waves - San Jose State UniversityElectromagnetic Waves EE142 Dr. Ray Kwok •reference:...