CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM...

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EM Harmonic Wave 8-1 Proprietary of Prof. Lee, Yeon Ho CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVE Maxwell’s equations Differential wave equation Traveling wave A harmonic wave An arbitrary wave A linear combination of harmonic waves. Superposition principle . Linear media, ( ε , μ are constant) Harmonic waves A constant amplitude Time-invariant Sinusoidal steady-state 9 Time-varying E and H Harmonic waves in 3D-space. ( ) cos t ± ω k r i , ( ) cos t ω ± k r i Maxwell’s eqs. + phase change, phase lead - phase change, phase lag 9 Subjects of this chapter - Propagation harmonic waves - Harmonic waves on an interface (Boundary condition) - Introduction of phasor Time-invariant part of the complex form of a harmonic wave. 8-1 THE PHASOR In a medium of 0 v = ρ , 0 = J and constant ε and μ Maxwell’s equations t ∇× = −μ H E (8-1a) t ∇× E H (8-1b) 0 = i E (8-1c) 0 = i H (8-1d) 9 Combining two curl equations 2 2 2 t = με E E (8-2) Its solution ( ) , Re j j t o t e = kr r E i E Uniform plane wave (8-3) Harmonic wave

Transcript of CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM...

Page 1: CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM Waves_a.pdf · EM Harmonic Wave 8-3 Proprietary of Prof. Lee, Yeon Ho The right side

EM Harmonic Wave 8-1 Proprietary of Prof. Lee, Yeon Ho

CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVE Maxwell’s equations → Differential wave equation → Traveling wave A harmonic wave An arbitrary wave → A linear combination of harmonic waves. ↑ Superposition principle. Linear media, ( ε , μ are constant) Harmonic waves → A constant amplitude ↑ Time-invariant Sinusoidal steady-state

Time-varying E and H → Harmonic waves in 3D-space. ↑ ( )cos t± ωk ri , ( )cos tω ± k ri Maxwell’s eqs. ↑ + phase change, phase lead - phase change, phase lag

Subjects of this chapter - Propagation harmonic waves

- Harmonic waves on an interface (Boundary condition) - Introduction of phasor ↑ Time-invariant part of the complex form of a harmonic wave. 8-1 THE PHASOR In a medium of 0v =ρ , 0=J and constant ε and μ Maxwell’s equations

t

∂∇ × = −μ

∂H

E (8-1a)

t

∂∇ × = ε

∂E

H (8-1b)

0∇ =iE (8-1c) 0∇ =iH (8-1d)

Combining two curl equations

2

22t

∂∇ = με

∂E

E (8-2)

Its solution ( ), Re j j t

ot e − + ω⎡ ⎤= ⎣ ⎦k rr E iE

↑ ↑ ↑ Uniform plane wave (8-3) Harmonic wave

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EM Harmonic Wave 8-2 Proprietary of Prof. Lee, Yeon Ho

where oE , vector complex amplitude k= ω μεk a , wavevector (8-4) Eq. (8-2) → k , but no ka ↑ From source or boundary condition

In a medium of finite extent The same differential wave eq. → Uniform plane waves Boundary condition → Linear combination (Different 'sk ) A wave on a boundary → A partial reflection → Total wave = incident wave + reflected wave

( ) ( ), Re i rj ji r j to ot e e e− − ω⎡ ⎤= +⎣ ⎦

k r k rr E Ei iE (8-5)

↑ Solution of wave equation if i r= = ω μεk k . In general form ( ) ( ), Re , , j tt x y z e ω⎡ ⎤= ⎣ ⎦r EE (8-6) ↑ Phasor, time invariant. It assumes sinusoidal steady-state condition. It shows relative phases, or time delays, in space. It is useful in combining multiple harmonic waves. It is valid in linear media.

Differential wave eq. for H has the same form as E ( ) ( ), Re , , j tt x y z e ω⎡ ⎤= ⎣ ⎦r HH (8-7) ↑ Phasor of H Example 8-1

Resolve a harmonic wave ( ) 31cos jk z

y oE k x e −=E a into two uniform plane waves and sketch their wavefronts.

Solution Using Euler formula, we rewrite

31 112

jk zjk x jk xy oE e e e −−⎡ ⎤= +⎣ ⎦E a

( ) ( )1 3 1 3

2 2

j k x k z j k x k zo oy yE Ee e− − + − += +

′ ′′≡ +

E a a

E E

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The right side is the sum of two uniform plane waves of wavevectors 1 3x zk k′ = − +k a a and 1 3x zk k′′ = +k a a , respectively. Both waves have the same wave number

2 21 3k k′ ′′= = +k k

The black(or blue) parallel lines represent plane wavefronts, viewed from the top, of the wave of ′k (or ′′k ). They correspond to phases 2 nπ , where n are integers. The dots on x axis are the points at which wavefronts of the same phase meet, and therefore the total electric field intensity has the maximum amplitude oE . The amplitude varies sinusoidally along x axis as is indicated by the term ( )1cos k x .

8-2 MAXWELL’S EQUATIONS IN PHASOR FORM Inserting Eqs. (8-6) and (8-7) in Eq. (8-1a)

( ) ( )Re , , Re , ,j t j tx y z e x y z et

ω ω∂⎡ ⎤ ⎡ ⎤∇ × = −μ⎣ ⎦ ⎣ ⎦∂E H (8-8)

→ ( ) ( )Re , , Re , ,j t j te x y z e j x y zω ω⎡ ⎤ ⎡ ⎤∇ × = −μ ω⎣ ⎦ ⎣ ⎦E H ↑ ↑ Curl, time derivative and real part are mutually exclusive Ignore Re and ( )exp j tω Faraday’s law in phasor form j∇ × = − ωμE H

Maxwell’s equations in phasor form with constant ε and μ j∇ × = − ωμE H (8-9a) j∇ × = ωεH E (8-9b) 0∇ =Ei (8-9c) 0∇ =Hi (8-9d) E and H are time-invariant complex vectors

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Combine two curl equations

2 j∇ ×∇ × = − ωμ∇ ×

= ω με

E HE

(8-10)

Using a vector identity, 2∇ ×∇ × = ∇∇ − ∇E E Ei , and Eq. (8-9c), 2 2 0k∇ + =E E : Vector Helmholtz’s equation (8-11)

2kvω π

= ω με = =λ

: Wavenumber (8-12)

where

ω , angular frequency μ , permeability ε , permittivity v , velocity λ , wavelength.

8-3 PLANE WAVE IN LOSSLESS DIELECTRIC

In Cartesian coordinates, vector Helmholtz’s equation

( ) ( )2 2 0x x y y z z x x y y z zE E E k E E E∇ + + + + + =a a a a a a (8-13) ↑ ↑ ↑ Mutually exclusive

2 2 0x xE k E∇ + = (8-14a) 2 2 0y yE k E∇ + = (8-14b) 2 2 0z zE k E∇ + = (8-14c)

Rewriting Eq. (8-14a)

2 2 22

2 2 2 0x x xx

E E Ek E

x y z∂ ∂ ∂

+ + + =∂ ∂ ∂

(8-15)

Its solution

( )ˆ ˆx y zj k x k y k z jx ox oxE E e E e± + + ±= = k ri (8-16)

↑ ↑ ↑ 2 2 2

k k x y zk k k k≡ = + +k a a , wavevector (8-17) Scalar complex amplitude Similarly

ˆ jy oyE E e −= k ri (8-18a)

ˆ jz ozE E e −= k ri (8-18b)

Helmholtz’s equations in Eq. (8-14) → The same k in Eqs. (8-16), (8-18a) and (8-18b) But no restriction on ka

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The same ka → Three solutions into a single wave

( )ˆ ˆ ˆ jx x y y z z ox x oy y oz zE E E E E E e −= + + = + + k rE a a a a a a i

or j

oe−= k rE E i (8-19)

where

ˆ ˆ ˆ

ˆ o ox x oy y oz z

jo E o E

E E E

E E e ϕ

= + +

= ≡

E a a a

a a (8-20)

↑ ↑ Scalar complex amplitude Vector complex amplitude

oE , amplitude ϕ , phase angle

joe

−= k rE E i , a uniform plane wave in 3D space. k , wavevector propagation direction and wavelength

oE , vector complex amplitude direction and amplitude of E r , position vector −k ri , time-phase of the field

oE → ˆo EE a

↑ ˆ

oxE , ˆoyE and ˆ

ozE have the same ϕ phase angle A single wave in an infinite medium → 0ϕ =

Transverse wave Inserting Eq. (8-19) in Eq. (8-9c)

( ) ( ) 0j jo oe j e− −∇ = ∇ = − =k r k rE E k Ei ii i i

or 0o =k Ei (8-21)

↑ Propagation direction ⊥ Oscillation direction

oE is in the wavefront

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H from Eq. (8-9a)

ˆ ˆ ˆ ˆ ˆ ˆ

x y z x y z

jx y z o

j j j j j jox oy oz ox oy oz

jk jk jk j ex y z

E e E e E e E e E e E e

j

− − − − − −

∂ ∂ ∂∇ × = = − − − = − ×

∂ ∂ ∂

= − ωμ

k r

k r k r k r k r k r k r

a a a a a a

E k E

H

i

i i i i i i

(8-22)

or

( )1k= ×

ηH a E (8-23)

μ

η =ε

: Intrinsic impedance (8-24)

Rewriting (8-23)

( )1 j jk o oe e− −= × ≡

ηk r k rH a E Hi i (8-25)

↑ Same form as E Inserting ˆ

o o EE=E a

( )ˆ

ˆoo k E o H

EH= × ≡

ηH a a a (8-26)

ˆ ˆ/o oE Hη = , in ohms [ ]Ω [ ]/ 120 377o o oη = μ ε = π ≅ Ω Lossless dielectrics → η is real → E and H are in phase Lossy dielectrics → η is complex → E and H are out of phase

oH ⊥ ( oE and k ), from Eq. (8-25) oE ⊥ k , from Eq. (8-21)

→ oE , oH and k are mutually orthogonal Transverse wave!

Real instantaneous expression ( ) ( ), cosE ot E t= ω − + ϕr a k riE (8-27a)

( ) ( ), cosoHE

t t= ω − + ϕη

r a k riH (8-27b)

Unit vectors k E H= ×a a a (8-28)

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Example 8-2

A plane wave, ( )1 jkzx oE j e −= +E a , has a frequency of 2 GHz and propagates in a dielectric

of 4rε = and 1rμ = . Find (a) E and H (b) wavelength

Solution (a) Rewriting the phasor

/42 j jkzx oE e eπ −=E a (8-29)

The real instantaneous expression

/4Re 2 2 cos4

j jkz j tx o x oE e e e E t kzπ − ω π⎛ ⎞⎡ ⎤= = ω − +⎜ ⎟⎣ ⎦ ⎝ ⎠a aE

Inserting Eq. (8-29) in Eq. (8-23)

( ) ( ) /4

/4

1 1 2

1 2

j jkzk z x o

j jkzy o

E e e

E e e

π −

π −

= × = ×η η

H a E a a

a

where k z=a a . Intrinsic impedance

188.5[ ]o

o r

μη = = Ω

ε ε.

Real instantaneous expression

/41Re 2 2 cos4

j jkz j t oy o y

EE e e e t kzπ − ω⎡ ⎤ π⎛ ⎞= = ω − +⎜ ⎟⎢ ⎥η η ⎝ ⎠⎣ ⎦a aH

Page 8: CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM Waves_a.pdf · EM Harmonic Wave 8-3 Proprietary of Prof. Lee, Yeon Ho The right side

EM Harmonic Wave 8-8 Proprietary of Prof. Lee, Yeon Ho

(b) From Eq. (8-12)

9

8

2 2 10 23 101/

2

r

o o

kω ε π × × ×

= ω εμ = =×μ ε

π=

λ

or

3 7.540m cmλ = =

8-4 POYNTING VECTOR AND POWER FLOW

Energy densities in static fields → 12D Ei and 1

2B Hi

Energy is stored in E and H of an electromagnetic wave. ↑ ↑ Same velocity and direction as the wave. Power delivery

Two curl equations

t

∂∇ × = −

∂B

E (8-30a)

t

∂∇ × = +

∂D

H J (8-30b)

A vector identity

( ) ( ) ( )∇ × = ∇ × − ∇ ×i i iE H H E E H (8-31) Inserting Eq. (8-30) in Eq. (8-31)

( )t t

∂ ∂−∇ × = +

∂ ∂i i i iD B

E H E E HJ H (8-32)

Assuming ε , μ and σ are constant

( ) ( )12 2t t t

∂ ε ∂ ∂= =

∂ ∂ ∂i i iD

E E E D E (8-33a)

( ) ( )12 2t t t

∂ μ ∂ ∂= =

∂ ∂ ∂i i iB

H H H B H (8-33b)

From Eqs. (8-32) and (8-33)

( ) ( ) ( )1 12 2t t

∂ ∂−∇ × = +

∂ ∂i i i iE H E D E B HHJ : Poynting’s theorem (8-34)

Integrating both sides over a volume and applying divergence theorem

( ) ( ) ( )1 12 2

d d d dt t∂ ∂

− × = +∂ ∂∫ ∫ ∫ ∫s

S V V Vv v vi i i iE H E D E B HHJ (8-35)

↑ ↑ ↑ ↑ ↑ Ohmic power-loss in V ↑ Increased energy in V per unit time. ↑ Net power flowing into V . Total instantaneous power flowing into V . (Integrand = power density)

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Poynting vector is defined by = ×S E H : 2/W m⎡ ⎤⎣ ⎦ (8-36) ↑ ⊥S E and H .

S k

Time-average power density S~ 2E , a nonlinear function of E → E and H cannot be expressed by phasors in S Using complex exponential and its complex conjugate.

( ) ( ) ( )

( ) ( )*

1cos21 ˆ ˆ 21 2

e ej jj t j j t jE o e E o E o

j j t j j tE o E o

j jo

E t E e e E e e

E e e E e e

e e

ϕ − ϕω − − ω +

− ω + − ω

− ω

⎡ ⎤= ω − + ϕ = +⎣ ⎦

⎡ ⎤= +⎣ ⎦

=

k r k r

k r k r

k r

a k r a a

a a

E

i i

i i

i

iE

*

*1 2

t j j to

j t j t

e e

e e

+ − ω

ω − ω

⎡ ⎤+⎣ ⎦

⎡ ⎤≡ +⎣ ⎦

k rE

E E

i (8-37a)

( ) *1cos2

j t j tH o hH t e eω − ω⎡ ⎤= ω − + ϕ = +⎣ ⎦a k r H HiH (8-37b)

where E , phasor oE , vector complex amplitude ˆ

oE , scalar complex amplitude oE , amplitude eϕ , phase angle * , complex conjugate. Inserting Eq. (8-37) in Eq. (8-36)

* *

2 * * 2 * *

1 12 21 4

j t j t j t j t

j t j t

e e e e

e e

ω − ω ω − ω

ω − ω

⎡ ⎤ ⎡ ⎤= + × +⎣ ⎦ ⎣ ⎦

⎡ ⎤= × + × + × + ×⎣ ⎦

S E E H H

E H E H E H E H (8-38)

Time average of S is defined by

/2

/2

1 T

Tdt

T −= ∫S S : T , temporal period of the wave

Time-average Poynting vector

*1 Re2

⎡ ⎤= ×⎣ ⎦S E H (8-39)

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Example 8-3 Given an electromagnetic wave, ( ) ( ), cosx ot E t kz= ω −r aE and

( ) ( ), cosy ot E t kzε= ω −

μr aH , find S using

(a) b and H (b) E and H Solution (a) From Eq. (8-36)

( )2 2cosz oE t kzε= × = ω −

μS aE H

Using trigonometry

( )2 1 1 cos 2 22z oE t kzε

= + ω −⎡ ⎤⎣ ⎦μS a

After time-average

2

2o

zE ε

S a (8-40)

(b) Phasors of E and H jkz

x oE e−=E a

jkzy oE e −ε

H a

From Eq. (8-39)

( )

( ) ( )

*

*

2

1 1Re Re2 2

Re2

jkz jkzx o y o

jkz jkzox y

E e E e

Ee e

− −

⎡ ⎤⎛ ⎞ε⎢ ⎥⎡ ⎤= × = × ⎜ ⎟⎣ ⎦ ⎜ ⎟μ⎢ ⎥⎝ ⎠⎣ ⎦

ε ⎡ ⎤= ×⎣ ⎦μ

S E H a a

a a

Rewriting it

2

2o

zE ε

S a (8-41)

The result is the same as Eq. (8-40).

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8-5 POLARIZATON OF PLANE WAVE Phasor of a uniform plane wave

( )expo j= −E E k ri ↑ oE ⊥ k from Maxwell No limits on oE and Eo

a ˆ

oxE , ˆoyE and ˆ

ozE of oE In phase → E oscillates with time along a straight line Out of phase → E changes magnitude and direction with time Orientation of E is called the polarization H is just given by Eq. (8-23).

8-5.1 Linear Polarization

A plane wave in z+ -direction → E should be in xy -plane ( )ˆ ˆjkz jkz

o ox x oy ye E E e− −= = +E E a a (8-42)

↑ ↑ Scalar complex amplitudes.

If ˆox oxE E= and ˆ

oy oyE E= , real values

( )

( ) ( )

Re

cos

jkz j tox x oy y

ox x oy y

E E e e

E E t kz

− ω⎡ ⎤= +⎣ ⎦

= + ω −

a a

a a

E (8-43)

↑ ↑ ↑ It oscillates with time along a line parallel to ox x oy yE E+a a ↑ Linearly polarized in the direction ox x oy yE E+a a . Linearly polarized along x -axis + Linearly polarized along y -axis

↑ ↑ Two component waves are in phase.

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Example 8-4 Given a linearly polarized wave, ( )3 jkz

o x yE e −= +E a a , propagating in free space, find (a) rotation angle of the polarization vector with respect to x -axis (b) time-average power density

Solution (a) The rotation angle

1 3tan 601

o− ⎛ ⎞θ = =⎜ ⎟⎜ ⎟

⎝ ⎠

(b) From Eq. (8-23)

( ) ( )3 3jkz jkzo oz o x y o y x

o o

E e E e− −ε ε= × + = −

μ μH a a a a a

From Eq. (8-39)

( ) ( )

( ) ( )

( )

*1 Re 3 32

3 32

1 32

jkz jkzoo x y o y x

o

o ox y y x

o

o oz

o

E e E e

E

E

− −⎡ ⎤⎧ ⎫ε⎪ ⎪⎢ ⎥= + × −⎨ ⎬

μ⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦

ε= + × −

μ

ε= +

μ

S a a a a

a a a a

a

The time-average power density is the sum of those of the waves that are polarized in x and y -directions, respectively.

8-5.2 Circular Polarization

y -component lags behind x -component by 90o in time dimension ( ) ( )/2jkz j jkz

o x o y o x o yE jE e E E e e− − π −= − = +E a a a a (8-44) where ˆ ˆ

ox oy oE E E= = . Instantaneous expression

( ) ( )cos cos /2x o y oE t kz E t kz= ω − + ω − − πa aE (8-45) At 0z =

( ) ( )( ) ( )

cos cos /2

cos sinx o y o

x o y o

E t E t

E t E t

= ω + ω − π

= ω + ω

a aa a

E (8-46)

At 0tω = , o xE= aE At /2tω = π , o yE= aE → Right-hand circularly polarized wave ↑ A circular locus in xy -plane

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Fig. 8-4, right-hand circular polarization y - comp. lags behind x -comp. by 90o in time phase y -comp. arrives at p at a later time than x -comp. by a quarter of the temporal period. y -comp. is displaced backward by a quarter wavelength with respect to the x -comp. The phase relationship is maintained throughout the space at all times.

y -comp. leads x -comp. by 90o in time phase → Left-hand circularly polarized wave

When the left fingers follow the rotation of the electric field vector, the left thumb points to the propagation direction of the wave.

( ) ( )/2jkz j jkz

o x o y o x o yE jE e E E e e− π −= + = +E a a a a (8-47) Instantaneous expression

( ) ( )cos cos /2x o y oE t kz E t kz= ω − + ω − + πa aE (8-48) At 0z =

( ) ( )( ) ( )

cos cos /2

cos sinx o y o

x o y o

E t E t

E t E t

= ω + ω + π

= ω − ω

a aa a

E (8-49)

At 0tω = , o xE= aE At /2tω = π , o yE= − aE → Clockwise rotation ↑ A circular locus in xy plane

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Fig. 8-5, left-hand circular polarization y - comp. leads x -comp. by 90o in time phase y -comp. arrives at p at an earlier time than x -comp. by a quarter of the temporal period. y -comp. is displaced forward by a quarter wavelength with respect to the x -comp.

Example 8-5

A linearly polarized wave can be expressed by the sum of two circularly polarized waves. Find two circularly polarized waves contained in (a) 1

jkzo xE e −=E a

(b) ( )2jkz

ox x oy yE E e −= +E a a Solution (a) By adding and subtracting ( )1/2 o yjE a

( ) ( )112

jkz jkzo x o x o y o x o yE e E jE E jE e− −⎡ ⎤= = − + +⎣ ⎦E a a a a a (8-50)

On the right side, the first term is a right-hand circularly polarized wave and the second is a left-hand circularly polarized wave. Thus, the answer is

( )12

jkzo x o yE jE e −−a a and ( )1

2jkz

o x o yE jE e −+a a

The linearly polarized wave in x -direction is the sum of two circularly polarized waves, with a half the amplitude of the linear polarization, which rotate in the opposite directions and meet each other at x -axis once in every cycle of their rotations.

(b) Magnitude of the amplitude 2 2

o ox oyE E E= + Phase angle of the amplitude

1tan oy

ox

EE

−θ =

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The polarization vector is parallel to a “reference line” that is rotated by θ with respect to x -axis. Referring to the result of (a), we can express the linearly polarized wave by the sum of two circularly polarized waves, which rotate in the opposite directions and meet at the reference line once in every cycle of their rotations. We add a phase θ to the right-hand circularly polarized wave in Eq. (8-44), and add a phase −θ to the left-hand circularly polarized wave in Eq. (8-47), so that their polarization vectors become parallel to the reference line at 0z = and 0t = , ( ) j jkz

o x o yE jE e eθ −−a a , (8-51a)

( ) j jkzo x o yE jE e e− θ −+a a , (8-51b)

Reducing the amplitude of the waves in Eq. (8-51) by half and adding two waves

( ) ( )21 12 2

j jkz j jkzo x o y o x o yE jE e e E jE e eθ − − θ −= − + +E a a a a (8-52)

which is certainly a sum of two circularly polarized waves Let us check the result in Eq. (8-52). Rearranging it

( ) ( )2 2 cos sin

j j j j jkzox y

jkzo x y

E e e j e e e

E e

θ − θ θ − θ −

⎡ ⎤= + − −⎣ ⎦

⎡ ⎤= θ + θ⎣ ⎦

E a a

a a

Since coso oxE Eθ = and sino oyE Eθ = , it is a linearly polarized wave in the direction that is rotated by θ with respect to x -axis.

8-5.3 Elliptical Polarization

( )j jkzox x oy yE E e eϕ −= +E a a : Elliptically polarized wave (8-53)

Linearly and circularly polarized waves are special cases of an elliptically polarized wave Instantaneous expression

( ) ( )cos cosox x oy yE t kz E t kz= ω − + ω − + ϕa aE (8-54)

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At 0z = plane

( )cosx oxE E t= ω (8-55a) ( )cosy oyE E t= ω + ϕ (8-55b)

For 0 /2< ϕ < π At 0tω = , x oxE E= and cosy oyE E= ϕ (E in first quadrant) At /2tω = π , 0xE = and siny oyE E= − ϕ (E parallel to y− -axis) ↑ E is rotating clockwise in xy -plane Left-hand elliptically polarized.

Elliptical locus by E in Eq. (8-54) New coordinate system, 'x and 'y ↑ ↑ Parallel to the major and minor axes Elliptical locus in general form

( ) ( )cos ' cos 'x ox oxE E t E t′ ′ ′= − ω + ϕ = ω + ϕ⎡ ⎤⎣ ⎦ (8-56a)

( ) ( )sin ' sin 'y oy oyE E t E t′ ′ ′= − ω + ϕ = − ω + ϕ⎡ ⎤⎣ ⎦ (8-56b) ↑ ccw rotation in 'x 'y -plane Coord. transformation of Eq. (8-55) to primed coord.

( ) ( )cos cos cos sinx ox oyE E t E t′ = ω θ + ω + ϕ θ (8-57a) ( ) ( )cos sin cos cosy ox oyE E t E t′ = − ω θ + ω + ϕ θ (8-57b)

By equating Eq. (8-57) with Eq. (8-56)

( )( ) ( )22

2 costan 2 ox oy

ox oy

E E

E E

ϕθ =

− (8-58)

→ tan /oy oxE Eθ = for 0ϕ = , linearly polarized wave.

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Example 8-6 Find time-average power density of a uniform plane wave, ( )j jkz

ox x oy yE E e eϕ −= +E a a , propagating in free space. Solution Vector complex amplitude j

o ox x oy yE E e ϕ= +E a a From Eq. (8-23)

( ) ( )1 1

j jkzk z ox x oy y

o ojkz

o

E E e e

e

ϕ −

= × = × +η η

H a E a a a

H

Vector complex amplitude

( )1 jo ox y oy x

o

E E e ϕ= −η

H a a

Time-average Poynting vector

( ) ( )

( )

* *

2 2

1 1Re Re2 21 Re

21

2

o o

j jox x oy y ox y oy x

o

ox oyo

E E e E E e

E E

ϕ − ϕ

⎡ ⎤ ⎡ ⎤= × = ×⎣ ⎦ ⎣ ⎦

⎡ ⎤= + × −⎣ ⎦η

= +η

S E H E H

a a a a

The power delivered by an elliptically-polarized wave is equal to the sum of the individual powers of two linearly-polarized waves.

8-6 PLANE WAVE IN LOSSY MEDIA

An electromagnetic wave in a lossy dielectric. → Induced electric dipoles → Power loss (Same freq. ω ) ↑ Damping force (Collision) Conduction current also causes power loss. In a lossy medium, ˆ j′ ′′ε = ε − ε ↑ Power loss Magnetic interaction is negligible → μ is real

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8-6.1 Lossy Dielectric ( )0σ = In a simple medium with no free charges and constant ε and μ

Maxwell’s equations j∇ × = − ωμE H (8-59a) ( )ˆ

j jj

′ ′′∇ × = ω ε − ε

≡ ωε

H EE

(8-59b)

0∇ =Ei (8-59c) 0∇ =Hi (8-59d)

Helmholtz’s equation

2 2ˆ 0k∇ + =E E (8-60) ↑ Complex wavenumber ( )ˆ ˆk j′ ′′= ω με = ω μ ε − ε (8-61) Plane wave solution

ˆkjk

oe−= a rE E i : ka by other means (8-62)

A wave in + z -direction

ˆ

jkzo

zo

ee

−γ

=

E EE

(8-63)

↑ Propagation constant

( )ˆ

jk j j

j

′ ′′γ = = ω μ ε − ε

≡ α + β (8-64)

where α , attenuation constant β , phase constant.

1/22

1 12

⎡ ⎤′ ′′με ε⎛ ⎞⎢ ⎥α = ω + −⎜ ⎟′ε⎢ ⎥⎝ ⎠⎣ ⎦ : Nepers per meter [Np/m] (8-65a)

1/22

1 12

⎡ ⎤′ ′′με ε⎛ ⎞⎢ ⎥β = ω + +⎜ ⎟′ε⎢ ⎥⎝ ⎠⎣ ⎦ : Radians per meter [rad/m] (8-65b)

Loss tangent is defined by

tan′′ε σ

ξ = =′ε ωε

: ξ , loss angle (8-66)

Inserting Eq. (8-64) in Eq. (8-63)

( )z j z z j zo o Ee e E e e−α − β −α − β= =E E a (8-67a)

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Instantaneous expression ( ) ( )cosz

o EE e t z−α= ω − βaE (8-67b) ↑ ↑

Attenuation. Phase variation

Phase difference is 2π between two points separated by λ 2βλ = π

Rewriting it

2πβ =

λ : β and λ depend on ε (8-68)

Phase velocity

pvω

(8-69)

For o o′με > μ ε β in medium > k in free space λ in medium < oλ in free space → pv is slower in medium

H from Eqs. (8-63) and Eq. (8-59a) ( ) ( )

z zo E z o EE e E e

j

−γ −γ∇ × = −γ ×

= − ωμ

a a aH

Rewriting it, with Eq. (8-64)

( ) ( )ˆˆ

z z j zoo z E z E

EE e e e−γ −α − βε

= × = ×μ η

H a a a a (8-70)

Complex intrinsic impedance, ˆ ˆˆ /o oE Hη =

ˆˆ jμ μ

η = =′ ′′ε ε − ε

(8-71)

Complex η̂ → H and E are out of phase

Example 8-7

From a wave, 0.3 /2z jzx oE e e− −=E a , given in a lossy dielectric of oμ = μ and 2 o′ε = ε ,

determine intrinsic impedance of the medium.

Solution Attenuation and phase constants

0.3α = 0.5β =

From Eq. (8-65)

2

2

2

1 1

1 1

′′ε⎛ ⎞+ −⎜ ⎟′ε⎛ ⎞α ⎝ ⎠=⎜ ⎟β⎝ ⎠ ′′ε⎛ ⎞+ +⎜ ⎟′ε⎝ ⎠

(8-72)

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Inserting α and β in Eq. (8-72), loss tangent

1.88′′ε=

′ε (8-73)

From Eq. (8-71), with 2 o′ε = ε and the loss tangent

( )1ˆ

2 1 /o

o jμ

η =ε ′′ ′− ε ε

0.54120 1 1832 1 1.88

jej

π= =

− (8-74)

H lags behind E by 0.54 radian. Note that α , β and η̂ are closely related to each other.

8-6.2 Lossy Dielectric ( )0σ ≠ Conduction current is the dominant source of power loss. In a dielectric with 0σ ≠ and no damping

j∇ × = − ωμE H (8-75a) ( )j j∇ × = + ωε = σ + ωεH J E E (8-75b)

0∇ =Ei (8-75c) 0∇ =Hi (8-75d)

E and H , generated outside → 0v∇ = ρ =Di , J is induced ↑ No source charge or current Rewriting Eq. (8-75b)

( ) ˆ

j j

j j j

σ⎛ ⎞∇ × = ω ε −⎜ ⎟ω⎝ ⎠′ ′′≡ ω ε − ε ≡ ωε

H E

E E (8-76)

ε̂ is newly defined ′ε = ε (8-77a)

σ′′ε =ω

(8-77b)

Plane wave solution

zoe

−γ=E E (8-78) Substituting Eq. (8-77) in Eqs. (8-64) and (8-65)

1j j j σ⎛ ⎞γ ≡ α + β = ω με −⎜ ⎟ωε⎝ ⎠ (8-79)

1/2

2

1 12

⎡ ⎤με σ⎛ ⎞⎢ ⎥α = ω + −⎜ ⎟ωε⎢ ⎥⎝ ⎠⎣ ⎦ (8-80a)

1/22

1 12

⎡ ⎤με σ⎛ ⎞⎢ ⎥β = ω + +⎜ ⎟ωε⎢ ⎥⎝ ⎠⎣ ⎦ (8-80b)

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Similarly

( )

1ˆ1 /j

μη =

ε − σ ωε (8-81)

Low-loss dielectric, ′′ ′ε << ε or / 1σ ωε << Applying binomial expansion

2112 8

j j

j

⎡ ⎤σ σ⎛ ⎞γ ≅ ω με − +⎢ ⎥⎜ ⎟ωε ωε⎝ ⎠⎢ ⎥⎣ ⎦≡ α + β

(8-82)

Binomial expansion in general form

( ) ( ) 211 1 ...

2!n n n

a na a−

+ = + + +

where 1a << . From Eq. (8-82)

2σ μ

α ≅ε

(8-83a) 211

8⎡ ⎤σ⎛ ⎞β ≅ ω με +⎢ ⎥⎜ ⎟ωε⎝ ⎠⎢ ⎥⎣ ⎦

(8-83b)

where

α ~ σ β ω με , perfect dielectric Similarly

ˆ 12jμ σ⎡ ⎤η ≅ +⎢ ⎥ε ωε⎣ ⎦

(8-84)

↑ Positive phase angle E leads H in time phase.

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Example 8-8

Given an electromagnetic wave, ( )0.21 2.2300 z j zx e − −=E a [ ]/V m , propagating in a medium of

ˆ 220 21jη = + [ ]Ω , find (a) S at 0.5z m= (b) time-average ohmic power-loss per unit volume at 0.5z m= at steady state

Solution (a) Magnetic field intensity

( ) ( )0.21 2.21 300ˆ 220 21

z j zz y e

j− −= × =

η +H a E a

Time average Poynting vector

( )

2 0.42*

0.42

1 1 300Re Re2 2 220 21

202.7

z

z

zz

ej

e

⎡ ⎤⎡ ⎤= × = ⎢ ⎥⎣ ⎦ −⎣ ⎦

=

S E H a

a (8-85)

At 0.5z m= ( ) 0.42 0.5 2202.7 164.3 /z ze W m− × ⎡ ⎤= = ⎣ ⎦S a a (b) Time averaging Poynting’s theorem in Eq. (8-34) at steady state −∇ × =i iE H E J or

*12

−∇ =S E Ji i (8-86)

where J is the real instantaneous value and J is the phasor. Evaluating the left side of Eq. (8-86) using Eq. (8-85)

( ) ( ) 0.42 0.42202.7 0.42 85.1z ze e− −−∇ = =Si At 0.5z m= , time-average ohmic power-loss per unit volume is 0.42 0.5 385.1 69.0 /e W m− × ⎡ ⎤−∇ = = ⎣ ⎦Si (8-87)

In the other method, the power loss can be calculated by the use of J First, from Eqs. (8-79) and (8-81)

j jγ σ⎛ ⎞= ωε −⎜ ⎟η ωε⎝ ⎠

From the equation we obtain

[ ]30.21 2.2Re Re 1.89 10 /ˆ 220 21

j S mj

−⎡ ⎤⎡ ⎤γ +σ = = = ×⎢ ⎥⎢ ⎥η +⎣ ⎦ ⎣ ⎦

At 0.5z m= ( )0.21 0.5 2.2 0.5300 j

x e − × − ×=E a ( ) ( )0.21 0.5 2.2 0.53300 1.89 10 j

x e − × − ×−= × ×J a

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Time-average ohmic power-loss

( )* 2 3 0.21 0.5 2 31 1 300 1.89 10 69.0 /2 2

e W m− − × × ⎡ ⎤= × × = ⎣ ⎦E Ji (8-88)

Two results in Eqs. (8-87) and (8-88) are identical. 8-6.3 Good Conductor

Good conductor → / 1σ ωε >> 1

2

jj j

j

σ +γ ≅ ω με − = ωμσ

ωε≡ α + β

: 1/2/2 /4 12

j j jj e e− π − π −⎡ ⎤− = = =⎣ ⎦ (8-89)

fα = β = π μσ (8-90) Plane wave in a good conductor

( )

z j zo

z f jz fo E

e e

E e e

−α − β

− π μσ − π μσ

=

=

E E

a (8-91a)

( ) ( )cosz f

o EE e t z f− π μσ= ω − π μσaE (8-91b)

↑ Attenuation

Depth of penetration or skin depth

1 1 1f

δ = = =α βπ μσ

: [m] (8-92)

From 2 /β = π λ

δ =π

(8-93)

Phase velocity in a good conductor

2pfv f π

= λ = ωδ =μσ

: 41.3 10 /pv m s= × at 1GHz in Cu (8-94)

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Example 8-9 Find the skin depth in copper of [ ]75.8 10 /S mσ = × as a function of frequency

Solution From Eq. (8-93), with oμ ≈ μ

7 7

1 1

4 10 5.8 10of f −δ = =

π μ σ π × π × × ×

The answer is

0.066f

δ = [m]

At a frequency of 1GHz, skin depth in copper is only 2.1 mμ .

From Eq. (8-81)

ˆ jωμη ≅

σ

or

( )1ˆ

j+η =

σδ : E leads H by 45o in time phase (8-95)

The real part of η̂

1sR =

σδ : Surface resistance per unit area (8-96)

Resistance of a good conductor of cross section w × δ and length l

Rw

=δσl (8-97)

Example 8-10

A plane wave ( ) ( )1 /j zo xE e − + δ=E a propagates into a good conduction of conductivitiy σ in

the region 0z ≥ . (a) Find time-average power loss in a volume 3 [ ]w m× × ∞l (b) Find total current, in phasor form, crossing the area 2 [ ]w m× ∞ in yz -plane (c) Assuming the current in (b) flows uniformly in a volume of cross section w × δ and

length l , find its resistance, and time-average power loss in the volume

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Solution (a) Time-average power loss

* *

0

2 2 /

0

2

1 1Re Re2 2

1 21 4

z

z

z zo z

o

P dv w dz

w E e dz

w E

=∞

=

=∞ − δ

=

⎡ ⎤⎡ ⎤= = σ⎣ ⎦ ⎢ ⎥⎣ ⎦

= σ

= σ δ

∫ ∫

J E E EV

i il

l

l

(8-98)

(b) Total current

( )1 /

0 0 0 1w z j z o

oy z z

w EI d wE e dz

j∞ =∞ − + δ

= = =

σ δ= σ = σ =

+∫ ∫ ∫E si (8-99)

(c) The conductor has a cross section wδ , length l and conductivity σ . The resistance

Rw

=δσl (8-100)

Time-average power loss, with Eqs. (8-99) and (8-100)

* 21 1Re2 4 oP R I I w E⎡ ⎤= = σ δ⎣ ⎦ l (8-101)

The results in Eqs. (8-98) and (8-101) are identical, from which we can assume that the current flows uniformly only in the region 0 z≤ ≤ δ , as far as the total power loss is concerned. We can also obtain the same power loss as Eq. (8-98) or (8-101) from the voltage oE l that is applied to the resistance R of Eq. (8-100).

Power density in a conductor Electric and magnetic field intensities

( )1 /j zx oE e

− + δE aZ

( ) ( )1 /1ˆ 1

z jk y oE ej

− + δσδ×

η +H a E aZ Z

Time-average Poynting vector

( ) ( )*/ / / / 2 2 /11 1Re Re

2 1 2 2z j z z j z z

x o y o z o

jE e e E e e E e

j− δ − δ − δ − δ − δ

⎡ ⎤ σδ +⎡ ⎤⎛ ⎞σδ⎢ ⎥= × = ⎢ ⎥⎜ ⎟+⎢ ⎥⎝ ⎠ ⎣ ⎦⎣ ⎦

S a a a

or

2 2 /

4z

z oE e − δσδ=S a : Attenuated (8-102)

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Example 8-11 Referring to Fig. 8-10, find the total power transmitted into the conductor through an area w × l of the conductor surface.

Solution

From Eq. (8-102), time-average power density at 0z = 2

4z oEσδ

=S a (8-103)

Time-average total power across the area

2

0 0

14

w

ox yP d E w

= == = σδ∫ ∫ S si

ll (8-104)

Note that ds is in za direction, which is in the same direction as S , not the outward surface normal. The power across the surface decays to zero at z = ∞ , as is shown in Eq. (8-102), implying that the transmitted power is dissipated all in the conductor. It is verified by the fact that the results in Eqs. (8-104) and (8-98) are identical.

8-7 PLANE WAVES AT BOUNDARY A uniform plane wave → Reflection Transmission ↑ Interface (Different ε and μ )

8-7.1 Normal Incidence of Plane Wave

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Incident wave polarized in x direction 1j zi i

x oE e− β=E a (8-105a)

1

1

ij zi o

yE e − β=η

H a (8-105b)

where 1β , phase constant 1η , intrinsic impedance i

oE , amplitude, assumed real The boundary condition → ( rk and tk ) za , uniform rE and tE at 0z = → ( rE and tE ) xa The reflected plane wave

1j zr rx oE e

β=E a (8-106a)

1

1

rj zr o

yE e β= −η

H a (8-106b)

↑ Traveling in z− direction. Same 1β and 1η as iE , same medium The transmitted plane wave

2j zt tx oE e

− β=E a (8-107a)

2

2

tj zt o

yEe − β=

ηH a (8-107b)

From the boundary condition at 0z = plane

( ) ( ) ( )0 0 0i r t+ =E E E (8-108a) ( ) ( ) ( )0 0 0i r t+ =H H H (8-108b)

Inserting Eqs. (8-105)-( 8-107)

i r to o oE E E+ = (8-109a)

1 1 2

i r to o oE E E− =

η η η (8-109b)

By solving the equation

2 1

2 1

r io oE E

η − η=η + η

(8-110a)

2

2 1

2t io oE E

η=η + η

(8-110b)

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We define 2 1

2 1

roio

EE

η − ηΓ = =

η + η : Reflection coefficient (8-111a)

2

2 1

2toio

EE

ητ = =

η + η : Transmission coefficient (8-111b)

↑ True only for normal incidence

Γ < 0 for 2 1η < η → ra = i−a 0Γ ≠ for 1 2η ≠ η → Mandatory reflection 0Γ = for 1 2η = η → No reflection Eq. (8-111) can be applied even to an interface between two lossy media → Complex 1η and 2η → Complex Γ and τ From Eq. (8-111) 1+ Γ = τ (8-112) ↑ Valid only for normal incidence

Example 8-12

For normal incidence of a uniform plane wave, prove ( )2 21 2/ 1Γ + η η τ = from the

conservation of energy Solution

Time-average power densities of the incident, reflected and transmitted waves

2*

1

1 1Re2 2

i i i iz oE⎡ ⎤= × =⎣ ⎦ η

S E H a (8-113a)

2 22

1 1

1 12 2

r r iz o z oE E= − = − Γ

η ηS a a (8-113b)

2 22

2 2

1 12 2

t t iz o z oE E= = τ

η ηS a a (8-113c)

From the conservation of energy

i r t= +S S S (8-114) Substituting Eq. (8-113)

2 21

2

Γ + τ =η

(8-115)

Two terms on the left side correspond to the reflected and the transmitted energies relative to the incident energy.

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EM Harmonic Wave 8-29 Proprietary of Prof. Lee, Yeon Ho

A. Standing Wave Ratio Total electric field intensity in medium 1

1 121

1

1

ˆ

j z j zi r ix o

x

E e e

E

− β β⎡ ⎤= + = + Γ⎣ ⎦≡

E E E a

a (8-116)

Γ is complex in general

je φΓ = Γ (8-117) Rewriting Eq. (8-116) with Eq. (8-117)

( )11 21

ˆ 1 j zj zioE E e e β +φ− β ⎡ ⎤= + Γ⎣ ⎦ (8-118)

The maximum amplitude and its location

( )1 maxˆ 1i

oE E= + Γ (8-119)

( )max1

1 22

z n= − φ + πβ

( )0, 1, 2, ...n = ± ± (8-120)

The minimum amplitude and its location

( )1 minˆ 1i

oE E= − Γ (8-121)

( )min1

1 22

z n= − φ + π + πβ

( )0, 1, 2, ...n = ± ± (8-122)

The intermaximal distance is 12 /2π β , a half wavelength of the wave in medium 1.

Rewriting Eq. (8-118) ( )

( ) ( )

11

1

21

/21

ˆ 1

1 2cos /2

j zj zio

j zi i jo o

E E e e

E e E e z

β +φ− β

− β φ

⎡ ⎤= − Γ + Γ + Γ⎣ ⎦= − Γ + Γ β + φ⎡ ⎤⎣ ⎦

(8-123)

Real instantaneous expression ( ) ( ) ( ) ( )1 1 11 cos 2 cos /2 cos /2i i

o oE t z E z t= − Γ ω − β + Γ β + φ ω + φE (8-124) ↑ ↑ Traveling wave Standing wave

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EM Harmonic Wave 8-30 Proprietary of Prof. Lee, Yeon Ho

Traveling wave, a blue colored line; standing wave, black colored line 0, /4tω = π and π for 0.5Γ = and 0φ =

E at 0, /4, /2, 3 /4,tω = π π π and π for 0.5Γ = and 0φ = . Black colored line, E ; blue colored line, amplitude of E as a function of position

The standing wave ratio is defined by 1 max

1 min

ˆ 1ˆ 1

ES

E+ Γ

= =− Γ

: Dimensionless (8-125)

Γ ranges from 0 to 1, and S ranges from 1 to ∞ .

Page 31: CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM Waves_a.pdf · EM Harmonic Wave 8-3 Proprietary of Prof. Lee, Yeon Ho The right side

EM Harmonic Wave 8-31 Proprietary of Prof. Lee, Yeon Ho

Example 8-13 A normally incident wave is reflected off an interface at 0z = plane. Measured in medium 1 are the standing wave ratio of 4, a distance of 0.2m between the first maximum of the electric field intensity and the interface, and a distance of 0.5m between two adjacent maxima. Find the intrinsic impedance 2η of medium 2.

Solution

The intermaximal distance is equal to a half wavelength of the wave in medium 1 2 0.5mλ = ×

or

12 2π

β = = πλ

.

The first maximum corresponds to 0n = in Eq. (8-120)

max1

1 0.22

z m= − φ = −β

Combining two equations 0.8φ = π (8-126)

From Eq. (8-125)

1 4 1 0.61 4 1

SS− −

Γ = = =+ +

(8-127)

The reflection coefficient, from Eqs. (8-126) and (8-127)

0.80.6 je πΓ = (8-128) Rewriting Eq. (8-111a)

2

1

11

η + Γ=

η − Γ (8-129)

Inserting Eq. (8-128) and 1 oη = η in Eq. (8-129)

0.8 0.60

0.8320.8 0.23

1 0.6 0.99 0.811 0.6 1.22

j jj

j jo

e e ee e

π

π −

η += = =

η −

The answer is [ ]0.83

2 305 jeη = Ω B. Interface Involving Perfect Conductor

Interface between a lossless medium and a perfect conductor ↑ ↑ 0z < 0z ≥ Perfect conductor → σ = ∞ → 0η = 1Γ = − and 0τ =

Page 32: CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM Waves_a.pdf · EM Harmonic Wave 8-3 Proprietary of Prof. Lee, Yeon Ho The right side

EM Harmonic Wave 8-32 Proprietary of Prof. Lee, Yeon Ho

Incident plane wave in medium 1 1j zi i

x oE e− β=E a : Polarized in x -direction (8-130a)

1

1

ij zi o

yEe − β=

ηH a (8-130b)

Reflected plane wave propagates in z− direction

1 1j z j zr i ix o x oE e E eβ β= Γ = −E a a (8-131a)

1

1

ij zr o

yEe β=

ηH a (8-131b)

where Amplitude of rE from r i

o oE E= Γ rH from Eq. (8-23).

E = 0 at conductor surface. Nodes with a period of /2λ . First node of H at /4λ from the conductor surface. The same period as E

Page 33: CHAPTER 8. ELECTROMAGNETIC HARMONIC WAVElab.icc.skku.ac.kr/~yeonlee/Electromagnetics/Class 8_EM Waves_a.pdf · EM Harmonic Wave 8-3 Proprietary of Prof. Lee, Yeon Ho The right side

EM Harmonic Wave 8-33 Proprietary of Prof. Lee, Yeon Ho

Total field in medium 1

( )

1 11

1 2 sin

j z j zi r ix o

ix o

E e e

j E z

− β β⎡ ⎤= + = −⎣ ⎦= − β

E E E a

a (8-132a)

( )

1 11

1

11

2 cos

ij z j zi r o

y

io

y

Ee e

Ez

− β β⎡ ⎤= + = +⎣ ⎦η

= βη

H H H a

a (8-132b)

Instantaneous expression ( ) ( )( ) ( )

1

1

2 sin cos /2

2 sin sin

ix o

ix o

E z t

E z t

β ω − π

= β ω

aa

ZE : Standing wave (8-133a)

( ) ( )11

2cos cos

io

xE

z t= β ωη

aH (8-133b)

Standing wave cannot transport energy in space, because the factor j− in Eq. (8-132a) makes *×E H be imaginary and results in 0=S .

Example 8-14

A right-hand circularly polarized wave, ( )i j zo x o yE jE e − β= −E a a , propagates in free space

and impinges on the surface of a perfect conductor at 0z = plane. Find (a) polarization vector of the reflected wave (b) surface current density on the conductor surface (c) E in free space

Solution (a) Intrinsic impedance of the perfect conductor 2 0η = Reflection coefficient 1Γ = − Reflected wave ( ) ( )r j z j z

o x o y o x o yE jE e E jE eβ β= Γ − = − +E a a a a (8-134) Real instantaneous expression for the reflected wave

( )

( )

( ) ( )

/2Re

cos cos2

cos sin

r j j z j to x o y

x o y o

x o y o

E e E e e

E t z E t z

E t z E t z

π β ω⎡ ⎤= − +⎣ ⎦π⎛ ⎞= − ω + β + ω + β +⎜ ⎟

⎝ ⎠= − ω + β − ω + β

a a

a a

a a

E

Let us check rE at 0z = plane

At 0tω = , rx oE= −aE

At /2tω = π , ry oE= −aE (8-135)

When the left fingers follow rE in Eq. (8-135), the left thumb points to the propagation direction of rE , or the z− direction. Since the x and y -components have the same amplitude, the reflected wave is left-hand circularly polarized.

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EM Harmonic Wave 8-34 Proprietary of Prof. Lee, Yeon Ho

If a right-hand circularly polarized wave is reflected, it becomes a left-hand circularly polarized wave, or vice versa.

(b) Magnetic field intensities of the incident and reflected waves

( ) ( )

( )

1 1

i i j zk z o x o y

o

j zoy x

o

E jE e

Ej e

− β

− β

= × = × −η η

= +η

H a E a a a

a a

( ) ( ) ( )

( )

1 1

r r j zk z o x o y

o

j zoy x

o

E jE e

Ej e

β

β

= × = − × − +η η

= +η

H a E a a a

a a

Total field at 0z = plane

( )2i r oy x

o

Ej= + = +

ηH H H a a

Surface current density

( ) ( )

( )

2

2

os n z y x

o

ox y

o

Ej

Ej

= × = − × +η

= −η

J a H a a a

a a

where na is the unit surface normal. (c) Total field in free space

( ) ( )( ) ( ) 2 sin 2 sin

i r j z j zo x o y o x o y

x o y o

E jE e E jE e

jE z E z

− β β= + = − + − +

= − β − β

E E E a a a a

a a

Instantaneous expression

( ) ( ) ( )

( ) ( ) ( )

/2Re Re 2 sin 2 sin

2 sin sin cos

j tj t j tx o y o

o x y

e E z e E z e

E z t t

ω −πω ω⎡ ⎤⎡ ⎤= = β − β⎣ ⎦ ⎣ ⎦⎡ ⎤= β ω − ω⎣ ⎦

E a a

a a

E