10.7 Power and the Poynting Vectorsite.iugaza.edu.ps/hkhozondar/files/2017/02/EMII_Chapter...(b) The...

3/24/2016

1

EELE 3332 – Electromagnetic IIChapter 10

Electromagnetic Wave PropagationProf. Hala J. El‐Khozondar

Islamic University of Gaza

Electrical Engineering Department

20161

Energy can be trasported from one point (where a transmitter is located)

to another point (with a receiver) by means of EM waves.

The rate of energy transportation can be obtined from

Maxwell's e

2

22

quations:

EUsing Maxwell equation: H E

Dotting both sides with E:

EE H E

But from vector identities: H E E H H E

E H H E H E

1H E H E ... (1)

2

t

Et

EE

t

2

10.7 Power and the Poynting Vector

3

2

22

2 22

2 22

Using Maxwell equation : E , Dotting both sides with H:

HH E H

2

1Substitute in equation (1): H E H E

2

1E H

2 2

1E H

2 2

H

t

H

t t

EE

t

H EE

t t

E HE

t

2 2 2

2 2 2

Take volume integral of both sides:

1 1E H

2 2

Applying the divergence theorem to the left hand side:

1 1E H

2 2

v v v

S v v

t

dv E H dv E dvt

dS E H dv E dvt

4

2 2 21 1E H

2 2S v v

dS E H dv E dvt

Power and the Poynting Vector

Total power leaving the volume

Rate of decrease in energy stored in electric and magnetic fields

Ohmicpower

dissipated

2the (Watts/m ) is defined as:

=E×H

It represents the instantaneous power density vector

associated with the EM field at a given point.

PoyntingVector

Poynting’s theorem

3/24/2016

2

5


Poynting theorem: states that the net power flowing out of a given

volume v is equal to the time rate of decrease in the energy stored

with v minus the ohmic losses.

Illustration of power balance for EM fields.

Note that is normal to both E and H and is therefore along the direction of propagation ak

=E×H

6

0

0

0

0

22

22

Assume that

E( , ) cos -

H( , ) cos -

( , )= E H = cos - cos -

( , )= cos cos 2 - 22

1since cosAcosB= cos cos

2

zx

zy

zz

zz

z t E e t z a

Ethen z t e t z a

Eand z t e t z t z a

Ez t e t z a

A B A B


0

0

0

*s s

22

22

The time-average Poynting vector ( ) over the period T=2 / is:

1( ) ( , )

It can also be found by:

1( ) Re E ×H

2

For ( , ) cos cos 2 - 22

( ) cos2

ave

T

ave

ave

zz

zave

z

z z t dtT

z

Ez t e t z a

Ez e

The total time-average power crossing a given surface S is given by:

S

z

ave ave

S

a

P d

7


0

2

2

0

*s s

2

( , , , ) Poynting vector

( , , ) time-average of Po

time-varying vector (watts/m )

( , , , ) E×H

(time-invariant vector) (watt

ynting

s/m )

1( , , , )

1Re E

vector

×H2

( )2

T

ave

ave

av

ave

e

x y z t

x y z t dtT

E

x y z t

x y z

z e

20

total time-average power through

cos E cos -

(scalar) wattsa surface

S

z zz x

ave av

e

e

v

S

a

a for E e t z a

P d

P

8


3/24/2016

3

9

7

r

2

In a nonmagnetic medium

E = 4 sin (2 10 0.8 ) V/m

Find

(a) ,

(b) The time-average power carried by the wave

(c) The total power crossing 100 cm of plane 2x + y = 5

zt x

a

Example 10.7

10

7

(a) Since =0 and /c, the medium is not free space but

a lossless medium.

0.8 , 2 10 , (nonmagnetic),

Hence

o o r

8

7

2

or

0.8 3 10 12

2 10

14.59

120120 . 10 98.7

12

o o r r

r

r

o

o r r

c

c

Example 10.7 ‐ Solution

22

x

22

x x x20

(b) sin ( )

1 16 81 mW/m

2 2 10

(c) On plane 2x + y = 5 (see Example 3.5 or 8.5),

2

5Hence the total power is

o

To

ave

x yn

av

Et x

Edt

T

P

E H a

a a a

a aa

3 4

5

2 . . 81 10 . 100 10

5

162 10 724.5 W

5

x ye ave ave n xdS S

a a

a a

11

Example 10.7 ‐ Solution

12

10.8 Reflection of a plane waveat normal incidence

When a plane wave from one medium meets a different medium, it

is partly reflected and partly transmitted.

The proportion of the incident wave that is reflected or transmitted

depends on the parameters (ε,μ,σ) of the two media involved.

Normal incidence (plane wave is normal to the boundary) and

oblique incidence will be studied.

3/24/2016

4

13

Reflection of a plane wave at normal incidence

Suppose a plane wave propagating along the +z direction is incident

normally on the boundary z=0 between medium 1 (z<0) characterised

by ε1,μ1,σ1 and medium 2 (z>0) characterised by ε2,μ2,σ2.

i i z(E ,H ) is traveling along +a in medium 1.

Assume the electric and megnetic filed (in phasor form) as follows:

Inciden

t Wave

14


1

1 1

0

00

1

E ( ) a ,

H ( ) a a

zis i x

z ziis i y y

z E e then

Ez H e e

0 is magnitude of

the incident electric

field at z=0

iE

z(E ,H ) is traveling along a in medium 1.

Reflected

Wav

e

r r

15


1

1 1

0

00

1

If E ( ) a ,

H ( ) a a

zrs r x

z zrrs r y y

z E e then

Ez H e e

0 is magnitude of

the reflected electric

field at z=0

rE

z(E ,H ) is traveling along +a in medium 2.

Transmitt

ed Wave

t t

16


2

2 2

0

00

2

If E ( ) a ,

H ( ) a a

zts t x

z ztts t y y

z E e then

Ez H e e

0 is magnitude of

the transmitted electric

field at z=0

iE

3/24/2016

5

1 1

2 2

Field in medium 1: E E E , H H H

Field in medium 2: E E , H H

Since the waves are transverse, E and H fields are entirely

tangential to the interface.

Applying the boundary

i r i r

t t

1t 2t 1t 2t

0 0 0 0 0

00 0

1 2

conditions at the interface 0:

(E =E and H =H )

:

E (0) E (0) E (0)

1H (0) H (0) H (0)

i r t i r t

ti r t i r

z

then

E E E

EE E

17


2 10 0

2 1

0 2 10 0

0 2 1

20 0

2 1

0 20 0

0 2 1

From the last two equations:

= ,Re or

2

2 = ,

flection Coeffici

or

ent

Transmission Coefficien t

r i

rr i

i

t i

tt i

i

E E

EE E

E

and E E

EE E

E

Note that:

1. 1+

2. Both and are dimensionless and may be complex.

( and are real for lossless media, and complex for lossy media)

3. 0 1

18


When medium 1 is a perfect dielectric (lossless , σ1=0), and

medium 2 is a perfect conductor (σ2=∞):

η2= 0 → Γ=-1 → τ=0

The wave is totally reflected and there is no transmitted wave

(E2 = 0).

The totally reflected wave combines with the incident wave to

form a standing wave.

A standing wave "stands" and does not travel; it consists of two

travelling waves (Ei and Er) of equal amplitudes but in opposite

directions. 19


0For conductor = 45

20

1 1

1 1

z z1s is rs i0 r0

01 1 1 1

0

z z1s i0

1s 0 1

1 1s

E =E +E = + a

But = = 1, =0, =0, =

E a

E 2 sin z a (since sin = )2

Thus E =R

The standing wave in medium 1 is:

e E ,

x

r

i

j jx

jA jA

i x

j t

E e E e

Ej

E

E e e

e eor jE A

j

e

1 0 1

01 1

1

E =2 sin z sin a

2Similarly, it can be shown that: H = cos z cos a

i x

iy

or E t

Et


3/24/2016

6

21


Standing waves E 2Eio sin 1z sin t ax. The curves 0, 1, 2, 3, 4, . . ., are, respectively, at times t 0, T/8, T/4, 3T/8, T/2, . . . ; 2/1.

22

Standing Waves Examples

Standing wave on a string

http://www.walter‐fendt.de/ph14e/stwaverefl.htm

Medium 1 : perfect dielectric 1=0

Medium 2: perfect dielectric 2=0

η1 and η2 are real and so are Γ and τ.

There is a standing wave in medium 1 but there is also a

transmitted wave in medium 2. (incident wave is partly reflected

and partly transmitted).

However, the incident and reflected waves have amplitudes that

are not equal in magnitude.

Two cases:

case 1: when η2 > η1

case 2: when η2 < η123


24

CASE 1

Medium 1 : perfect dielectric 1=0, Medium 2: perfect dielectric 2=0

2 12 1

2 1

0

If , , 0,

0

and are real

j oe

1 1

1 1

1

1

2

21

E E E ( )

(1 )

E 1

j z j zs is rs oi

j z j zoi

j zs oi

E e e

E e e

E e

121 1 max

1 max 1max

1 max 1

E is maximum when 1 E 1

2 0,2 ,4 ,6 ... 0,1,2,3

or 0, ,2 ,3 ,... 2

j zoie E

z n nz n

z

121 1 min

1 min 1 min

1min

1

E is minimum when 1 E 1

3 52 ,3 ,5 ... or , , ...

2 2 2(2 1) (2 1)

0,1,2,32 4

j zoie E

z z

n nz n

3/24/2016

7

25

CASE 2

Medium 1 : perfect dielectric 1=0, Medium 2: perfect dielectric 2=0

2 12 1

2 1

If , , 0,

180

and are real

j oe

1 1

1 1

1

1

2

21

E E E ( )

(1 )

E 1

j z j zs is rs oi

j z j zoi

j zs oi

E e e

E e e

E e

121 1 min

1 min 1min

1 min 1

E is minimum when 1 E 1

2 0,2 ,4 ,6 ... 0,1,2,3

or 0, ,2 ,3 ,... 2

j zoie E

z nnz n

z

121 1 max

1 max 1 max

1max

1

E is maximum when 1 E 1

3 52 ,3 ,5 ... or , , ...

2 2 2(2 1)(2 1)

0,1,2,32 4

j zoie E

z z

nnz n

26

Standing waves due to reflection at an interface between two lossless media; 2/1.

• Measures the amount of reflections, the more reflections, the larger the standing wave that is formed.

• The ratio of |E1|max to |E1|min

or

Since 0 |≤ |Γ|≤1, it follows that 1 ≤s ≤∞.

* When Γ=0, s=1, no reflection, total transmission.

* When |Γ|=1, s=∞, no transmission, total reflection.

s is dimensionless, expressed in decibels (dB) as: s dB=20log10 s27

1

1

min1

max1

min1

max1

H

H

E

Es

1

1

s

s

Standing Wave Ratio, SWR

8

1 8

1

2

2 1

2

10 1

3 10 3120

4 .(4) 4

3

2

o

o o r r

o ro

o r

c

c

Solution :

28

8x

o o

r r

In freespace (z 0),a plane wave with

=10 cos(10 t )a mA/m

is incident normally on a lossless medium ( =2 , =8 ) in region z 0.

Determine the reflected wave , and th

i z

H

H E t te transmitted wave , H E

Example 10.8

3/24/2016

8

29

81 x

81 i

1

Given that =10 cos(10 t ) we expect that

= cos(10 t )

where and = =10

Hence, = 10 cos(

i

i io E

Ei Hi ki x z y io io o

i o

z

z

E H

H a

E E a

a a a a a a

E 81

2 1

2 1

8

r

10 t ) mV/m

2 1 1Now = = ,

2 3 3

10 1Thus cos 10 t + mV/m

3 3

from which we easily obtain as

y

ro o oro io

io o o

r o y

z

EE E

E

z

a

E a

H

810 1 cos 10 t + mA/m

3 3r x z

H a

Example 10.8 – solution continued

82

Similarly,

4 4 1 or

3 3

Thus

cos 10 t +

where .Hence,

t

toto io

io

t to E

Et Ei y

E E E

E

E z

E a

a a a

8

8

40 4 cos 10 t mV/m

3 3

from which we obtain

20 4 cos 10 t mA/m

3 3

t o y

t x

z

z

E a

H a

30

Example 10.8 – solution continued

31

x y

Given a uniform plane wave in air as

(a) Find

(b) If the wave encounters

=40 cos( t

a perfectly conducting plate norm

) + 30 sin(

al

to the z axis at z

t ) V/m

= 0, fi

i

i

z z E a a

H

r rnd the reflected wave and .

(c) What are the total and fields for z 0 ?

(d) Calculate the time-average Poynting vectors

for z 0 and z 0.

E H

E H

Example 10.9

1 2

x 2 y

(a) This is similar to the problem in Example 10.3.

We may treat the wave as consisting of tw

=40 cos( t ) , = 30 sin( t

o waves and where

At

) i i

i iz z

E E

E

Solutio

a

n

a E

r

1

11

1 2

1

atmospheric pr

essure, air has = 1.0006 1.

Thus air

cos( t )

may be regarded as free space.

Let

40 1

120 3

1

i i i

i

i

Hi o

oi o

o

zH

EH

H H H

H a

1 y

Hence = cos( t )1

3

1H k E z x y

i z

a a a a a a

H a32

Example 10.9 ‐solution

3/24/2016

9

33


2

2

22

2 2

Similarly,

where

30 1

= sin( t )

Hence

120 4

i i o H

H k E z y x

i oi o

o

H

E H

z

H a

a a a a a a

2 x

1 2 x y

= sin( t )

and

sin( t ) + cos( t )

1

4

1 1

4 3This problem can also be solved using Method 2 of Example 10.

mA

3

/m

.

i

i i i

z

z z

H a

H H H a a

22 1

2

(b) Since medium 2 is perfectly conducting,

1 <<

that is 1 , = 0

showing that the incident and fields are totally reflected.

= =

Hence, =ro io io

r

E E E

E H

E x y

1 1

40 cos( t ) 30 sin( t ) V/m

1 1 H = cos( t ) sin( t ) A/m

3 4(c) The total fields in air

and

can be shown to be standing wave.

The total

r y x

i r i r

z z

z z

a a

a a

E E E H H H

2 2fields in the conductor are 0 , 0.t t E E H H 34


35


22 21

1 k z z1

2 2 2 2z z

2 22

2 k z2 2

(d) For z 0,

| | 1[ ]

2 2

1= 40 30 40 30 =0

240

For z 0,

| |0

2 2

because the whole incident power is reflected.

save io ro

o

s toave

EE E

E E

a a a

a a

a a

36

• Wave arrives at an angle.

• Assume lossless media.

• Uniform plane wave in general form

• For lossless unbounded media, k =

( )

2 2 2 2 2

( , ) cos( ) Re[ ]

ˆ ˆ ˆ position vector

ˆ ˆ ˆ wave number or propagation vector

j to o

x y z

x x y y z z

x y z

E t E t E e

xa ya za

k a k a k a

k k k k

k rr k r

r

k

Oblique incidence

3/24/2016

10

zy

z=0

Medium 1 : 1 , 1Medium 2 : 2, 2

r

i

t

kr

ki

kt

kix

kiz

an

θi is angle of incidence.The plane defined by propagation vector k and a unit normal

vector an to the boundary is called plane of incidence. 37

Oblique incidencex

38

1 1 1

2 2 2

1

1

E cos( )

E cos( )

E cos( )

cos

sin

i io ix iy iz

r ro rx ry rz

t to tx ty tz

i r

t

ix i

iz i

E k x k y k z t

E k x k y k z t

E k x k y k z t

where k k

k

k

k

ki =β1

kix

kizi

Oblique incidence

It's defined as E is || to incidence plane (E‐field lies in the xz‐plane)

39

Parallel Polarization

1

1

1

1

sin cos

sin cos

1

sin cos

sin cos

1

E (cos sin )

H

E (cos sin )

H

i i

i i

r r

r r

j x zis io i x i z

j x ziois y

j x zrs ro r x r z

j x zrors y

E e

Ee

E e

Ee

a a

a

a a

a

40


3/24/2016

11

2

2

sin cos

sin cos

2

E (cos sin )

H

t t

t t

j x zts to t x t z

j x ztots y

E e

Ee

a a

a

41


42


21 1

21 1

sinsin sin

sinsin sin

1 1 2

Tangential components of E and H should be continuous

at the boundary z=0,

(cos ) (cos ) (cos )

Th

ti r

ti i

j xj x j xio i ro r to t

j xj x j xio ro to

E e E e E e

E E Ee e e

1 1 2

1

2

e exponential terms must be equal for the previous equations

to be valid: sin sin sin

(Incidence angle = reflection angle)

sin

sin

i r t

i r

t

or

1 1 1

2 2 2

(snell's law)i

n

n

43

1 1 2

2 1|| ||

2 1

,

cos cos cos (x-components of E)

(y-component of H)

cos cos ,

cos cos

io i ro r to t

io ro to

ro t iro io

io t i

Hence

E E E

E E E

EE E

E

Reflection coefficient

Transmission coefficient 2|| ||

2 1

|| ||

2 cos ,

cos cos

coswhere (1 )

cos

to ito io

io t i

i

t

EE E

E


44

• defined as the incidence angle at which the reflection coefficient is 0 (all transmission).

2 1 ||||

2 1 ||

2 1 ||

2 2 2 22 1 ||

1 1||

2 2

2 1 2 2 1|| 2

1 2

By setting :

cos cos0

cos cos

cos cos

1 sin 1 sin

sin , and

sin

1 ( / ) sin

1 ( / )

i B

t B

t B

t B

t B

ti B

i

B

or

Since

Parallel Polarization ‐ Brewster angle, B

3/24/2016

12

45

Perpendicular Polarization

In this case, the E field is perpendicular to the plane of incidence (the xz‐plane)

46

1

1

1

1

sin cos

sin cos

1

sin cos

sin cos

1

E

H ( cos sin )

E

H (cos sin )

i i

i i

r r

r r

j x zis io y

j x ziois i x i z

j x zrs ro y

j x zrors r x r z

E e

Ee

E e

Ee

a

a a

a

a a


47


2

2

sin cos

sin cos

2

E

H ( cos sin )

t t

t t

j x zts to y

j x ztots t x t z

E e

Ee

a

a a

(cos sin )t x t z a a

48

i

1 1 2

components of E and H should be continuous

at the boundary z=0, and by setting = :

(y-component of E)

cos cos (x-component of H)

r

io ro to

io ro toi t

Tangential

E E E

E E E

Reflection coeffi 2 1

2 1

2

2 1

cos cos ,

cos cos

2 cos ,

cos cos

where 1

ro i tro io

io i t

to ito io

io i t

EE E

E

EE E

E

cient

Transmission coefficient


3/24/2016

13

49

• For no reflection (total transmission):

2 1

2 1

2 1

2 2 2 22 1

1 1

2 2

2 2 1 1 22

1 2

By setting :

cos cos0

cos cos

cos cos

1 sin 1 sin

sin , and

sin

1 ( / ) sin

1 ( / )

i B

B t

B t

B t

B t

ti B

i

B

or

Since

Perpendicular Polarization ‐ Brewster angle

50

Summary

Property Normal Incidence

Perpendicular Parallel

Reflection coefficient

Transmission coefficient

Relation

it

it

coscos

coscos

12

12||

ti

i

coscos

cos2

12

2

ti

ti

coscos

coscos

12

12

2

2 1

2

it

i

coscos

cos2

12

2||

12

12

1 1 t

i

cos

cos1 ||||

51

j(0.866y+0.5z)s x

An EM wave travels in free space with the

electric field component

= 100e V/m

Determine

(a) and

(b) The magnetic

fiel

d compone

(

nt

c

E a

) The time average power in the wave

Example 10.10

52

j( )js o o x

2 2 2

(a) Comparing the given E with

= e = e

it is clear that

0 , 0.866 , 0.5

(0 8 6

. 6

x y zk x k y k z

x y z

x y z

E

k k k

k k k k

k.rE E a

2 2

8

) (0.5) 1

But in free space,

2

Hence, 3 10 rad/s

2 2 6.283 m

o ok =c

kc

k


3/24/2016

14

53

j(0.866 0.5 )

s

2 2

js x

j(0.866 0.5 )s y z

(b) the corresponding magnetic field is given by

0.866a 0.5a0.866 0.5

0.866 0.50.866 0.5

100 e ; = 120

(0.132 0.23 ) e A/m

(c) The tim

y z

k s

y zk y z

y z

y z

a

a

EH

a a

a aH a

H a a

22*

s s k y z

y z

e average power is

1001 = Re (0.866 +0.5 )

2 2 2 120

=11.49 + 6.631 W/m2

oavg

E

E H a a a

a a


54

Example 10.11

r

y

A uniform plane wave in air with

= 8 cos V/m

is incident on a dielectric slab (z 0) with 1 , 2.5 , =0.

Find

(a) The polarization of the wave

(b) The angle o

f

( t 4 )

n

3

i

r

x z

E a

cidence

(c) The reflected field

(d) The transmitted field

E

H

55

8

(a) From the incident field, it is evident that the propagation vector is

4 3 5

Hence, =5c=15 10 rad/s

A unit vector normal to the interface (z = 0)

i x z i o okc

E

k a a

i

is .

The plane containing and is y = constant, which is the xz-plane,

the plane of incidence. Since is normal to this plane, we have

perpendicular polarization (similar to Figure 10.17).

z

z

a

k a

E

Example 10.11 ‐ solution

i i

i i

n

i

4(b) from the figure, tan 53.13

3

Alternatively, we can obtain from the fact that

is the angle between and , that is,

cos .

4 3

5

oix

iz

k n

x z

k

k

k a

a a

a a

i

3.

5

or 53.13

z

o

a

56


3/24/2016

15

57

i y

(c) Let cos( . )

which is similar to form to the given . The unit vector is chosen in

view of the fact that the tangential component of must be continuous

at the interface. From t

r ro r yE t E k r a

E a

E

he Figure:

sin , cos

But = and = = 5

because both and are in

the same medium. Hence

4 3

r rx x rz z

rx r r rz r r

r i r i

r i

r x z

k k k k

k k

k k

k

k k a k a

a a


58

0

1 11

2 2 2

2 1

2 1

0 21 0 2

2

To find , we need . From Snell's law

sin 53.13sin sin sin

2.5

or 30.39

cos cos

cos cos

377where 377 , 238.4

2.5

r t

o

t i i

ot

ro i t

io i t

r

o r

ro

io

E

cn

n c

E

E

E

E

8

238.4cos53.13 377cos30.390.389

238.4cos53.13 377cos30.39

Hence, 0.389(8) 3.112

3.112cos(15 10 4 3 ) V/m

o o

o o

ro io

r y

E E

t x z

E a


59

8

2 2 2 2 2 8

(d) Similarly, let the transmitted electric field be

cos( . )

15 10where 1 2.5 7.906

3 10From the Figure ,

sin =4

cos 6.819

4 6.819

No

t to t y

t r r

tx t t

tz t t

t x z

E t

kc

k k

k k

E k r a

k a a

2

2 1

tice that

2 cos

cos cos

2 238.4cos53.130.611

238.4cos53.13 377cos30.39

ix rx tx

to i

io i t

o

o o

k = k = k

E

E


60

8

t t

tt

2

The same result could be obtained from the relation =1+ . Hence,

0.611 8 4.888

4.88cos(15 10 4 6.819 )

From , is easily obtained as

4 6.8194.888 cos

7.906(238.4)t

to io

t y

k x zy

E E

t x z

E a

E H

a E a aH a

8t

( . )

( 17.69 10.37 )cos(15 10 4 6.819 ) mA/m

r

x z

t

t x z

k r

H a a


10.7 Power and the Poynting Vectorsite.iugaza.edu.ps/hkhozondar/files/2017/02/EMII_Chapter...(b) The...

Documents

Transcript of 10.7 Power and the Poynting Vectorsite.iugaza.edu.ps/hkhozondar/files/2017/02/EMII_Chapter...(b) The...