1

Propagation of Electro-magnetic Waves

Dr. M.A.Motawea

In this chapter we shall apply Maxwell’s equations to introduce the fundamental theory of wave

motion.

The uniform plane wave represents one of the simplest applications of Maxwell’s equations, and yet it

illustrates the principles behind the propagation of energy.

We shall introduce the velocity of propagation, u, wavelength. , wave impedance, , phase

constant, , attenuation constant, , and the use of Poynting theorem in finding the power density.

Finally we shall consider reflection and transmission of a uniform plane wave at the boundary between

two different media.

The use of stand wave ratio and input impedance will prepare us to consider many of the practical

problems of the guided transmission of power and information in the following chapter.

Wave motion in free space:

As we indicated in our discussion of B.C. in the previous studies, the solution of Maxwell’s equations

without the application of any B.C. at all represents a very special type of problems.

Although we restrict our attention to a solution in Cartesian coodt , it may seen even then that we are

solving several different problems as we consider various special cases.

Solutions are obtained 1st for good conductor, then for perfect dielectrics, next for lossy dielectrics,

and finally for free space conditions, we do this to take advantages of the approximation that are

applicable to each special case and to emphasize the special characteristics of wave propagation in

these media.

A- Wave Equation:

Maxwell’s equations may be written in terms of E and H : tje

HjBt

XE

(1)

EjEDt

EXH

(2)

2

by taking curl eqn 1:

XHjXBt

XEX

(3)

Substitute eqn 2 into eq

n 3 gives:

XHjXBt

XEX

)().( 2 EjEjEE

EjjE )(2 (4)

EE 22 (5)

Where: )(2 jj (6)

Then , propagation constant:

jjj )( (7)

Where : attenuation constant

wave number

Eqn 5 is wave equation for electric field, also the wave eq

n for magnetic field is:

HH 22 (8)

- ( attenuation constant)& ( wave number):

Eqn(7)

2 :

2222 2 jj (9)

3

From eqn (9) we see that: 222 (10)

2 (11)

Eqn(10)

2 + Eq

n(11)

2 :

222224224224 42

]1[)(22

2224222

22

2222 1

(12)

Eqn(10) + Eq

n(12) :

2

22

222 12

]11[222

222

2

1

22

2

]11[2

(13)

Eqn(12) - Eq

n(10) :

2

22

222 12

4

]11[2 22

2

(14)

,, parameters of wave eqn (5), (8).

In cartesian coordinate system :

EEz

Ey

Ex

E xxx

2

2

2

2

2

2

22

i.e. xx EEz

2

2

2

(15)

Also, yy HHz

2

2

2

direction of propagation , z direction. E in x-direction, H in y-direction

Remember this shape:

directionZ .

ZHE yx ,, direction of travel are mutually orthogonal

The solution of wave eqn (15): x

ztj

mx aeeEtzE ),(

x

ztj

mx

ztj

mx aeeEaeeEtzE ),( (16)

Forward direction backward direction

5

plane wave solutions, depend on the properties of any medium ( ,, ).

Also,

y

ztj

my aeeHtzH ),(

y

ztj

my

ztj

my aeeHaeeHtzH ),( (16)

- Intrinsic impedance of the medium:

From Faraday’s Law:

HjXE

E in x-direction, H in y-direction

y

zyx

zyx

Hj

EEE

zyx

aaa

, yyx HjaEz

,

y

z

m

z

m HjeEeEz

)( ,

)( z

m

z

m

z

m

z

m eHeHjeEeE

,

By equating both sides, then

mm HjE ,

mm HjE

j

H

E

m

m

,

j

H

E

m

m

j

,

j

(forward &backward direction)

where is intrinsic impedance of the medium,

6

j

j

jj

j

)( (17)

EH

- Phase velocity, propagation constant:

We have:

x

ztj

mx

ztj

mx aeeEaeeEtzE ),(

Take forward direction only then,

)(

),(

ztjz

m

tjzjz

mx

ztj

m

eeE

eeeEaeeEtzE

conszt , dt

dz , v ,

v

phase velocity of the wave (18)

v propagation constant (19)

B- Types of media:

EI c , EjEt

I d

,

d

c

I

I

50

for perfect conducting medium

7

50

1

for perfect dielectric medium

1- Perfect conducting medium ; Skin depth:

Materials are classified as good conductors when 50

, in the range of

practical frequencies.

Propagation constant:

)1()(

jjjj

)4/sin4/(cos4/2/ jeejj jj

jj 22

2

(20)

Intrinsic impedance:

4/2/

)/1(

je

jj

j

j

j

j

jXRjj

22)4/sin4/(cos

2 XR (21)

It is seen that for all conductors the E and H waves are attenuated.

Numerical examples will show that, this is a very rapid attenuation.

At each fixed location H is out of time phase with E by 450 or 4/ rad.

Once again assume E in ax and propagation in az , the field equations are:

8

x

ztjz aeeEtzE )(

0),(

y

ztjz aeeE

tzH )(0),(

The term of attenuation is Ae z

eAdB 10log20

2

4

2/2

22

ff

-skin depth or depth of penetration

f

1

(22)

2

4.

2

eeef

dBeAdB 5.54log20 2 very high

So, no wave can propagate in good conductor.

velocity of propagation v,

24

2/2

2 f

f

fv

2- Perfect dielectric:

)02.0(,50

1

0 , ,

9

jjjjj 1)( 2

Then, 0 ,

Also, phase velocity(v):

0)( vztdt

d ,

1v

Wavelength, :

f

122

Intrinsic impedance:

j

j

Field equations: ),( tzEx , ),( tzH y

x

ztj

mx

ztj

m

x

zjtj

mx

zjtj

mx

ztj

mx

ztj

mx

aeEaeE

aeeEaeeEaeeEaeeEtzE

)()(

),(

y

ztj

my

ztj

m

y

zjtj

my

zjtj

my

ztj

my

ztj

my

aeHaeH

aeeHaeeHaeeHaeeHtzH

)()(

),(

10

because of the term of attenuation is negligible( 0 )

3- Free space:

Free space is nothing more than the perfect dielectric for which

mH /10*4 7

0

mF /1036

1 9

0

also, 0

Maxwell’s equations in this case :

Ht

Bt

XE

0

Et

Dt

XH

0

Et

EE2

2

00

2).(

Etv

Et

E2

2

22

2

00

2 1

Etv

E2

2

2

2 1

Uniform plane wave propagated in z-direction for E&H

Htv

H2

2

2

2 1

v --- velocity of light (3*10

8 m/s)

The field components of E &H are function of (z,t) only, i.e. 2

2

2

2

2

2

2

22

zzyx

Then, the plane wave becomes:

11

Etvz 2

2

22

2 1

H

tvz 2

2

22

2 1

plane wave equation for E&H

x

ztj

mx aeeEtzE ),(

x

ztj

m

ztj

mx aeeEeeEtzE )(),(

x

ztjz

m

ztjz

m

x

tjzjz

m

tjzjz

m

aeeEeeE

aeeeEeeeEtzE

)(

)(),(

)()(

Forward direction backward direction

H(z,t) is similar to E(z,t)

v

00 , v..velocity of light, Sometime denoted by k

f

vv

22

0

0

0 120

j

j intrinsic impedance of free space

Example: if a wave x

kztj

m aeEtzE )(),( is propagated in free space with

Erms=10v/m, f=900 MHz, Find: H, k, ,vph ,P

Solution: mAEH m /37.0/ 0

68.18

10*3

10*900*2228

6

c

fk

smv ph /10*31 8

00

wattaaaaE

EXHPRp zzzz 6.26

5

12

10

2)(.

2

1

0

2*

12

Poynting Vector & power considerations

In order to find the power in a uniform plane wave, it is necessary to develop a power

theorem for the electromagnetic field known as Poynting theorem. It was originally

postulated in 1884 by an English physicist, John H. Poynting.

Let us begin with Maxwell’s equations:

Bt

XE

B

tHXEH

.. 1

Dt

JXH

D

tEJEXHE

... 2

By subtracting Eqn 2 - Eq

n 1 , yield

Bt

HDt

EJEXEHXHE

..... 3

By using vector identity,

XEHXHEEXH ..).(

Which may be proved by expansion in Cartesian coordinate

Then, Eqn 3 becomes,

Bt

HDt

EJEEXH

...).( 4

By integrating Eqn 4 over volume V bounded by a closed surface S

dVBt

HdVDt

EJdVEdVEXHV V VV

...).( 5

Apply Divergence theorem:

S V

DdVdSD ..

13

Then, Eqn 5 becomes,

dVt

BHdV

t

DEJdVEdSEXH

V VVS

...).(

6

a b c

a- 2.. EEEEJ Energy dissipated /time

b- Et

EEt

EEtt

EE

t

DE

2)().(.. 2

But we have, t

EE

t

DEE

tEE

t

2

2

2

2

1. E

tt

DE

t

E

t

EE

t

DE

2

2

1.. Electric energy/m

3

c- t

HH

tHB

tH

2

2

1.. Magnetic energy /m

3

Eq 6 becomes:

dV

t

BHdV

t

DEJdVEdSEXH

V VVS

...).(

14

dV

t

HdV

t

EdVEdSEXH

V VVS

222

22.

VSVV

dVEdSEXHdVt

HdV

t

E 222

).()22

(

Energy dissipated as heat/time

(ohmic term)

Rate of energy entering the volume from outside

P(EXH) poynting vector

م/المعدل اللحظى لسريان الطاقة) 2

)

Densities of energy stored in the electric & magnetic field

(-ve sign means : decreasing in stored energy) EXH dS

dVDVV

EXHP

)Re(2

1 *EXHPav

Complex power in circuit analysis: *

2

1VIS ,where

*Re2

1VIP .

For plane waves, the direction of energy flow is the direction of propagation.

Thus the poynting vector offers a useful, coordinate-free way of specifying the direction

of propagation , or of determining the directions of the fields if the direction of

15

propagation is known. This can be particularly valuable where incident , transmitted,

and reflected waves are being examined.

For THV fields:

)(

00

tjjtj eEeeEE )(

00

tjjtj eHeeHH

Instantaneous Power:

)]cos()[cos(.. 00 ttXHEHEXRPRPPinst

dtT

XHEdtPT

P

TT

inaver )]cos()2[(cos2

111

0

00

0

).(2

1]

2

1.[

.2

1)][cos(

1

2

1

*

00

)(

0000

EXHRPeeXHeeERP

eRPXHETT

XHEP

jtjtjj

j

aver

Poynting vector = average power density

Poynting vector represents the directional energy flux density (the rate of energy

transfer per unit area ) of an electromagnetic field.

S=EXH [watt/m2]

16

Reflection of normally incident plane wave from planer interface

General medium:

We have from eqn 16 that:

x

ztj

mx

ztj

mx aeeEaeeEtzE ),(

y

ztj

my

ztj

my aeeHaeeHtzH ),(

Forward &backward components

j

j

- Incident wave:

x

ztj

mi aeeEE 1 y

ztjmy

ztj

mi aeeE

aeeHH 11

1

- Reflected wave:

x

ztj

mr aeeEE 1

y

ztjmy

ztj

mr aeeE

aeeHH 11

1

- Transmitted wave:

x

ztj

mt aeeTEE 2 y

ztjmt aee

ETH 2

2

X

Z

Y

transmittedincident

reflected

Hi

Hr

Ht

Ei Et

Er

medium 1 medium 2

17

-----reflection coefficient T -----transmission coefficient

Apply B.C. at the interface between 2 media:

1- 21 tgttgt EE i.e trxrefinc EEE

x

ztj

mx

ztj

mx

ztj

m aeeTEaeeEaeeE 211

zzz

Teee 211 ,

at 0z

T1 17

2- 21 tgttgt HH i.e trxrefinc HHH

trxrefinc HHH

y

ztjmy

ztjmy

ztjm aeeE

TaeeE

aeeE

211

211

zzz

eTee 211

211

111

at z=0

211

111

T

18

By solving eq 17, 18

211

1)1(

11

,

21

11

)( 121212

18

12

12

12

2

12

12 21

T

- at no reflection, 0 12 T=1

- when 1st medium is air ,

120

0

0

01

22

22

j

j

- when 1st medium is air & 2

nd medium is perfect dielectric,

120

0

0

01

0

2

2

2

2

2

22

r

r

j

j

1

1

2

2

2

2

12

12

r

r

r

r

,

1

222

2

2

2

2

00

2

2

0

2

2

12

2

r

r

r

r

r

r

r

r

T

Pi = incident power density

zm

zm

y

ztjmx

ztj

miii aE

aE

aeeE

XaeeERPXHERPP

2402][

2

1][

2

12

0

2

1

* 11

Pt= transmitted power density

zm

y

ztjmx

ztj

mttt aET

aeeE

XTaeeTERPXHERPP2

22

2

*

2][

2

1][

2

122

Pr = reflected power density

zm

y

ztjmx

ztj

mrrr aE

aeeE

XaeeERPXHERPP1

22

1

*

2][

2

1][

2

111

19

- when 1st medium is air &2

nd medium is good conductor,

120

0

0

01

jXRjj

jf

jj

eejj

j

j jj

222

2

2

2

2

2

2

4/

2

22/

2

2

2

2

2

2

22

22

)1(

)1()2

1

2

1()4/sin4/(cos

2

XR ’ )1(

2

*

2 j

2

11*

2

j

For air

Pi = incident power density

zm

zm

zm

y

ztjmx

ztj

miii

aE

aE

aE

RPaeeE

XaeeERPXHERPP

2402

][2

1][

2

1][

2

1

2

0

2

1

2

1

* 11

Pt= transmitted power density

z

zm

z

zm

yzjzm

z

zjz

my

ztjm

x

ztj

mttt

aeE

TaeE

RPT

aeeE

XTaeeTERPaeeE

XTaeeTERPXHERPP

222

][2

1][

2

1][

2

1

22

222222

2

*

2

2

22

*

2

2

2

*

22

*

PEC

no field

air

2

medium 1 medium 2

20

- when 1st medium is air&2

nd is perfect electric conductor(PEC)

12001 PEC 0,,,, 2

10

0

0

0

12

12

0

2

21

2

T wave is totally reflected

Total field in air region:

reflectedincidenttotal EEE

x

ztj

mx

zjtj

mx

ztj

mi aeEaeeEaeeEE)( 111

,

x

ztj

mx

ztj

mr aeEaeeEE)( 11

xmx

tjzjzj

mx

zjtjzjtj

m

x

zjtj

mx

zjtj

mx

zjtj

mx

zjtj

mtotal

azjEaeeeEaeeeeE

aeeEaeeEaeeEaeeEE

)sin(2)()( 11111

1111

xmx

tjzjzj

mtotal azjEaeeeEzE )sin(2)()( 111

reflectedincidenttotal HHH

y

ztjmy

ztj

mi aeeE

aeeHH 11

1

, y

ztjmy

ztj

mr aeeE

aeeHH 11

1

y

tjzzmtotal

y

ztjmy

ztjmy

ztjmy

ztjmtotal

aeeeE

H

aeeE

aeeE

aeeE

aeeE

H

)( 11

1111

1

1111

ym

y

tjzzmtotal za

Eaeee

EzH 1

01

cos2

)()( 11

j

eekz

jkzjkz

2)sin(

,

2)cos(

jkzjkz eekz

21

- SWR

Incident wave:

x

ztj

mi aeEE)( 1

x

ztj

mx

ztj

mr aeEaeEE)()( 11

xm

x

tjzjzj

m

tazjE

aeeeEtzE

sin)sin(2

)(),(

1

11

The ratio of the maximum to minimum amplitudes

is called SWR

1

1)(SSWR 1

1 if reflected amplitude = incident amplitude

All the energy incident is reflected, S

PEC

no field

air

2

medium 1 medium 2

lossless

totally reflected

Z

X

22

Microwave Engineering

Sheet #2

(2/11/2015)

Plane Wave equation

Q1: For electric circuit , V=V0 I=I0 . Find: ½ RP [ VI*]

Q2: An EM field E = ( E0ax + 2 E0ay) , E0 =10 V/m ,

k/ωµ=1/120π Find: magnetic field ?

Q3: An EM field E = E0 ax . Find H &P ?

Q4: If H =H0 ay, Find E? H0= 20 A/m , k/ωµ=1/120π

Q5: If E = E0 ay , Find: H?

Q6 : An EM wave has its E = ( E1ax + 2 E2ay) , assume free Space , E1 =2

V/m E2 =1 V/m k/ωµ=1/120π Find : magnetic field, poynting vector, electric

energy stored, magnetic energy stored .

Q7: If a wave x

kztj

m aeEtzE )(),( is propagated in free space with Erms=10v/m,

f=900 MHz, Find: H, k, , vph , P.

Q8: Consider a uniform plane wave in homogeneous medium with Ex , Hy. Show

that

j

j

Q9: x

ztjz aeAeE )( , y

ztjz aeeA

H )(

, Find: Poynting vector P?

Q10: Show that the following field vector in free space satisfy Maxwell’s equations,

xaztEE )cos(0 , yaztE

H )sin(0

23

Good luck, Dr. M.A.Motawea