propagation of electro-magnetic waves - delta...
TRANSCRIPT
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