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1
3
2
6
8
6
(a)
10 1 1 1822.782
loss tangent=tan =1822.78
(b) 1822.78( ) 1822.78(2 5 10 2 ) 1.02 S/m
1.02(c) 2 2 3.227 10
2 5 10
o
c o oj j j
4
2
(d) 1 1 10 Np/m2
(e) intrinsic impedance: 14 45
(1 )
o
j
5
r
o
1/4 1/42 2
11
(a) 30 60
loss tangent=tan = 3
(b) nonmagnetic: 1
// | | 240
1 31
1.1 10
dielectric constant= 1.24
o o
o r
r
6
12
12
2
6 3
(c) tan = 3 3 3 (1.24) 18.8 10
1.24 18.8 10
(d) 1 12
(1.24 ) =(2 10 ) 1 3 1 =16.45 10 Np/m
2
o
c o
o o
j j
7
: is a unit vector in the direction of propagation
-Both (E&H) are in a plane that is normal(transerve)
to direction of wave propagation. This is called TEM waves.
-The direction of E-filed is
k E H
k
a a a
a
called the polarization of TEM
8
skin depth
The skin depth is a measure of the depth to which an EM wave can penetrate the
medium.
The distance δ, shown in Figure through which the wave amplitude decreases by
a factor e-l (about 37%) is called skin depth or penetration depth of the medium.
1
o oE e E e
1
valid for any material medium
9
For good conductors
1 1 = ,
2
1
2 f
skin depth decreases with increase in frequency. Thus, E and H
can hardly propagate through good conductors.
10
The phenomenon whereby field intensity in a conductor rapidly decreases is
known as skin effect.
Skin effect
hollow tubular conductors are used instead of solid conductors.
11
21resistance, ( /m ) real part of
{This is the resistance of a unit width and unit length of the conductor}
,
For a conductor wire
w=2 a
of radius a :
dc
c
c
s
s
ac
a
d
RS
skin R
R =w
R w
w
R
R
2
2
( a )(2 a)
(2 a
at high frequencies,
) (2)
a
2
( )
ac dc
aa
S
a R R
12
6
r
1
for
(a) n
exampl
onmagne
e: let
tic:
40 , 18% of =40*0.18 7.2
reduced by 7.2=40 7.2 32.8
: reduc
1 , =2 10 10 , 24
(1 0.18)
(0.82) ln
ed
(0.82) 0.19 4
% 40*
8
18
o
o
o
o
o
o o
o
o
E E
E
or E by
E E e
E E e
(1 0.18) 32.8
13
2
22
2
tan(2 )= tan(48) 1.11
1 12
1 1.11 12
3.6
tan(2 ) 1.11 1.11( ) 0.0022 S/m
1 1 0.445 rad/sec2
propagation constant:
o r o
r
= +j =0.1984+j0.445
14
22
22
2
2
tan(2 )= tan(48) 1.11
1 11 12
1 11 12
1 1.11 10.19840.445 rad/sec
1 1.11 1
propagation constant: = +j =0.1984+j0.445
Other sol.
15
2(b) = 14m
1(c) 5 m
(d) 1.11( ) 0.0022 S/m
16
6
2
26
66
(a)
49
(2 100 10 )(80 )
1 12
80 = (2 100 10 ) 1 9 1 42 rad/m
2
2 100 1015 10 m/s
42
o
o o
u
u
17
2
26
2(b) 0.15 m
(c) 1 12
80 = (2 100 10 ) 1 9 1 37.5 Np/m
2
10.0266 m
(d) intrinsic impedance: 14 41.83
(1 )
o o
o
j
18
8
8
8
2
2
2(a) =0.1 Np/m , 10 2
22 10 sec
10
2(b) =
1 1 0.1 0.09552
1 1 2.097 rad/m2
2= 3 m
fT
T
19
0.1 8
(c) intrinsic impedance:
(1 )
= 188 2.72 4 (1 0.0955)
| | 12(188) 2256
2.72
, ,
2256 sin( 10 2.097 2.7
oo
o
o o
o
k E H k y H x E z
y
j
j
E H
E e t y
a a a a a a a a a
z2) , +ve E leads H
(d) E leads H by 2.72o
a
20
6
8 8
6
1
6
2
(a) (freespace) \\\\ , , 0 , 0
2 =
2 100.0209 rad/m
3 10 3 10
2= 300 m
(b) At z=0 E =10 cos(2 10 )
300 At z= / 4 75 E =10 cos(2 10 (0.0209 75))
4
o o
o o
y
y
t
t
a
a
6
2
oo
6
E =10 cos(2 10 1.5)
E 10( ) =377 H
377 377
10 H= cos(2 10 0.0209 )( )
377
y
x
t
c
t z
a
a
21
8
o o
H y
8
(a)
(b) , 0 , 0
6 6 (2 10 ) 80.88
( )
41.88 | |
(1 )
E = | | H (25)(41.88) 1047.2
,
E=1047.2 sin(2 10 6 )
k x
o
o r
k x
k E H E z
z
c
j
t x
a a
a a a a
a a a a a
a
22
64
6
6
10(a) 4 10
(5 )(2 10 10 )
(b) 750 , 5 , 0 , 0
(2 10 10 ) (5 )(750 ) 12.8 rad/m
20.49 m
( ) we want to find the phase (rad) for =2m
rad12.8
o
o o
o o
lossless
c
phasem
2m=25.66 rad
(d) 4617
(1 )j
23
8
88
E H k z x y
3
o o
(a) 0 , 0
6 (2 10 ) ( )( ) 5.7
2 2 10(b) 1.25 m , u= 1.257 10 m/s
5
377(c) 157.9
(1 )
(d)
(e) E = | | H (157.9)(30 10 ) 4.737
E=4.
r o o r
o
r o rj
a a a a a a
8737 sin(2 10 5 ) V/myt x a
24
8 8
D
8 2
D
D D
dD dE(f) J = =ε (ε)(2 10 )4.737 cos(2 10 5 )
dt dt
J 0.15 cos(2 10 5 ) A/m
: H J J J
y
y
t x
t x
or
a
a
25
9
o1o1
o2o2
H1 k 1
H2 k 2
9
(a) 0 , 0 , 1
8 (10 ) ( )( ) 5.75
377 157.8
5.75(1 )
(d)
E 50H = 0.3168
| | 157.8
E 40H = 0.253
| | 157.8
H=0.3168 cos(10
r
r o o r
o
r o
E x y z
E x z y
j
t
a a a a a a
a a a a a a
98 ) +0.253 sin(10 8 )( ) A/mz yx t x a a
26
2
2
77
(1 0.6)
(0.4) 2 ln(0.4) 0.458 Np/m
1 = 2.18 m
2 103.9 10 m/s
1.6
o o
o o
E E e
E E e
u
27
7 3 2
6
6 7
7 6 3
2
600(a) 2.287
(5.8 10 ) (1.2 10 )
1 1 1(b) 6.61 10 m
(100 10 ) (5.8 10 )
600= 207.9
(5.8 10 )(6.61 10 )(2 1.2 10 )
(c) (2 )2
1
dc
o
sac
dc ac
RS
f
RR =
w w
aR R S w a a
S w
f
2
2
1 412.13 kHz
2 4
a af
f a
28
7 3 2
67
7 3 3
600(a) 2.287
(5.8 10 ) (1.2 10 )
1 1 1(b) 0.1203 mm (small)
10( ) (3.5 10 )
2
a=12mm not 9mm w=2
40= 0.126
(3.5 10 )(0.1203 10 )(2 12 10 )
dc
o
sac
RS
f
a
RR =
w w
29
1 1
2 2 2
2 is shorter than
f
f
30
7
9 7
7 6
1 1 15.88 10 m
(12 10 ) (6.1 10 )
5 5(5.88 10 ) 3 10 m
of
thickness
31
Poynting's 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 within V minus
the conduction losses.
Poynting's theorem
32
33
2
k(1) Time varying Poynting vector : ( ) (W/m )E H a
(2) is called poyinting=pointing vector since it points in the
direction of wave propagation
2* 2
k
(3) the time average of the Poynting vector
1 = Re[ ] cos
2 2 | |
koavg
EE H e
a
avg
(4) total time average power through a surface(it is scalar in Watt):
P = .dSavg
34
Poynting vector time varying watt/mk
( , , , )
= ( )
x y z t
E H
a
22
k
( , , )
cos 2 | |
avg
ko
x y z
Ee
aThe time average
of the Poynting vector time invariant watt/m
Total time average
power through a surface scalar watt avg P = .dSavg
35
8
k
3
8
2 8
2 8
(a)
6 6 (2 10 ) 8.2
131.56 , 08.2
(b)
( , , , )= ( )
| | (131.56)(30 10 ) 4
4 cos(2 10 6 )( )
(4)(3)cos (2 10 6 )( )
0.1124cos (2 10 6 )
o r
o
o
o o
z
z y
x y z t E H
E H
E t x
E H t x
t x
a
a
a a
a2 W/mx
36
22
k
20 2
x x
3 2
avg
0 0
(c) ( , , ) cos 2 | |
(4) cos0 0.06 W/m
2(131.65)
P = .dS= 0.06 0.36 watt
koavg
avg
Ex y z e
e
dydz
a
a a
37
8
8
(a)
22 2.82 10 rad/sec
177.6
(40 / ) 0.255
177.6
0.255sin(2.82 10 2 )
oo
H K E z
EH
H t z
a a a a a a
a
38
k
2 8
2 8 2
2
22
k
2
0 2
z z2
av
(b)
( , , , )= ( )
40 0.225( )( )sin (2.82 10 2 )( )
9sin (2.82 10 2 ) W/m
( ) ( , , ) cos 2 | |
40( )
4.5 cos 0 W/m
2(177.6)
P
z
koavg
x y z t E H
t z
t z
Ec x y z e
e
a
a a
a
a
a a
2 0.03 2 0.03
g 2
0 0 0 0
4.5 4.5= .dS= 11.455 Wattavg d d d d