1
School of Electrical and Computer Engineering, Cornell University
ECE 303: Electromagnetic Fields and Waves
Fall 2007
Homework 6
Due on Oct. 05, 2007 by 5:00 PM
Reading Assignments:
i) Review the lecture notes.
ii) Review sections 1.5, 3.33.6 of the paperback book
Electromagnetic Waves
.
Special Note:
Graders have been instructed to take off points (as much as 50%) if proper units are not
included in your answers. You must specify the correct units with your numerical answers.
Problem 6.1: (Conductive Media)
Consider a plane wave in a conductive medium given by the expression:
( )
z
jk
o
e
E
x
r
E
−
=
ˆ
r
r
The medium has a conductivity
σ
and a dielectric permittivity
ε
. As explained in the lectures, the wave
decays because the wave looses energy due to dissipation in the medium. The Poynting theorem is:
(
)
(
)
(
)
[
]
(
)
(
)
t
r
E
t
r
J
t
r
W
t
r
W
t
t
r
S
m
e
,
.
,
,
,
,
.
r
r
r
r
r
r
r
r
+
+
∂
∂
=
∇
−
The timeaverage version of the Poynting theorem for timeharmonic fields is (as in homework 4):
(
)
(
)
(
)
t
r
E
t
r
J
t
r
S
,
.
,
,
.
r
r
r
r
r
r
=
∇
−
Using phasors this becomes:
( )
[
]
( )
( )
[
]
r
E
r
J
r
S
r
r
r
r
r
r
*
.
Re
2
1
.
Re
2
1
=
∇
−
This is saying that if a wave is slowly loosing energy, and so the Poynting vector is a function of position,
then there must be dissipation going on. Verify the above relation for the plane wave propagating in a
conductive medium given above (without making any “lossy dielectric” or “imperfect metal”
approximations). You need to evaluate the left and right sides of the equation separately and show that
they are equal.
Hint
: First show (without making any approximations) that for any conductive medium:
σ
µ
ω
o
k
k
=
'
'
'
2
.
Problem 6.2: (Ground Penetrating Radar  GPR)
A conductor is considered a good conductor for the frequency
ω
of interest if the loss tangent
ε
ω
σ
is
much greater than unity (i.e.
1
>>
ε
ω
σ
) and a bad conductor if
1
<<
ε
ω
σ
. These conditions can also
be stated in terms of the
dielectric relaxation time
σ
ε
τ
=
d
:
Good conductor:
1
<<
d
τ
ω
(This is also the “imperfect metal” case discussed in the lecture notes)
Bad conductor:
1
>>
d
τ
ω
(This is also the “lossy dielectric” case discussed in the lecture notes)
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2
Solid ground has a conductivity of approximately
3
10
5
−
×
S/m and a permittivity
ε
equal to
o
ε
10
. You
need to design a ground penetrating radar (GPR). You have at our disposal two sources of
electromagnetic radiation – one at a frequency of 10 kHz and the other one at a frequency of 100 MHz.
a) Figure out if ground is a good or a bad conductor at each of the two frequencies: 10 kHz and 100 MHz
(don’t forget that
f
π
ω
2
=
).
b) Figure out the penetration depth of electromagnetic radiation in ground at the two frequencies: 10 kHz
and 100 MHz. Take penetration depth to be the depth at which the
time average power
in the
electromagnetic radiation going into the ground drops to 1% of its value at the surface.
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 Fall '06
 RANA
 Polarization, Electromagnet, Hertz, electromagnetic wave, circularly polarized output

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