Lecture 8
•Traveling Electromagnetic Waves
•Using Ampere’s Law and Faraday’s
Law to derive the EM Wave Equation
•Solution to the Wave Equation
•Wave velocity “vp”
•Intrinsic impedance of medium “η”
•Waves in Lossy (Conductive) Media
•Skin Depth “δ”
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Lecture 8 EM Wave (KEH)
1
Electromagnetic Wave
•
•
•
•
If a sinusoidal current i(t) is set up in a vertical wire (antenna), Ampere’s Circuital Law
predicts that a circular time-varying H field lines will be set up (horizontally) around the
wire.
Faraday’s Law predicts that circulating time-varying E field lines will be set up (vertically
linking) around these varying H field lines
Ampere’s Circuital Law predicts that circulating H field lines will be set up horizontally
around these varying E field lines, etc.
A propagating electromagnetic wave results with H oriented horizontally (H = Hyiy) and
E oriented vertically (E = Exix)
x
H
i(t)
+
H
H
z
y
E
E
E
sin(ωt)
-
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Lecture 8 EM Wave (KEH)
2
If the observer is far enough away from the wire antenna, the cylindrical wave
front becomes “locally plane”. A “snapshot” of the sinusoidal traveling wave
that results, with B oriented horizontally along y direction, E oriented along the x
direction, and the wave traveling in the z direction, is shown below:
y
x
Fig. 8-1 Locally Plane EM Wave
dz
z
dz
x
This is called a “uniform plane wave” since Hy and Ex depend only on z and t
and NOT on x or y. Thus there is no field variation in the z = constant plane.
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Lecture 8 EM Wave (KEH)
3
Fig. 8-2. Two views of Traveling EM Plane Wave:
(a) xz plane showing E field (b) yz plane showing B (or H) field
x
h
zx
dz
y
h
zx
dz
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Lecture 8 EM Wave (KEH)
4
Applying Faraday’s Law of Induction around shaded contour in (a)

 Ex dl

d
 
dt
(8-1)
C
Note there is NO contribution around top and bottom of the closed
rectangular loop “C”, which has width “dz” and height “h”, since E = Exix
is perpendicular to top and bottom loop segments. Thus the left-hand
side of (8-1) may be written as:

 Ex dl

C

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 Ex  dEx h   Ex h
  Hy   dz h
=>
d

dt
Lecture 8 EM Wave (KEH)
dEx  h
  dz h 
d
Hy
dt
5
Faraday's Law =>
dEx  h
d
  dz h  Hy
dt
dEx
=>
dz
 
dHy
dt
Both Hy and Ex are functions of both time (t) and space (z). When
evaluating dEx/dz we must assume t is constant because Figure(a) is
a instantaneous snapshot at a fixed time. Likewise, when evaluating
dHy/dt, we must assume that z is a constant, since we are calculating
the time rate of change of Hy at a fixed location. To mathematically
take this into account, we replace the derivatives in the above
equation by partial derivatives:
E x
z
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 
H y
(8-2)
t
Lecture 8 EM Wave (KEH)
6
Fig. 8-2. Two views of Traveling EM Plane Wave:
(a) xz plane showing E field (b) yz plane showing B (or H) field
x
h
zx
dz
y
h
zx
dz
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Lecture 8 EM Wave (KEH)
7
Applying Ampere’s Circuital Law to the shaded area in (b)

 Hy dy

d 
   Ex ds
dt 
C
S
We assume a
nonconducting medium
(σ = 0) so the
conduction current term
in Ampere’s Circuital
Law has been omitted
(8-3)
Left Side of Ampere's Circuital Law

 Hy dy

 Hy  dHy  h  Hy  h
h  dHy
C
Right Side of Ampere's Circuital Law
d 
   Ex ds
dt 

dEx
dt
  h  dz
Assuming Ex is
uniform over surface S
S
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Lecture 8 EM Wave (KEH)
8
Thus Ampere's Circuital Law yields
h  dHy

dEx
dt
  h  dz
=>
Hy
z
E x
Recalling Equation (8-2):
 
z
 
Ex
(8-4)
t
H y
(8-2)
t
Differentiating (8-2) with respect to z, exchanging the order of differentiation, and
substituting in (8-4) yields
2
 Ex
2
 
z
  Hy 


z  t 
2
=>
 Ex
2
z
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 
2
  
 Ex
t
2
  Hy 


t  z 
 

  

t 
t 
This the “free
space” E-field
wave equation!
Lecture 8 EM Wave (KEH)
Ex 
(8-5)
9
In similar fashion, we may find a wave equation for the magnetic
field “Hy” by differentiating (8-4) with respect to distance (z),
changing the order of differentiation, and substituting in (8-2).
2
  Hy
2
z
Ex 

  

dz 
t 
2
2
=>
  Hy
2
z
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Ex 

  

dt 
z 
  
d Hy
dt
dHy 
 
    

dt  
dt 
2
  
d Hy
dt
2
Free Space H-field
wave equation
2
Lecture 8 EM Wave (KEH)
(8-6)
10
Solution to Plane Wave Equation
The solution to (8-5) is given by
E x  z  t
f  t  z    g  t  z  
(8 – 7)
Where f and g are arbitrary functions of the space-time arguments
 tgzt z  
z gand
E x Ez xtz  tf  t f zt 
Thus the wave equation has an infinite number of possible solutions!
The specific shape of f and g corresponding to a particular situation is
dictated by the wave shape of the source current, i(t), that flows in the
antenna wire.
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Lecture 8 EM Wave (KEH)
11
Verification of Solution to Plane Wave Equation
E x  z  t
f  t  z
   g  t  z
u
For notational convenience, let
Thus
Ex
f  u  g  w
 
Start with the
proposed solution
t  z  and
w
t  z 
where u and w are functions of z and t
Using chain differentiation:
dEx
dz
2
d Ex
2
dz
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df du dg dw



du dz dw dz
=>
d f  du 
df d u d g  dw 
dg  d w 
  


  
 2
2  dz 
2
2
du
dz
dw


du
dz
dw
 dz 
2
2
2
2
Lecture 8 EM Wave (KEH)
2
2
12
du
But note:
dz
2
d Ex
2
dz
 
 d 2f d 2g 
  


2
2
dw 
 du
dw
dz
2

2
2
dz
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 d 2f d 2g 
  


2
2
dt 
 dt
d w
2
2
dz
0
dz
 d2f  du  2 d2g  dw  2 
  
  

 
2  dt 
2  dt 
dw
 du

(=1)
d Ex
2
d u
(=1)
2
 
d Ex
dt
2
Lecture 8 EM Wave (KEH)
This is the right-hand
side of the wave
equation (8 – 5), so
we have proved that
this proposed solution
(8 - 7) is indeed valid!
13
Interpretation of solution to plane wave equation
The general function
f  t  z  
(be it a sinusoid, a rectangular pulse, a triangular pulse, etc.) represents a
traveling wave moving in the positive z direction, and will thus be called the “+”
wave. To illustrate this, imagine that we desire to “follow” a certain point on the
waveform. At time t1, let us imagine this point on the waveform is at position
z1. Then at a later time t2, this point moves to position z2. Since we are
following a specific point (or specific value of f), we require

f t 1  z 1  


f t 2  z 2  

Since f is arbitrary, we must, in turn, require that
t 1  z 1  
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t 2  z 2  
Lecture 8 EM Wave (KEH)
14
Thus the velocity of that point (and hence the velocity of the entire traveling wave)
must be given by
vp
z2  z1
t2t1
z
t
1
(8 – 8)

For an electromagnetic wave in free space
10
9
F
 0 
36   m
vp_air 
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1
 0  0
 0  4    10
7 H
vp_air  3  10
Lecture 8 EM Wave (KEH)

m
8m
s
This is the
standard
value of the
“speed of
light”
15
In glass
 r  5.0
vp_glass 
1
 r  0   r  0
In distilled water
vp_water 
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 r  1.0
 r  80.0
1
 r  0   r  0
vp_glass  1.342  10
8m
s
 r  1.0
vp_water  3.354  10
Lecture 8 EM Wave (KEH)
7m
s
16
Likewise, the function
g  t  z  
Represents a wave traveling in the –z direction with the same speed:

g t1  z1  
t1  z1  
vp
z
t


g t2  z2  

t2  z2  
z2  z1
t2  t1
1

(8 – 9)
The superposition (sum) of functions f and g represents the most general
solution to the wave equation, where the f function represents a forward
moving wave component away from the radiating source, and the presence
of the g function represents a reflected wave moving back toward the
radiating source. If no reflections are present, the g function will be zero.
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Lecture 8 EM Wave (KEH)
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Finding Hy from Ex: Intrinsic Impedance “η”
We can find a corresponding traveling wave solution for Hy by substituting
(8 - 7), which is our solution for Ex into either (8 - 4) or (8 – 2).
Ex
z
Ex
z
Ex
z
 
Hy
(8-2)
t
f  t  z       t  z     g  t  z       t  z    



z
z
  t  z    
  t  z    
f  t  z    
  t  z    
 
g  t  z

  
  t  z
  
  
  
 
 
Hy
t
Hy
t
Since we are applying this equation at a fixed point in space, z = constant, so
t
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  t  z    
  t  z    
Lecture 8 EM Wave (KEH)
18
Multiplying through by δt and then integrating both sides yields
Hy
1

f  t  z
   
1

 g  t  z
  
where



(8 – 10)
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Lecture 8 EM Wave (KEH)
19
Note that the (+z) progressing Ex wave component “f” (= Ex+) must be
scaled (divided) by +η to turn it into the corresponding Hy (+z)
progressing wave. In similar fashion, the (-z) progressing Ex wave
component “g” (= Ex-) must be divided by –η to turn it into the
corresponding Hy (-z) progressing wave.
Since Ex+ / η = Hy+ and Ex- / -η = HyThe units of η must be (V/m) / (A/m) = Ohms. Thus η is called the
“intrinsic impedance” of the medium. Note that η sets the ratio between
the positive-moving Ex and Hy waves. Likewise, -η sets the ratio
between the negative-moving Ex and Hy waves.
For a wave in free space or air,
0 
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0
0
 0  376.991 
Lecture 8 EM Wave (KEH)
20
Sinusoidal EM Waves
•
•
•
The shape of the f and g functions is determined by the wave shape of i(t)
that flows in the radiating source (antenna).
A sinusoidally time-varying source makes f and g take on a sinusoidal form,
while a digital pulse generating source makes f and g take on the form of a
rectangular pulse.
For a sinusoidal source, the resulting EM wave moving in the +z direction is:
Ex
A  cos    t    z




A  cos    t 

 

 z
f  t  z    
f  t 

z 
v p 
WHERE
ω = angular frequency in radians/second of Ex at a fixed point in space (where z = constant)
β = “phase constant” in radians/meter
= rate at which phase of wave changes with distance in a “snapshot” of Ex at a fixed instant
of time (where t = constant)
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Lecture 8 EM Wave (KEH)
21
Ex
A  cos    t    z
Note: for a fixed time (t = constant => a “snapshot” of Ex), the
traveling wave will vary through one complete sinusoidal cycle when
βz changes through 2π radians. Thus the length of one complete
cycle of the wave, frozen at an instant of time, is called the
“wavelength” (λ), where
λ = 2π / β
Note the velocity of propagation of the wave is
vp
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1

 

Also, the phase
constant is
given by
Lecture 8 EM Wave (KEH)

   
22
Example:Consider the EM wave Ex = 2cos(4πt – πz) => ω = 4π r/s, β = π r/m
Note that this implies
λ = 2π/ β = 2 m (=>one complete Ex(z) cycle every 2 meters for fixed time “snapshot”)
vp = ω/β = 4 m/s (Every ¼ second, the wave moves 1 meter in the +z direction)
t  0
2
2  cos  4    t  z
A
0
λ=2m
2
0
0.5
1
1.5
1
t 
4.
2
2.5
3
z
2
A
2  cos  4    t  z
0
2
2
t 
4.
0
0.5
1
1.5
2
2.5
3
z
2
2  cos  4    t  z
A
0
2
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Note that from the fixedtime “snapshots” of Ex(z)
shown here that the
wavelength of the
propagating wave (the
length of one complete
cycle of Ex(z)) is λ = 2m,
and that reference point
“A” travels at 4 m/s, or
1 m every ¼ second, as
predicted above.
0
0.5
1
1.5
2
z
2.5
3
Lecture 8 EM Wave (KEH)
23
Uniform Plane Waves in Lossy (Electrically Conductive) Medium
• Of special interest in EMC problems is how plane EM waves attenuate as
they travel through various conductive media, such as sea water, moist earth,
or even various conductors, such as copper or aluminum.
• In such materials, the electric field component (Ex) in an EM wave sets up
conduction current according to the microscopic form of Ohm’s Law, Jx =
σEx.
• This induced conduction current serves to partially “short out” the E field
component of the EM wave, and thus attenuates (gradually reduces the
strength) of the EM wave as it propagates through the conductive medium.
• This attenuation is sometimes undesirable (such as when transmitting a cell
phone signal through a metal wall of a building, or a transmitting a
submerged submarine’s radio communication signal through sea water).
• But this attenuation is often desirable in RF shielding applications, such as
the use of copper mesh to screen out interfering radio waves in a laboratory’s
“screen room”, or the use of shielded coaxial cable in a cable TV distribution
system, where outer shield on the coaxial cable must be made thick enough
and of the right kind of material in order to successfully shield the
transmitted cable signal from radiating out of the cable and interfering with
other wireless service. At the same time, this coaxial shield must prevent the
emissions from wireless services from entering the cable TV system!
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Lecture 8 EM Wave (KEH)
24
Sinusoidal Steady-State Lossy EM Plane Wave Equation
Faraday’s Law applied around the closed contour of Fig. 8-2(a) does not change
if the medium is conductive, and we still end up with Equation (8-2)
2
E x
z
 
H y
2  cos  4t
   t  z
(8 – 2)
0
But when Ampere’s Circuital Law is applied around the closed contour of Fig.
2
0
0.5 involving
1 J (=1.5
8-2(b), we must now include the conduction current
term
σE)
that was formerly omitted.
z

 Hy dy

C

 Hy dy

C
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2
2.

d 
   Ex ds   Jx ds
dt 

S
S

d 
   Ex ds     Ex ds
dt 

S
S
Lecture 8 EM Wave (KEH)
25
Working as before to derive the plane wave equation for a lossy medium:
Left Side of Ampere's Circuital Law

 Hy dy

 Hy  dHy  h  Hy  h
h  dHy
C
Right Side of Ampere's Circuital Law


 Ex
d
   Ex ds  
ds
σEx
dt 
 

S

dEx
dt
  h  dz    Ex   h  dz
Note the additional
conduction current
term now present!
S
Thus Ampere's Circuital Law yields
h  dHy
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 dEx

   Ex   h  dz
 
 dt

=>
Hy
z
 
Lecture 8 EM Wave (KEH)
Ex
t
   Ex
(8 - 11)
26
Combining the Faraday’s Law result (8-2) with the new
Ampere’s Circuital Law result (8-11)
  Hy 
2
 Ex
2


z  t 
 
z
2
=>
  Ex
2
z
 
  Hy 
  Ex
t
2

    
   Ex
t 
t



t  z 
2
  

   
Ex
Ex
t
(8-12)
Plane EM wave equation for conductive
medium. Notice the additional “loss”
term that is now present.
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Lecture 8 EM Wave (KEH)
27
Sinusoidal Steady State Solution to the Lossy Wave Equation
The wave equation (8-12) has been rewritten to remind us that Ex is a “time-domain” function of
time and space (t and z)
  E x  t  z
2
  E x  t  z
2
  
2
z
t
2
   
E x  t  z
t
(8-12)
Let us solve the wave equation assuming sinusoidal steady state excitation. This
permits us to use phasor representation. In phasor form, the lossy wave equation
becomes:
2
 E x
2
z
2
      E x  j        E x
(8-13)
Note that Ex is now taken to be the complex-valued phasor representation of the
time-domain expression for the electric field, Ex(t,z):
E x  t  z
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
Re e
j  t
Lecture 8 EM Wave (KEH)

E x
(8-14)
28
We begin by postulating a possible solution to this second-order sinusoidal steady
state (phasor) differential equation that varies exponentially with distance:
A e
Ex
 z
Substituting this proposed solution into the phasor form of the wave equation
(8-13) yields
2
 A e
 z
2
      A  e
 z
 j    A e
 z
Dividing both sides by Aeγz yields the “characteristic equation”

2
2
      j      
Solving for the value of γ that will permit the proposed solution to satisfy the
wave equation yields two, NOT ONE!, complex-valued solutions!
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29
j      j   

 j      j   
Thus, the complete solution must be the superposition (sum) of
both of these solutions:
Ex
A e
where

 z
 B e
z
j      j   
(8-15a)
(8-15b)
Note
defined
in terms
source
frequency,
 is
 , and
Note that
γ, the
propagation
constant,of
tellsthe
us how
the sinusoidal
steady-state
propagates
through the
the wave
medium
properties,
, medium.
 , andIt.is a function of the source
frequency, ω, and the properties of the medium through which it is propagating:
The medium properties are: conductivity σ, permittivity, ε, and permeability, μ
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30
Physical interpretation of lossy wave equation solution
Let us define γ as the complex-valued “propagation constant”, and we shall call its real
part α and its imaginary part β. We will find that α will be called the “attenuation
constant”, and it dictates how fast the wave’s amplitude exponentially decays (attenuates)
with distance as it propagates through the lossy medium. Likewise, we shall call “β” the
“phase constant” (just as we have named it already for the lossless case), and it will
indicate the rate at which the phase of the wave changes with distance at a fixed time.

j      j   
  j 
(8-16)
Now let us convert the phasor solution back to the time domain by
multiplying by exp(jωt) and taking the real part:
 j t E x
E x  t  z
Re e
E x  t  z
j  t 
   j    z
  j   z  

Re  e
 A e
 B e

Using Euler’s identity, ejθ = cos θ + j sin θ we find that
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Lecture 8 EM Wave (KEH)
31
Wave propagating in the +z direction
whose amplitude attenuates with
distance traveled.
E x  t  z
A e
  z
Wave propagating in the –z
direction whose amplitude
attenuates with distance traveled.
 cos    t    z  B e
 z
 cos    t    z
(8-17)
Thus we find that the solution for a plane EM wave in a
conductive (lossy) medium consists of a wave moving in the
“+z” direction and a wave moving in the –z direction.
However, now these two traveling waves no longer have a
constant amplitude as they move, but rather they each decay
exponentially with the distance traveled, as indicated by the
e-αz term in the (+z) wave and the eαz term for the (-z) wave.
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32
Intrinsic Impedance of lossy medium
To find the solution for Hy, we may (as for the lossless case) substitute our
solution for Ex (8-15a) into (8-2) which is recalled below:
E x
 
z
H y
(8-2)
t
Putting this into phasor form,
Ex
  j    H y
z
Solving for Hy and substituting Equation (8-15a)
Hy
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1
j  

E x
dz
  A  e
  z
  B e
 z
j    
Lecture 8 EM Wave (KEH)
33
Recalling from (8-15b)
j      j   

Substituting this into our expression for Hy
Hy
A e
  z


B e
 z
Where


j  
  j  
For the lossy medium, note that the intrinsic impedance (for the sinusoidal steady state) is
complex-valued, and may be written in polar form as |η|exp(jθn) This implies in the time
domain there is a small phase shift “θn” imparted to the +z progressing and the –z
progressing waves due to the phase angle of the intrinsic impedance of the lossy medium.
(8 – 18)
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34
Approximations for Good Dielectric
|
• Definition of a “Good Dielectric” --- a
medium where the ratio of the magnitude
of conduction current to displacement
current is much less than unity:
|Conduction Current|/|Displacement Current|
= |σE|/|jωD| = σE/(ωεE) = σ/(ωε) << 1
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
  j 
j      j   
j      1  j 

 
Thus for a good conductor,
where
dielectric, where

 

j    
<< 1
j 
Thus, in a good dielectric, to a high degree of approximation,

   
and  is approximately 0 => wave propagates with little attenuation
The velocity of the wave is
vp



   
1
 
Thus a good dielectric closely resembles a lossless medium.
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Intrinsic Impedance of Good Dielectric

j  

  j  

1

1  j

 
Thus for a good conductor,
dielectric where

<< 1
 
To a high degree of
approximation:
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


Lecture 8 EM Wave (KEH)
Same result as for the
lossless case!
37
Approximations for “Good Conductor”
• Definition of a “Good conductor” --- a
medium where the ratio of the magnitude
of conduction current to displacement
current is much greater than unity:
|Conduction Current|/|Displacement Current|
= |σE|/|jωD| = σE/(ωεE) = σ/(ωε) >> 1
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  

j     1  j 

 

j      j   

For a good conductor
 


<< 1 =>
j   
(approximately)
j
But if we write "j" in polar form as
e
j
Then we see that
j
e

2
 j  
 2
e 

2
1
2
j
e

4
Therefore, to a high degree of approximation, in
a conductive medium:

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 
      exp  j  
 4
  
  1  j
2
Lecture 8 EM Wave (KEH)
  j
39
Thus for a good conductor, there is considerable
loss, since


  
2
(8 – 19)
Also note the phase angle between the time varying
Ex and Hy field at a point in the good conductor is
45 degrees.
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40
Skin Depth in a Good Conductor
• The shielding ability of a good conductor (such as a
copper or aluminum sheet) is indicated by its skin depth
or depth of penetration, which is the depth to which the
Ex and Hy fields penetrate the conductor before they are
reduced in intensity by a factor of exp(-1) = 37%.
• From our solution to the lossy wave equation in the
sinusoidal steady state (8-17), we found that
Ex  t  z
A e
  z
 cos    t    z  B e
 z
 cos    t    z
Thus, the (+z) progressing wave will attenuate as it travels
through the lossy medium in the +z direction:
A e
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  z
 cos    t    z
Lecture 8 EM Wave (KEH)
41
Substituting our expression for the attenuation constant, α
A e
  z
 cos    t    z
A e

  
z
2
 cos    t    z
Thus the Ex field will attenuate by exp(-1) = 37% as it travels
a distance
1
2
1


  
 f  
δ is called the skin depth. It decreases with increasing source frequency f,
permeability of the medium μ, and conductivity of the medium σ.
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Note the wave’s phase velocity in a lossy medium is still given by
vp = ω / β
But vp is no longer = 1/sqrt(με),
which is only valid for lossless or
good dielectric materials
In a good conductor, the wave velocity is given by
vp
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


  
2
Lecture 8 EM Wave (KEH)
2 
 
43