NUS/ECE
EE2011
Plane Wave Propagation in Lossy Media
1 Plane Waves in Lossy Media
In a lossy medium, J = σE, ρ ≠ 0, σ ≠ 0.
Maxwell’s equations:
Usually,if damping loss is small :
ε ' = ε 0ε r
∇ × E = - jωμH
⎛
σ ⎞
∇ × H = σE + jωD = jω⎜⎜ ε '+
⎟⎟E = jωε c E
jω ⎠
⎝
σ
σ ⎞
⎛
ε c = ε ' − j = ε ' ⎜1 − j
⎟ = complex permittivity
ω
ωε ' ⎠
⎝
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1
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
σ
= loss tangent = tanδ c
ωε '
The complex permittivity can be written as:
ε c = ε 0ε rc
⎛
σ ⎞
⎟⎟ = complex relative permittivity
ε rc = ⎜⎜ ε ' r − j
ωε 0 ⎠
⎝
Usually ε ' = ε
Current terms:
σE = conduction current
r
r
jωD = jωε ' E = displacement current
Note that conduction current and displacement current
are out of phase by π/2.
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Helmholtz’s equation:
∇ 2E + kc2E = 0
kc = ω με c
a complex number
By replacing k with kc, all the previous results derived
for lossless media are applicable to lossy media. But
since kc is a complex number, the plane wave will
experience loss when it propagates.
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
The solutions for the Helmholtz equation in lossy media
are same as those for lossless media if k is replaced by kc.
E x ( z ) = E0± e ∓ jkc z
= E0± e ∓γz
= E0± e −αz e ∓ jβz
where γ = jkc = α + jβ
= complex propagation constant.
Note that solutions with the +α constant have been
discarded as they imply waves with increasing
amplitudes which is impossible.
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4
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
α = attenuation constant
2
⎞
⎛
με ' ⎜
⎛ σ ⎞
− 1⎟
=ω
1+ ⎜
⎟
⎟
ωε ' ⎠
2 ⎜
⎝
⎠
⎝
(Np/m)
β = propagation constant
2
⎞
⎛
με ' ⎜
⎛ σ ⎞
⎟
1+ ⎜
1
=ω
+
⎟
⎟
⎜
ωε ' ⎠
2
⎝
⎠
⎝
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5
(rad/m)
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
As in the lossless case, the general form of a plane wave
in a lossy medium is:
− jk c ⋅r
jφ − jk c (kˆ c ⋅r )
k c = kcx xˆ + kcy yˆ + kcz zˆ = kc kˆ c
E = E0e
= eˆ E0 e e
H = H0e
− jk c ⋅r
r = xxˆ + yyˆ + zzˆ
φ = initial phase
E0 jφ − jkc ( kˆ c ⋅r )
ˆ
=h
e e
ηc
kˆ c = eˆ × hˆ
hˆ = kˆ × eˆ
E ⊥ H ⊥ kc
1 ˆ
H = k c × E,
ηc
c
E = η c H × kˆ c
eˆ = hˆ × kˆ c
ηc = complex intrinsic impedance
=
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6
μ
= ηr + jηi
εc
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
The real and imaginary parts of ηc can be derived as:
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μ0
ηr =
ε0
⎛
⎜
1 ⎜
⎜
2ε ' r ⎜
⎜⎜
⎝
⎞
⎟
⎛ σ ⎞
1+ ⎜
⎟ +1 ⎟
⎝ ωε ' ⎠
⎟
2
⎟
⎛ σ ⎞
1+ ⎜
⎟
⎟
'
ωε
⎟
⎠
⎝
⎠
Ω
μ0
ηi =
ε0
⎛
⎜
1 ⎜
⎜
2ε ' r ⎜
⎜⎜
⎝
⎞
⎟
⎛ σ ⎞
1+ ⎜
⎟ −1 ⎟
⎝ ωε ' ⎠
⎟
2
⎟
⎛ σ ⎞
1+ ⎜
⎟
⎟
'
ωε
⎟
⎝
⎠
⎠
Ω
2
2
7
ηc = ηr + jηi
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
2 Approximation Analysis for Two Special Cases
The formulas for kc and ηc are complicated. For some
special cases depending on the loss tangent σ/ωε’,
approximation formulas can be derived. We study two
special cases in which the loss tangent is either much
greater than 1 or much smaller than 1.
σ
>> 1
ωε '
σ
<< 1
ωε '
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⇒ a good conductor
⇒ a low - loss dielectric
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
2.1 Good conductors
Condition:
σ
loss tangent =
>> 1
ωε '
Then,
σ
⎛
⎞
for
example
>
100
⎜
⎟
ωε '
⎠
⎝
γ = α + jβ ≈ (1 + j ) πfμσ
α = β ≈ πfμσ
μ
α
ηc =
≈ (1 + j )
εc
σ
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9
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
In a good conductor, the intrinsic impedance ηc is a
complex number, meaning that the electric and
magnetic fields are not in phase as in the case of a
lossless medium.
2ω
ω
u p = phase velocity = ≈
β
μσ
2π
2π
π
λ = wavelength =
≈
=2
β
fμσ
πfμσ
Note that H is still ⊥ to E as shown on next page.
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
H and E inside a good conductor
See animation “Plane Wave Simulator”
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
In a good conductor, because of the attenuation
constant α, the wave amplitude becomes smaller when
it propagates. The distance δ through which the
amplitude of a travelling plane wave decreases by a
factor of e-1 = 0.368 = 36.8% is called the skin depth
or depth of penetration of the conductor.
Field amplitude
E0
E0e-1
0
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δ
2δ …….
12
z
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Ex (z = δ )
Ex (z = 0 )
=
E0+ e −αδ e − jβδ
E0+
= e −αδ = e −1
αδ = 1
δ=
1
α
For copper, σ = 5.8 × 107 S/m and μr = 1.
6.61 × 10
δ= =
=
α
ωμ0 μ rσ
f
1
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2
13
−2
at 60Hz, δ = 8.5 × 10-3 m
at 1MHz, δ =6.6 × 10-5 m
at 30GHz, δ = 3.8 × 10-7 m
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Skin depth of some common materials
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14
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
2.2 Low-loss dielectrics
1/ 2
σ ⎞
⎛
γ = α + jβ = jkc = jω με ' ⎜1 − j
⎟
ωε ' ⎠
⎝
Condition:
Binomial expansion (when x << 1) :
σ
loss tangent =
<< 1
ωε '
(1 + x )n ≈ 1 + nx + 1 n(n − 1)x 2 +
2
σ
⎛
⎞
≤ 0.01⎟
⎜ for example
ωε '
⎝
⎠
Then,
2
⎡
ε '' 1 ⎛ ε '' ⎞ ⎤
γ ≈ jω με ' ⎢1 − j
+ ⎜ ⎟ ⎥
2ε ' 8 ⎝ ε ' ⎠ ⎥⎦
⎢⎣
σ
ε '' = ,
ω
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ε '' σ
=
= loss tangent
ε ' ωε '
15
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Hence,
α≈
ωε ' '
2
μ
,
ε'
⎡ 1 ⎛ ε' ' ⎞2 ⎤
β ≈ ω με ' ⎢1 + ⎜ ⎟ ⎥
⎢⎣ 8 ⎝ ε ' ⎠ ⎥⎦
μ⎛
ε' ' ⎞
ηc ≈
⎟
⎜1 + j
ε' ⎝
2ε ' ⎠
ω
up = ≈
β
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2
⎡
1
1 ⎛ ε' ' ⎞ ⎤
⎢1 − ⎜ ⎟ ⎥
με ' ⎢⎣ 8 ⎝ ε ' ⎠ ⎥⎦
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
2
⎡
2π
1
1 ⎛ ε' ' ⎞ ⎤
≈
λ = wavelength =
⎢1 − ⎜ ⎟ ⎥
β
f με' ⎢⎣ 8 ⎝ ε ' ⎠ ⎥⎦
A skin depth δ can be similarly defined as in the good
conductor case:
δ=
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1
α
=
2
σ
17
ε 0ε r
μ0 μ r
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Example 1
The electric field intensity of a linearly polarised uniform
plane wave propagating in the +z direction in seawater is
E = xˆ 100cos(107 πt ) at z = 0. The constitutive parameters of
seawater are εr = 72, μr = 1, and σ = 4 S/m.
(a) Determine the attenuation constant, intrinsic impedance,
phase velocity, wavelength, and skin depth.
(b) Write expressions for H(z,t) and E(z,t).
(c) Find the distance z1 at which the amplitude of the
electric field is 1% of its value at z = 0.
(d) Compute the skin depth at a frequency of 1 GHz.
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Solutions
(a) ω = 10 7 π rad/s ⇒ f = 5×106 Hz
σ
≈ 200 >> 1 . We may therefore approximate
Here
ωε’
seawater as a good conductor at this frequency.
α = β = π f μ r μ 0 σ = 8.89 Np/m or rad/m
πfμ r μ0
π
ηc = (1 + j )
=
(1 + j ) = π e jπ / 4 Ω
σ
2
ω
u p = = 3.53 × 106 m/s
β
2π
λ=
= 0.707 m
β
δ = 1/ α = 0.112 m
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19
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
(b) Phasor fields:
jφ − jkc ( kˆ c ⋅r )
In a lossy medium, E(r ) = eˆ E0e e
eˆ = unit vector of E(r )
jφ −α ( kˆ c ⋅r ) − jβ ( kˆ c ⋅r )
= eˆ E0e e
Here,
e
eˆ = xˆ , φ = 0, E0 = 100, α = β = 8.89, kˆ c ⋅ r = z
Therefore,
E( z ) = xˆ 100 e −8.89 z e − j 8.89 z
1 ˆ
H (r ) = k c × E
ηc
100 −8.89 z − j 8.89 z
100 −8.89 z − j (8.89 z +π / 4 )
⇒ H ( z ) = yˆ
e
e
= yˆ
e
e
jπ / 4
πe
π
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20
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Instantaneous fields:
[
E( z , t ) = Re E( z ) e jωt
[
= Re xˆ 100 e
]
−8.89 z
e
− j 8.89 z
e
j107 π t
]
= xˆ 100 e −8.89 z cos(107 π t − 8.89 z )
[
H( z , t ) = Re H( z ) e jωt
]
⎡ 100 −8.89 z − j ( 8.89 z +π / 4 ) j107 π t ⎤
= Re ⎢ yˆ
e
e
e
⎥⎦
⎣ π
100 −8.89 z
= yˆ
e
cos(107 π t − 8.89 z − π / 4)
π
(b) exp (− αz1 ) = 0.01 ⇒ z1 = 0.518 m
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
ω = 2π × 109 , σ = 4, μ = μ0 = 4π × 10−7 , ε ' = ε 0ε r = 8.854 × 10−12 × 72
Using the general formulas for α and β on page 5, we have :
2
⎛
⎞
με ' ⎜
⎛ σ ⎞
1+ ⎜
∴α = ω
− 1⎟ = 80.837 (Np/m)
⎟
⎟
2 ⎜
ωε ' ⎠
⎝
⎝
⎠
2
⎛
⎞
με ' ⎜
⎛ σ ⎞
1+ ⎜
+ 1⎟ = 195.35 (rad/m)
β =ω
⎟
⎟
2 ⎜
ωε ' ⎠
⎝
⎝
⎠
Hence at 1 GHz, α = 80.837 Np/m and δ = 1/α = 12.37 mm.
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
3 Power Flow and Poynting Vector
The cross product of E and H has the dimension of power
per unit area. It is called the Poynting vector, S and it
represents the power carried by the electromagnetic field
through a unit area (see Note 1* below). The direction of
the Poynting vector indicates the direction of power flow.
Instantaneous Poynting vector:
S = E( x , y , z , t ) × H ( x , y , z , t )
= Re{E( x, y , z )e jωt }× Re{H ( x, y , z )e jωt }
(W/m 2 )
(*Note 1: for a rigorous derivation of the Poynting vector, pls see Intensive Reading notes, section 2.5.)
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Average Poynting vector (see Intensive Reading notes, section 2.5):
T 2
{
}
1
1
*
2
(
)
(
)
S av = lim
S
dt
=
Re
E
x
,
y
,
z
×
H
x
,
y
,
z
(W/m
)
∫
T →∞ T
2
−T 2
The magnitude of Sav gives the average power density
(per unit area) of the EM wave.
Vector magnitude, not absolute value
3.1 In a lossless medium
Pav = Sav
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1
= Re {E × H*}
2
1
η
2
2
(W/m 2 )
=
E0 = H 0
2η
2
24
1ˆ
H = k×E
η
E = ηH × kˆ
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
3.2 In a lossy medium
jφ − jk c ⋅r
jφ −α (kˆ c ⋅r ) − jβ (kˆ c ⋅r )
E = E0e e
= eˆ E0 e e
e
E0 jφ −α (kˆ c ⋅r ) − jβ (kˆ c ⋅r )
− jk c ⋅r
ˆ
H = H 0e
=h
e e
e
ηc
Pav = Sav
1
= Re {E × H*}
2
2
⎧
1
⎪ ˆ E0 −2α (kˆ c ⋅r ) ⎫⎪
e
= Re ⎨k c
⎬
ηc
2
⎪⎩
⎪⎭
⎧1⎫
1 ˆ
2 −2α ( kˆ c ⋅r )
= k c E0 e
Re ⎨ ⎬
2
⎩ηc ⎭
⎧1⎫
1
2 −2α ( kˆ c ⋅r )
= E0 e
Re ⎨ ⎬ (W/m 2 )
2
⎩ηc ⎭
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Example 2
For the plane wave in Example 1, find the power densities at
distances of skin depth z = δ and z = 0.
Solutions
From Example 1:
E( z ) = xˆ 100 e −8.89 z e − j 8.89 z
100 −8.89 z − j (8.89 z +π / 4 )
H ( z ) = yˆ
e
e
π
skin depth δ = 0.112 m
η c = π e jπ / 4 Ω
α = 8.89 Np/m
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26
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Power density at skin depth δ:
δ
⎧1⎫
1
2 −2α ( kˆ c ⋅r )
Pδ = E0 e
Re ⎨ ⎬
2
⎩ηc ⎭
⎧ 1 ⎫
1
2 −2×8.89×0.112
= 100 e
Re ⎨ jπ /4 ⎬
2
⎩π e ⎭
= 153.63 (W/m 2 )
Power density at z = 0:
⎧ 1 ⎫
1
2
P0 = 100 Re ⎨ jπ /4 ⎬
2
⎩π e ⎭
= 1125.4 (W/m 2 )
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27
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Example 3
Compute the average power density Pav of a uniform
sinusoidal plane wave propagating in air which has the
following expression for the instantaneous magnetic field:
1
⎛ 1
H( x, z , t ) = ⎜ −
xˆ +
20π
⎝ 15π
⎞
zˆ ⎟ cos(ωt − 6 x − 8 z ) A/m
⎠
Solutions
In phasor form:
1
⎛ 1 ˆ
H (r ) = H ( x , z ) = ⎜ −
x+
20π
⎝ 15π
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28
ˆz ⎞⎟ e − j ( 6 x +8 z )
⎠
A/m
Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
H(r ) = hˆ H 0 e
hˆ = unit vector of H(r )
− j k ( kˆ ⋅r )
2
2
1 1
1
1
⎛ 1 ⎞ ⎛ 1 ⎞
H0 = ⎜ −
+⎜
=
+ 2 =
⎟
⎟
2
π 15
12π
20
⎝ 15π ⎠ ⎝ 20π ⎠
ˆh = ⎛ − 1 xˆ + 1 zˆ ⎞ / 1 = (−0.8, 0, 0.6)
⎜ 15π
⎟ 12π
20
π
⎝
⎠
k (kˆ ⋅ r ) = k x x + k y y + k z z = 6 x + 8 z
⇒ k x = 6, k y = 0, k z = 8
k = k x2 + k z2 = 10
kˆ = (k x ,0, k z ) k = (0.6,0,0.8)
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Plane Wave Propagation in Lossy Media
NUS/ECE
EE2011
Using, E(r ) = −η0 kˆ × H
− j ( 6 x +8 z )
ˆ
E(r ) = E( x, z ) = y 10 e
[
V/m
]
1
Sav = Re E(r ) × H * (r )
2
1 ⎡
1
1
⎛
− j ( 6 x +8 z )
xˆ +
= Re ⎢ yˆ 10 e
×⎜−
2 ⎣
20π
⎝ 15π
⎞
j ( 6 x +8 z ) ⎤
zˆ ⎟ e
⎥
⎠
⎦
1 ⎡ 10
10 ⎤
xˆ +
zˆ ⎥
= Re ⎢
15π ⎦
2 ⎣ 20π
5
1
1
2
=
xˆ +
zˆ (W/m ) ∴ Pav = Sav =
12π
4π
3π
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30
(W/m 2 )
Plane Wave Propagation in Lossy Media