Problem 7.19 Ignoring reflection at the air–soil boundary, if the amplitude of a
3-GHz incident wave is 10 V/m at the surface of a wet soil medium, at what depth will
it be down to 1 mV/m? Wet soil is characterized by µr = 1, εr = 9, and σ = 5 × 10−4
S/m.
Solution:
E(z) = E0 e−α z = 10e−α z ,
5 × 10−4 × 36π
σ
=
= 3.32 × 10−4 .
ωε
2π × 3 × 109 × 10−9 × 9
Hence, medium is a low-loss dielectric.
r
σ µ
σ 120π
5 × 10−4 × 120π
√
= 0.032 (Np/m),
α=
= · √ =
2 ε
2
εr
2× 9
10−3 = 10e−0.032z , ln 10−4 = −0.032z,
z = 287.82 m.
Problem 8.4 A 200-MHz, left-hand circularly polarized plane wave with an electric
field modulus of 5 V/m is normally incident in air upon a dielectric medium with
εr = 4, and occupies the region defined by z ≥ 0.
(a) Write an expression for the electric field phasor of the incident wave, given that
the field is a positive maximum at z = 0 and t = 0.
(b) Calculate the reflection and transmission coefficients.
(c) Write expressions for the electric field phasors of the reflected wave, the
transmitted wave, and the total field in the region z ≤ 0.
(d) Determine the percentages of the incident average power reflected by the
boundary and transmitted into the second medium.
Solution:
(a)
ω
2π × 2 × 108 4π
=
=
rad/m,
c
3 × 108
3
8π
ω
ω√
4π √
k2 =
εr2 =
4=
=
rad/m.
up2
c
3
3
k1 =
LHC wave:
e i = a0 (x̂ + ŷe jπ /2 )e− jkz = a0 (x̂ + jŷ)e− jkz ,
E
Ei (z,t) = x̂a0 cos(ω t − kz) − ŷa0 sin(ω t − kz),
|Ei | = [a20 cos2 (ω t − kz) + a20 sin2 (ω t − kz)]1/2 = a0 = 5
Hence,
e i = 5(x̂ + jŷ)e− j4π z/3
E
(b)
η1 = η0 = 120π
(Ω),
(V/m).
(V/m).
η0
η0
η2 = √ =
= 60π
εr
2
(Ω).
Equations (8.8a) and (8.9) give
Γ=
−60
η2 − η1 60π − 120π
1
=
=
=− ,
η2 + η1 60π + 120π
180
3
τ = 1+Γ =
(c)
e r = 5Γ(x̂ + jŷ)e jk1 z = − 5 (x̂ + jŷ)e j4π z/3 (V/m),
E
3
10
t
−
jk
z
2
e = 5τ (x̂ + jŷ)e
E
= (x̂ + jŷ)e− j8π z/3 (V/m),
3
2
.
3
(d)
·
¸
1 j4π z/3
i
r
− j4π z/3
e
e
e
E1 = E + E = 5(x̂ + jŷ) e
− e
3
(V/m).
100
= 11.11%,
9
µ ¶2
120π
2
2 η1
% of transmitted power = 100 × |τ |
= 88.89%.
×
= 100 ×
η2
3
60π
% of reflected power = 100 × |Γ|2 =
Problem 7.16 Dry soil is characterized by εr = 2.5, µr = 1, and σ = 10−4 (S/m).
At each of the following frequencies, determine if dry soil may be considered a good
conductor, a quasi-conductor, or a low-loss dielectric, and then calculate α , β , λ , µp ,
and ηc :
(a) 60 Hz
(b) 1 kHz
(c) 1 MHz
(d) 1 GHz
Solution: εr = 2.5, µr = 1, σ = 10−4 S/m.
f→
ε ′′
σ
=
′
ε
ωε
σ
=
2π f εr ε0
Type of medium
α (Np/m)
β (rad/m)
λ (m)
up (m/s)
ηc (Ω)
60 Hz
1 kHz
1 MHz
1 GHz
1.2 × 104
720
0.72
7.2 × 10−4
Good conductor
Good conductor
Quasi-conductor
Low-loss dielectric
1.54 × 10−4
6.28 × 10−4
1.13 × 10−2
1.19 × 10−2
104
180
0.19
1.8 × 108
1.9 × 108
1.54 × 10−4
6.28 × 10−4
2.45 × 106
107
4.08 × 104
1.54(1 + j)
6.28(1 + j)
3.49 × 10−2
204.28 + j65.89
33.14
238.27
Problem 7.17 In a medium characterized by εr = 9, µr = 1, and σ = 0.1 S/m,
determine the phase angle by which the magnetic field leads the electric field at
100 MHz.
Solution: The phase angle by which the magnetic field leads the electric field is −θη
where θη is the phase angle of ηc .
0.1 × 36π
σ
=
= 2.
ωε
2π × 108 × 10−9 × 9
Hence, quasi-conductor.
ηc =
r
µ
¶
µ
¶−1/2
µ
ε ′′ −1/2 120π
σ
=
1
−
j
1
−
j
√
ε′
ε′
εr
ωε0 εr
◦
= 125.67(1 − j2)−1/2 = 71.49 + j44.18 = 84.04∠31.72 .
Therefore θη = 31.72◦ .
Since H = (1/ηc )k̂ × E, H leads E by −θη , or by −31.72◦ . In other words, H lags
E by 31.72◦ .
Problem 7.18 Generate a plot for the skin depth δs versus frequency for seawater
for the range from 1 kHz to 10 GHz (use log-log scales). The constitutive parameters
of seawater are µr = 1, εr = 80, and σ = 4 S/m.
Solution:
δs =

µε ′
s
1
1
 1+
= 
α
ω
2
µ
ε ′′
ε′
¶2
ω = 2π f ,
−1/2
− 1
,
εr
80
80
= 2 =
,
2
c
c
(3 × 108 )2
ε ′′
σ
σ
72
4 × 36π
=
=
=
× 109 .
=
ε′
ωε
ωε0 εr 2π f × 10−9 × 80 80 f
µε ′ = µ0 ε0 εr =
See Fig. P7.18 for plot of δs versus frequency.
Skin depth vs. frequency for seawater
1
10
0
Skin depth (m)
10
−1
10
−2
10
−3
10
−2
10
−1
10
0
1
10
10
Frequency (MHz)
2
10
3
10
4
10
Figure P7.18: Skin depth versus frequency for seawater.
Problem 8.1 A plane wave in air with an electric field amplitude of 20 V/m is
incident normally upon the surface of a lossless, nonmagnetic medium with εr = 25.
Determine the following:
(a) The reflection and transmission coefficients.
(b) The standing-wave ratio in the air medium.
(c) The average power densities of the incident, reflected, and transmitted waves.
Solution:
(a)
η1 = η0 = 120π
(Ω),
η0
120π
= 24π
η2 = √ =
εr
5
(Ω).
From Eqs. (8.8a) and (8.9),
−96
η2 − η1 24π − 120π
=
=
= −0.67,
η2 + η1 24π + 120π
144
τ = 1 + Γ = 1 − 0.67 = 0.33.
Γ=
(b)
S=
1 + |Γ| 1 + 0.67
=
= 5.
1 − |Γ| 1 − 0.67
(c) According to Eqs. (8.19) and (8.20),
|E0i |2
400
=
= 0.52 W/m2 ,
2η0
2 × 120π
r
i
Sav
= |Γ|2 Sav
= (0.67)2 × 0.52 = 0.24 W/m2 ,
i
Sav
=
t
Sav
= |τ |2
|E0i |2
η1 i
120π
× 0.52 = 0.28 W/m2 .
= |τ |2 Sav
= (0.33)2 ×
2η2
η2
24π
Problem 8.2 A plane wave traveling in medium 1 with εr1 = 2.25 is normally
incident upon medium 2 with εr2 = 4. Both media are made of nonmagnetic, nonconducting materials. If the electric field of the incident wave is given by
Ei = ŷ8 cos(6π × 109t − 30π x) (V/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each of
the two media.
(b) Determine the average power densities of the incident, reflected and
transmitted waves.
Solution:
(a)
Ei = ŷ 8 cos(6π × 109t − 30π x) (V/m),
η0
η0
η0
377
=
η1 = √ = √
=
= 251.33 Ω,
εr1
1.5
1.5
2.25
377
η0
η0
η2 = √ = √ =
= 188.5 Ω,
εr2
2
4
η2 − η1 1/2 − 1/1.5
Γ=
= −0.143,
=
η2 + η1 1/2 + 1/1.5
τ = 1 + Γ = 1 − 0.143 = 0.857,
Er = ΓEi = −1.14 ŷ cos(6π × 109t + 30π x) (V/m).
Note that the coefficient of x is positive, denoting the fact that Er belongs to a wave
traveling in −x-direction.
E1 = Ei + Er = ŷ [8 cos(6π × 109t − 30π x) − 1.14 cos(6π × 109t + 30π x)] (A/m),
8
Hi = ẑ cos(6π × 109t − 30π x) = ẑ 31.83 cos(6π × 109t − 30π x) (mA/m),
η1
1.14
cos(6π × 109t + 30π x) = ẑ 4.54 cos(6π × 109t + 30π x) (mA/m),
Hr = ẑ
η1
H1 = Hi + Hr
= ẑ [31.83 cos(6π × 109t − 30π x) + 4.54 cos(6π × 109t + 30π x)] (mA/m).
√
√
Since k1 = ω µε1 and k2 = ω µε2 ,
r
r
ε2
4
k1 =
k2 =
30π = 40π (rad/m),
ε1
2.25
E2 = Et = ŷ 8τ cos(6π × 109t − 40π x) = ŷ 6.86 cos(6π × 109t − 40π x) (V/m),
8τ
H2 = Ht = ẑ cos(6π × 109t − 40π x) = ẑ 36.38 cos(6π × 109t − 40π x) (mA/m).
η2
(b)
82
64
= x̂ 127.3 (mW/m2 ),
=
2η1 2 × 251.33
Srav = −|Γ|2 Siav = −x̂ (0.143)2 × 0.127 = −x̂ 2.6 (mW/m2 ),
Siav = x̂
|E0t |2
2η2
(8)2
(0.86)2 64
= x̂ 124.7 (mW/m2 ).
= x̂ τ 2
= x̂
2η2
2 × 188.5
Stav =
Within calculation error, Siav + Srav = Stav .
Problem 8.3 A plane wave traveling in a medium with εr1 = 9 is normally incident
upon a second medium with εr2 = 4. Both media are made of nonmagnetic, nonconducting materials. If the magnetic field of the incident plane wave is given by
Hi = ẑ 2 cos(2π × 109t − ky) (A/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each of
the two media.
(b) Determine the average power densities of the incident, reflected, and
transmitted waves.
Solution:
(a) In medium 1,
3 × 108
c
up = √ = √
= 1 × 108 (m/s),
εr1
9
ω
2π × 109
=
= 20π (rad/m),
k1 =
up
1 × 108
Hi = ẑ 2 cos(2π × 109t − 20π y) (A/m),
377
η0
η1 = √ =
= 125.67 Ω,
εr1
3
η0
377
= 188.5 Ω,
η2 = √ =
εr2
2
Ei = −x̂ 2η1 cos(2π × 109t − 20π y)
= −x̂ 251.34 cos(2π × 109t − 20π y) (V/m),
η2 − η1 188.5 − 125.67
= 0.2,
=
η2 + η1 188.5 + 125.67
τ = 1 + Γ = 1.2,
Γ=
Er = −x̂ 251.34 × 0.2 cos(2π × 109t + 20π y)
= −x̂ 50.27 cos(2π × 109t + 20π y) (V/m),
50.27
cos(2π × 109t + 20π y)
Hr = −ẑ
η1
= −ẑ 0.4 cos(2π × 109t + 20π y) (A/m),
E1 = Ei + Er
= −x̂ [25.134 cos(2π × 109t − 20π y) + 50.27 cos(2π × 109t + 20π y)] (V/m),
H1 = Hi + Hr = ẑ [2 cos(2π × 109t − 20π y) − 0.4 cos(2π × 109t + 20π y)] (A/m).
In medium 2,
r
r
40π
4
× 20π =
(rad/m),
9
3
µ
¶
40π y
9
t
E2 = E = −x̂ 251.34 τ cos 2π × 10 t −
3
¶
µ
40π y
9
(V/m),
= −x̂ 301.61 cos 2π × 10 t −
3
¶
µ
40π y
301.61
H2 = Ht = ẑ
cos 2π × 109t −
η2
3
¶
µ
40π y
(A/m).
= ẑ 1.6 cos 2π × 109t −
3
k2 =
ε2
k1 =
ε1
(b)
|E0 |2
(251.34)2
= ŷ 251.34 (W/m2 ),
= ŷ
2η1
2 × 125.67
Srav = −ŷ |Γ|2 (251.34) = ŷ 10.05 (W/m2 ),
Siav = ŷ
Stav = ŷ (251.34 − 10.05) = ŷ 241.29 (W/m2 ).
Problem 8.4 A 200-MHz, left-hand circularly polarized plane wave with an electric
field modulus of 5 V/m is normally incident in air upon a dielectric medium with
εr = 4, and occupies the region defined by z ≥ 0.
(a) Write an expression for the electric field phasor of the incident wave, given that
the field is a positive maximum at z = 0 and t = 0.
(b) Calculate the reflection and transmission coefficients.
(c) Write expressions for the electric field phasors of the reflected wave, the
transmitted wave, and the total field in the region z ≤ 0.
(d) Determine the percentages of the incident average power reflected by the
boundary and transmitted into the second medium.
Solution:
(a)
ω
2π × 2 × 108 4π
=
=
rad/m,
c
3 × 108
3
8π
ω
ω√
4π √
k2 =
εr2 =
4=
=
rad/m.
up2
c
3
3
k1 =
LHC wave:
e i = a0 (x̂ + ŷe jπ /2 )e− jkz = a0 (x̂ + jŷ)e− jkz ,
E
Ei (z,t) = x̂a0 cos(ω t − kz) − ŷa0 sin(ω t − kz),
|Ei | = [a20 cos2 (ω t − kz) + a20 sin2 (ω t − kz)]1/2 = a0 = 5
Hence,
e i = 5(x̂ + jŷ)e− j4π z/3
E
(b)
η1 = η0 = 120π
(Ω),
(V/m).
(V/m).
η0
η0
η2 = √ =
= 60π
εr
2
(Ω).
Equations (8.8a) and (8.9) give
Γ=
−60
η2 − η1 60π − 120π
1
=
=
=− ,
η2 + η1 60π + 120π
180
3
τ = 1+Γ =
(c)
e r = 5Γ(x̂ + jŷ)e jk1 z = − 5 (x̂ + jŷ)e j4π z/3 (V/m),
E
3
10
t
−
jk
z
2
e = 5τ (x̂ + jŷ)e
E
= (x̂ + jŷ)e− j8π z/3 (V/m),
3
2
.
3
(d)
·
¸
1 j4π z/3
i
r
− j4π z/3
e
e
e
E1 = E + E = 5(x̂ + jŷ) e
− e
3
(V/m).
100
= 11.11%,
9
µ ¶2
120π
2
2 η1
% of transmitted power = 100 × |τ |
= 88.89%.
×
= 100 ×
η2
3
60π
% of reflected power = 100 × |Γ|2 =