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Problem 7.19 Ignoring reflection at the air–soil boundary, if the amplitude of a3-GHz incident wave is 10 V/m at the surface of a wet soil medium, at what depth willit be down to 1 mV/m? Wet soil is characterized byµr = 1, εr = 9, andσ = 5×10−4
S/m.
Solution:
E(z) = E0e−αz = 10e−αz,
σωε
=5×10−4×36π
2π ×3×109×10−9×9= 3.32×10−4.
Hence, medium is a low-loss dielectric.
α =σ2
√µε
=σ2· 120π√
εr=
5×10−4×120π2×
√9
= 0.032 (Np/m),
10−3 = 10e−0.032z, ln10−4 = −0.032z,
z = 287.82 m.
Problem 8.4 A 200-MHz, left-hand circularly polarized plane wave with an electricfield modulus of 5 V/m is normally incident in air upon a dielectric medium withεr = 4, and occupies the region defined byz ≥ 0.
(a) Write an expression for the electric field phasor of the incident wave, given thatthe field is a positive maximum atz = 0 andt = 0.
(b) Calculate the reflection and transmission coefficients.
(c) Write expressions for the electric field phasors of the reflected wave, thetransmitted wave, and the total field in the regionz ≤ 0.
(d) Determine the percentages of the incident average power reflected by theboundary and transmitted into the second medium.
Solution:(a)
k1 =ωc
=2π×2×108
3×108 =4π3
rad/m,
k2 =ωup2
=ωc√
εr2 =4π3
√4 =
8π3
rad/m.
LHC wave:
Ei = a0(x+ ye jπ/2)e− jkz = a0(x+ jy)e− jkz,
Ei(z, t) = xa0cos(ωt − kz)− ya0sin(ωt − kz),
|Ei | = [a20cos2(ωt − kz)+a2
0sin2(ωt − kz)]1/2 = a0 = 5 (V/m).
Hence,Ei = 5(x+ jy)e− j4πz/3 (V/m).
(b)
η1 = η0 = 120π (Ω), η2 =η0√
εr=
η0
2= 60π (Ω).
Equations (8.8a) and (8.9) give
Γ =η2−η1
η2 +η1=
60π−120π60π+120π
=−60180
= −13
, τ = 1+Γ =23
.
(c)
Er = 5Γ(x+ jy)e jk1z = −53(x+ jy)e j4πz/3 (V/m),
Et = 5τ (x+ jy)e− jk2z =103
(x+ jy)e− j8πz/3 (V/m),
E1 = Ei + Er = 5(x+ jy)
[e− j4πz/3− 1
3e j4πz/3
](V/m).
(d)
% of reflected power= 100×|Γ|2 =1009
= 11.11%,
% of transmitted power= 100×|τ |2η1
η2= 100×
(23
)2
× 120π60π
= 88.89%.
Problem 7.16 Dry soil is characterized byεr = 2.5, µr = 1, andσ = 10−4 (S/m).At each of the following frequencies, determine if dry soil may be considered a goodconductor, a quasi-conductor, or a low-loss dielectric, and then calculateα , β , λ , µp,andηc:
(a) 60 Hz
(b) 1 kHz
(c) 1 MHz
(d) 1 GHz
Solution: εr = 2.5, µr = 1, σ = 10−4 S/m.
f → 60 Hz 1 kHz 1 MHz 1 GHz
ε ′′
ε ′=
σωε
=σ
2π f εrε0
1.2×104 720 0.72 7.2×10−4
Type of medium Good conductor Good conductor Quasi-conductor Low-loss dielectric
α (Np/m) 1.54×10−4 6.28×10−4 1.13×10−2 1.19×10−2
β (rad/m) 1.54×10−4 6.28×10−4 3.49×10−2 33.14
λ (m) 4.08×104 104 180 0.19
up (m/s) 2.45×106 107 1.8×108 1.9×108
ηc (Ω) 1.54(1+ j) 6.28(1+ j) 204.28+ j65.89 238.27
Problem 7.17 In a medium characterized byεr = 9, µr = 1, andσ = 0.1 S/m,determine the phase angle by which the magnetic field leads the electric field at100 MHz.
Solution: The phase angle by which the magnetic field leads the electric field is−θηwhereθη is the phase angle ofηc.
σωε
=0.1×36π
2π ×108×10−9×9= 2.
Hence, quasi-conductor.
ηc =
√µε ′
(1− j
ε ′′
ε ′
)−1/2
=120π√
εr
(1− j
σωε0εr
)−1/2
= 125.67(1− j2)−1/2 = 71.49+ j44.18= 84.04∠31.72 .
Thereforeθη = 31.72.SinceH = (1/ηc)k×E, H leadsE by−θη , or by−31.72. In other words,H lags
E by 31.72.
Problem 7.18 Generate a plot for the skin depthδs versus frequency for seawaterfor the range from 1 kHz to 10 GHz (use log-log scales). The constitutiveparametersof seawater areµr = 1, εr = 80, andσ = 4 S/m.
Solution:
δs =1α
=1ω
µε ′
2
√
1+
(ε ′′
ε ′
)2
−1
−1/2
,
ω = 2π f ,
µε ′ = µ0ε0εr =εr
c2 =80c2 =
80(3×108)2 ,
ε ′′
ε ′ =σ
ωε=
σωε0εr
=4×36π
2π f ×10−9×80=
7280f
×109.
See Fig. P7.18 for plot ofδs versus frequency.
10−3
10−2
10−1
100
101
102
103
104
10−2
10−1
100
101
Skin depth vs. frequency for seawater
Frequency (MHz)
Ski
n de
pth
(m)
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 isincident normally upon the surface of a lossless, nonmagnetic medium withεr = 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), η2 =η0√
εr=
120π5
= 24π (Ω).
From Eqs. (8.8a) and (8.9),
Γ =η2−η1
η2 +η1=
24π−120π24π+120π
=−96144
= −0.67,
τ = 1+Γ = 1−0.67= 0.33.
(b)
S =1+ |Γ|1−|Γ| =
1+0.671−0.67
= 5.
(c) According to Eqs. (8.19) and (8.20),
Siav =
|E i0|2
2η0=
4002×120π
= 0.52 W/m2,
Srav = |Γ|2Si
av = (0.67)2×0.52= 0.24 W/m2,
Stav = |τ |2 |E
i0|2
2η2= |τ |2η1
η2Si
av = (0.33)2× 120π24π
×0.52= 0.28 W/m2.
Problem 8.2 A plane wave traveling in medium 1 withεr1 = 2.25 is normallyincident upon medium 2 withεr2 = 4. Both media are made of nonmagnetic, non-conducting materials. If the electric field of the incident wave is given by
Ei = y8cos(6π×109t −30πx) (V/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each ofthe two media.
(b) Determine the average power densities of the incident, reflected andtransmitted waves.
Solution:(a)
Ei = y8cos(6π×109t −30πx) (V/m),
η1 =η0√εr1
=η0√2.25
=η0
1.5=
3771.5
= 251.33Ω,
η2 =η0√εr2
=η0√
4=
3772
= 188.5 Ω,
Γ =η2−η1
η2 +η1=
1/2−1/1.51/2+1/1.5
= −0.143,
τ = 1+Γ = 1−0.143= 0.857,
Er = ΓEi = −1.14ycos(6π×109t +30πx) (V/m).
Note that the coefficient ofx is positive, denoting the fact thatEr belongs to a wavetraveling in−x-direction.
E1 = Ei +Er = y [8cos(6π×109t −30πx)−1.14cos(6π×109t +30πx)] (A/m),
Hi = z8η1
cos(6π×109t −30πx) = z31.83cos(6π×109t −30πx) (mA/m),
Hr = z1.14η1
cos(6π×109t +30πx) = z4.54cos(6π×109t +30πx) (mA/m),
H1 = Hi +Hr
= z [31.83cos(6π×109t −30πx)+4.54cos(6π×109t +30πx)] (mA/m).
Sincek1 = ω√µε1 andk2 = ω√µε2 ,
k2 =
√ε2
ε1k1 =
√4
2.2530π = 40π (rad/m),
E2 = Et = y8τ cos(6π×109t −40πx) = y6.86cos(6π×109t −40πx) (V/m),
H2 = Ht = z8τη2
cos(6π×109t −40πx) = z36.38cos(6π×109t −40πx) (mA/m).
(b)
Siav = x
82
2η1=
642×251.33
= x127.3 (mW/m2),
Srav = −|Γ|2Si
av = −x(0.143)2×0.127= −x2.6 (mW/m2),
Stav =
|E t0|2
2η2
= xτ 2(8)2
2η2= x
(0.86)2642×188.5
= x124.7 (mW/m2).
Within calculation error,Siav+Sr
av = Stav.
Problem 8.3 A plane wave traveling in a medium withεr1 = 9 is normally incidentupon a second medium withεr2 = 4. Both media are made of nonmagnetic, non-conducting materials. If the magnetic field of the incident plane wave is given by
Hi = z2cos(2π×109t − ky) (A/m).
(a) Obtain time-domain expressions for the electric and magnetic fields in each ofthe two media.
(b) Determine the average power densities of the incident, reflected, andtransmitted waves.
Solution:(a) In medium 1,
up =c√εr1
=3×108√
9= 1×108 (m/s),
k1 =ωup
=2π×109
1×108 = 20π (rad/m),
Hi = z2cos(2π×109t −20πy) (A/m),
η1 =η0√εr1
=3773
= 125.67Ω,
η2 =η0√εr2
=3772
= 188.5 Ω,
Ei = −x2η1cos(2π×109t −20πy)
= −x251.34cos(2π×109t −20πy) (V/m),
Γ =η2−η1
η2 +η1=
188.5−125.67188.5+125.67
= 0.2,
τ = 1+Γ = 1.2,
Er = −x251.34×0.2cos(2π×109t +20πy)
= −x50.27cos(2π×109t +20πy) (V/m),
Hr = −z50.27
η1cos(2π×109t +20πy)
= −z0.4cos(2π×109t +20πy) (A/m),
E1 = Ei +Er
= −x [25.134cos(2π×109t −20πy)+50.27cos(2π×109t +20πy)] (V/m),
H1 = Hi +Hr = z [2cos(2π×109t −20πy)−0.4cos(2π×109t +20πy)] (A/m).
In medium 2,
k2 =
√ε2
ε1k1 =
√49×20π =
40π3
(rad/m),
E2 = Et = −x251.34τ cos
(2π×109t − 40πy
3
)
= −x301.61cos
(2π×109t − 40πy
3
)(V/m),
H2 = Ht = z301.61
η2cos
(2π×109t − 40πy
3
)
= z1.6cos
(2π×109t − 40πy
3
)(A/m).
(b)
Siav = y
|E0|22η1
= y(251.34)2
2×125.67= y251.34 (W/m2),
Srav = −y |Γ|2(251.34) = y10.05 (W/m2),
Stav = y(251.34−10.05) = y241.29 (W/m2).
Problem 8.4 A 200-MHz, left-hand circularly polarized plane wave with an electricfield modulus of 5 V/m is normally incident in air upon a dielectric medium withεr = 4, and occupies the region defined byz ≥ 0.
(a) Write an expression for the electric field phasor of the incident wave, given thatthe field is a positive maximum atz = 0 andt = 0.
(b) Calculate the reflection and transmission coefficients.
(c) Write expressions for the electric field phasors of the reflected wave, thetransmitted wave, and the total field in the regionz ≤ 0.
(d) Determine the percentages of the incident average power reflected by theboundary and transmitted into the second medium.
Solution:(a)
k1 =ωc
=2π×2×108
3×108 =4π3
rad/m,
k2 =ωup2
=ωc√
εr2 =4π3
√4 =
8π3
rad/m.
LHC wave:
Ei = a0(x+ ye jπ/2)e− jkz = a0(x+ jy)e− jkz,
Ei(z, t) = xa0cos(ωt − kz)− ya0sin(ωt − kz),
|Ei | = [a20cos2(ωt − kz)+a2
0sin2(ωt − kz)]1/2 = a0 = 5 (V/m).
Hence,Ei = 5(x+ jy)e− j4πz/3 (V/m).
(b)
η1 = η0 = 120π (Ω), η2 =η0√
εr=
η0
2= 60π (Ω).
Equations (8.8a) and (8.9) give
Γ =η2−η1
η2 +η1=
60π−120π60π+120π
=−60180
= −13
, τ = 1+Γ =23
.
(c)
Er = 5Γ(x+ jy)e jk1z = −53(x+ jy)e j4πz/3 (V/m),
Et = 5τ (x+ jy)e− jk2z =103
(x+ jy)e− j8πz/3 (V/m),
E1 = Ei + Er = 5(x+ jy)
[e− j4πz/3− 1
3e j4πz/3
](V/m).
(d)
% of reflected power= 100×|Γ|2 =1009
= 11.11%,
% of transmitted power= 100×|τ |2η1
η2= 100×
(23
)2
× 120π60π
= 88.89%.
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