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%.