Problem #1 A linearly polarized plane wave propagates in free space and has its E-field lying along the x-direction. Its propagation vector 1K makes an angle of 30 degrees with the z axis and in the quadrant y>0, z>0 the wave propagates away from the origin. Assume that the magnitude of its electric field is Eo and its phase is zero at the origin.

1) Write down an expression for the electric field. If the electric field is pointing in the x-direction, the wave vector must be in the yz-plane and if it makes an positive angle of 30º with the z-axis, we can write

30sinF30cosF01 yzkk

Therefore our electric field is given by

30sin30cos001

001 FF yjkzjkrkj eeExeExE

2) Write down an expression for the magnetic field. Using Faraday’s Law

HjE

0

yjkzjkyz

yjkzjkyz

x

yzyz eezkykEeezjkyjk

jE

Ezyxzyx

jH FFFF

00

FFF

0

0

0

0

0

and in terms of the angle of propagation

30sin30cos

0

00

0030sinF30cosF yjkzjk eezykEH

where we can simplify the coefficient to get

00

0

0

00

0

0 1

k

leaving us with

30sin30cos

0

0 0030sinF30cosF yjkzjk eezyEH

Page 2

A second linearly polarized plane wave is also propagating in free space and has its E-field lying along the x-direction as well. Its propagation vector 2K however makes an angle of 150 degrees with the z axis and in the quadrant y>0, z<0 this wave propagates toward the origin. Assume that its electric field magnitude is also Eo with a phase of zero at the origin.

3) Write down an expression for the electric field of this second plane wave. Similarly to part 1) but with a new angle and propagating inward so we add a negative sign to the front

30sinF30cosF150sinF150cosF 002 yzkyzkk

and the electric field becomes

30sin30cos002

002 FF yjkzjkrkj eeExeExE

4) Write down an expression for the total electric field. The total electric field is simply the super-position of the first an second field

30sinF30cosF150sinF150cosF 002 yzkyzkk

and the electric field becomes

30sin30sin30cos021

000F yjkyjkzjk eeeExEEE

which we can simplify to

30sincos2F 030cos

00 ykeExE zjk

5) Determine the loci of the points where the total electric field vanishes. The electric field vanishes when the cosine is zero, which occurs when the argument satisfies 2130sin0 nyk where n is an integer

This occurs when ... ,

25 ,

23 ,

221212

30sin21

00

nnk

kny

The loci of the points is the set of planes normal to the y-axis, separated by a wavelength, and passing through 2y .

Page 3

8-20

The values from problem 8-12 are used to give numerical solutions.

a) For r = 80, µr = 1, = 4, and ! = 104, we have ! = 5.65 103, which is

much greater than 1. Hence, seawater is a good conductor. For a good conductor, we have↵ =

pfµ = 1

. Hence,

=1p

fµ= 6.31 m

b) For a good conductor, we have ↵ = =p

fµ = 1 . Hence, adding the attenuation

and phase terms, we have:

H(t, z) = ay

H0e z

cos(104t z

)

Next, we calculate the intrinsic impedance for a good conductor:

c = (1 + j)↵

=

p2

ej

4

Finally, we have:

E(t, z) = caz

H = ax

H0

p2

e

z cos(104t z

+

4)

2

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 by µr = 1, εr = 9, and σ = 5×10−4S/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

=5×10−4×120π

2×√9

= 0.032 (Np/m),

10−3 = 10e−0.032z, ln10−4 = −0.032z,z= 287.82 m.

Problem 7.21 Based on wave attenuation and reflection measurements conductedat 1 MHz, it was determined that the intrinsic impedance of a certain medium is28.1∠45 (Ω) and the skin depth is 2 m. Determine the following:(a) The conductivity of the material.(b) The wavelength in the medium.(c) The phase velocity.

Solution:(a) Since the phase angle of ηc is 45, the material is a good conductor. Hence,

ηc = (1+ j)ασ

= 28.1e j45

= 28.1cos45 + j28.1sin45,

or ασ

= 28.1cos45 = 19.87.

Since α = 1/δs = 1/2= 0.5 Np/m,

σ =α

19.87=

0.519.87

= 2.52×10−2 S/m.

(b) Since α = β for a good conductor, and α = 0.5, it follows that β = 0.5.Therefore,

λ =2πβ

=2π0.5

= 4π = 12.57 m.

(c) up = fλ = 106×12.57= 1.26×107 m/s.

Problem 7.22 The electric field of a plane wave propagating in a nonmagneticmedium is given by

E= z25e−30x cos(2π×109t−40x) (V/m)

Obtain the corresponding expression for H.

Solution: From the given expression for E,

ω = 2π×109 (rad/s),α = 30 (Np/m),β = 40 (rad/m).

From (7.65a) and (7.65b),

α2−β 2 = −ω2µε ′ = −ω2µ0ε0ε ′r = −ω2

c2ε ′r ,

2αβ = ω2µε ′′ =ω2

c2ε ′′r .

Using the above values for ω , α , and β , we obtain the following:

ε ′r = 1.6,ε ′′r = 5.47.

ηc =

!µε ′

"1− j

ε ′′

ε ′

#−1/2

=η0$ε ′r

"1− j

ε ′′rε ′r

#−1/2=377√1.6

"1− j

5.471.6

#−1/2= 157.9e j36.85

(Ω).

%E= z25e−30xe− j40x,

%H=1ηck××× %E=

1157.9e j36.85

x××× z25e−30xe− j40x = −y0.16e−30xe−40xe− j36.85 ,

H= Re%He jωt = −y0.16e−30x cos(2π×109t−40x−36.85) (A/m).

Problem 7.24 In a nonmagnetic, lossy, dielectric medium, a 300-MHz plane waveis characterized by the magnetic field phasor

H= (x− j4z)e−2ye− j9y (A/m)

Obtain time-domain expressions for the electric and magnetic field vectors.

Solution:!E= −ηck××× !H.

To find ηc, we need ε ′ and ε ′′. From the given expression for !H,

α = 2 (Np/m),β = 9 (rad/m).

Also, we are given than f = 300 MHz= 3×108 Hz. From (7.65a),

α2−β 2 = −ω2µε ′,

4−81= −(2π×3×108)2×4π×10−7× ε ′r×10−9

36π,

whose solution givesε ′r = 1.95.

Similarly, from (7.65b),

2αβ = ω2µε ′′,

2×2×9= (2π×3×108)2×4π×10−7× ε ′′r ×10−9

36π,

which gives

ε ′′r = 0.91.

ηc =

"µε ′

#1− j

ε ′′

ε ′

$−1/2

=η0%ε ′r

#1− j

0.911.95

$−1/2=

377√1.95

(0.93+ j0.21) = 256.9e j12.6.

Hence,

!E= −256.9e j12.6y××× (x− j4z)e−2ye− j9y

= (x j4+ z)256.9e−2ye− j9ye j12.6

= (x4e jπ/2+ z)256.9e−2ye− j9ye j12.6,

E= Re!Ee jωt= x1.03×103e−2y cos(ωt−9y+102.6)

+ z256.9e−2y cos(ωt−9y+12.6) (V/m),

H= Re!He jωt= Re(x+ j4z)e−2ye− j9ye jωt= xe−2y cos(ωt−9y)+ z4e−2y sin(ωt−9y) (A/m).

Problem 7.29 The electric-field phasor of a uniform plane wave travelingdownward in water is given by

!E= x5e−0.2ze− j0.2z (V/m)

where z is the downward direction and z= 0 is the water surface. If σ = 4 S/m,(a) Obtain an expression for the average power density.(b) Determine the attenuation rate.(c) Determine the depth at which the power density has been reduced by 40 dB.

Solution:(a) Since α = β = 0.2, the medium is a good conductor.

ηc = (1+ j)ασ

= (1+ j)0.24

= (1+ j)0.05= 0.0707e j45

(Ω).

From Eq. (7.109),

Sav = z|E0|2

2|ηc|e−2αz cosθη = z

252×0.0707

e−0.4z cos45 = z125e−0.4z (W/m2).

(b) A= −8.68αz= −8.68×0.2z= −1.74z (dB).(c) 40 dB is equivalent to 10−4. Hence,

10−4 = e−2αz = e−0.4z, ln(10−4) = −0.4z,

or z= 23.03 m.

Problem 7.32 At microwave frequencies, the power density considered safe forhuman exposure is 1 (mW/cm2). A radar radiates a wave with an electric fieldamplitude E that decays with distance as E(R) = (3,000/R) (V/m), where R is thedistance in meters. What is the radius of the unsafe region?

Solution:

Sav =|E(R)|2

2η0, 1 (mW/cm2)= 10−3 W/cm2 = 10 W/m2,

10=

!3×103

R

"2×

12×120π

=1.2×104

R2,

R=

!1.2×104

10

"1/2= 34.64 m.

Problem 7.33 Consider the imaginary rectangular box shown in Fig. P7.33.(a) Determine the net power flux P(t) entering the box due to a plane wave in air

given byE= xE0 cos(ωt− ky) (V/m)

(b) Determine the net time-average power entering the box.

a

b

c

x

y

z

Figure P7.33: Imaginary rectangular box of Problems 7.33 and 7.34.

Solution:(a)

E= xE0 cos(ωt− ky),

H= −zE0η0cos(ωt− ky).

S(t) = E×××H= yE20η0cos2(ωt− ky),

P(t) = S(t) A|y=0−S(t) A|y=b =E20η0

ac[cos2ωt− cos2(ωt− kb)].

(b)

Pav =1T

! T

0P(t) dt.

where T = 2π/ω .

Pav =E20acη0

"ω2π

! 2π/ω

0[cos2ωt− cos2(ωt− kb)] dt

#= 0.

Net average energy entering the box is zero, which is as expected since the box is ina lossless medium (air).

Problem set 9

Problem 10

!"#

$!"#$= −'()

$*"#+, = 0.$ = 2000/2

3 r = 0.01 S/m

f = 4 GHz

Solution:

a)

3c = 3′ – j 3"

3′ = 3r 30 = 38.9×10-12 F/m and 3"=σ/ω = 3.98×10-13 F/m

σ/ω 3′ = 7.79 +:/;.<=+> ?@A.+BC.D×97FGH I/;.

= 0.0102 <<1

We can use low loss formulae

J ≅ LMNO3′[1 − L3"

2 3R + 18

U3"3RV

<

]

J ≅ 0.898 + L 176.04 2\9

b)

β= Im(γ)= 176.04 rad/m

λ=2π/β = 35.8 mm

α= Re(γ) = 0.898 Np/m

eαy = 10-6/20

(0.898 Np/m) y = -0.691

y = -769 mm => d = 769 mm

d/λ = (769 mm)/(35.8 mm) = 21.5

c)

We are given $*"#+, = 0, ^ = 0.$ = *7'(_ = 200 '(_ 0/2. For RHCP, the z-component of electric field must lag the x-component by 90 degrees. Therefore,

*"#+,. = 200`a)'(_ − L 200 `a)'(A

*"#+,. = 200 `7.CDC) `b9cd.7>) '(_ − L 200 `7.CDC) `b9cd.7>)'(A (Phasor form)

*"#+,.`bef = 200 `7.CDC) `b+efg9cd.7>). '(_ + 200 `7.CDC) `b+efg9cd.7>)\=<

. '(A

*"#+,, ^. = h`[*"#+,.`bef]

*"#+,, ^. = 200 `7.CDC)ijk+M^ + 176.04,.'(_ + 200 `7.CDC) klm+M^ + 176.04,.'(A

d)

n""#+,. =9

op'(q × *7

""""rs`a)

where here, '(q = −'() and *7""""r = 200'(_ − L200'(A

and t ≅ uv

wx y1 + Lw"

< wxz = 179.6 + L0.92 = 179.8 ∠0.29°

n""r+,. =9

9c9cD.C ∠ 7.<D°

'(_ '() '(A

0 −1 0200 0 −L200

`a) = 9

9cD.C∠7.<D°+L200'(_ + 200'(A.`7.CDC)`b9cd.7>)

n""r+,. = +1.11∠89.7° '(_ + 1.11∠ − 0.29° '(A.`7.CDC)`b9cd.7>) (phasor form)

n""r+,, ^. = +1.11 cos+M^ + 176.04, + 89.7°. '(_ + 1.11cos +M^ + 176.04, − 0.29°. '(A.`7.CDC) (real-time form)

Real-time form:

Ez

Ex !"#

z

x

-y

RHCP