EE541 Homew ork #4, #5, and #6 · Balanis 5.17 6 4.5 Balanis 5.21 7 4.6 Balanis 5.28 1.2 Homew ork...

EE541 Homework #4, #5, and #6

Robert W. Schuler

December 5, 2002

This paper collects Rob Schuler's answers to homework problems assigned before the second examinationin EE541. The purpose of type setting these assignments is two fold. The �rst point is to learn how totypeset equations of this nature using LATEX. The second purpose is to produce an electronic form that willbe easy to keep for years after the course.

1 Numbering of Assigned Problems

The sections are numbered sequentially by the LYX editor starting from 1. Problems are numbered at the�rst tier title using a scheme of Assignment-dot-problem. For example Problem 1.1 is the �rst problem ofthe �rst homework assignment and Problem 2.3 is the third problem of the second homework assignment.

1.1 Homework #4

Homework #4 consists of six problems assigned from Chapter 5 of the textbook: 5.1, 5.5, 5.16, 5.17, 5.21,and 5.28.

Section in this Paper Problem Number Assigned Number

2 4.1 Balanis 5.13 4.2 Balanis 5.54 4.3 Balanis 5.165 4.4 Balanis 5.176 4.5 Balanis 5.217 4.6 Balanis 5.28

1.2 Homework #5

Homework #5 consists of a single derivation assigned in class. Speci�cally to derive the equations for theTM modes of a circular wave guide of cross section a, using cylindrical coordinates.


8 5.1 Class Notes

1

1.3 Homework #6

Homework #6 consists of three problems assigned in class and three problems assigned from Chapter 6 ofthe textbook: 6.1, 6.17, and 6.21.


9 6.1 Class Notes 110 6.2 Class Notes 211 6.3 Class Notes 312 6.4 Balanis 6.113 6.5 Balanis 6.1714 6.6 Balanis 6.2015 6.6 Balanis 6.21

2

2 Problem 4.1 (Balanis 5.1)

2.1 Statement

A uniform plane wave traveling in a dielectric medium with "r = 4 and �r = 1 is incident normally upon afree-space medium. If the incident electric �eld is given by

fEi =cay2� 10�3e�j�z

write:(a) The corresponding incident magnetic �eld.(b) The re�ection and transmission coe�cients.(c) The re�ected and transmitted electric and magnetic �elds.(d) The incident, re�ected, and transmitted power densities.

y z

x

"r = 4

�r = 1

"0; �0

2.2 Solution

First we calculate the intrinsic impedance � for the dielectric medium:

� =

r�

"

=

r1� �04� "0

:=

r1� 4� � 10�7

4� 8:85� 10�12

:=

r�

8:85� 10�5

:= 188:4

Part (a) fH i

H i = bai � 1

�fEi

where bai is a unit vector in the direction of wave propagation (the direction of incidence in this case). In

this case, bai = baz and the direction of fEi is cay, which means that fH i must be in the �cax direction sincebaz �cay = �cax . To verify this, note that the direction of fEi (cay ) crossed with the direction of fH i (�cax )yields the direction of propagation baz. This allows us to directly write:

fH i = �cax�2� 10�3

�

�e�j�z

= �cax�2� 10�3

188:4

�e�j�z

= �cax �10:6� 10�6�e�j�z

3

Part (b) R and T

R =�2 � �1�2 + �1

=377� 188:4

377 + 188:4= 0:334

T =2�2

�2 + �1

=2(377)

377 + 188:4= 1:334

Part (c) Re�ected and Transmitted eE and eH. In this case fEi, fEr and fEt are all polarized in the

same direction (cay). This means that fH i and fHt will be in the �cax direction, but fHr will be in the +caxdirection (So that eE � eH will point in the direction of wave propagation, which is � baz for the re�ectedwave). It is also important to consider what happens to � as we go from the dielectric to free-space. Recallthat the wave constant � = !

p�" or

� = !p�r�0"r"0

= !p�r"r

p�0"0

= �0p�r"r

where �0 =�p�r"r

= 0:5� is the wave constant in free space. From this we can write directly:

fEr = RfEi= (0:334)cay2� 10�3e�j�z

= cay0:668� 10�3e�j�z

fEt = e�0:5�jzTfEi= (1:334)cay2� 10�3e�j0:5�z

= cay2:668� 10�3e�j0:5�z

fHr = �RfH i

= (0:334)cax �10:6� 10�6�e�j�z

= cax �3:54� 10�6�e�j�z

fHt = baz � 1

�2fEt

= �cax 1

3772:668� 10�3e�j0:5�z

= �cax �7:08� 10�6�e�j0:5�z

4

Part (d) Power Densities Recall that the time average power density = 12RE

� eE � fH��and the in-

stantaneous power density =�!E ��!H . Since all of the �elds are stated in the frequency domain, we only �nd

the time average power densities here.

transmitted time average power density =1

2RE

�fEt �gHt��

=1

2RE

�cay2:668� 10�3e�j0:5�z ��cax �7:08� 10�6�ej0:5�z

�=

1

2RE

0@��cax cay baz0 2:668� 10�3e�j0:5�z 0

� �7:08� 10�6�ej0:5�z 0 0

��1A

=1

2RE

��7:08� 10�6

�ej0:5�z baz2:668� 10�3e�j0:5�z

�=

1

2RE

��1:89� 10�8

� baz�= 9:44� 10�9 baz

incident time average power density =1

2RE

�fEi �gH i��

=1

2RE

�cay2� 10�3e�j�z ��cax �10:6� 10�6�ej�z

�=

1

2RE

0@��cax cay baz0 2� 10�3e�j�z 0

� �10:6� 10�6�ej�z 0 0

��1A

=1

2RE

��10:6� 10�6

�ej�z baz2� 10�3e�j�z

�=

1

2RE

��2:12� 10�8

� baz�= 1:06� 10�8 baz

reflected time average power density =1

2RE

�fEr �gHr��

=1

2RE

�cay0:668� 10�3e�j�z ��cax �3:54� 10�6�e+j�z

�=

1

2RE

0@��cax cay baz0 0:668� 10�3e�j�z 0

� �3:54� 10�6�ej�z 0 0

��1A

=1

2RE

��3:54� 10�6

�ej�z baz0:668� 10�3e�j�z

�=

1

2RE

��2:36� 10�9

� baz�= 1:18� 10�9 baz

NOTE: the re�ected plus the transmitted average power densities equal the incident average powerdensity:

0:18� 10�8 + 0:944� 10�8 = 1:12� 10�8 � 1:06� 10�8

5


3.1 Statement

A time-harmonic electromagnetic wave traveling in free space is incident normally upon a perfect conductingplanar surface, as shown in Figure P5-5. Assuming the incident electric �eld is given by

fEi =caxE0e�j�0z

�nd (a) the re�ected electric �eld, (b) the incident and re�ected magnetic �elds, and (c) the current

density eJs induced on the conducting surface.

Reflected

Incident

y z

x

"0; �0 � =1

3.2 Solution

The �rst step in this problem is to calculate the Re�ection and Transmission Coe�cients. In this case, wemust account for the in�nite conductivity in the P.E.C. when calculating �2.

�1 = 377

�2 =

sj!�

� + j!"

=

sj!�

1+ j!"

= 0

So

R =�2 � �1�2 + �1

=0� 377

0 + 377= �1

T =2�2

�2 + �1

=2(0)

0 + 377= 0

6

Part (a) re�ected electric �eld The re�ected electric �eld can be found using equation 5-48a on page207 of Balanis:

fEr = cax�bE0e+�1ze+j�1z

= �caxE0e+j�0z

Part (b) incident and re�ected magnetic �elds These �elds can be written directly from the incidentand re�ected electric �elds by simply dividing by the intrinsic impedance �, which is approximately 377 in free space. The orientation of eH will be perpendicular to both the direction of the wave propagation andthe orientation of eE, and can be found using the right-hand rule.

fH i =cay E0

377e�j�0z

fHr =cay E0

377e+j�0z

Part (c) induced current Recall that eJs = bn� eH . In this case, bn = � baz andgHtotal = fH i + fHr

= cay E0

377

�e+j�0z + e�j�0z

�gHtotal

��z=0

= cay 2E0

377

eJs = bn� eH= � baz �cay 2E0

377

= cax 2E0

377Amps

7


4.1 Statement

A perpendicularly polarized plane wave traveling in a dielectric medium with relative permittivity of 9 isobliquely incident on another dielectric with a relative permittivity of 4. Assuming that the permeabilitiesof both media are the same, �nd the incident angle (measured from the normal to the interface) that resultsin total re�ection.

4.2 Solution

This problem is asking us to �nd the critical angle for the two given media.The �rst step is to visualize the situation:

�t"r = 9

"r = 4�i

Here �i is the angle of incidence and �t is the angle of the transmitted wave. We know that the transmittedwave will �move� in the direction indicated because the permittivity in the second medium is less than thatin the �rst ("2 < "1). In other words, we know that �t will always be greater than �i. Using equation 5-36on page 197 of Balanis:

�i � �c = sin�1

�r"2"1

�We see that the critical angle is:

�c = sin�1

�r"2"1

�= sin�1

�r4"09"0

�= sin�1 (:666)

= :729rad

= 41:8Æ

This means that for any incident angle greater than 0.729 radians or 41.8 degrees, all of the energy ofthe wave shall be re�ected back into the �rst medium.

8


5.1 Statement

Calculate the Brewster and critical angles for a parallel polarized wave when the plane interface is (a) waterto air ("r of water is 81), (b) air to water, and (c) high density glass to air ("r of glass is 9).

5.2 Solution

Recall that the Brewster angle is that angle of incidence (from the normal) that results in total transmissionof a wave from the �rst medium into the second. When both media have the same permeability (as is truein all 3 cases in this problem), the Brewster angle can be found by equation 5-33b on page 195 of Balanis:

�i = �B = tan�1

�r"2"1

�Also, recall equation 5-36 on page 197 of Balanis for �nding the critical angle:

�i � �c = sin�1

�r"2"1

�NOTE: in the case of total transmission, �i must exactly equal the Brewster angle �B , which is di�erent

than the physical relationship for the critical angle in which �i must equal or be greater than the criticalangle �c.

Part (a) water to air

�B = tan�1

�r"081"0

�= tan�1

�1

9

�= 0:111rad

= 6:34Æ

�c = sin�1

�r"081"0

�= sin�1

�1

9

�= 0:111:rad

= 6:38Æ

Part (b) air to water

�B = tan�1

�r81"0"0

�= tan�1 (9)

= 1:46rad

= 83:66Æ

�c = sin�1

�r81"0"0

�= sin�1 (9)

=�

2� j2:887rad

= 90Æ � j2:887There is no physically realizable critical angle when going from air to water!

9

Part (c) glass to air

�B = tan�1

�r"09"0

�= tan�1

�1

3

�= 0:322rad

= 18:43Æ

�c = sin�1

�r"09"0

�= sin�1

�1

3

�= 0:340rad

= 19:47Æ

10


6.1 Statement

The heights above the earth of a transmitter and receiver are, respectively, 100 and 10 m, as shown inFigure P5-21. Assuming that the transmitter radiates both perpendicular and parallel polarizations, howfar apart (in meters) should the transmitter and receiver be placed so that the re�ected wave has no parallelpolarization? Assume that the re�ecting medium is a lossless �at earth with a dielectric constant of 16.

6.2 Solution

From page 195 of Balanis, �The incidence angle �i, as given by (5-31a) or (5-33), which reduces the re�ectioncoe�cient for parallel polarization to zero, is referred to as the Brewster angle �B . It should be noted thatwhen �1 = �2, the incidence Brewster angle �i = �B of (5-33) exists only if the polarization of the wave isparallel (vertical).�

For completeness:

sin �i =

r"2"1��2�1

"2"1� "1"2

(5-31a)

�i = �B = sin�1�q

"2"1+"2

�(5-33)

From this we can see that if we pick the distance between the transmitter and receiver such that thewave strikes the ground at the Brewster angle �B , then we know that the parallel part of the wave shall becompletely transmitted into the earth, while the vertical part of the wave will be re�ected (at least partially).

Brewster Angle First we calculate the Brewster angle using equation 5-33:

�B = sin�1

�r"2

"1 + "2

�= sin�1

�r16"0

"0 + 16"0

�= sin�1

r16

17

!= sin�1 (:970)

= 1:33rad

= 76:0Æ

11

Distance s Using elementary trigonometry, we know that the distance d1 from the transmitter to thepoint on the ground where the wave hits the ground at the Brewster angle can be found by

tan �B =d1

100md1 = 100m tan(1:33)

= 407:23m

We also know from Snell that the angle of incidence equals the angle of re�ection for the vertical part ofthe wave, so the distance d2 from the point on the ground from which the wave hits to the receiver can befound by

tan �B =d210m

d2 = 10m tan(1:33)

= 40:723m

The total distance s between the transmitter and receiver is then

s = d1 + d2 = 407:23m+ 40:723m:= 448m

100

10d2d1

�B�B

�B

12


7.1 Statement

At large observation distances the �eld radiated by a satellite antenna which is attempting to communicatewith a submerged submarine is locally TEM (also assume uniform plane wave), as shown in Figure P5-28.Assuming the incident electric �eld before it impinges on the water is 1mV=m and the submarine is directlybelow the satellite, �nd at 1MHz:

(a) The intensity of the re�ected�!E �eld.

(b) The SWR created in air.(c) The incident and re�ected power densities.

(d) The intensity of the transmitted�!E �eld.

(e) The intensity of the transmitted power density.(f) The depth d (in meters) of the submarine where the intensity of the transmitted electric �eld is 0:368

of its value immediately after it enters the water.(g) The depth (in meters) of the submarine so that the distance from the surface of the ocean to the

submarine is 20� (� in water).(h) The time (in seconds) it takes the wave to travel from the surface of the ocean to the submarine at

a depth of 100m.(i) The velocity of the wave in water to that in air (v=vo).

7.2 Solution

First, since E0 = 1mV = 1� 10�3V , and � = !c = 2�106

3�108 = 23� � 10�2we can directly write the equation for

the incident�!E �eld:

fEi = E0e�j�z

= 10�3e�j2

3��10�2z

Next we calculate the coe�cients of re�ection and transmission.In air, �1

:= 377.

In this case, the Sea Water has a conductivity of 1 S/m, so:

�2 =

sj!�

� + j!"

=

sj2� � 106�0

1 + j2� � 106"0 (81)

:= 377

sj2� � 106

1 + j162� � 106

13

= 377

sj2� � 106

1 + j162� � 106� 1� j162� � 106

1� j162� � 106

= 377

sj2� � 106 + 2(162) (� � 106)

2

1 + (162� � 106)2

= 377

rj2� + 3:2� 109

1 + 2:59� 1011

=377

5� 105

p3:2� 109 + j2�

= 7:54� 10�4�5:7� 104 + j5:5� 10�5

�= 42:7 + j4:2� 10�8

NOTE: In this case, the real part is much much bigger than the imaginary part�we can consider �2 = 43.

R =�2 � �1�2 + �1

=43� 377

43 + 377= �0:8

T =2�2

�2 + �1

=2(43)

43 + 377= 0:2

Part (a) Re�ected Now we can �nd the re�ected eE �eld by multiplying the incident eE �eld by R.

fEr = RfEi= (�0:8)10�3e�j

2

3��10�2z

= 8� 10�4e�j2

3��10�2z

Part (b) SWR In this case "2 > "1, so we can use equation 5-9b on page 185 of Balanis to calculate thestanding wave ratio (SWR):

SWR =

r"2"1

=

r81

1= 9

Part (c) Incident and Re�ected Power Densities Using equation 5-6a on page 183 of Balanis, theincident power density is:

Siav =1

2Re�fEi �gH i�

�= baz jE0j2

2�1

= baz 10�6

754

= baz1:326� 10�9

14

Using equation 5-6b on page 183 of Balanis, the re�ected power density is:

Srav =1

2Re�fEr �gHr�

�= � baz ��b��2 jE0j2

2�1

= � baz (�0:8)2 10�6

754

= � baz8:49� 10�10

Part (d) Transmitted We can �nd the transmitted eE �eld by multiplying the incident eE �eld by T andaccounting for the attenuation constant �.

�:=

�

2

r�

"

= 0:5 (377)

r1

81= 20:9

fEr = TfEie�21z

= (0:2)10�3e�jz2

3��10�2

e�21z

= 2� 10�4e�(21+j2

3��10�2)z

Part (e) Transmitted Power Density Using equation 5-51c on page 207 of Balanis, the transmittedpower density is:

Stav = baz ��T b��2 jE0j22

e�2�2zRe

�1

��2

�= baz (0:2)2 10�6

2Re

�1

43

�= (:04)(1:16� 10�8)

= baz4:7� 10�10

Part (f) Depth

0:368 = e�z21

ln(0:368) = z21

z =�21

ln(0:368)

z = 21meters

Part (g) Wave Length The wave length in water can be found by looking at the wave length in air andthe wave number in water.

!p�" = � =

2�

�or

� =2�

�

In water: �w = 2� � 106p�0"081 = 9�air = 9 2

3� � 10�2 = 6� � 10�2 , so the wave length in water is

15

� =2�

�w

=2�

6� � 10�2

=1

3� 102 meters

20� = 20(33:333) = 666:67meters

Part (h) Wave Velocity in Water The speed of the wave is found by:

c =!

�

=2� � 106

6� � 10�2

=1

3� 108

m

s

So the time for a wave to travel 100 meters is:

t =d

c

=100m

0:333� 108m=s

= 300� 10�8 seconds

Part (i) Relative Velocity

13 � 108

3� 108=

1

9

16

8 Problem (Class Notes)

8.1 Statement

Given a circular waveguide with cross section de�ned by radius a, �nd the TMz modes.NOTE: The answer will involve Bessel Functions.

8.2 Solution

The TMz modes will be expressed in cylindrical coordinates matching the geometry shown in the following�gure.

��

��

z

y

x�

�

�Transverse magnetic modes (often also known as transverse magnetic �elds) are �eld con�gurationswhose magnetic �eld components lie in a plane that is transverse to a given direction. That direction isoften chosen to be the path of wave propagation. For example, if the desired �elds are TM to z (TMz),this implies that Hz = 0. The other two magnetic �eld components (Hx and Hy) and the three electric �eldcomponents (Ex, Ey, and Ez) may or may not all exist.

By examining (6-43) and (6-51) it is evident that to derive the �eld expressions that are TM to a given

direction, independent of the coordinate system, it is su�cient to let the vector potential eA have only acomponent in the direction in which the �elds are desired to be TM. The remaining components of eA as wellas those of eF are set equal to zero.� (Balanis, p. 269).

eA = bazAz (�; �; z)eF = 0

Once the vector potentials are known, we can �nd the eE and eH �elds using the equations (6-70) on page273 of Balanis, for TMz Cylindrical Coordinates:

E� = �j 1!�"

@2Az@�@z H� =

1�1�@Az@�

E� = �j 1!�"

1�@2Az@�@z H� = � 1

�@@�

Ez = �j 1!�"

�@2

@z + �2�Az Hz = 0

The vector potential eA must satisfy the vector wave equation, so we end up with:

r2Az (�; �; z) + �2Az (�; �; z) = 0

For a circular waveguide, Az takes the form:

17

Az (�; �; z) = [A1Jm (��) +B1Ym (��)]

� [A2 cos (m�) +B2 sin (m�)]

� �A3e

�j�zz +B3e+j�zz

��2 = �2� + �2z

Next we look at the boundary conditions:

- eE = 0 when � = a.

- The �elds must be �nite everywhere.

- The �elds must repeat every 2� radians in �.

Since, Ym(0) =1, B1 = 0, otherwise we violate the second boundary condition.If we assume that the wave only propagates in the +z direction, then B3 = 0.We can collect constants�Amn = A1A3

So, now we can write:

A+z (�; �; z) = AmnJm (��) [A2 cos (m�) +B2 sin (m�)] e

�j�zz

Now we can look at fEz :fEz (�; �; z) = �j 1

!�"

�@2

@z+ �2

�Az

= �j 1

!�"

�@2

@z+ �2

�AmnJm (��) [A2 cos (m�) +B2 sin (m�)] e

�j�zz

= �j 1

!�"

�@2

@z

�AmnJm (��) [A2 cos (m�) + B2 sin (m�)] e

�j�zz

+ �j 1

!�"

��2�AmnJm (��) [A2 cos (m�) +B2 sin (m�)] e

�j�zz

= �j 1

!�"

��2 + (j�z)

2�AmnJm (��) [A2 cos (m�) +B2 sin (m�)] e

�j�zz

= �j �2�

!�"AmnJm (��) [A2 cos (m�) +B2 sin (m�)] e

�j�zz

Since eE = 0 when � = a, we can say:

�j �2�

!�"AmnJm (��a) [A2 cos (m�) +B2 sin (m�)] e

�j�zz = 0

This requires that (see equation 9-25 on page 478 of Balanis):

Jm (��a) = 0) ��a = Xmn ) �� =Xmn

a

Xmn is the nth zero (n = 1, 2, 3, . . .) of the Bessel function Jm of the �rst kind of order m (m = 0, 1,2, 3, . . .).

This leads to a complex de�nition for �z (see equation 9-26 on page 478 of Balanis):

(�z)mn =

8>><>>:q�2 � �Xmn

a

�2for � > ��

0 for � = �c = ��

�jq�

Xmn

a

�2 � �2 for � < ��

18

9 Problem (Class Notes 1)

9.1 Statement

Calculate the re�ection coe�cient and percent of incident energy re�ected when a uniform plane waveis normally incident on a Plexiglas radome (assume a in�nite �at plane) of thickness 3/8 inch, relativepermittivity 2.8, with free space on both sides. Frequency corresponds to free-space wavelength of 20 cm.

9.2 Solution

Recall that:

R =�12 + �23e

�j2�2d

1 + �12�23e�j2�2d

�12 =�2 � �1�2 + �1

�23 =�3 � �2�3 + �2

In this case

�1 = 377

�2 =

r�0"02:8

= 377

r1

2:8= 225

�3 = 377

�12 =225� 377

225 + 377= �0:25

�23 =377� 225

377 + 225= +0:25

! =2�c

�

=2�(3� 108)

:2

= 3� � 109

d =3

8inch

= :375inch

= 0:0095meter

19

�2 = !p�0"o2:8

= 3� � 109p4� � 10�7 � 8:85� 10�12 � 2:8

= 52:6

So

R =�0:25+ 0:25e�j2(52:6)(:0095)

1 + (�0:25)(0:25)e�j2(52:6)(:0095)

=�0:25 + 0:25e�j

1 + 0:0625e�j

=�0:25 + 0:25 (cos 1� j sin 1)1 + 0:0625 (cos 1� j sin 1)

=�0:25 + 0:25(0:54� j0:84)1 + 0:0625(0:54� j0:84)

=�0:115� j0:211:034� j0:0525

= �0:101� j0:208

Applying the principle of equation 5-6b on page 182 and equation 5-51 on page 207 of Balanis, thepercentage of incident energy re�ected is found by squaring the magnitude of R:

jRj2 = (0:101)2 + (0:208)2 = 0:053 = 5:3%

20


10.1 Statement

What refractive index and what thickness do you need to make a quarter wave anti-re�ection coating betweenair and silicon at 10 GHz? The relative permittivity of silicon is 3.5.

10.2 Solution

Recall that:

R =�12 + �23e

�j2�2d

1 + �12�23e�j2�2d

�12 =�2 � �1�2 + �1

�23 =�3 � �2�3 + �2

For the slab to be anti-re�ective, R must equal 0, which happens when the numerator equals 0.

R = 0 = �12 + �23e�j2�2d

For f0 = 10GHz, d = �24 , then

2�2djf=10GHz = 2

�2�

�2

��24

�= �

e�j� = �1

So we get (see example 5-10 on pages 225-226 of Balanis for the algebra):

R = 0 = �12 + �23e�j2�2d

= �12 � �23

0 =�2 � �1�2 + �1

� �3 � �2�3 + �2

�2 =p�1�2

�1 = 377

�3 =

r�0"03:5

= 377

r1

3:5= 202

�2 =p377� 202

= 275:6

This means that the dielectric constant of the slab is

275:6 = 377

r1

"r"r = 1:870

21

Index of Refraction =

1p�0"01p

�0"0"r

=p"r

=p1:87

= 1:37

We also know that2�

�= � = !

p"�

2�

�= 20� � 109

p1:87� 8:85� 10�12 � 4� � 10�7

= 45:6

� = 2:2cm

d =�

4

=2:2cm

4= 0:55cm

22


11.1 Statement

Compare the currents that would be required between a half-wave dipole (0:5�) antenna and a small dipoleof length 0:05� to produce 100 W of radiated power from each.

11.2 Solution

The power radiated by a half-wave dipole can be found from the equation:

Prad = I2Rrad

= I273:1

So it will take:

100 = I273:1

I =

r100

73:1= 1:17 amps

to radiate 100 W of power.The power radiated by a hertzian dipole can be found from the equation (assuming free-space and far-

�eld):

Pe = ��

3

��I`��2 �1� j

(kr)3

�= 377

�

3

��I:05��2 �1� j

(kr)3

�' 394:8(0:05I)2

= 0:987I2

So it will take:

100 = :987I2

I =

r100

:987= 10:07 amps

to radiate 100 W of power.It takes 10 times more current to radiate the same power from the smaller dipole!

23

12 Problem (Balanis 6.1)

12.1 Statement

If fHe = j!"r��e, where �e is the Hertzian potential, show that:(a) r2�e + �2�e = j (1=!") eJ(b) fEe = �2�e +r (r � �e)(c) �e = �j (1=!�") eA

12.2 Solution

NOTE: This is based on the discussion in section 6.2 on pages 256 and 257 of Balanis.Substitute the given equation into Maxwell's curl equation:

r� fEe = �j!�fHe

reduces it to

r� fEe = �j!� (j!"r��e) = !2"�r��e

which can also be written as

r�hfEe � �2�ei = 0

From the vector identity

r� (�r�e) = 0

we can transform the previous equation to

fEe � �2�e = �r�eor

fEe = �r�e + �2�e (1)

�e represents an arbitrary electric scalar potential that is a function of position.Taking the curl of both sides of the given equation and using the vector identiy

r�r��e = r(r ��e)�r2�e

leads to

r��

1

j!"fHe

�= r(r � �e)�r2�e

For a homogeneous medium, this reduces to

1

j!"r� fHe = r(r ��e)�r2�e

Equating Maxwell's equation

r� fHe = eJ + j!"fEetransforms the previous equation to

24

1

j!"

� eJ + j!"fEe� = r(r � �e)�r2�eeJj!"

+ fEe = r(r � �e)�r2�e

Now we substitute the previous value for fEe to get:

eJj!"

+ fEe = r(r ��e)�r2�eeJj!"

�r�e + �2�e = r(r ��e)�r2�e

r2�e + �2�e = �eJ

j!"+r(r � �e) +r�e

r2�e + �2�e =j

!"eJ +r (r ��e + �e)

So far, only the curl of �e has been speci�ed. This means that we are free to specify its divergence. Tomake life easy and to satisfy the identity of Part a of the problem, we choose:

r ��e = ��e (2)

Part (a) This reduces the previous equation to:

r2�e + �2�e =j

!"eJ

which demonstrates that (a) is true.

Part (b) Inserting (2) into (1) gives us:

Ee = �r�e + �2�e

= r (r � �e) + �2�e

= �2�e +r (r � �e)which demonstrates that (b) is true.

Part (c) Finally, if we equate equation (6-4a) on pate 256 of Balanis with the given equation in thisproblem, we see that:

1

�r�A =gHA = fHe = j!"r��e

r� 1

�A = r� j!"�e

1

�A = j!"�e

1

�!"A = j�e

�e =�j!�"

A

which demonstrates (c) is true.

25


13.1 Statement

Show that for observations made at very large distance (�r � 1) the electric and magnetic �elds of Example6-3 reduce to:

E� = j��Ie`e

�j�r

4�rsin �

H� ' E��

Er ' 0

E� = Hr = H� = 0

13.2 Solution

Example 6-3 de�nes the electric and magnetic �elds for a very thin linear electric current element of veryshort length and a constant current to be:

Er = � Ie` cos �2�r2

�1 + 1

j�r

�e�j�r Hr = 0

E� = j� �Ie` sin �4�r

h1 + 1

j�r � 1(�r)2

ie�j�r H� = 0

E� = 0 H� = j �Ie` sin �4�r

�1 + 1

j�r

�e�j�r

When �r � 1 terms that have �r in the denominator will become negligible. Terms with any power(greater than 1) of r in the denominator will also become zero. These terms are highlighted in bold below.

Er = � Ie` cos �2�r2

�1 + 1

j�r

�e�j�r Hr = 0

E� = j� �Ie` sin �4�r

�1 + 1

j�r� 1

(�r)2

�e�j�r H� = 0

E� = 0 H� = j �Ie` sin �4�r

�1 + 1

j�r

�e�j�r

Zeroing these terms out, yields:

Er = 0 Hr = 0

E� = j� �Ie` sin �4�r e�j�r H� = 0

E� = 0 H� = j �Ie` sin �4�r e�j�r = E��

26


14.1 Statement

The current distribution on a very thin wire dipole antenna of overall length ` is given by

Ie =

� bazI0 sin �� `2 � z0�� 0 � z0 � `2bazI0 sin �� `2 + z0

�� `2 � z0 � 0

where I0 is a constant. Representing the distance R of (6-112) by the far-�eld approximations of (6-112a)through (6-112b), derive the far-zone electric and magnetic �elds radiated by the dipole using (6-97a) andthe far-�eld formulations of Section 6.7.

For completeness:

R =�r2 + (r0)2 � 2rr0 cos

� 12 (6-112)

R = r � r0 cos for phase variations (6-112a)

R = r for amplitude variations (6-112b)eA = �4�

Rc Ie (x

0; y0; z0) e�j�R

R dl0 (6-97)

14.2 Solution

Fitting (6-112a) and (6-112b) into (6-97) and using the Calculus shown on slides 13 through 15 of Lecture#8 gives:

eA = baz �4�

Zc

I0 sin

��

�`

2� jz0j

��e�j�(r�r

0 cos

rdz0

= baz �

4�rI0e

�j�rZ + `

2

� `2

sin

��

�`

2� jz0j

��ej�r

0 cos dz0

= baz �

4�rI0e

�j�r"2�cos�� `2 cos

�� cos�� `2��

� sin2

#

Now we can use the short-cut equations for far-�eld approximation from slide 17 of Lecture #8 (see alsoequations 6-101 on page 281 of Balanis).

fEr ' 0 fHr ' 0fE� ' �j!A� fH� ' j!� A�fE� ' �j!A� fH� ' � j!� A�

This requires converting eA into spherical coordinates. Since we have only an baz component, this meansequation (II-12a) on page 924 of Balanis reduces to:

Ar = Az cos �

A� = �Az sin �A� = 0

fEr ' 0 fHr ' 0fE� ' 0 fH� ' 0fE� ' j! sin � �4�r I0e

�j�r�2[cos(� `

2cos )�cos(� `

2 )]� sin2

� fH� ' j!� sin � �

4�r I0e�j�r

�2[cos(� `

2cos )�cos(� `

2 )]� sin2

�

27


15.1 Statement

Show that the radiated far-zone electric and magnetic �elds derived in Problem 6.20 reduce for a half-wavelength dipole (` = �=2) to

E� ' j�I0e

�j�r

2�r

"cos��2 cos �

�sin �

#

H� ' E��

Er ' E� ' Hr ' H� ' 0

15.2 Solution

Recall from the solution to problem 6.20 (see section 14 of this paper):

fEr ' 0 fHr ' 0fE� ' 0 fH� ' 0fE� ' j! sin � �4�r I0e

�j�r�2[cos(� `

2cos )�cos(� `

2 )]� sin2

� fH� ' j!� sin � �

4�r I0e�j�r

�2[cos(� `

2cos )�cos(� `

2 )]� sin2

�From this, we can see by direct observation that the last two lines of the stated problem are true. That

is:

H� ' E��

Er ' E� ' Hr ' H� ' 0

This leaves us with proving that

E� ' j�I0e

�j�r

2�r

"cos��2 cos �

�sin �

#when

` =�

2

fE� ' j! sin ��

4�rI0e

�j�r

242hcos�� (�=2)

2 cos �� cos

�� (�=2)

2

�i� sin2

35' j! sin �

�

4�rI0e

�j�r"2�cos�2��4 cos

�� cos�2��4

�� sin2

#

' j! sin ��

4�rI0e

�j�r"2�cos��2 cos

�� cos��2

�� sin2

#

' j! sin ��

4�rI0e

�j�r"2�cos��2 cos

�� sin2

#

At this point we note that = � = azimuthal angle.

28

fE� ' j! sin ��

4�rI0e

�j�r"2�cos��2 cos �

�� sin2 �

#

' j!�

2��rI0e

�j�r"cos��2 cos �

�sin �

#

' j!�

�

I0e�j�r

2�r

"cos��2 cos �

�sin �

#

' j!�

!p�"

I0e�j�r

2�r

"cos��2 cos �

�sin �

#

' j�p�"

�"

I0e�j�r

2�r

"cos��2 cos �

�sin �

#

' jp�"

"

I0e�j�r

2�r

"cos��2 cos �

�sin �

#

' jp"p"

r�

"

I0e�j�r

2�r

"cos��2 cos �

�sin �

#

' j�I0e

�j�r

2�r

"cos��2 cos �

�sin �

#

29

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