Final Exam Antenna 2006

D-ITET Antennas and Propagation

Student-No.:.....................................................................

Name: ......................................................................

Address: ......................................................................

......................................................................

Antennas and PropagationFall 2006

October 12, 2006, 09:00 am – 12:00 noon

Dr. Ch. Fumeaux, Prof. Dr. R. Vahldieck

This exam consists of 6 problems. The total number of pages is 26, including

the cover page. You have 3 hours to solve the problems. The maximum

possible number of points is 70.

Please note:

This is an open book exam.

Attach this page as the front page of your solution booklet.

All the calculations should be shown in the solution booklet to justify the solutions.

Please, do not use pens with red ink.

Do not forget to write your name on each solution sheet.

Please, put your student card (LEGI) on the table.

Possible further references of general interest will be written on the blackboard during

the examination.

Problem Points Initials

1

2

3

4

5

6

Total

— 1 / 26 —

Masters

olution

D-ITET Antennas and Propagation October 12, 2006

Problem 1 (12 Points)

A 60° perfectly conducting corner reflector is used in a radar tracking system, as shown in

the figure below. The conducting walls of the reflector can be considered infinitely long. A

/ 2l dipole, located at a distance d from the vertex, is used as a feed element.

4 Points a) Find the equivalent array of dipoles by determining a system of images properly placed

in the absence of the reflector plates. Explain your procedure in a few words.

Note: The total number of dipoles n is determined by 360 / na = ° .

6 Points b) Derive the array factor AF in the xy -plane for the configuration found in a). Start with

determining the path delays of the array elements.

After algebraic transformations, the normalized array factor nAF of b) can be written as

( ) 1 1 34sin cos cos cos cos sin

2 2 2nAF kd kd kdf f f fé ùæ öæ ö æ ö= -ê úç ÷ç ÷ ç ÷ ç ÷è ø è øê úè øë û

2 Points c) A requirement for the radar system is that the antenna has a zero in the radiation pattern

along the symmetry axis ( x -axis). Find all possible feed-to-vertex spacings d , which

fulfill this requirement.

— 2 / 26 —

PEC


Solution 1

a)

A system of images has to be found, which when properly placed in the absence of the

reflector plates, form an array that yields the same field within the space formed by the

reflector plates as the actual system.

According to image theory, the polarity of the images changes by 180° with respect to the

“real” element. In the first step 1) the feed is mirrored with respect to the upper plate. In the

second step 2), both feed and image 2 are mirrored with respect to the lower plate, resulting

in image 5 and image 6. In a third step 3), the image 3 and image 4 are found by mirroring

with respect to the upper plate.

b)

The path delay between the element n and the center of the array can be calculated by the

projection

— 3 / 26 —


( )cos n n rd d a ag = ×r r

,

as can be seen in the figure above. Here the unit vector( ) ( )cos ( 1) sin ( 1)n x ya n a n aa a= - + -

r r r

points towards the element n , and 60a = ° is given by the geometry of the reflector

antenna. The radial unit vector in spherical coordinates in the xy -plane ( 90q = ° ) directed

towards the observation point is given bycos sinr x ya a af f= +

r r r.

Now either the angles ng can be determined, and therewith cos ng :( 1)n ng a f= - - .

Or the vectors pointing toward the elements can be set up:

1 xa a=r r

2

1 3cos sin

2 2x y x ya a a a aa a= + = +r r r r r

( ) ( )3

1 3cos 2 sin 2

2 2x y x ya a a a aa a= + = - +r r r r r

( ) ( )4 cos 3 sin 3x y xa a a aa a= + = -r r r r

( ) ( )5

1 3cos 4 sin 4

2 2x y x ya a a a aa a= + = - -r r r r r

( ) ( )6

1 3cos 5 sin 5

2 2x y x ya a a a aa a= + = -r r r r r

With the far-field approximations for the amplitude term

nr r»

and for the phase term

cosnr r d g» -

the AF can be determined.

( )1 1 1cos cos sin

cos

r x x yr r d r da a r d a a a

r d

g f f

f

= - = - × = - × +

= -

r r r r r

( )2 2

1 3cos cos sin

2 2

1 3cos sin

2 2

x y x yr r d r d a a a a

r d

g f f

f f

æ ö= - = - + × +ç ÷ç ÷

è øæ ö

= - +ç ÷ç ÷è ø

r r r r

3

1 3cos sin

2 2r r d f f

æ ö= - - +ç ÷ç ÷

è ø( )4 cosr r d f= - -

5

1 3cos sin

2 2r r d f f

æ ö= - - -ç ÷ç ÷

è ø

— 4 / 26 —


6

1 3cos sin

2 2r r d f f

æ ö= - -ç ÷ç ÷

è ø

And therefore, with alternating sign of the amplitude (image theory, see figure above)

( ) ( )

( )

1 3 1 3exp cos exp cos sin exp cos sin

2 2 2 2

1 3 1 3exp cos exp cos sin exp cos sin

2 2 2 2

AF jkd jkd jkd

jkd jkd jkd

f f f f f f

f f f f f

æ ö æ öæ ö æ ö= - + + - +ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷è ø è øè ø è ø

æ ö æ öæ ö æ ö- - + - - - -ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷è ø è øè ø è ø

or

( ) ( ) 1 3 1 32 sin cos sin cos sin sin cos sin

2 2 2 2AF j kd kd kdf f f f f f

é ùæ ö æ öæ ö æ ö= - + + - +ê úç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷ê úè ø è øè ø è øë û

with ( )sin sin cos cos sinx y x y x y± = ± and a normalization

( ) ( )

( )

3 1 3 12sin cos 2sin sin cos cos 2cos sin sin cos

2 2 2 2

3 1 3 12sin sin cos cos 2cos sin sin cos

2 2 2 2

32 sin cos 2cos sin s

2

nAF kd kd kd kd kd

kd kd kd kd

kd kd

f f f f f f

f f f f

f f

æ ö æ öæ ö æ ö= - -ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷è ø è øè ø è øæ ö æ öæ ö æ ö+ -ç ÷ ç ÷ç ÷ ç ÷ç ÷ ç ÷è ø è øè ø è ø

æ ö= - ç ÷ç ÷

è ø

1in cos

2kd f

é ùæ öê úç ÷

è øê úë ûand finally with sin 2 2sin cosx x x=

( ) 1 1 34sin cos cos cos cos sin

2 2 2nAF kd kd kdf f f fé ùæ öæ ö æ ö= -ê úç ÷ç ÷ ç ÷ ç ÷è ø è øê úè øë û

c)

There should be a zero on the x -axis ( 0f = ° ):

( )!1 1

0 4sin cos 1 02 2nAF kd kdf é ùæ ö æ ö= ° = - =ç ÷ ç ÷ê úè ø è øë û

Thus, there are two possibilities to fulfill the requirement:

i)1

sin 02

kdæ ö =ç ÷è ø

or ii)1

cos 12

kdæ ö =ç ÷è ø

2

kdmp= , 0, 1, 2,m = ± ± K 2

2

kdmp= , 0, 1, 2,m = ± ± K

d ml= , 0, 1, 2,m = ± ± K 2d ml= , 0, 1, 2,m = ± ± K

Together, the following distances are possible in order to fulfill the requirementd ml= , 0, 1, 2,m = ± ± K

— 5 / 26 —



On a highway, car 1 detects car 2 (radar cross section 23ms = ) in a distance 1 2y y+ (

1 15y = m, 2 6y = m) in front of it with the help of a radar system (transmit power 1tP = W,

frequency 24radarf = GHz, linear polarization in z -direction). Assume that the radar system

is omnidirectional in xy -plane with 1radarG = . The two cars drive on two different lanes in a

distance 1 6x = m.

In the present instantaneous situation on the highway, the signal can propagate on two

different ways from car 1 to car 2 as shown in the figure below:

way 1: line of sight,

way 2: between car 1 and car 2, the signal undergoes a total reflection from a

truck driving on the right lane ( 2 4x = m; reflection coefficient 1G = - ).

Note: Ground reflection can be neglected because of a narrow pattern of the radar system in

elevation.

2 Points a) List the 4 main propagation paths of the radar signal from car 1 to car 2 and back to

car 1.

4 Points b) Determine the power received by the radar system on each of those paths.

4 Points c) Determine the effective area of the radar antenna. Find the amplitudes of the received

electric fields for each of the propagation paths.

2 Points d) Determine the phases of the received electric fields on each of those paths.

2 Points e) How can the total amplitude of the electric field received by the radar system be

computed? Give one equation without developing the overall solution.

— 6 / 26 —


Solution 2

a)

The four main propagation paths are:

Path 1: On way 1 to car 2, and back on way 1 to car 1.




b)

Radar range equation2

1 24 4t r

r tGGP P

RRls

p pæ ö÷ç ÷= ç ÷ç ÷çè ø

As can be seen from the figure above, path 1 has the length

( ) 221 1 1 2 21.8403d x y y= + + = m

Path 2 has the length

2 25.2389d = m

with2 2

21 1 2 1( ) 18.0278d x x y= + + = m2 2

22 2 2 7.2111d x y= + = m.

The following numbers are given:

1tP = W, and thus , dB 30tP = dBm,

1tG = , and thus , dB 0tG = dB,23ms = ,

2 1G = , and thus dB 0G = dB

24f = GHz, and thus 312.49 10l -= × m.

and

2dB

4 53.83psl

æ ö÷ç =÷ç ÷çè ø dB

— 7 / 26 —


Path 12 42

2 121 2 2

1 1

4 1.037 104 4 4t

r t t tGP P P G

d dl ps ls

p p l p-

æ ö æ öæ ö÷ ÷ç ç÷ç÷ ÷= = = ×ç ç÷ç÷ ÷÷çç ç÷ ÷è øç çè ø è øW

with4

1 dB

173.674 dlp

æ ö÷ç ÷ = -ç ÷ç ÷çè ødB

( ) ( )( ) 421, dB , dB , dB 1dB dB

4 / 2 / 4 89.84r t tP P G dps l l p= + + + = - dBm

Path 24

4 2 132 2

2

4 5.813 104r t tP P Gd

ps ll p

-æ öæ ö ÷ç÷ç ÷= G = ×ç÷ç ÷÷ç ç ÷è ø çè ø

where 1G = - is the reflection coefficient of the total reflection from the truck. Since the

reflection coefficient is defined over the amplitude, and the radar signal is two times

reflected by the truck (on the way there and on the way back), it is considered with a power

of four in the radar range equation.

With4

2 dB

176.194 dlp


and thus

2, dB 92.36rP = - dBm

Path 3 and Path 4

( )

222 2 13

3 4 221 2

4 7.763 104r r t tP P P Gd d

ps ll p

-æ öæ ö ÷ç÷ç ÷= = G = ×ç÷ç ÷÷ç ç ÷è ø çè ø

with

( )

21

21 2 dB

174.934 d dlp


3, dB 91.1rP = - dBm

c)

In general, the amplitude is connected to the power by2 2

2 4r

r r r r

EP W A Gl

h pæ ö÷ç= × = × ÷ç ÷÷çè ø

and therewith 2 rr

r

PEAh=

— 8 / 26 —


The effective aperture is2

5 21.24 10 m4r rA Glp

-= = ×

or 2

, dB 10 log 49.05994r rA Glp

æ ö÷ç= = -÷ç ÷÷çè ødB

Path 1

311 2 7.93 10rr

r

PEAh -= = × V/m

or

1 dB42.0127rE = - dB

Path 2

322 2 5.94 10rr

r

PEAh -= = ×

or

2 dB44.5252rE = - dB

Path 3 and path 4

313 4 2 6.86 10rr r

r

PE EAh -= = = ×

or

3,4 dB43.2690rE = - dB

d)

The phase on each propagation path is given by:

Path 14

1 12 2 2.1971 10 313.8862r dpjl

= = × °@

Path 24

2 22 2 2.5390 10 4.6902r dpjl

= = × °@

Path 3 and path 4

( ) 43 4 1 2

2 2.3681 10 339.2882r r d dpj jl

= = + = × °@

e)1 2 3

1 2 32r r rj j jtot r r rE E e E e E ej j j- - -= + -

— 9 / 26 —


The minus sign is due to the phase jump caused by the total reflection at the truck.

NOTE:

The total received power is2 2

2 4tot

tot t

EP Gl

h p= ×

With the following approximation for the amplitude

( )1 2 1 2D d d d d» » » + the amplitude of the total electric field can be written as2

22 21 3 22

1 2 exp( ) 2 exp( ) exp( )4tot t tE P G j j jD

h s j j jp

æ ö÷ç= + G + G÷ç ÷çè ø

and therewith

( ) ( )

422 2

1 3 22

422 2

3 1 2 12

4 exp( ) 2 exp( ) exp( )4

4 1 2 exp( ) exp( )4

tot t t

t t

P P G j j jD

P G j jD

ps l j j jl pps l j j j jl p

æ ö÷ç= + G + G÷ç ÷çè øæ ö÷ç= + G - + G -÷ç ÷çè ø

Since ( )3 1 20.5j j j= +

( )4

422 12

4 1 exp( /2)4tot t tP P G jD

ps l j jl p

æ ö÷ç= + G -÷ç ÷çè ø ,

see exam Spring 2006 Problem 2b).

— 10 / 26 —



A lossless antenna A is CW circularly polarized. Its far field radiation is described by the

electric field with magnitude

( )0 sin sin cos cos for 0 and 02 6 2

0 elsewhere

jkr

A

eE

r

E

p p pp q f q f p-ì é ùæ ö× + £ £ £ £ï ç ÷ê úè øë ûï

ï= íïïïî

4 Points a) Find the direction ( 0 0,q f ) of maximum radiation intensity of antenna A.

4 Points b) Determine the half-power beamwidths (HPBWs) in the elevation plane ( 0f f= ) and the

plane perpendicular to it ( 0q q= ).

1 Point c) Estimate the directivity of antenna A using the appropriate approximation formula.

A dipole antenna is positioned in the far field region of antenna A, in its maximum radiation

direction. This is illustrated in the Figure below. The dipole is perpendicular to the

maximum radiation direction of the antenna A, and lies in the elevation plane.

2 Points d) Determine the polarization loss factor (PLF) of this antenna system.

1 Point e) Find the maximum effective aperture of antenna A, if the operating frequency is

150 MHz. Use the estimated value of the directivity.

— 11 / 26 —

Antenna A

Dipole


Solution 3

a) The radiation intensity of antenna A can be obtained from the electric field as

( ) ( )2

2 21, , ,

2 2A A A

rU E r Eq f q f

h h= =

( )20

1sin sin cos cos , 0 and 0

2 2 6 2AU Ep p pp q f q f p

hé ùæ ö= × + £ £ £ £ç ÷ê úè øë û

The maximum radiation direction is given by

( )

( )

0 0

0 0

sin sin cos cos 12 6

sin sin 1and cos cos 12 6

p pp q f

p pp q f

é ùæ ö× + =ç ÷ê úè øë ûé ùæ öÞ = + =ç ÷ê úè øë û

( )0

0

0

0

sin sin 1and 02

sin2

1sin

2

6

pp q q

pp q

q

pq

= £ £

Þ =

Þ =

Þ =

0

0

0 0

0

0

cos cos 12 6

cos 02 6

cos 0 and 06

(the only solution)6 2

3

p pf

p pf

pf f p

p pf

pf

é ùæ ö+ =ç ÷ê úè øë ûæ öÞ + =ç ÷è ø

æ öÞ + = £ £ç ÷è ø

Þ + =

Þ =

Thus, the direction of maximum radiation is determined by 0 030 and 60q f= =o o .b) Elevation plane 0 60f f= = o

Normalized radiation intensity is given by

( )

( )

1 0

1

sin sin cos cos2 6

sin sin

An

An

U

U

p pp q f

p q

é ùæ ö= × +ç ÷ê úè øë û=

Half-power angle hpq can by obtained by

— 12 / 26 —


( )1

1 2

1sin sin

25

sin and6 6

1 5sin and

6 6

9.6 and 56.44

An hp

hp

hp

hp hp

U p q

p pp q

q

q q

= =

Þ =

Þ =

Þ = =o o

elevation 2 1HPBW 46.84hp hpq q= - = o

The normal plane 0 30q q= = o

2 cos cos2 6AnUp pfé ùæ ö= +ç ÷ê úè øë û

1 2

1cos cos

2 6 2

cos2 6 3

2cos and 0

6 3

0.8411and 2.3rad60.3175rad and 1.7764 rad

hp

hp

hp

hp

hp hp

p pf

p p pf

pf f p

pf

f f

é ùæ ö+ =ç ÷ê úè øë ûæ öÞ + = ±ç ÷è ø

æ öÞ + = ± £ £ç ÷è ø

Þ + = ± ±

Þ = =Or in degrees

1 2

30 48.19 and 131.81

18.19 and 101.81

hp

hp hp

f

f f

Þ + = ± ±

Þ = =

o o o

o o

perpendicular 2 1HPBW 83.62hp hpf f= - = o

c)As demonstrated in the script (pg. 2.18), Kraus’ approximation formula works better for wider beams. Thus, we can estimate the directivity using

0elevation perpendicular

0

41,253

HPBW HPBW

41,25310.53 10.23dB

46.84 83.62

A

A

D

D

»×

» = =×

d)Elevation plane 60f = o is shown in the figure below

— 13 / 26 —


Antenna A in transmitting mode radiates in rer

direction. It is CW circularly polarized. Thus, its wave vector is given by

2A

e jeq fr-

=r r

r

The dipole antenna is linearly polarized. Its polarization vector is given by

D e ey qr = =r r r

Thus, the polarization loss factor can be obtained as2

2 1PLF

22TD A

e jee q fqr r

-= × = × =

r rr r r

e)Antenna A is lossless. Thus, its maximum effective aperture is given by

2

_ 04A em AA Dlp

æ ö= ç ÷è ø

150MHz 2mf l= Þ =2

2_

210.53 3.36m

4A emAp

æ ö= =ç ÷è ø

— 14 / 26 —



A linearly polarized plane wave with amplitude 0E is incident upon an aperture cut into a

perfectly electric conducting ground plane of infinite extent. The wave polarization is

parallel as shown in the figure below. The aperture has dimensions a and b . Assume that

the field over the aperture is given by the incident field.

4 Points a) Determine the E and H fields over the aperture.

2 Points b) Find the sources of the aperture’s equivalent system.

3 Points c) Find the far zone spherical components of the fields for 0x > .

— 15 / 26 —

Aperture: front viewAperture: side view


Solution 4

a)Incident electric and magnetic fields are given by

( ) ( )

( )

0 0

0 0

sin cos0 0 0

sin cos0

sin cos jk y xix y

jk y xiz

E E e e e

EH e e

f f

f f

f f

h

-

-

= + ×

= ×

r r

r

In the aperture, 0x = . Thus, the fields over the aperture are

( ) 0

0

sin0 0 0

sin0

sin cos jk yax y

jk yaz

E E e e e

EH e e

f

f

f f

h

×

×

= + ×

= ×

r r

r

b)The given aperture can be replaced by an equivalent system as shown below.

The equivalent magnetic source is

( ) 0

0

sin0 0 0

sin0 0

ˆ2 2 sin cos

2 cos

jk yas x x y

jk ys z

M n E e E e e e

M e E e

f

f

f f

f

×

×

= - ´ = - ´ + ×

= - ×

r r r rr r

c)

0N Nq f= =and

'cossin ' and 0jkrz

S

L M e ds Lyq fq += - × =òò

— 16 / 26 —

0

S

S

J

M =

r

r

0

ˆ2

S

aS

J

M n E

=

= - ´

r

r rˆ

ˆ

aS

aS

J n H

M n E

= ´

= - ´

r r

r r

0

0

S

S

J

M

=

=

r

r0

S

S

J

M =

r

r

0

0

S

S

J

M

=

=

r

r

0

S

S

J

M =

r

r

0

S

S

J

M =

r

r

ˆ

ˆ

aS

aS

J n H

M n E

= ´

= - ´

r r

r r


( ) ( )'cos ' ' ' sin cos sin sin cos

'cos 'sin sin 'cos

r y z x y zr r e y e z e e e e

r y z

y q f q f q

y q f q

= × = + × + +

= +

r r r r r r r

( )

( )

( )

0

0

'sin sin 'cos

'sin sin 'cos'sin0 0

2 2' sin sin sin 'cos

0 0

2 2

0 0

sin ' '

2 cos sin ' '

2 cos sin ' '

sin sin2 cos sin

jk y zz

S

jk y zjk y

S

a bjky jkz

a b

L M e dy dz

E e e dy dz

E e dy e dz

Y ZE a b

Y Z

q f qq

q f qf

q f f q

q

f q

f q

f q

+ +

+ +×

+ + +

- -

= - ×

= × ×

= ×

é ù é ù= × ×ê ú ê úë û ë û

òò

òò

ò ò

Where

( )0sin sin sin2

cos2

kaY

kbZ

q f f

q

= +

=

Thus

0 0

4

sin sincos sin

2

0

jkr

jkr

keE j L

r

ke Y ZE jE ab

r Y Z

EH

E H

f q

f

fq

q f

p

f qp

h

-

-

=

=

= -

= =

— 17 / 26 —



Two 2l dipoles are crossed above a perfect electric ground plane as shown in the figure

below. The height at which the antennas are mounted is: 0.5mh = . The operating frequency

of this antenna system is 300 MHz.

For this antenna system above the ground plane

2 Points a) Sketch the antennas, their images, with respect to the location of the ground plane and

explain the position of the images.

2 Points b) Find the far-zone E-field radiated by each, vertical and horizontal dipole over the

ground plane separately.

2 Points c) Find the total, far-zone E-field ( ,E Eq f -components) radiated by the whole system in the yz -plane above the ground. The two dipoles are excited with the same phase.

2 Points d) What type of polarization (linear, circular, or elliptical) does the field in the yz -plane

have? An intuitive explanation is sufficient.

2 Points e) At what distance from the antenna are the field expressions obtained in c) valid?

— 18 / 26 —


Solution 5

a)

The operating frequency is 300 MHz, thus the wavelength is = 1 m and h = /2.

The given antenna system above the ground plane can be represented using image theory as:

The images are oriented as shown in the figure because the tangential component of the E-field on the boundary has to be zero.

b)

E-field of the vertical /2 dipole is given by

10

,11

cos cos2

2 sin

jkRv I e

E jRq

p qh

p q

-æ ö÷ç ÷ç ÷çè ø=

The electric field radiated by the image of the vertical dipole is given by

20

,22

cos cos2

2 sin

jkRv I e

E jRq

p qh

p q


The total E-field radiated by the vertical dipole is given by

,1 ,2v v vE E Eq q q= +

R1 and R2 are distances from the antenna and the image to some point P in the far field, respectively. Generally, we can write:

— 19 / 26 —


1 2

1

2

for amplitude terms

R cosfor phase terms

R cos

R R r

r h

r h

qq

= =ü= - ïïýï= + ïþ

Thus, the total E-field radiated by the vertical dipole is given by

( )0

cos cos2

2cos cos2 sin

jkrv I e

E j khrq

p qh q

p q

-æ ö÷ç ÷ç ÷çè ø é ù= ë û

Since h=/2, the E-field radiated by the vertical dipole is:

( )0

cos cos2

2cos cos2 sin

jkrv I e

E jrq

p qh p q

p q


(1)

The horizontal dipole above the ground plane radiates E-field given by

10

,11

cos cos2

2 sin

jkRh I e

E jRy

p yh

p y


And its image

20

,22

cos cos2

2 sin

jkRh I e

E jRy

p yh

p y

-æ ö÷ç ÷ç ÷çè ø= -

where -direction is shown in the figure above. The relation between angles and angles

and is given by:

2 2

cos sin sin

sin 1 sin sin

y q f

y q f

=

= -Similarly to the case of the vertical dipole, the total E-field radiated by the horizontal dipole is found as

( )0

cos cos2

2 sin cos2 sin

jkrh I e

E j jry

p yh p q

p y


(2)

c)The total E-field of the antenna system is obtained by adding (1) and (2). The yz-plane with the antenna system is shown in the figure below.

— 20 / 26 —


In yz-plane 90f o= , thus

2 2

cos sin sin sin

sin 1 sin sin cos

y q f q

y q f q

= =

= - =and

( )0

cos sin2

2 sin cos2 cos

jkrh I e

E j jry

p qh p q

p q


From the sketch above, one can see that in Q1 and Q3

in Q2 and Q4

a a

a ay q

y q

r rr r

= -=

Therefore, in Q1 and Q3, the total field is:

( ) ( )

( )

0 0

0

cos cos cos sin2 2

2cos cos 2 sin cos2 sin 2 cos

cos cos cos sin2 2

cos cos sin cosin cos

v h

jkr jkr

jkr

E E a E a

I e I eE j a j j a

r r

I eE j j

r

q q y q

q q

p pq qh p q h p q

p q p q

p pq qh p q p

p q q

r r

r r- -

-

= × - ×æ ö æ ö÷ ÷ç ç÷ ÷ç ç÷ ÷ç çè ø è øé ù é ù= × - ×ë û ë û

æ ö æ ö÷ ÷ç ç÷ ÷ç ç÷ ÷ç çè ø è ø= - ( )s aqq ré ùê úê ú

×ê úê úê úê úë û

and in Q2 and Q4, the total field is

( ) ( )0

cos cos cos sin2 2

cos cos sin cossin cos

jkrI eE j j a

r q

p pq qh p q p q

p q qr-

é ùæ ö æ ö÷ ÷ç çê ú÷ ÷ç ç÷ ÷ç çê úè ø è ø= + ×ê úê úê úê úë û

d)The field in the yz-plane is linearly polarized.

e)

— 21 / 26 —


The expressions for the E-field obtained in c) and d) are valid for the far field region, which starts at 22FFR D l= . In order to determine the size of the antenna (D), the whole antenna system, including the original antennas and their images, has to be considered.

3

2 22

D hl l= + =4

we have2

2

32

2 924.5m

2FF

DR

l

ll l

æ ö÷ç ÷ç ÷çè ø= = = =

— 22 / 26 —



A rectangular patch antenna, shown in figure below, with the dimensions 6cma = and

3cmb = is realized on a substrate with 2.2re = and 1h = mm.

For this antenna system

2 Points a) Calculate the fundamental resonance (lowest resonant frequency). Neglect the fringing

fields!

3 Points b) For the first four (in terms of frequency) resonant modes, calculate the resonance

frequency. Neglect the fringing fields!

The feeding network consists of two microstrip lines connected at the center of two

neighboring sides of the patch antenna as shown in the figure below.

- continued on the next page -

— 23 / 26 —

b

a

z

yx

a

x

yz b

z

yx


5 Points c) One of the first four modes calculated CAN NOT be exited with this feeding structure.

i. Point out the mode.

ii. Explain the problem of the excitation.

iii. Find a simple solution to feed this mode.

iv. Is this mode useful for use in a radiating application?

In figure below, a quadratic patch is given with the dimensions 3a b= = cm. The two

connected feeding lines have a width 0 1.6w = mm.

3 Points a) Calculate the length difference between the two feeding lines in order to achieve

circular polarization of the patch antenna at the lowest resonant frequency. Take the

fringing fields into account!

— 24 / 26 —

b

b

z

yx


Solution 6

a)The fundamental mode is 010

xTM and the resonance frequency can be calculated as follows.

æ ö æ ö÷ ÷ç ç= +÷ ÷ç ç÷ ÷è ø è ø

Þ = =

2 20

,0

0,010

2

1.6092

r npr

rr

c n pfa b

cf GHza

p pp e

eb)First four Modes are 010

xTM , 001xTM , 020

xTM , 011xTM .

æ ö÷ç= ÷ç ÷è ø

Þ = =

æ ö÷ç= ÷ç ÷è øÞ = =

æ ö÷ç= =÷ç ÷çè øæ ö æ ö÷ ÷ç ç= + =÷ ÷ç ç÷ ÷è ø è øæ öç= ççè ø

20

,010

0,010

20

,001

0,001

20

,020

2 20

,011

0,030

2

1.6852

23.37

22 3.37

2

3.7682

32

rr

rr

rr

rr

rr

rr

rr

cfa

cf GHzacf

bcf GHzacf GHz

acf GHz

a bcf

a

pp e

ep

p e

ep

p ep p

p ep

p e÷ =÷÷

æ ö÷ç= = =÷ç ÷çè ø

2

20

,040 ,002

5.055

2 6.742r r

r

GHz

cf f GHzbp

p e

c)The 011

xTM can NOT be exited by this port configuration because the mode exhibits zero E-field at the tapping position.The mode can be feed by a feeding line at the corner of the patch antenna. Still the mode is not useful for any radiation application because the slots have opposite current distributions. Therefore the radiation is nearly zero because the different contributions cancel each other.d)

-é ù+ -= + + =ê úê úë û

12

,1 1 1 12 2.107

2 2r r

reff patchha

e ee

( )( )

+ +D =- +

D =

( 0.3) / 0.2640.412

( 0.258) / 0.80.527

reff

reff

a hLh a hL mm

ee

= = =+ D

0

, 0 ,

1 3.3272 2( 2 )rceff reff patch o reff patch

cf GHzL a Le m e e

-é ù+ -= + + =ê úê úë û

12

,1 1 1 12 1.806

2 2r r

reff linehw

e ee

— 25 / 26 —


The circular polarization is achieved by a phase difference of 90°, which translate into a

length difference of = = 1.6784eff

diffl cml

.

— 26 / 26 —

Final Exam Antenna 2006

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