Final Exam Antenna 20051

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Student-No.:... Student-No.:... Name: ... Name: ... Address: ... Address: ... ... ...

Antennas and Propagation

Spring 2005

March 17, 2005, 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 19, including

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

possible number of points is 67.

Please note:

Please note: •

• This is anThis is an open bookopen book exam.exam. •

• Attach this page as the front page of your solution booklet.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.All the calculations should be shown in the solution booklet to justify the solutions. •

• Please, do not use pens with red ink.Please, do not use pens with red ink. •

• Do not forget toDo not forget to write your namewrite your name onon eacheach solution sheet.solution sheet. •

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

• Possible further references of general interest will be written on the Possible further references of general interest will be written on the blackboard duringblackboard during the examination.

the examination.

Problem

Problem Points Points InitialsInitials 1 1 2 2 3 3 4 4 5 5 6 6

(2)

Problem 1 (10 Points)

Assume

Assume a a receiver receiver is is located located km km from from a a 50 50 W W transmitter. transmitter. The The receiver receiver andand transmitter are mounted on 2.5 m and 5 m high posts, respectively. The carrier frequency is transmitter are mounted on 2.5 m and 5 m high posts, respectively. The carrier frequency is 900

900 MHz. MHz. The The receiver receiver and and transmitter transmitter antennas antennas have have gains gains and and ,, respectively. respectively.

10

d d

=

rr

11

G G == G G _tt ==

22

5 5 mm 2.5 m 2.5 m ground ground d d == 10 km10 km transmitter transmitter receiver receiver 2 Points

2 Points a) a) Find free Find free space space loss anloss and received received powed power if r if reflections reflections from earth from earth are negare neglected.lected. 2 Points

2 Points b) b) FinFind d addadditiitionaonal l powpower er losloss s ((LL_ref_ref) ) in in dB dB due due to to refrefleclectiotions ns frofrom m the the grogroundund.. 3 Points

3 Points c) c) Assume Assume a wall poa wall positioned in sitioned in between between the transmitter athe transmitter and receiver, nd receiver, 8 km 8 km away from away from thethe transmitter. Calculate the height of the wall so that

transmitter. Calculate the height of the wall so that the power loss due to the knife-edgethe power loss due to the knife-edge dif

diffrafractiction ion is ths the sae same ame as ths the pe poweower lor loss ss due due to rto refleflectection ion frofrom Eam Earth rth ((LL_ref_ref) in ) in b).b). 3 Points

3 Points d) d) What would What would be the be the maximum hmaximum height of eight of the wall sthe wall so that tho that the power e power loss due loss due to theto the diffraction is negligible? Can such a wall be built?

(3)

Solution 1

a) a)

990000 M

MH

Hz

z

c c

00..333333 m

m

f f f f λ λ

=

⇒

=

Free space loss: Free space loss:

( ( ))

22

( (

))

22 LOS LOS ₂₂ 11 11 LOS LOS

0.333

1 1 22

44

4 4 110

0 1100

11..4400772

2 110

0 110088..551 d

1 dB

B

r r t t L L G G G G d d L L λ λ π π ππ − −

=

⋅

⋅⋅

⋅ ⋅

⋅⋅

=

⋅

=

−

Received Power Received Power L LOOSSr r LLOOSS 10 10 LOSr LOSr

77..00336

6 110

0 W

W

9911..5 d

5 dB

BW

W

6611..5 d

5 dB

Bm

m

t t P P P P LL P P −−

=

⋅⋅

=

⋅

=

−

=

−

or or L LOOSSr r t t LLOOSS

5500 W

W

1166..99889dB

9dBW

W

1166..99889

9 110088..551

1 9911..5dB

5dBW

W

t t P P P P P P LL

=

= +

+

=

−

=

−

b) b)

The power loss due to the reflection from the

The power loss due to the reflection from the ground isground is

(

( )

) (

( ))

( (

))

2 2 22 ref ref 22 ref ref ref ref

2

22

1

1 eex

xp

p

2

22 0.00222

0.00222

26.535dB

t t rr tt rr t t r r h h h h h h h h L L jjkk kk d d d d h h h h L L d d L L π π λ λ

=

= +

+

Γ

⋅

≈

≈ ⋅⋅

=

⋅ ⋅

=

= −

−

The plane earth loss (PEL) is The plane earth loss (PEL) is

P

PEEL L LLOOS S rreef f

135.045dB

L

(4)

c) c)

Power loss due to the diffraction is t

Power loss due to the diffraction is the same as the power loss due to the reflection, thushe same as the power loss due to the reflection, thus d

diifff f rreef f

26.535dB

L

=

LL

=

= −

−

From the graph knife-edge diffraction

From the graph knife-edge diffraction vs. parametervs. parameter υ,υ,we see thatwe see that υυ > 2.4. Thus:> 2.4. Thus:

( (

))

diff diff

0.225

0.225 2200 lloog

g

44..7788

L L υυ υ υ

=

⇒

=

( ( )) ( ( )) ( ( )) 1 1 22 1 1 22 1 1 22 1 1 22 ttoot t rr

22

4.78

4.78 78.1m

78.1m

2 8

2 800000

0 22000000

0.

0.33

333 3 80

8000

00 20

2000

00 80.5m

80.5m

d d d d h h h h d d d d d d d d d d d d h h h h h h h h υ υ υ υ λ λ λ λ

+

=

⇒

⇒ =

=

+

=

+

⋅ ⋅

⋅⋅

=

= +

+

=

d) d)

The diffraction loss is negligible if most of the 1

The diffraction loss is negligible if most of the 1stst Fresnel zone is clear (uncovered by theFresnel zone is clear (uncovered by the wall).

wall).

More specifically, the Fresnel zone clearance condition has to

More specifically, the Fresnel zone clearance condition has to be satisfied, i.e.be satisfied, i.e.

ttoot t rr

00..8

8 1133..006633

10.563m

h h h h h h h h υ υ

=

= −

−

⇒

⇒ =

= −

−

=

= +

+ =

= −

−

Therefore, such wall cannot be built. Therefore, such wall cannot be built.

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Problem 2 (13 Points)

An infinitesimal dipole of length

An infinitesimal dipole of length l l is placed at a distance s is placed at a distance s from a ground plane and at anfrom a ground plane and at an angle of 45 degrees from the vertical axis, as shown in the figure below. The dipole lies in angle of 45 degrees from the vertical axis, as shown in the figure below. The dipole lies in the

the yz yz -plane.-plane.

s s y y x x 45° 45° 45° 45° normal normal 3 Points

3 Points a) a) Determine the Determine the location and location and direction of direction of the image the image source, whsource, which can ich can be used be used to accounto accountt for reflections of the ground plane. Your answer should be in a form

for reflections of the ground plane. Your answer should be in a form of a very clearof a very clear sketch.

sketch. 6 Points

6 Points b) b) With the With the coordinate scoordinate system giveystem given in the n in the figure, find figure, find the exprethe expression for ssion for the total the total far fieldfar field electric field in the 2nd quadrant of the

electric field in the 2nd quadrant of the yz yz -plane (the shaded region).-plane (the shaded region). 4 Points

4 Points c) c) Find Find the the smallest smallest non-zero non-zero distancedistance s s for which the total field for which the total field obtained in b) is zeroobtained in b) is zero along the normal.

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Solution 2

a) a)

The image dipole is shown in the figure below. Tangential field at the ground plane is The image dipole is shown in the figure below. Tangential field at the ground plane is zero.zero.

45 45 45°45° normal normal E Eqq11 E Eqq22 b) b)

The original dipole and the image form an array as shown in the

The original dipole and the image form an array as shown in the sketch belowsketch below

s s y y x x 45° 45° 45° 45° 45°45° normal normal r r11 r r₂₂ 45 45_°-_°-_q_q q q y y z’ z’ ( ( )) ( ( )) 11 22 1 1 22

ccos

os 45

45 phase

phase

co

cos

s 45

45 magnitude

magnitude

r r r r s s r r r r s s r r r r r r θθ θθ

=

= −

− ⋅

⋅

−

_⎫⎫⎪⎪

⎪⎪

⎬⎬

⎪⎪

=

= +

+ ⋅

⋅

−

_⎪⎪⎭⎭

=

    (*) (*)

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The field produced by the original dipole is The field produced by the original dipole is

11 00 1 1 11 11

sin

44

jkr jkr kI le kI le E E EE jj aa r r θ θ η η θθ θθ π π − −

=

⋅⋅

    __

The image dipole is directed along the z’ direction and produces field in

The image dipole is directed along the z’ direction and produces field in ψψdirection, defineddirection, defined on the graph above. Thus the image dipole field is given as

on the graph above. Thus the image dipole field is given as

22 00 2 2 22 22

sin

44

jkr jkr kI le kI le E E EE jj aa r r ψ ψ η η ψψ ψψ π π − −

=

⋅⋅

    __ where where ( ( )) 22 22 22 22 ''

ssiin

n

ψ ψ

=

= −

1 1 ccoos

−

s

ψ ψ

=

= −

1

1 − ⋅

aazz

⋅

aa rr

=

= −

1

1 − ⋅

aayy

⋅

a ar r

=

= −

1 ssiin

1 −

n ssiin

θθ

n

φφ

        Thus Thus ( ( )) 22 22 00 2 2 22 22

1

1 ssiin s

n siin

n

44

jkr jkr kI le kI le E E EE jj aa r r ψ ψ η η θθ ψψ π π − −

=

−

φφ

⋅⋅

    __

In the second quadrant of the yz-plane we have

In the second quadrant of the yz-plane we have φφ

=

= °

270 ,

°

a a ψ ψ

=

−

a a θθ

    Thus: Thus: ( ( )) 22 22 00 22 2 2 22 22 00 2 2 22 22

1

1 ssiin

n

44 cos

cos

44

jkr jkr jkr jkr kI le kI le E E EE jj aa r r kI le kI le E E E E jj aa r r ψ ψ θθ ψ ψ θθ η η θθ π π η η θθ π π − − − −

=

−

⋅⋅

=

−

⋅⋅

    __     __

The total field is equal to t

The total field is equal to the sum of the two fields, i.e.he sum of the two fields, i.e.

1 1 22 0 0 00 ttoot t 1 1 22 1 1 22

ssiin

n

ccooss

4

44

jjkkr r jjkkr r

kkI I lle e kkI I lle e E E EE EE jj jj r r r r θθ η η θ θ ηη π π ππ − − −− a a θθ

⎡ ⎡

⎤⎤

⎢ ⎢

⎥⎥

=

+

=

−

⋅⋅

⎢ ⎢

⎥⎥

⎣ ⎣

⎦⎦

    __ Expresing

Expresing r r ₁₁ and and r r ₂₂ in in terms terms of of r r (*), we obtain the total field in the 2(*), we obtain the total field in the 2ndnd quadrant asquadrant as

(

( )) (( ))

00 ccoos s 445 5 ccoos s 4455 tot

tot

₄₄

ssiin

n

ccooss

jkr jkr jjkks s jjkks s kI le kI le E E jj ee ee r r θ θ θθ θθ η η θθ π π − − + + −− −− −− _θθ _a_a

⎡ ⎡

⎤⎤

=

_{⎣ ⎣}

 

−



_⎦⎦

⋅⋅

  __ c) c)

Along the normal, we have

Along the normal, we have θθ

=

45

 Thus, the field can

Thus, the field can be expressed asbe expressed as

( ( )) 00 tot tot 00 tot tot

sin45

44

22

2

2 ssiin

n

4

22

jkr jkr jjkks s jjkks s jkr jkr kI le kI le E E jj ee ee r r kI le kI le E E jj kkss aa r r θθ θθ η η π π η η π π − − + + −− − −

⎡ ⎡

⎤⎤

=

_{⎣ ⎣}

−

_⎦⎦

=

= −

−

⋅⋅

  a a

⋅⋅

  __   __ So So E E _tot_tot

= ⇐

=

0

0 ⇐ =

kks s n

=

n n ππ

,

n

=

0,,11,,22,,...

0

 

11

n

=

= ⇒

⇒

smallest non-zero lengthsmallest non-zero length s s isis

22

s s

=

λλ

(8)

Problem 3 (12 Points)

Given are three radiation patterns that are taken from rectangular horn antennas. The horn Given are three radiation patterns that are taken from rectangular horn antennas. The horn antennas

antennas are are fed fed by by a a rectangular rectangular waveguide waveguide WR90 WR90 (dimension (dimension mm mm andand mm) operated in the dominant TE

mm) operated in the dominant TE

22.86

a a

=

10.16

bb

=

1010 mode mode at at f f ==

11

GHz.GHz. E-Plane E-Plane H-Plane H-Plane (1) (2) (3) (1) (2) (3) 2 Points

2 Points a) a) Relate each Relate each of the of the three radiation three radiation pattern to pattern to a type a type of rectanguof rectangular horn alar horn antenna. Givntenna. Give thee the physical explanation of your choice!

physical explanation of your choice! 2 Points

2 Points b) b) Which effect Which effect is responsis responsible for large ible for large back raback radiation of diation of the above the above patterns? Whpatterns? What wouldat would you recommend in order to

you recommend in order to decrease the back-side radiation?decrease the back-side radiation? 2 Points

2 Points c) c) Estimate Estimate the the dimensions dimensions and and of of a a pyramidal pyramidal horn horn aperture aperture antenna antenna required required toto exhibit

exhibit a a gain gain of of dBi dBi at at GHz if GHz if a a typical typical aperture aperture efficiency efficiency of of 50% 50% isis assumed.

assumed. The The side side length length ratio ratio is is ..

11 a a bb₁₁ 00

17

G G == f f ==

11

1 1

/

11

22

a a bb == 2 Points

2 Points d) d) Determine the Determine the maximum pomaximum power that cwer that can be an be received by received by the antennthe antenna designea designed in cd in c) if it) if it is

is illuminated illuminated by by a a plane plane wave wave of of the the frequency frequency GHz with GHz with an an amplitude amplitude of of mV/m. mV/m.

11

f f == 00

30

E E == 2 Points

2 Points e) e) Determine Determine the maxthe maximum power imum power that can that can be recbe received by eived by the feethe feeding wavding waveguide eguide (without(without the horn antenna) if it is

the horn antenna) if it is illuminated by the same plane wave as in d).illuminated by the same plane wave as in d). 2 Points

2 Points f) f) Comparing Comparing the apthe aperture efficienerture efficiencies cies of a of a horn ahorn antenna ntenna and and a waa waveguide, veguide, why why areare waveguides, having the same dimension as the aperture of

waveguides, having the same dimension as the aperture of a horn antenna, not used asa horn antenna, not used as radiating elements? Find an

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Solution 3

a) a)

Pattern

Pattern (1) (1) H-plane H-plane rectangular rectangular horn horn antennaantenna Pattern

Pattern (2) (2) pyramidal pyramidal horn horn antennaantenna Pattern

Pattern (3) (3) E-plane E-plane rectangular rectangular horn horn antennaantenna

The larger the aperture in one direction, the narrower the beam becomes in that direction The larger the aperture in one direction, the narrower the beam becomes in that direction (until a certain limit).

(until a certain limit). The pattern of the pyramidal horn, that has a flaring The pattern of the pyramidal horn, that has a flaring in both directions,in both directions, is a combination of both the E-plane, and the H-plane pattern.

is a combination of both the E-plane, and the H-plane pattern.

b) b)

The back-radiation is caused by diffraction at the horn aperture edges. Solutions to decrease The back-radiation is caused by diffraction at the horn aperture edges. Solutions to decrease the back-side radiation are:

the back-side radiation are: 1.

1. Corrugated horns, where a similar boundary condition is enforced for both, theCorrugated horns, where a similar boundary condition is enforced for both, the E E

and the

and the H H -field, i.e. both fields exhibit a tapered field distribution. Additionally to a-field, i.e. both fields exhibit a tapered field distribution. Additionally to a reduced back-side radiation, the pattern becomes rotationally symmetric.

reduced back-side radiation, the pattern becomes rotationally symmetric. 2.

2. Aperture matched horns, where a curved surface section is added to the outside of Aperture matched horns, where a curved surface section is added to the outside of the aperture edges. This decreases diffraction, that occurs at the sharp edges of a the the aperture edges. This decreases diffraction, that occurs at the sharp edges of a the aperture of a regular horn and additionally provides a smooth transition from the aperture of a regular horn and additionally provides a smooth transition from the horn to the f

horn to the free-space impedance.ree-space impedance.

c) c)

The connection between antenna gain and effective aperture size at a certain frequency is The connection between antenna gain and effective aperture size at a certain frequency is given by given by 00

44

e e G G ππAA λ λ

=

If

If an an aperture aperture efficiency efficiency of of is is assumed, assumed, the the physical physical aperture aperture size size can can bebe determined determined ap ap

0.5

η η == 0 a 0 ₂₂ app ₂₂ 11

4

4 4 11

22

G G π π ηη A A ππ a a λ λ λλ

=

⋅⋅

bb11 22 1 1 1 1

22

00

44

a a b b λλ G G π π

⋅ ⋅

=

(10)

The

The gain gain is is given given as as dBi, dBi, which which is is corresponding corresponding to to and and the the wavelengthwavelength at at GHz GHz is is mm.mm. 00

17

G G == G G ₀₀ ==

50

11

f f == λλ

=

27.25

Therefore the physical aperture is Therefore the physical aperture is

( (

33

))

22 22 1 1 11

27

27.2 .25

5 10

10 11000

0 00..000066 m

m

44

a a bb π π − −

⋅⋅

⋅ ⋅ =

=

with

with the the given given ratio ratio of of a a ₁₁

//

bb₁₁ ==

22

this this givesgives a a 11

=

2 0

2 0..000066 m

⋅ ⋅

m

=

110099..5555 m

mm

m

, and thus, and thus ..

11

54.775mm

bb

=

At

(11)

d) d) The

The maximal maximal received received power power is is given given by by , , where where the the effective effective aperture aperture isis known known e e P P = = A A W ⋅⋅W 22

11 0.006m

0.006m

22

e e A

A

=

= ⋅⋅

and the power density of the incident wave isand the power density of the incident wave is

22 66 00 66 2 2 22 w w

1

1 1 99000

0 110

0 W

W

1.

1.2 2 10

10

2

2 2 112200

m

E E W W Z Z ππ − − − −

⋅⋅

=

≈

=

⋅⋅

W

The maximal received power thus is The maximal received power thus is

66 horn horn

1

1 1 99000

0 1100

00..00006

6 W

W 33..66 n

nW

W

2

2 2 112200

P P π π − −

⋅⋅

=

⋅ ⋅

=

.. e) e) Because of the TE

Because of the TE1010 field distribution in the feeding waveguide, the aperture efficiency isfield distribution in the feeding waveguide, the aperture efficiency is

determined as

determined as ηη_ap_ap

88

₂₂

0.81

π π

=

and thereforeand therefore

22 66

00 3 3 33

w

waavveegguuiidde e aapp

w w

1

1 99000

0 1100

00..881

1 2222..886

6 110

0 1100..116

6 110

0 W

W

2

2 112200

0.225nW

E E P P a a bb Z Z η η π π − − − − −−

⋅⋅

=

⋅

=

⋅

⋅⋅

=

f) f)

The larger dimensions (in terms of wavelength) of such a waveguide enable the support of The larger dimensions (in terms of wavelength) of such a waveguide enable the support of higher order modes in the waveguide. Higher order modes exhibit different group velocities higher order modes in the waveguide. Higher order modes exhibit different group velocities which results in a

(12)

Problem 4 (11 Points)

You

You have have the the choice choice of of two two dielectric dielectric substrates, substrates, one one with with , , the the other other withwith .

. Both Both have have a a thickness thickness of of mm. mm. The The microstrip microstrip antenna antenna should should be be designeddesigned to

to radiate radiate at at the the frequency frequency GHz.GHz.

r1 r1

1.5

εε == r2 r2

44

εε == h h

=

1.5

rr

22

f f == L L W W h h w w 00 e err w w 11 L L W W h h w w 00 e err (a) (b) (a) (b) 2 Points

2 Points a) a) Which dieWhich dielectric subslectric substrate would trate would you preyou prefer as fer as a dielea dielectric layer ctric layer for a for a microstrip amicrostrip antennantenna shown in figure (a) and why?

shown in figure (a) and why? 4 Points

4 Points b) b) Determine Determine the the lengthlength LL and the widthand the widthW W of a linearly polarized patch on the aboveof a linearly polarized patch on the above chosen substrate with the help of the transmission line model.

chosen substrate with the help of the transmission line model. 2 Points

2 Points c) c) Redesign Redesign the pathe patch negletch neglecting fringing cting fringing fields. Ffields. Find the ind the error of error of frequency frequency shift that shift that isis arising from this neglect.

arising from this neglect. 3 Points

3 Points d) d) With With the the matching matching network network (microstrip (microstrip line line 1: 1: width width mm mm and and lengthlength at

at GHz) GHz) shown shown in in figure figure (b), (b), the the microstrip microstrip patch patch is is matched matched to to the the microstripmicrostrip line

line 0 0 with with the the width width . . Give Give the the characteristic characteristic impedance impedance of of the the line line 0. 0. Assume Assume therethere is

is no no mutual mutual coupling coupling between between the the slots slots of of the the microstrip microstrip patch patch and and that that isis valid. valid. 11

1.62

w w == λλ_gg

// 44

rr

22

f f == 00 w w 00 W W  λλ

(13)

Solution 4

a) a) The

The material material with with the the lower lower permittivity permittivity εε_r_r ==

1.5

is to is to choose choose because because of of several several reasons:reasons: 1.

1. Radiation efficiencyRadiation efficiency rraad d radrad ttoott

ttoot t rraadd tot tot

11

P P Q Q P P Q Q η η

=

Q Q Q

Q where the radiation Q-factor iswhere the radiation Q-factor is

proportional

proportional to to . . The The smaller smaller (and (and thus thus ), ), the the lower lower the the losses losses andand the higher the r

the higher the radiation efficiency.adiation efficiency. rad rad r r Q Q ∼∼ εε Q Q _rad_rad εεr r 2. 2. BandwidthBandwidth

BW

0 0 ttoott , and , and

11

f f f f

=

Q Q   rr

11 BW

BW

εε ∼

∼ respectively. respectively. The The lower lower , , thethe higher the bandwidth of the

higher the bandwidth of the antenna.antenna.

r r

εε

3.

3. Directivity Directivity is is higher higher for for lower lower εεr r , see , see graph graph on on slide slide 8.44 8.44 (of (of lecture lecture notes notes 2004).2004).

b) b)

The width

The widthW W can be determined bycan be determined by 00

r r rr

22

67.036

2

11

c c W W f f εε

=

+

mm.mm.

The effective permittivity is The effective permittivity is

1 2 1 2 r r rr eff eff

1

11

1 1 112

2 11..4477

2

22

h h W W ε ε εε εε − −

+

−

− ⎡ ⎡

⎤⎤

=

+

_{⎢ ⎢}

+

_⎥⎥

=

⎣ ⎣

⎦⎦

The length extension caused by fringing effects can

The length extension caused by fringing effects can be calculatedbe calculated ( ( ))

( (

))

( ( ))

( (

))

eff eff eff eff

00..3

3 00..226644

00..44112

2 00..889911

00..22558

8 00..88

W W h h L L h h W W h h εε εε

+

=

= ⋅ ⋅

=

−

+

  mmmm

The total length of the patch is therefore The total length of the patch is therefore

00 r r eefff f

2

2 5599..999922

22

c c L L LL f f εε

=

−



=

mmmm c) c)

Without effective permittivity and fringing fields the length of a

Without effective permittivity and fringing fields the length of a patch is determined bypatch is determined by 00 r r rr

*

6611..119955

2

22

c c L L f f λ λ εε

=

mmmm

In reality, the resonant frequency for this case would be In reality, the resonant frequency for this case would be

(

))

00 eff eff

*

11..996622

2

2 *

* 22

r r c c f f L L LL εε

=

+

 GHz.GHz.

(14)

d) d)

The input impedance of the microstrip patch can be found by (no mutual coupling) The input impedance of the microstrip patch can be found by (no mutual coupling)

Ω

withwith 22 11 00

11

2.222

90

W W G G λ λ

⎛

_⎞⎞⎟⎟

⎜⎜

=

_⎜⎜

_⎟⎟

=

⎜⎜⎝ ⎝ ⎠⎠

mS.mS. in in 11

11

225

Z Z G G

=

A

A quarter-wavelength quarter-wavelength transformer transformer with with the the width width is is used used to to match match the the patch patch with with thethe impedance

impedance to to the the microstrip microstrip line line with with the the impedance impedance . . The The impedance impedance of of thisthis transformer has to be transformer has to be 11 w w in in Z Z Z Z ₀₀ 1 1 00 Z

Z

=

Z Z

⋅⋅

Z Z inin which can be found from slide 8.26which can be found from slide 8.26

22 22 22

ttaan

n(

( ))

ttaan

n(

( ))

C C C C C C Y Y jjY Y LL Y Y Y Y Y Y jjY Y LL β β β β

+

=

+



 _{, here that is}_{, here that is} inin

in in 11 0 0 11 1 1 11

ttaan

n(

(

))

ttaan

n(

(

))

Y Y jjY Y LL Y Y Y Y Y Y jjY Y LL β β β β

+

=

+

11 _and_and and

and thus thus ..

11

// 22

L L β β == ππ 11

ttaan

n(

(

β β LL

))

→ → ∞∞ Hence Hence iinn iinn 22 1 1 11 0 0 11 Y Y Y Y Y Y Y Y Y Y Y Y

=

or or Z Z ₁₁22

=

Z Z ₀₀

⋅⋅

Z Z _in_in..

( (

))

11 eff1 eff1

120

106.1

106.1 11..33993 0

3 0..66667 l

7 ln

n

11..444444

Z Z W W W W h h h h π π εε

≈

=

⎡ ⎡

⎤⎤

+

⎢ ⎢

⎥⎥

⎣ ⎣

⎦⎦

Ω

with with 1 2 1 2 r r rr eff1 eff1 11

1

11

1 1 112

2 11..3322

2

22

h h w w ε ε εε εε − −

+

−

− ⎡ ⎡

⎤⎤

=

+

⎢ ⎢

+

⎥⎥

=

⎢ ⎢

⎥⎥

⎣ ⎣

⎦⎦

..

Thus, the patch is matched to a Thus, the patch is matched to a

22 11 00 in in

50

Z Z Z Z Z Z

=

Ω

line.line.

(15)

Problem 5 (11 Points)

Design an ordinary end-fire array with its maximum radiation directed towards 180 degrees Design an ordinary end-fire array with its maximum radiation directed towards 180 degrees and half-power

and half-power beamwidth (HPBW) of beamwidth (HPBW) of 30 30 degrees. The spacing degrees. The spacing between the between the elementselements is

is λλ

// 44

, , and and the the array’s array’s length length is is much much larger larger than than the the spacing.spacing.

3 Points

3 Points a) a) Determine nDetermine number oumber of elemenf elements and ts and progressive progressive phase phase shift betweeshift between the n the elements elements (in(in degrees).

degrees). 2 Points

2 Points b) b) Estimate Estimate the the array’s array’s directivity.directivity. 2 Points

2 Points c) c) Redesign Redesign the array the array in order in order to increase to increase its directivity its directivity for the for the end-fire radend-fire radiation. Theiation. The number of elements, spacing between them and amplitude uniformity

number of elements, spacing between them and amplitude uniformity should stay theshould stay the same.

same. 2 Points

2 Points d) d) How How much much can can you you increase increase directivity directivity maximally?maximally? 2 Points

2 Points e) e) Could you Could you use the use the same msame method to ethod to increase direincrease directivity of the ctivity of the original array original array (designed (designed inin a)),

(16)

Solution 5

a) a)

Ordinary end-fire array has maximum and

Ordinary end-fire array has maximum and 180 degrees if 180 degrees if

22

90

4

22

kd kd π π λ λ ππ β β λ λ

=

= =

= ⋅

⋅ =

= =

=



HPBW of an ordinary end fire array is

HPBW of an ordinary end fire array is given bygiven by

( (

))

11

1.391

1.391 H

HP

PB

BW

W

h h

2 2 ccoos 1

s 1

dN dN λ λ π π − −

=

Θ

Θ =

= ⋅

⋅

−

Thus, the total number of elements is Thus, the total number of elements is

( (

))

1.391

52

1 1 ccooss

h h

₂₂

N N d d λ λ π π

=

Θ

⎡⎡ −

−

⎢ ⎢

⎥⎥

⎣ ⎣

⎦⎦

⎤⎤

elements. elements. b) b)

Directivity of an ordinary end-fire array is given as Directivity of an ordinary end-fire array is given as

( ( ))

00

4

552 2 1177..1166 d

dB

B

d d D D N N λ λ

=

c) c)

To improve the directivity, given everything else staying the same, the progressive phase To improve the directivity, given everything else staying the same, the progressive phase shift should be calculated from

shift should be calculated from

11..6633 rraad

d

9933..4466

2

52

kd kd N N π π π π ππ β β

=

= +

+ =

= +

+ =

=



The obtained array is Hansen-Woodyard end-fire array with

The obtained array is Hansen-Woodyard end-fire array with maximum beam at 180 degrees.maximum beam at 180 degrees.

d) d)

Directivity of a Hansen-Woodyard array shows improvement of 1.789 times or 2.526 dB Directivity of a Hansen-Woodyard array shows improvement of 1.789 times or 2.526 dB over the ordinary end-fire array.

over the ordinary end-fire array.

e) e)

No, because for the Hansen-Woodyard array to have the maximum in the desired direction No, because for the Hansen-Woodyard array to have the maximum in the desired direction the

the spacing spacing between between the the elements elements have have to to be be around around . . For For the the spacing spacing of of , , the the sideside lobes would have larger maxima than the main lobe.

lobes would have larger maxima than the main lobe.

// 44

λ

(17)

Problem 6 (10 Points)

A

A generator generator with with impedance impedance and and supply supply power power dBm dBm isis feeding a transmitting antenna. The transmitting antenna is a lossless, half-wavelength feeding a transmitting antenna. The transmitting antenna is a lossless, half-wavelength dipole

dipole and and is is oriented oriented according according to to the the figure. figure. An An airplane airplane (RCS (RCS of of , , heightheight km, the dimensions of the airplane are much smaller than

km, the dimensions of the airplane are much smaller than h h ) is illuminated by the) is illuminated by the antenna

antenna under under an an angle angle of of and and scatters scatters the the electromagnetic electromagnetic wave. wave. Due Due to to thethe scattering, the polarization of the incident wave is turned by

scattering, the polarization of the incident wave is turned by φφ in thein the xy xy -plane. In-plane. In an

an angle angle of of , , a a pyramidal pyramidal horn horn antenna antenna (gain (gain dBi) dBi) is is employed employed as as aa receiver. The orientation and the polarization of the horn antenna are depicted in the figure. receiver. The orientation and the polarization of the horn antenna are depicted in the figure.

(( G G

550

0 2200

Z Z

=

+

j j ))

Ω

= =  S S

40

P P == 22

20 20 m

m

σ σ

=

33

h h

=

tt

30

θθ ==  ss

30

rr

45

θθ ==  G G ₀₀ ==

23

airplane airplane t trraannssmmiitttteer r rreecceeiivveerr d d tt h h d d rr q q_t_t  30°30° qq_r_r  4545°° E E ® ® x x y y z z 2 Points

2 Points a) a) For For which which frequency frequency is is the the power power at at the the receiver receiver higher, higher, for for GHz GHz oror GHz if the transmitter in both

GHz if the transmitter in both cases is a half-wavelength dipole? What is thecases is a half-wavelength dipole? What is the difference of the received power between both

difference of the received power between both cases in dB?cases in dB?

11

88

f f == 22

16

f f == 4 Points

4 Points b) b) Calculate Calculate the the power power that that is is captured captured by by the the airplane airplane for for a a carrier carrier frequency frequency of of GHz. GHz. cc P P

88

f f == 4 Points

(18)

Solution 6

a) a) The

The received received power power in in a a free-space free-space propagation propagation is is proportional proportional to to . . Thus, Thus, the the higherhigher the frequency, the higher are the free-space losses. Doubling the frequency, the received the frequency, the higher are the free-space losses. Doubling the frequency, the received po

powewer dr dececrereasases es by by a fa facactotor or of 4 (f 4 ( dB dB rerespspecectitivevelyly).).

22 rr P P ∼∼ λλ

66 



b) b)

The input impedance of a half-wavelength dipole is known as

The input impedance of a half-wavelength dipole is known as Z Z _in_in

=

( (

773

3 +

+

j j

4422..55

))

Ω

. Thus. Thus the generator is not matched to the dipole, resulting in a return loss of the generator is not matched to the dipole, resulting in a return loss of

in in GG in in GG

0.233

Z Z Z Z Z Z Z Z

−

Γ

Γ =

=

+

..

The distance between the transmitter and the airplane is The distance between the transmitter and the airplane is

tt

33 / c

/ coos 3

s 30

0 3300000 /

0 /

m

3344664 m

4 m

22

d d

=

h h 

=

..

The directivity (lossless half-wavelength dipole) is given The directivity (lossless half-wavelength dipole) is given byby

( ( )) 22 00 33 33 22 22 rad rad 00

sin

(

)

₈₈

ssiin

n

4

44

2.435

88

I I U U D D P P I I η η θθ θ θ _π_π θθ θ θ π π ππ η η π π

=

≈

=

.. In

In direction direction of of the the airplane airplane θθ_tt ==

30

 this givesthis gives ( ( ))

33 tt tt

sin

4

00

2.435

D D θθ

=

θθ

=

= .205

.205

.. The power of the generator is

The power of the generator is

{ { }} { { }} ( ( )) 22 S S GG 00 00 GG rraadd 22 0 0 G G rraadd

1

11 R

Re

e

Ree

R

2

22

11

22

A A P P UU II II ZZ RR I I R R RR ∗∗

=

⋅

=

+

=

+

X X S S 22 00 G G rara

22

P P I I R R RR

=

+

_d_d rad rad 22 t t rraad d 0 0 rraad d SS G G rraadd

11

5.93

22

R R P P PP II RR PP R R RR

=

+

W.W.

For the captured power this means For the captured power this means

( ( )) tt c c t t tt ₂₂ tt

161.6

44

P P P P W W D D d d σ σ σ σ θθ π π

=

nW.nW.

(19)

c) c)

For the received power the polarization has to be taken into account. For the received power the polarization has to be taken into account.

Lecture notes, slide 2.26 (SS 2004): The PLF can be described using the angle between two Lecture notes, slide 2.26 (SS 2004): The PLF can be described using the angle between two unit vectors

unit vectors

P

PL

LF

F

=

  ρ ρ _w_w

⋅ ⋅

ρρ  _aa 22

ccooss 330

0

 22

=

00..7755

..

OR: OR: y y ’’ x x ’’ ® ® r rww ® ® r raa 30° 30°

Using the coordinate system shown in the figure above, the polarization vector of the Using the coordinate system shown in the figure above, the polarization vector of the scattered wave is

scattered wave is _w_w

3

_{' '}

11

_''

2

a a x x

22

a a y y ρ

ρ

=



+

 and and of of the the receiving receiving antenna antenna _''. . Thus Thus thethe polarization loss factor is

polarization loss factor is

aa a a x x ρ ρ

=

    22 w w aa

P

PL

LF

F

=

  ρ ρ

⋅ ⋅

ρρ 

=

00..7755

. The gain of the antenna is. The gain of the antenna is ..

00

223d

3dB

Bi

i

220000

G

=



The distance between the airplane and the

The distance between the airplane and the receiverreceiver

rr

/ c

/ coos 4

s 45

5 3300000

0 2 m

2 m

4422443 m

3 m

d

=

h h 

=

⋅

=

The received power therefore is determined by The received power therefore is determined by

22 cc 1818 r r c c 00 ₂₂ rr

G

G P

PL

LF

F

1111..995

5 1100

44

e e P P P P A A W W d d λ λ π π ππ − −

=

⋅

=

⋅

=

⋅⋅

W.W.