Final Exam Antenna 20051

D-ITET Antennas and Propagation Student-No.:.....................................................................

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

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:

• 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 / 19 —

D-ITET Antennas and Propagation March 17, 2005

Problem 1 (10 Points)

Assume a receiver is located km from a 50 W transmitter. The receiver and

transmitter are mounted on 2.5 m and 5 m high posts, respectively. The carrier frequency is

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

respectively.

10d =

r 1G = t 2G =

5 m

2.5 m

ground d 10 km

transmitter

receiver

2 Points a) Find free space loss and received power if reflections from earth are neglected.

2 Points b) Find additional power loss ( ) in dB due to reflections from the ground. refL

3 Points c) Assume a wall positioned in between the transmitter and receiver, 8 km away from the

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

diffraction is the same as the power loss due to reflection from Earth ( ) in b). refL

3 Points d) What would be the maximum height of the wall so that the power loss due to the

diffraction is negligible? Can such a wall be built?

— 2 / 19 —


Solution 1 a)

900MHz 0.333mc

ff

λ= ⇒ = =

Free space loss:

( ) ( )2 2

LOS 2

11LOS

0.3331 2

4 4 10 10

1.4072 10 108.51dB

r tL G Gd

L

λπ π

−

= = ⋅ ⋅⋅ ⋅

= ⋅ = −

Received Power LOSr LOS

10LOSr 7.036 10 W 91.5dBW 61.5dBm

tP P L

P −

= ⋅

= ⋅ = − = −

or

LOSr t LOS

50W 16.989dBW

16.989 108.51 91.5dBW

tP

P P L

= =

= + = − = −

b)

The power loss due to the reflection from the ground is

( ) ( )( )

2 2

ref

2

ref

ref

2 21 exp

2 20.00222

26.535dB

t r t r

t r

h h h hL jk k

d d

h hL

dL

πλ

= + Γ ⋅ ≈ ⋅

= ⋅ =

= −

The plane earth loss (PEL) is PEL LOS ref 135.045dBL L L= + = −

— 3 / 19 —


c)

Power loss due to the diffraction is the same as the power loss due to the reflection, thus diff ref 26.535dBL L= = −

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

( )diff0.225

20 log 4.78L υυ

= ⇒ =

( )

( )

( )

1 2

1 2 1 2

1 2

tot r

22

4.7878.1m

2 8000 20000.333 8000 2000

80.5m

d dh h

d d d dd d

h

h h h

υυ

λλ

+= ⇒ =

+

= =+

⋅ ⋅= + =

d)

The diffraction loss is negligible if most of the 1st Fresnel zone is clear (uncovered by the

wall).

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

tot r

0.8 13.063

10.563m

h

h h h

υ = − ⇒ = −

= + = −

Therefore, such wall cannot be built.

— 4 / 19 —



An infinitesimal dipole of length l is placed at a distance s from a ground plane and at an

angle of 45 degrees from the vertical axis, as shown in the figure below. The dipole lies in

the yz -plane.

s

yz

x

45°45°

normal

3 Points a) Determine the location and direction of the image source, which can be used to account

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

sketch.

6 Points b) With the coordinate system given in the figure, find the expression for the total far field

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

4 Points c) Find the smallest non-zero distance s for which the total field obtained in b) is zero

along the normal.

— 5 / 19 —


Solution 2

a)

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

45¡ 45°

normal

Eq1

Eq2

b)

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

s

yz

x

45°45° 45°

normal

r

r1r1

r2r2

q

45°-q

q

y z’ ( )

( )

1

2

1 2

cos 45phase

cos 45

magnitude

r r s

r r s

r r r

θ

θ

= − ⋅ − ⎫⎪⎪⎬⎪= + ⋅ − ⎪⎭= =

(*)

— 6 / 19 —


The field produced by the original dipole is 1

01 1

1sin

4

jkrkI leE E j a

rθ θη θπ

−= = ⋅

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

on the graph above. Thus the image dipole field is given as 2

02 2

2sin

4

jkrkI leE E j a

rψ ψη ψπ

−= = ⋅

where

( )22 22'sin 1 cos 1 1 1 sin sinz r y ra a a aψ ψ θ= − = − ⋅ = − ⋅ = − φ

Thus

( )2

202 2

21 sin sin

4

jkrkI leE E j a

rψ ψη θπ

−= = − ⋅φ

In the second quadrant of the yz-plane we have 270 , a aψ θφ = ° = −

Thus:

( )2

2

0 22 2

2

02 2

2

1 sin4

cos4

jkr

jkr

kI leE E j a

r

kI leE E j a

r

ψ θ

ψ θ

η θπ

η θπ

−

−

= = − − ⋅

= = − ⋅

The total field is equal to the sum of the two fields, i.e. 1 2

0 0tot 1 2

1 2sin cos

4 4

jkr jkrkI le kI leE E E j j

r r θη θ ηπ π

− −aθ

⎡ ⎤⎢ ⎥= + = − ⋅⎢ ⎥⎣ ⎦

Expresing and in terms of r (*), we obtain the total field in the 21r 2rnd quadrant as

( ) ( )0 cos 45 cos 45tot sin cos

4

jkrjks jkskI le

E j e er

θ θθη θ

π

−+ − − − aθ⎡ ⎤= − ⋅⎣ ⎦

c) Along the normal, we have 45θ =

Thus, the field can be expressed as

( )

0tot

0tot

sin 454

22 sin

4 2

jkrjks jks

jkr

kI leE j e e

rkI le

E j ks ar

θ

θ

ηπ

ηπ

−+ −

−

⎡ ⎤= −⎣ ⎦

= − ⋅

a⋅

So tot 0 , 0,1,2,...E ks n nπ= ⇐ = =

1n = ⇒ smallest non-zero length s is 2

sλ

=

— 7 / 19 —



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

antennas are fed by a rectangular waveguide WR90 (dimension mm and

mm) operated in the dominant TE

22.86a =

10.16b = 10 mode at GHz. 11f =

E-Plane

H-Plane

(1) (2) (3)

2 Points a) Relate each of the three radiation pattern to a type of rectangular horn antenna. Give the

physical explanation of your choice!

2 Points b) Which effect is responsible for large back radiation of the above patterns? What would

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

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

exhibit a gain of dBi at GHz if a typical aperture efficiency of 50% is

assumed. The side length ratio is .

1a 1b

0 17G = 11f =

1 1/ 2a b =

2 Points d) Determine the maximum power that can be received by the antenna designed in c) if it

is illuminated by a plane wave of the frequency GHz with an amplitude of

mV/m.

11f =

0 30E =

2 Points e) Determine the maximum power that can be received by the feeding waveguide (without

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

2 Points f) Comparing the aperture efficiencies of a horn antenna and a waveguide, why are

waveguides, having the same dimension as the aperture of a horn antenna, not used as

radiating elements? Find an explanation.

— 8 / 19 —


Solution 3

a)

Pattern (1) H-plane rectangular horn antenna

Pattern (2) pyramidal horn antenna

Pattern (3) E-plane rectangular horn antenna

The larger the aperture in one direction, the narrower the beam becomes in that direction

(until a certain limit). The pattern of the pyramidal horn, that has a flaring in both directions,

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

b)

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

the back-side radiation are:

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

and the H -field, i.e. both fields exhibit a tapered field distribution. Additionally to a

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

2. 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

aperture of a regular horn and additionally provides a smooth transition from the

horn to the free-space impedance.

c)

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

given by

04

eG Aπλ

=

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

ap 0.5η =

0 ap 12 24 4 1

2G A a

π πη

λ λ= = 1b⋅

2

1 1 024

a b Gλπ

⋅ =

— 9 / 19 —


The gain is given as dBi, which is corresponding to and the wavelength

at GHz is mm. 0 17G = 0 50G =

11f = 27.25λ =

Therefore the physical aperture is

( )23 21 1

27.25 10100 0.006m

4a b

π

−⋅⋅ = =

with the given ratio of this gives 1 1/a b = 2 1 2 0.006m 109.55mma = ⋅ = , and thus

. 1 54.775mmb =

At the given frequency this corresponds to . 21 1 8a b λ⋅ ≈

— 10 / 19 —


d) The maximal received power is given by , where the effective aperture is

known

eP A W= ⋅

210.006m

2eA = ⋅ and the power density of the incident wave is 2 6

0 62 2

w

1 1 900 10 W1.2 10

2 2 120 m mE

WZ π

−−⋅

= ≈ = ⋅W

The maximal received power thus is 6

horn1 1 900 100.006 W 3.6nW2 2 120

Pπ

−⋅= ⋅ = .

e) Because of the TE10 field distribution in the feeding waveguide, the aperture efficiency is

determined as ap 28

0.81ηπ

= = and therefore 2 6

0 3 3waveguide ap

w

1 1 900 100.81 22.86 10 10.16 10 W

2 2 1200.225nW

EP a b

Zη

π

−− − ⋅

= ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅

=

f) 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

which results in a declination of signal transmission and antenna excitation.

— 11 / 19 —



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

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

to radiate at the frequency GHz.

r1 1.5ε =

r2 4ε = 1.5h =

r 2f =

L

W

h

w0

er

w1

LW

h

w0

er

(a) (b)

2 Points a) Which dielectric substrate would you prefer as a dielectric layer for a microstrip antenna

shown in figure (a) and why?

4 Points b) Determine the length L and the width W of a linearly polarized patch on the above

chosen substrate with the help of the transmission line model.

2 Points c) Redesign the patch neglecting fringing fields. Find the error of frequency shift that is

arising from this neglect.

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

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

line 0 with the width . Give the characteristic impedance of the line 0. Assume there

is no mutual coupling between the slots of the microstrip patch and that is

valid.

1 1.62w = g / 4λ

r 2f =

0w

0W λ

— 12 / 19 —


Solution 4 a) The material with the lower permittivity is to choose because of several reasons: 1.5rε =

1. Radiation efficiency rad totrad

tot radtot

1

1P QP Q

η = = =QQ

where the radiation Q-factor is

proportional to . The smaller (and thus ), the lower the losses and the higher the radiation efficiency.

rad rQ ε∼ radQ rε

2. Bandwidth BW0 tot

, and 1ff Q

=r

1BW

ε∼ respectively. The lower , the

higher the bandwidth of the antenna. rε

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

b)

The width W can be determined by 0

r r

267.036

2 1c

Wf ε

= =+

mm.

The effective permittivity is 1 2

r reff

1 11 12 1.47

2 2hW

ε εε

−+ − ⎡ ⎤= + + =⎢ ⎥⎣ ⎦

The length extension caused by fringing effects can be calculated

( )( )( )( )eff

eff

0.3 0.2640.412 0.891

0.258 0.8

WhL h Wh

ε

ε

+ += ⋅ =

− +mm

The total length of the patch is therefore 0

r eff2 59.992

2c

L Lf ε

= − = mm

c)

Without effective permittivity and fringing fields the length of a patch is determined by 0

r r* 61.1952 2c

Lf

λε

= = = mm

In reality, the resonant frequency for this case would be

( )0

eff* 1.9622 * 2r

cf

L L ε= =

+GHz.

— 13 / 19 —


d)

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

Ω with 2

10

12.222

90W

Gλ

⎛ ⎞⎟⎜= =⎟⎜ ⎟⎜⎝ ⎠mS. in

1

1225Z

G= =

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

impedance to the microstrip line with the impedance . The impedance of this

transformer has to be

1w

inZ 0Z

1 0Z Z Z= ⋅ in which can be found from slide 8.26

2

22

tan( )tan( )

CCC

Y jY LY Y

Y jY Lββ

+=

+, here that is in

in

10 1

1 1

tan( )tan( )

Y jY LY Y

Y jY Lββ

+=

+1 and

and thus .

1 /2Lβ π=

1tan( )Lβ → ∞ Hence

in in

21 1

0 1Y Y

Y YY Y

= = or . in21 0Z Z Z= ⋅

( )1

eff1

120106.1

1.393 0.667 ln 1.444Z W W

h h

π

ε≈ =

⎡ ⎤+ + +⎢ ⎥⎣ ⎦

Ω

with 1 2

r reff1

1

1 11 12 1.32

2 2hw

ε εε

−+ − ⎡ ⎤= + + =⎢ ⎥

⎢ ⎥⎣ ⎦.

Thus, the patch is matched to a 21

0in

50Z

ZZ

= = Ω line.

— 14 / 19 —



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

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

is , and the array’s length is much larger than the spacing. / 4λ

3 Points a) Determine number of elements and progressive phase shift between the elements (in

degrees).

2 Points b) Estimate the array’s directivity.

2 Points c) Redesign the array in order to increase its directivity for the end-fire radiation. The

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

same.

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

2 Points e) Could you use the same method to increase directivity of the original array (designed in

a)), if the spacing between the elements was ? Why? /2λ

— 15 / 19 —


Solution 5

a)

Ordinary end-fire array has maximum and 180 degrees if 2

904 2

kdπ λ π

βλ

= = ⋅ = =

HPBW of an ordinary end fire array is given by

( )1 1.391HPBW 2 cos 1h dN

λπ

−= Θ = ⋅ −

Thus, the total number of elements is

( )1.391

521 cos 2

hN

d

λ

π= =Θ⎡ −⎢ ⎥⎣ ⎦

⎤ elements.

b)

Directivity of an ordinary end-fire array is given as

( )0 4 52 17.16dBd

D Nλ

= = =

c)

To improve the directivity, given everything else staying the same, the progressive phase

shift should be calculated from

1.63 rad 93.462 52

kdNπ π π

β = + = + = =

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

d)

Directivity of a Hansen-Woodyard array shows improvement of 1.789 times or 2.526 dB

over the ordinary end-fire array.

e)

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

the spacing between the elements have to be around . For the spacing of , the side

lobes would have larger maxima than the main lobe.

/ 4λ /2λ

— 16 / 19 —



A generator with impedance and supply power dBm is

feeding a transmitting antenna. The transmitting antenna is a lossless, half-wavelength

dipole and is oriented according to the figure. An airplane (RCS of , height

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

antenna under an angle of and scatters the electromagnetic wave. Due to the

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

an angle of , a pyramidal horn antenna (gain dBi) is employed as a

receiver. The orientation and the polarization of the horn antenna are depicted in the figure.

(G 50 20Z j= + ) Ω

=

S 40P =

220mσ =

3h =

t 30θ =

s 30

r 45θ = 0 23G =

airplane

transmitter receiver

dt drh

q = 30°t qr = °45

E®x

y

z

2 Points a) For which frequency is the power at the receiver higher, for GHz or

GHz if the transmitter in both cases is a half-wavelength dipole? What is the

difference of the received power between both cases in dB?

1 8f =

2 16f =

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

GHz. cP

8f =

4 Points c) Determine the power received by the horn antenna. rP

— 17 / 19 —


Solution 6

a)

The received power in a free-space propagation is proportional to . Thus, the higher

the frequency, the higher are the free-space losses. Doubling the frequency, the received

power decreases by a factor of 4 ( dB respectively).

2rP λ∼

6

b) The input impedance of a half-wavelength dipole is known as ( )in 73 42.5Z j= + Ω . Thus

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

in G0.233

Z ZZ Z

−Γ = =

+.

The distance between the transmitter and the airplane is

t3

/ cos 30 3000/ m 3464m2

d h= = = .

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

( )

20 3

32

2rad 0

sin( ) sin84 4 42.435

2.4358

IU

DP I

η θθ θπθ π πη

π

= ≈ = .

In direction of the airplane this gives t 30θ = ( )3t

tsin4 02.435

Dθ

θ = = .205 .

The power of the generator is

( )

2S G 0 0 G rad

20 G rad

1 1Re Re2 212

AP U I I Z R

I R R

∗= ⋅ = + +

= +

X

S2

0G ra

2P

IR R

=+ d

rad2

t rad 0 rad SG rad

15.93

2R

P P I R PR R

= = = =+

W.

For the captured power this means

( ) tc t t 2

t161.6

4P

P W Dd

σ σ θπ

= = = nW.

— 18 / 19 —


c) 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

unit vectors 2 2w aPLF cos 30 0.75ρ ρ= ⋅ = .

OR: y’

x’

®rw

®ra

30°

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

scattered wave is w ' '3 12 2x ya aρ = + ' and of the receiving antenna . Thus the

polarization loss factor is

a xaρ =

2w aPLF 0.75ρ ρ= ⋅ = . The gain of the antenna is

. 0 23dBi 200G =

The distance between the airplane and the receiver r / cos 45 3000 2m 4243md h= = ⋅ =

The received power therefore is determined by 2

c 18r c 0 2

rG PLF 11.95 10

4 4eP

P A Wd

λπ π

−= ⋅ = ⋅ = ⋅ W.

— 19 / 19 —

Final Exam Antenna 20051

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