final exam antenna 2005

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

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

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

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

Antennas and Propagation Fall 2005

October 13, 2005, 09:00 am – 12:00 noon

Dr. C. Fumeaux, Prof. Dr. R. Vahldieck

This exam consists of 6 problems. The total number of pages is 18,, including the cover page. You have 3 hours to solve the problems. The maximum possible number of points is 71. 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 / 18 —

D-ITET Antennas and Propagation October 13, 2005

Problem 1 (7 Points)

Given is a WR 90 ( a = 22.86mm, b = 10.16mm) waveguide that is operated at f = 11GHz

in the dominant TE10 mode. The waveguide is used to feed three different types of horn

antennas.

2

b

a1

1b1

a a1

b1

a) b) c)

2 Points a) Design an H-plane sectoral horn with the maximum directivity dBi.

Determine the aperture dimensions ( , b ) and the horn length . H 12.12D =

1a 2ρ

2 Points b) What is the directivity of an E-plane sectoral horn with the same horn length

and the same length of the longer aperture side as the antenna in

question a).

ED

2ρ = 1ρ 11b a=

1 Point c) Explain shortly (one or two sentences) the difference in directivity between the two

horns from question a) and b).

2 Points d) Determine the directivity of a pyramidal horn with the aperture dimensions

and ! PD

1a 1b

Horn aperture ( )a1

No

rmal

ized

dir

ecti

vit

y(

)

bD

H

0 4 8 12 16 20 24 280

20

40

60

80

100

120

140

H-plane sectoral horn

No

rmal

ized

dir

ecti

vit

y(

)

aD

E

Horn aperture ( )b1 0 4 8 12 16 20 24 28

0

20

40

60

80

100

120

E-plane sectoral horn

— 2 / 18 —


Solution 1

a) The given directivity is dB, or H 12.12D = DH = 101.212 = 16.3 respectively. Using the

graph on slide 7.69 with the parameter

lb

DH= 43.724 gives r 2

= 10l and a . 1= 5.5l

b) The dimension of the E-plane horn is b1

= 5.5l . With the graph on slide 7.62 and

r 1= 10l this gives

la

DE= 33.045 and the gain is DE

= 27.7173 or DE= 14.427 dB.

c) The E-plane sectoral horn has a larger aperture as the H-plane horn, and the flare angles

and of both horns are the same. Therefore a larger directivity of the E-plane horn is

expected. eψ hψ

d) The directivity of a pyramidal horn is given by D

P=

pl 2

32abD

ED

H= 141.8485 or

DP= 21.518 dB.

— 3 / 18 —



Given is a linear array consisting of 21 slots cut in one wall of a WR 90 ( mm,

mm) waveguide. The coupling between the slots is negligible. The slots are

equally spaced with a distance of mm. The waveguide is operated at GHz in

the dominant TE

22.86a =

10.16b =

20d = 10f =

10 mode.

Note: The propagation constant is ( )2210 k

aπ

β = − with wave number 02 /k π λ=

(or: the guided wavelength is ( )2g 0 01 /2aλ λ λ= − ).

a

b

dl

x

y

z

1 Point a) Find the length l of the slots in order that the array is resonant at GHz.

Assume as a simplified model, that the slots are very thin and fed in the center.

10f =

3 Points b) The direction of the main lobe is not broadside. Explain why! Determine the direction

of the main lobe.

4 Points c) Determine the half-power beamwidth of the array factor for the main lobe.

Note: Use sin / 1/ 2x x = for . 1.391x =

3 Points d) Do grating lobes exist? If yes, in which direction(s) do they occur and how can they be

suppressed?

4 Points e) The main lobe should be brought closer to broadside to . How can this be

accomplished? Include numerical results. 0 130θ =

— 4 / 18 —


Solution 2 a) To be resonant, the slot should be around half a wavelength at the operating frequency, thus 00.5 15

cl

f= ≈ mm.

b) The progressive phase of the elements is corresponding to the phase difference of the

guided wave between two slots in the waveguide: g

2d

ξ πλ

= where the guided wavelength

is ( )2g 0 01 /2 39.707aλ λ λ= − = mm. The direction of the main lobe occurs, where

( )1sin 0

2Ψ = , or 1

2nπΨ = ± or 0

gcos n

nkdλ λ

θλ

= − 0 . Thus, with (main lobe) this

gives

0n =

0 00

gcos 0.755

2 dλ λ

θ ξπ λ

= − = − = − or . 0 139.026θ =

c) The half-power point of the main lobe occurs when ( )sin 12AF

22

N

N

Ψ= =

Ψ, thus

( )hcos 1.3912 2N N

kd θ ξΨ = + = and ( )01h

2.782cos 136.336

2 d Nλ

θ ξπ

− ⎡ ⎤= − + =⎢ ⎥⎣ ⎦

.

The half-power beamwidth is found by h 0 h2 5θ θΘ = − = .38 .

d) The first grating lobe occurs at 12

πΨ = ± ( ) and thus 1n =

( )0 0 01

g

0.74395cos 2

2.253972 d dλ λ λ

θ ξ ππ λ

⎧⎪⎪= − ± = − ± = ⎨⎪−⎪⎩.

This means there is a grating lobe at . The grating lobes can be avoided, if the

spacing between the slots is minimized. The smallest spacing if obviously

mm and therefore is always valid. In order to prevent grating lobes,

the following condition has to be fullfilled:

1 41.93θ =

d

15d l∗ = = g dλ >

0 01

gcos 1

dλ λ

θλ ∗= − ± > .

This gives two solutions 0

0 g

17.08mm

122.37 mm1 /d

λλ λ

∗⎧⎪⎪< = ⎨⎪⎪⎩∓

. The distance between the slots

has to be changed to 15 in order to suppress grating lobes. mm 17.08mmd∗< <

— 5 / 18 —


e) The main lobe should be shifted to . This can be accomplished by changing the

dimension of the waveguide to fullfill

0 130θ∗ =0*

0 *g

cosλ

θλ

= − :

( )( )20 0

0 02g 0 0

cos 1 /21 /2

aa

λ λθ λ

λ λ λ∗ ∗

∗ ∗= − = − = − −

−

or 001 cos sin

2aλ

θ∗∗ = − = 0θ∗ . Therefore the dimension a of the waveguide has to be

changed to 0

019.57

2 sina

λθ

∗∗= = mm. As an alternative, the frequency could be changed to

GHz. 8.56f ∗ =

— 6 / 18 —



A small, lossless dipole antenna, with length l and triangular current distribution is fed by a

transmission line. The antenna’s operation frequency is 150 MHz and the wire radius

is mm. The input reactance of the antenna can be approximated by

50Ω

0.5a =

( )( )in

ln 2 1120

tan

l aX j

lπ λ

−= − ⋅ .

The antenna is matched to the feeding transmission line by means of a series inductor L

and an ideal transformer with turns ratio 1 2 10n N N= = . and are the numbers of

turns on the feeding line's and antenna's sides, respectively. 1N 2N

3 Points a) Draw the equivalent circuit of the antenna and the matching network.

3 Points b) Find the length l of the antenna.

2 Points c) Find the value of the series inductor L .

The antenna is placed above a perfect MAGNETIC ground plane at height h , as shown in

the Figure below.

magnetic wall

h

I0

2 Points d) Sketch the antenna and its image. Indicate the polarity of the image.

3 Points e) Find the total E - and -fields above the ground plane. H

3 Points f) Calculate the minimum height h for which the array factor of the total field is

maximum at q = 60° . In this case, where are the zeros of the field pattern?

Note: The total field can be written as a product of the field of the small antenna and an

array factor.

— 7 / 18 —


Solution 3 a) The operation frequency is MHz, thus m. 150f = 2λ =The equivalent circuit is given in the figure below:

Rin

Xin

LN1N2

Z = 500

b) The input impedance of the antenna is

( )( )

ln 2 1120

tanin in in in

l aZ R X R j

lπ λ

−= + = − ⋅

The series inductor cancels the antenna reactance. Thus, the resistance at the antenna’s side

of the transformer is . Since this resistance is matched to by the transformer, we have inR 0Z2

0 1

2100 0.5in

in

Z NR

R N⎛ ⎞⎟⎜= = ⇒ =⎟⎜ ⎟⎜⎝ ⎠

Ω

For the small antenna, the input impedance is equal to its radiation impedance, thus,

( )2

220 10cm20in r

lR R l

λπ

λ= = ⇒ = =

c) Since,

( )( )

ln 2 1120 2731.5

tanin

l aX j j

lπ λ

−= − ⋅ = − ⋅ Ω

the series inductor needed for cancel reactance is given by

2731.5

2.9 H

inj L X

L

L

ω

ωµ

= −

Ω=

=

— 8 / 18 —


d) The antenna and its image are sketched in the figure below

magnetic wall

h

h

P

H1

H2

z

x

y

The image is oriented as shown in the figure, because the tangential component of the

magnetic field has to be zero at the perfect magnetic ground plane. This is also given in

lecture notes, pg. 5.20. e) E-field generated in point P by the original antenna is

10

11

sin8

jkRkI leE j

Rθ η θπ

−=

while the E-field radiated by the image is

( )2 2

0 02

2 2sin sin

8 8

jkR jkRkI le kI leE j j

R Rθ η π θ ηπ π

− −= − − = − θ

where and are distances from the antenna and the image to point P , respectively. The total field is obtained by adding the and , and using

1R 2R

1E θ 2E θ

1 2

1

2

for amplitude terms

R cosfor phase terms

R cos

R R r

r h

r h

θ

θ

= =

= − ⎫⎪⎪⎬⎪= + ⎪⎭

The obtained total field above the ground plane

( )[ ]0 sin 2 sin cos8

jkrI kI le

E j j khrθ η θ

π

−= θ (1)

( )[ ]0 sin 2 sin cos8

jkrI kI le

H j j khrφ θ θ

π

−=

— 9 / 18 —


f) The array factor is give by

( )AF 2 sin cosj kh θ=

It has its maximum when ( ) 60sin cos 1kh θθ = ° =

Thus,

,2

h n nλ

λ= + ∈ Z

The smallest positive solution is obtained for . Thus, 0n =

1m2

h hλ

π= = =k

The zeros of the field pattern are obtained from

( )

sin 0 0

sin cos 0 0 & 90 & 180

θ θ

π θ θ θ θ

= ⇒ = °

= ⇒ = ° = ° = °

— 10 / 18 —



A square patch antenna is designed on 10 mm thick RT Duroid 5880 substrate with

. The dimensions of the patch are mm. 2.25rε = 24.5W L= =

3 Points a) What is the effective length of the antenna?

1 Point b) Find the resonant frequency of the dominant TM010 mode.

2 Points c) The fringe factor is defined as the ratio of the resonant frequency and the would-be

resonant frequency if the fringing effects were not taken into account. Calculate the

fringe factor of the patch antenna.

3 Points d) Describe three ways to modify this patch antenna to make it circularly polarized.

2 Points e) Describe the modes excited in a circularly polarized patch antenna, i.e. how many

modes are there and what is their phase difference?

— 11 / 18 —


Solution 4 24.5W L= = mm, and mm. 2.25rε = 10h =

a)

For 1W h > , the effective permittivity is given by: 1/21 1

1 122 2

1.8824

r reff

eff

hW

ε εε

ε

−+ − ⎡ ⎤= + +⎢ ⎥⎣ ⎦

=

L∆ , the distance by which each end of the patch is effectively extended because of the

fringing is given by: ( )( )( )( )

0.3 0.2640.412

0.258 0.8

4.6mm

eff

eff

WL hWh h

L

ε

ε

+ +∆= ⋅

− +

∆ =

Thus, the effective length of the patch antenna is

2

33.745mm

eff

eff

L L

L

= + ⋅∆

=

L

b) The resonant frequency of the dominant mode is given by

2rceff eff

cf

L ε= and 0

cf

λ =

3.2378GHzrcf = and 0 92.595mmλ =

c) The fringe factor or the length reduction factor is defined by the ratio of the resonant

frequency and the would be resonant frequency if the fringing effects were not taken into

account. rc

r

fq

f=

where

4.0789GHz2r

r

cf

L ε= =

Thus, 0.7938q =

— 12 / 18 —


d) A square patch can be made circularly polarized by:

• Feeding it at two adjacent edges using a 90° hybrid.

• Cutting right- or left-hand thin slots in the center of the patch.

• Trimming the opposite corners of the patch.

A rectangular and near-square patch can be made circularly polarized if:

• The lengths of the patch’s edges are related by ( )11t

L W Q= + .

• The patch is single-fed on the diagonal at one of its corners. e)

For a patch antenna to be circularly polarized, two orthogonal modes have to be excited in

it.

— 13 / 18 —



For a satellite link budget calculation, assume the following data:

The uplink/downlink distances are km. The uplink/downlink frequencies

are GHz and GHz. The diameters of the earth and satellite antennas are 15 m

and m with 60% aperture efficiencies. The earth antenna transmits power of

kW and the satellite transponder (the amplifier chain of the satellite) gain is

dB.

36000u dr r= =

6uf = 4df =

0.5

1TEP =

90G =

Note: Boltzmann's constant J/K 231.38 10k −= ⋅

2 Points a) Calculate the up and down free space losses.

3 Points b) Determine the gain of the antennas.

5 Points c) How much power in dBW is received at the earth ground station?

The receiving satellite antenna is looking down at an earth temperature of 30 K and has a

noisy receiver of effective noise temperature of 2 K. The receiving earth antenna is

looking up at a sky temperature of K and uses a high-gain LNA amplifier of K (feed

line losses may be ignored). The bandwidth is MHz.

0

700

50 80

30

3 Points d) Calculate the system noise temperatures and system noise powers of the satellite and

ground receivers (the connection lines are lossless). Give the values in dBW!

2 Points e) For the calculation in d), give the Signal to Noise Ratio SN in dB. R /P N=

— 14 / 18 —


Solution 5 a) The uplink and downlink wavelengths are m and m, corresponding 0.05uλ = 0.075dλ =

to 6 and 4 GHz. The up and down free-space gains and losses are:

( ) ( )2 2

;u dfu fdG G

R Rλ λπ π

= =4 4

199.13dB

195.61dB

fu

fd

G

G

= −

= −

b)

Effective area of the dish antenna 2

4apd

A eπ

=

The antenna gains ( )2

apd

G eπλ

= are calculated to be:

57.27 dB, 27.72dB

24.20dB, 53.75dB

TE RS

TS RE

G G

G G

= =

= =

c)

With , the EIRP of the transmitting earth antenna will be: 1kW 30dBWTEP = =

30 57.27 87.27 dBWEIRPP = + = .

The power received by the satellite will be

. 87.27 199.13 27.72 84.14dBWRSP = − + = −

After boosting this up by the transponder gain of90 , the power transmitted down to dB

the receiving earth antenna will be: . 90 84.14 5.86dBWTSP = − =

The EIRP of the transmitting satellite antenna will be ( )dB 5.86 24.20 30.06dBWTS TSP G = + = .

The downlink power received by the earth antenna will be:

30.06 195.61 53.75 111.80dBWREP = − + = −

— 15 / 18 —


d)

The system noise temperatures are: 300 2700 3000K, 34.77 dBK

50 80 130K, 21.14 dBK

RS RS

RE RE

T T

T T

= + = =

= + = =

The bandwidth is . 30MHz 10dB 10 log (30 106) 74.77 dBHzB = ⋅ =

Using the Boltzmann's constant k in dB, , we calculate the receiver

system noise powers in dB, using :

dB 228.6dBk = −

dB dB dBN k T B= + +288.6 34.77 74.77 119.06dBW

288.6 21.14 74.77 132.69dBW

RS

RE

N

N

= − + + = −

= − + + = −

e) SNR 84.14 119.06 34.92dB

SNR 111.8 132.69 20.89dB

u RS RS

d RE RE

P N

P N

= − = − + =

= − = − + =

— 16 / 18 —



Two lossless ( ) antennas are operating at 10 GHz. Their maximum effective

aperture at this frequency is .

1cde =3 22.265 10 m−⋅

T R

LHCPlinear

z

2 Points a) Find the directivity and gain of the antennas.

2 Points b) The E-field vector of the transmitting antenna T is forming an angle with the

-axis. The receiving antenna R is left-hand circular polarized (LHCP).

Calculate the polarization loss factor.

45

x

These Antennas are now used as repeaters at 10 GHz to relay television signals into a valley.

Two repeaters are separated in distance by 10 km. For an acceptable signal quality the power

received at the repeater must be greater then 10 nW.

3 Points c) Determine the minimum power that should be used for transmitting.

— 17 / 18 —


Solution 6

a)

9

3 2

2

0.03m10

2.265 m

431.625; 15dB

em

em

A

A G

G D G

λ

λπ

8

−

3 ⋅ 10= =

⋅ 10= ⋅ 10

=

= = =

b)

2 1ˆ ˆPLF 3dB

2t rρ ρ= ⋅ = = −

c)

Friis Transmission equation:

( )2

2

2

11

ˆ ˆ4

10nW

1ˆ ˆ 3dB

215dB

2.85 10351W

rt r t r

t

r

t r

t r

rt

t

PG G

P RP

G G

PP

P

λρ ρ

π

ρ ρ

−

= ⋅

≥

⋅ = − =

= =

=⋅

⇒ ≥

— 18 / 18 —

final exam antenna 2005

Documents

points d

spacing d

distance of d

hplane horn

eplane horn

points b

d guided wave

points e