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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 —
D-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
Problem 2 (15 Points)
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 —
D-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
Problem 3 (16 Points)
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 —
D-ITET Antennas and Propagation October 13, 2005
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-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
Problem 4 (11 Points)
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 —
D-ITET Antennas and Propagation October 13, 2005
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-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
Problem 5 (15 Points)
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 —
D-ITET Antennas and Propagation October 13, 2005
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-ITET Antennas and Propagation October 13, 2005
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 —
D-ITET Antennas and Propagation October 13, 2005
Problem 6 (7 Points)
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 —
D-ITET Antennas and Propagation October 13, 2005
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 —