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Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 1.1 :
If a transmitter produces 50 W of power, express the transmit power in units of
a) dBm,
b) dBW.
Note: For the unit dBmW, the abbreviation dBm is commonly used.
Transmitter power is 50WtP = .
a)[ ]
[ ]3,dBm
mW10 log 10 log 50 10 47.0dBm
1mW
t
t
PP
¹ ̼ Í= Õ = Õ Õ =¼ ͼ ͽ Î
b)[ ]
[ ],dBW
W10 log 10 log 50 17.0dBW
1W
t
t
PP
¹ ̼ Í= Õ = Õ =¼ ͼ ͽ Î
Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 1.2 :
If 50 watts are transmitted by a unity gain antenna at 900 MHz carrier frequency, find the
received power (assuming unity gain receiver antenna) in [dBm] and [W] at a free space
distance of:
a) 100 m from antenna,
b) 10 km from antenna.
Power density as a function of distance:24
tPpowerS
area Rπ= =
Unity-gain / isotropic receive antenna described by its effective area: 2
4isotropicA
λ
π=
Received power:
2
4rec tP S A P
R
λ
π
à Ô= ⋅ = Ä Õ
Å Ö
a)
received power = 3.5 x 10–6
W ≅ –24.5 dBm
b)
received power = 3.5 x 10–10
W ≅ –64.5 dBm
Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 1.3 :
Assume an air-filled metallic rectangular waveguide of cross-section 22.86 mm x 10.16 mm.
Calculate
a) the monomode frequency range,
b) the guided wavelength at 10 GHz,
c) group velocity and phase velocity at 10 GHz.
a)
From [FuK II, 4.39], with a = 22.86 mm : , 10
16557
2c TEf MHz
a µε= =
Similarly, fcTE20 = 13114 MHz and fcTE01 = 14753 MHz. Therefore, the monomode
frequency range is limited by the cutoff frequencies of modes TE10 and TE20, respectively.
b)
From [FuK II, 4.26], considering TE10 and β=2π/λg :
2 22
g c a
π ω π
λ
Ã Ô Ã Ô= −Ä Õ Ä Õ
Å Ö Å Ö , λg = 39.7 mm
c)
From [FuK II, 4.7, 4.9] : 8
03.97 10 1.32mphase g s
v f x cλ= = = ⋅
28
02.27 10 0.76mgroup s
phase
cv x c
v= = = ⋅
Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 1.4 :
A hypothetical isotropic antenna is radiating in free space. At a distance of 100 m from the
antenna, the total electric field is measured to be 5 V/m.
a) Find the power density at this location.
b) Determine the total power radiated by the antenna.
a) power density (vectorial!) 2 2
* 2rad
1 50.03315 W/m
2 2 2(120 )r r r
EW E H a a a
h p¹ Ì= fl = = =¼ ͽ Î
f f f f f f
b) radiating power2
2rad rad
0 0
2
2 2rad
0 0
rad
0.03315 sin
0.03315 (100) sin 2 0.03315 100 2
4166.67 W
S
P W dS r d d
P d d
P
p p
f q
p p
q q f
q q f p
= =
= Õ = Õ Õ Õ
= Õ Õ Õ = Õ Õ Õ
=
ÄÄ Ä Ä
Ä Ä
•
Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 1.5 :
A dipole of length 3λ/2 is resonant at f = 150 MHz. Calculate its mechanical length
a) in air,
b) in water (εrel = 81).
a)
0
r r
cfλ
ε µ= with λ=2m, the dipole length in air is 3 m.
b)
with λ=2/9 m, the dipole length in air is 1/3 m = 0.33 m.
Solutions 2 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 2.4 :
The power radiated by a lossless antenna is 10 Watts. The directional characteristics of the
antenna are represented by the radiation intensity of
30 cos (W/sr) 0 /2, 0 2U B q q p f p= ‘ ‘ ‘ ‘
a) Find the maximum power density (in watts per square meter) at a distance of 1000 m
(assume far field distance). Specify the angle where this occurs.
b) Find the directivity of the antenna (dimensionless and in dB).
c) Calculate the half-power beamwidth (HPBW).
d) Find the first-null beamwidth (FNBW).
a)
Prad
= U sinθ ⋅ dθ ⋅ dφθ =0
π2
Ðφ =0
2π
Ð = B0
⋅ cos3θ ⋅ sinθ ⋅ dθ ⋅ dφθ =0
π2
Ðφ =0
2π
Ð
Prad
= 2π B0
cos4 θ
4
Ã
ÅÄÔ
ÖÕ0
π
2
=π
2⋅ B
0= 10 µ B
0=
20
π= 6.3662
å 3 3
0( ) cos 6.3662cosU Bθ θ θ= =
W =U
r 2=
6.3662
r 2cos3θ =
6.3662
10002cos3θ = 6.3662 × 10−6 ⋅ cos3θ
Wmax
= Wcosθ =1
= 6.3662 × 10−6 Watts/m2 @ θ = 0
b)
max0
4 4 6.36628 9.03
10rad
UD dB
P
π π ⋅= = = =
c)
The radiation half-power occurs when :
1
3 13
1
1 1cos cos 37.5 0.21
2 2θ θ
θ θ π−
=
à Ô= µ = = ° =Ä Õ
Å Ö
The half power beamwidth is given as : 1 12 0.42 74.9d θ πΘ = = = ° . The pattern is
independent on φ, thus HPBW in a plane at right angle to the first one is Θ
2d= Θ
1d. The
directivity can be estimated using an approximate formula (see lecture slides) as
0
1 2
41,253 41,2537.35 8.66
74.9 74.9d d
D dB= = = =Θ Θ ⋅
. This compares quite well with the result of (b).
d)
! cos3θ = 0
θ =θ2
µ θ2
=π
2= 90o µ FNBW = 180°
Solutions 2 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011
Problem 2.5 :
The normalized radiation intensity of a given antenna is given by:
1. 1 sin sinU q f= Õ
2. 22 sin sinU q f= Õ
3. 23 sin sinU q f= Õ
The intensity exists only in 0 180 , 0 180q f‘ ‘ ‘ ‘c c c c region, and is zero elsewhere.
a) Find the exact directivity (dimensionless and in dB).
b) Find the Azimuthal and elevation plane half-power beamwidths (in degrees).
c) Find the directivity by using approximate formulas.
a)
max1,2,3
,1,2,3
4
rad
UD
P
π= where Umax = 1, and it occurs when θ = φ = π / 2 (in all 3 cases 1,2,3)
2
,1
0 0 0 0
sin sin sin 22
radP U d d d d
π π π π
φ θ φ θ
πθ θ φ φ φ θ θ π
= = = =
= = ⋅ = ⋅ =Ð Ð Ð Ð , 1
4 (1)4 6.02D dB
π
π= = =
2 2
,2
0 0 0 0
sin sin sin2 2
radP U d d d d
π π π π
φ θ φ θ
π πθ θ φ φ φ θ θ
= = = =
= = ⋅ = ⋅Ð Ð Ð Ð , 2
165.09 7.07D dB
π= = =
3
,3
0 0 0 0
4sin sin sin 2
3radP U d d d d
π π π π
φ θ φ θ
θ θ φ φ φ θ θ= = = =
= = ⋅ = ⋅Ð Ð Ð Ð , 3
34.71 6.73
2D dB
π= = =
using
b)
The half-power beamwidths are equal to
( )0 1 0
1, 2 90 sin (1/ 2) 120azimuthHPBW −= ⋅ − = ( )0 1 0
1, 2 90 sin (1/ 2) 120elevationHPBW −= ⋅ − =
( )0 1 0
2, 2 90 sin (1/ 2) 90azimuthHPBW −= ⋅ − = ( )0 1 0
2, 2 90 sin (1/ 2) 120elevationHPBW −= ⋅ − =
( )0 1 0
3, 2 90 sin (1/ 2) 120azimuthHPBW −= ⋅ − = ( )0 1 0
3, 2 90 sin (1/ 2) 90elevationHPBW −= ⋅ − =
c)
Directivity using approx. formulas ( 1,2,3
41253
azimuth elevation
D ≈Θ ⋅Θ
, 1,2,3 2 2
72815
azimuth elevation
D ≈Θ + Θ
) :
This gives for case 1 (HPBWs of 120º,120º) : D ≈ 2.86 = 4.6 dB and D ≈ 2.53 = 4.0 dB
It gives for cases 2,3 (HPBWs of 90º,120º) : D ≈ 3.82 = 5.8 dB and D ≈ 3.24 = 5.1 dB
Smith Chart Problems
! The % ! length line shown has a characteristic impedance of 5%6 and isterminated with a load impedance of ZL 7 5 8 j956!
:a; Locate zL 7ZL
Z
7 % 8 j% 5 on the Smith chart!
See the point plotted on the Smith chart!
:b; What is the impedance at % 7 % !@
Since we want to move away from the load :i!e!C toward thegenerator;C read read % %DE! on the wavelengths towardgenerator scale and add % 7 % ! to obtain % DE! on thewavelengths toward generator scale! A radial linefrom the center of the chart intersects the constant reGecHtion coeIcient magnitude circle at z 7 % KL 8 j LL HenceZ 7 zZ 7 5%:% KL 8 j LL; 7 N 8 NE6!
:c; What is the VSWR on the line@
Find VSWR 7 zmax 7 K on the horizontal line to the rightof the chartTs center! Or use the SWR scale on the chart!
:d; What is VL@
From the reflection coefficient scale below the chartCWnd jVLj 7 % L55! From the angle of reflection coef3ficient scale on the perimeter of the chartC Wnd the angle ofVL 7 9X 5 ! Hence VL 7 % L55ej!"#"$
!
:e; What is V at % 7 % ! from the load@
Note that jVj 7 jVLj 7 % L55! Read the angle of the reGectioncoeIcient from the angle of reflection coefficientscale as 55 % ! Hence V 7 % L55ej$$"
!
Lz 0.1 0.5j= +
0.1λ=!
0 1
0.074 . . .T Gλ0.174 . . .T Gλ
z 0.38 1.88j= +
maxVSWR 13z= =
0
L 0.855126.5Γ =
0
LAng 126.5Γ =
0Ang 55.0Γ =
00.855 55.0Γ =
L 0.855Γ = Γ =
VSWR 13=
LZLZ 5 25 [ ]j= + Ω0Z 50= Ω
0.1λ=!
0 1
Problem 1
! A transmission line has Z . /!01 ZL . zL . 0! j0! 2!
3a4 What is z at $ .%
6. 0! 7%8
From the chart1 read 0!6<=% from the wavelengths to+ward generator scale! Add 0! 7% to obtain 0!=/=% onthe wavelengths toward generator scale! This is noton the chart1 but since it repeats every half wavelength1 itis the same as 0!=/=% 0!700% . 0! /=%! Drawing a raHdial line from the center of the chart1 we Ind an intersecHtion with the constant reJection coeKcient magnitude circleat z . Z . !7 L j !7!
3b4 What is the VSWR on the line8
From the intersection of the constant reJection coeKcient cirHcle with the right hand side of the horizontal axis1 read VSWR. zmax . 7!R!
3c4 How far from the load is the Irst voltage minimum8 The Irst currentminimum8
The voltage minimum occurs at zmin which is at a distance of0!700% 0!6<=% . 0!0RR% from the load! Or read this distancedirectly on the wavelengths toward load scale!The current minimum occurs at zmax which is a quarter of awavelength farther down the line or at 0!0RR%L0! 7% . 0! VR%from the load!
Lz 0.2 0.2j= −
0.25λ=!
0.467 . . .T Gλ
0.217 . . .T Gλ
maxVSWR 5.3z= =
VSWR 5.3=
LZL LZ z 0.2 0.2 [ / ]j= = − Ω Ω0Z 1.0= Ω
0.25λ=!
Problem 2
z Z 2.5 2.5j= = +
0.033λ
minz
! The air()lled two(wire line has a characteristic impedance of 567 and isoperated at f 8 GHz! The load is ZL 8 <66 = j>67!
?a@ For the line aboveD )nd zL on the chart!
The normalized load is zL 8ZLZ
8<66 = j>6
568 E$6 = j6$F!
See the Smith chart for location of point!
?b@ What is the line impedance E$5 cm from the loadI
Note that % 8c
f8
<6! cmKsec
<6" Hz8 <6 cm$ Since we are
going to move toward the generator ?away from the load@D atthe normalized load positionD )rst read 6$E<N% on the wave$lengths toward generator scale! Then add E!5 cmK<6cm 8 6$E5% to this value to obtain 6$>PN% on the wave$lengths toward generator scale! A radial line fromthe center at this point intersects the constant reRection co(eScient magnitude circle at z 8 6$> ! j6$<ND so Z 8 zZ 856?6$> = j6$<N@ 8 E<$5! jF$57!
?c@ What is the VSWR on the lineI
From the intersection of the reRection coeScient circle andthe horizontal axis on the right hand side of the chartD readVSWR 8 E!>! Or use the SWR scale below the chart!
Solutions 9 – microstrip patch antennas Antennas and Propagation, Frühjahrssemester 2011
Microstrip transmission line
Problem 9.1 :
A microstrip line is used as a feed line to a microstrip patch. The substrate of the line is
alumina (εrel = 10) while the dimensions of the line are w0 / h = 1.2.
a) Determine the effective dielectric constant εreff and
b) the characteristic impedance Z0 of the line.
c) Calculate the reflection coefficient if a line of the computed characteristic
impedance is connected to a 50 Ω line.
d) Calculate the respective lengths of half-wavelength long lines of impedance Z0
on alumina, of a line in free-space, and of a line fully embedded in alumina halfspace,
at f = 1.5 GHz.
a)
For 0 / 1.2w h = and 10re = , the effective dielectric constant can be computed as 1
2
0
1 11 12 6.86
2 2r r
reff
h
w
e ee
-+ - ¹ Ì= + + =¼ Í
¼ ͽ Î.
b)
For a microstrip line with 0 / 1w h > the characteristic impedance is given by:
( )0 0
120
1.393 0.667 ln 1.444
44.414
c
reff
Z w w
h h
p
e=
¹ Ì+ + +¼ ͽ Î= W
c)
The reflection coefficient from a 50W line is
50 44.4140.059 24.6dB
50 44.414
-G = = -
+5 .
d)
The free-space wavelength is λ0 = 200 mm. Using 0eff
reff
λλ
ε= ,
for the 44.4 Ω line on alumina, / 2 38.2 mmeffλ = ;
for a line in free-space, / 2 100mmeffλ = ;
for a line embedded in alumina, / 2 31.6mmeffλ = .
Solutions 9 – microstrip patch antennas Antennas and Propagation, Frühjahrssemester 2011
Microstrip patch antenna
Problem 9.4 :
Design a patch antenna working at the frequency 1.575 GHz and etched on Duroid 6010
substrate ( 10re = ) of thickness 0.508h = mm.
a) Determine W and L ,
b) Determine the inset length 1L needed for matching of antenna to the microstrip
feed line (50W )
a)
For an efficient radiator, a practical width that leads to good radiation efficiency is:
0
0
240.58mm
2 1r
cW
f e= =
+
The effective dielectric constant is
1/21 11 12 9.696
2 2r r
reff
h
W
e ee
-+ - ¹ Ì= + + =¼ ͽ Î
Because of the fringing effects, one has to take into account that the antenna looks
greater than its physical dimensions of 2 LD :
( )( )( )( )
0.3 0.2640.412 0.2202mm
0.258 0.8
reff
reff
W
hL h
W
h
e
e
+ +D = =
- +
So the real physical length will be
0
0
2 30.124mm2 reff
cL L
f e= - D = .
L1
Solutions 9 – microstrip patch antennas Antennas and Propagation, Frühjahrssemester 2011
b)
For the matching:
( )21 1
1
1( ) cos
2inR L LG L
p‚
where 1( )inR L has to match the microstrip line impedance 50W .
11 2 2
sin2 cos ( )
490.326S120 120
i
XX XS XI XG
p p
- + + += = =
0 1.3396X kW= = and 0 0.19ml = .
1
11019.73
2inRG
= = W
11
50cos 0.4289
1019.73
LL L
p- ² Éʳ= =ʳ ʳµ Ë
or if the approximate expression is used (with 0/ 0.2132W l = , 0/ 0.0027h l = ):
26
10
1505.05 10
90
WG
l-² Éʳ= = Õʳ ʳµ Ë
1
1990
2inRG
‚ = W
11
50cos 0.4278
990
LL L
p- ² Éʳ= =ʳ ʳµ Ë
D-ITET-IFH Antennas and Propagation - Final exam 13th August 2009
-7-
Solution to Problem 4
a) Distance divided by speed of light; i.e., 41000 seconds, or, 11 hours 23 minutes.
b) Friis transmission equation,
2
4r t t rP PG G
R
λ
π
à Ô= Ä Õ
Å Ö
Gain of a dish, , 2
4r tG efficiency area
π
λ= ⋅ ⋅
Transmit antenna gain = 2769 = 34.4 dBi
Receive antenna gain = 1’807’309 = 62.6 dBi
Received power = 3.14 x 10-20
W = -165 dBm
c) calculate first null beamwidth αfnbw from feed aperture directivity,
22
19.5 88a
D dBiπ
λ
à Ô= = = Ä Õ
Å Ö , a = 205 mm , a / λ = 1.5
the plot in (6.37) reads αfnbw = 48…50º (alternatively, the asymptotic formula
on (6.36) states αfnbw = 69.9º / (a / λ) = 46.6º, but this formula holds only if a>>λ).
From geometry it follows that distance = 340…360 cm for αfnbw = 47…50º.
dish
aperture
first null beamwidth
46cm
274cm
distance d
D-ITET-IFH Antennas and Propagation - Final exam 13th August 2009
-8-
d) this requires much larger feed aperture, leading to shadowing and therefore,
gain reduction (an uniform aperture with 29 dBi gain would have 20 % of the
area of the main dish).
Note: ray-optics assumption, such that the feed is in the focal point and the
waves leaving the aperture are forming (at the beginning) a parallel, that is,
non-divergent, beam.
e) corrugated horn
f) Half-power beamwidth of the large dish: the aperture distribution is as shown
in (6.86), that is, circular symmetric. The best approximation for the HPBW is
the H-plane HPBW of the TE11-aperture distribution in groundplane.
The E-plane HPBW is too narrow, as it equals those of a uniform distribution,
which cannot be realized on a dish aperture.
0370.145
/
o
HPBWa λ
= =
If 360º rotation occur in 24 hours, then 0.072º rotation takes 17.4 seconds.
Note: observation point is very far away.
g) non-constant field distribution (main factor), blocking by feed and mechanical
support structures, spillover at edge, roughness and shape imperfections,
conductive loss.
Solutions 11 – wave propagation Antennas and Propagation, Frühjahrssemester 2011
Propagation close to earth
Problem 11.1 :
A mobile phone is located 5 km away from a base station. It uses a vertical /4 monopole
antenna with a gain of 2.55 dB to receive cellular radio signals. The free space E-field at
1 km from the transmitter is measured to be 310 V/m. The carrier frequency used for this
system is 900f MHz.
a) Find the length and effective aperture of the receiving antenna.
b) Find the received power at the mobile using the 2-ray ground reflection model
assuming the height of the transmitting antenna is 50 m, and the receiving antenna is
1.5 m above ground.
a)
At frequency of 900 MHz, the wavelength is 0.333 m. The length of antenna
is /4 8.33L cm. The antenna gain can be expressed in terms of the effective
aperture as follows
2
4 eAG
where gain is given as 2.55dB 1.8G . Thus,
22
20.333 1.80.0159m
4 4e
GA .
b)
Transmitter-receiver separation distance is 5d km. The field at the receiver end is
a sum of the direct and the reflected rays. Thus, we have
' '' ' ''tot
jkd jkd jkd jkdd r d dE E e E e E e E e
where dE and rE are the magnitudes of the direct and reflected field components,
respectively. From here, we can find the magnitude of the total field as
' ''tot
'' '1
1
jkd jkdd d
jk d dd
jkd
E E e E e
E e
E e
If d is large compared to rh and th we have
2
'' ' r th hd dd
Solutions 11 – wave propagation Antennas and Propagation, Frühjahrssemester 2011
Thus, with 1
2
tot 1 1 2 sin ,2
r th hjkjk d
d d dE E e E e E
where k . Since free-space E-field is reversely proportional to the distance
d we can write
0 0d
E dE
d,
where 0E is the field at distance 0 1d km. Thus,
0 0tot 2 sin
2
E dE
d
For r th h d one can write
0 0 0 0tot
3
6
2 22 sin
2
2 10 1 1000 2 (50) (1.5)
5000 0.333 5000
113.1 10 V/m
r dE d E d h hE
d d d
The received power is
262tot 13113.1 10
0.0159 2.7 10 W2 120 240r e
EP A
In dB: ,dB 125.68 dBW 95.68 dBmrP
Solutions 11 – wave propagation Antennas and Propagation, Frühjahrssemester 2011
Edge diffraction and Fresnel zones
Problem 11.2 :
Compute the diffraction loss for the scenario shown in the figure below. Assume 1/3m,
1 1d km, 2 1d km, and
a) h = 25 m,
b) h = 0,
c) h = –25 m.
T R
a
h
d1 d2
Compare your answers using values from the graph showing the knife-edge diffraction in
function of the diffraction parameter , as well as the approximate solution given by the
equation on slide 8.68. For each of these cases identify the Fresnel zone within which the tip
of the obstruction lies.
a)
h = 25 m. The Fresnel diffraction parameter is obtained as
1 2
1 2
22.74
d dh
d d
From the figure (p.8.64) the diffraction loss is obtained as 22 dB. Using the numerical
approximation
0.225
(dB) 20 log 21.7 dBdL
The path length difference is given by
21 2
1 2
( )0.625m
2
h d d
d d, and we get
23.75n .
Therefore the tip of the knife edge completely blocks the first three Fresnel zones.
b)
h = 0. From the figure the diffraction loss is obtained as 6 dB. Using the numerical
approximation
Solutions 11 – wave propagation Antennas and Propagation, Frühjahrssemester 2011
0.95(dB) 20 log 0.5 6dBdL e
With 0h m and 0 the tip of the knife edge lies in the middle of the first
Fresnel zone.
c)
h = –25 m. The knife edge lies below LOS and 2.74 , 3.75n . From the figure
a diffraction loss of 1 dB is obtained. Using the numerical approximation
(dB) 0dBdL . The tip of the edge is in the 4th
Fresnel zone, leaving the first three
Fresnel zones unobstructed.
Solutions 11 – wave propagation Antennas and Propagation, Frühjahrssemester 2011
Path loss / Propagation close to earth
Problem 11.3 :
A base station transmits a power of 10 W into a feeder cable with a loss of 10 dB. The
transmit antenna has a gain of 12 dBd (dBd refers to a /2 dipole) in the direction of a
mobile receiver, with antenna gain 0 dBd and feeder loss 2 dB. The mobile receiver has a
sensitivity of –104 dBm.
a) Determine the maximum acceptable path loss.
b) Calculate the maximum range of the communication system, assuming 1 1.5h m,
2 30h m, 900f MHz and that propagation takes place over a plane earth.
c) How does this range change if the base station height is doubled?
Quantity Value in
original units
Value in
consistent units
TP : transmitted power 10 W 10 dBW
TG : gain of transmitting antenna 12 dBd 14.15 dBi
RG : gain of receiving antenna 0 dBd 2.15 dBi
RP : received power –104 dBm –134 dBW
TL : transmitter feeder loss 10 dB 10 dB
RL : receiver feeder loss 2 dB 2 dB
a)
The acceptable path loss is
,dB ,dB ,dB ,dB ,dB ,dB 148.3dBT T R R T RL P G G P L L
b)
The plane earth loss is given by 4
PEL 2 21 2
148.3 dBr
Lh h
. Therefore, 34 kmr .
c)
Doubling one heigt reduces the plain earth loss by a factor of four. The distance
inceases by the fourth root of this ratio, thus, by a factor of 1.41. Therefore,
48 kmr .