final exam antenna 2007
TRANSCRIPT
D-ITET Antennas and Propagation September 7 2007 Student-No helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip
Name helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip
Address helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip
helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip
Antennas and Propagation Fall 2007
September 7 2007 0900 am ndash 1200 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 86 Please note
bull This is an open book exam
bull Attach this page as the front page of your solution booklet
bull All the calculations should be shown in the solution booklet to justify the solutions
bull Please do not use pens with red ink
bull Do not forget to write your name on each solution sheet
bull Please put your student card (LEGI) on the table
bull 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
mdash 1 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 1 (16 Points) An important and direct communication channel is being established between the ETZ building of ETH and the community of Aesch ZH on the other side of Uetliberg The figure not drawn to scale shows the distances between transmitter receiver and obstacle and their respective elevations The communication frequency is 166 MHz
ETZ 479 m
Uetliberg 871 m
5555 m4000 m
Aesch 540 m
α
4 Points a) Using a knife-edge diffraction model determine h the excess height above line of sight
and α the pitch angle
1 Point b) Determine the excess path length Δ
3 Points c) The peak of the mountain comes closest to which Fresnel ellipsoid
2 Points d) Evaluate the Fresnel-Kirchoff diffraction parameter ν
2 Points e) What is the approximate knife-edge diffraction gain in dB
4 Points f) Is it possible to achieve half-power transmission (compared to the obstacle free case) by
adjusting the transmission frequency Explain If not how many meters would we
need to remove from Uetliberg to achieve half-power transmission If so at which
transmission wavelength(s) would this be possible
mdash 2 19 mdash
D-ITET Antennas and Propagation September 7 2007
61 h
γ β
x
Solution 1 The center leg now has a height 871-479 = 392 Angles β and γ are calculated as As α=β+γ we find α = 877deg h can be found by using congruent triangles to find the length of the section of the center leg below the line of sight and subtracting it from the total length of the center leg b) The diffracted path length is The excess path length is then the diffracted path length minus the ground distance or Δ = 958229 ndash 9555 = 2729 m c) Fresnel ellipsoids represent locations where the excessive path length is constant and an integer multiple of half wavelengths For 166 MHz we find a wavelength of 1807 m d) The Fresnel-Kirchoff diffraction parameter can be obtained from the geometric simplification e) The corresponding diffraction gain is -21 dB which can be obtained by computing the Fresnel integrals or read directly from a knife-edge diffraction gain plot as given in the notes It is additionally possible to use the approximation formula given in the lecture handouts f) It is not possible to achieve half-power transmission by adjusting the frequency The half-power point corresponds to -6 dB on the diffraction gain plot or a diffraction parameter of 0 As seen in the formula in part d) the only way to get ν to approach 0 is to increase the transmission wavelength to infinity Alternatively we can dig the mountain until its height matches the line of sight This would require digging out 35653 meters as calculated in part a)
mdash 3 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 2 (16 Points) A three element dipole array is arranged along the y-axis as shown in the figure below
xφ
y
λ
The dipoles conduct current in the +z direction ( ) or the ndashz direction ( ) as indicated and the current amplitudes are equal They are separated by a distance λ
3 Points a) Determine the azimuthal (ie x-y plane) array factor as a function of φ and λ
3 Points b) At what angle in degrees does the first major lobe peak occur
3 Points c) At what angle in degrees does the first null occur
4 Points d) Sketch a polar plot of the antenna power pattern Make sure to include all lobes
(including minor and grating lobes) Clearly indicate the angular orientation of each
lobe
3 Points e) Determine the gain ratio relating the largest lobe and the smallest lobe
G(major)G(minor)
mdash 4 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 2 a) The array factor for the field is given by b) The first peak occurs when the array factor is maximum in magnitude which happens when the cosine term of the array factor is equal to -1 c) The first null occurs when the cosine term equals one half d) The power pattern is proportional to the array factor (and is not influenced by the dipole pattern which is uniform in the xy plane) Because the array factor is descriptive of the electric field the power is proportional to e) The maximum peak occurs at 30 degrees with a normalized amplitude of 9 The minor lobes as seen in the sketch of d) occur at 0 90 180 and 270 degrees and all have an equal normalized amplitude of 1 The gain ratio G(major)G(minor) = 91 = 9
mdash 5 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 3 (12 Points) The electric field of a plane wave propagating in air is given by the phasor
((5 2 ) 4 expx y )E j e e jkz⎡ ⎤= + minus⎣ ⎦ where is the wavenumber and k λ the wavelength The
plane wave is incident upon a RHCP helical antenna of directivity radiation resistance and loss resistance
0 10 dBiD =90 radR = Ω 10 LR = Ω The antenna is perfectly matched and
the frequency of operation is 3 GHzf =
4 Points a) Determine the polarization state of the incident wave (linear circular elliptical
polarization right- or left handed)
3 Points b) If according to Maxwellrsquos equations the magnetic field component of a plane wave is
given by k EHωμtimes
= ω being the angular frequency find the time-averaged power
density of the plane wave
3 Points c) Determine the polarization factor PLF for the described arrangement
2 Points d) Determine the power received by the antenna
mdash 6 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Problem 1 (16 Points) An important and direct communication channel is being established between the ETZ building of ETH and the community of Aesch ZH on the other side of Uetliberg The figure not drawn to scale shows the distances between transmitter receiver and obstacle and their respective elevations The communication frequency is 166 MHz
ETZ 479 m
Uetliberg 871 m
5555 m4000 m
Aesch 540 m
α
4 Points a) Using a knife-edge diffraction model determine h the excess height above line of sight
and α the pitch angle
1 Point b) Determine the excess path length Δ
3 Points c) The peak of the mountain comes closest to which Fresnel ellipsoid
2 Points d) Evaluate the Fresnel-Kirchoff diffraction parameter ν
2 Points e) What is the approximate knife-edge diffraction gain in dB
4 Points f) Is it possible to achieve half-power transmission (compared to the obstacle free case) by
adjusting the transmission frequency Explain If not how many meters would we
need to remove from Uetliberg to achieve half-power transmission If so at which
transmission wavelength(s) would this be possible
mdash 2 19 mdash
D-ITET Antennas and Propagation September 7 2007
61 h
γ β
x
Solution 1 The center leg now has a height 871-479 = 392 Angles β and γ are calculated as As α=β+γ we find α = 877deg h can be found by using congruent triangles to find the length of the section of the center leg below the line of sight and subtracting it from the total length of the center leg b) The diffracted path length is The excess path length is then the diffracted path length minus the ground distance or Δ = 958229 ndash 9555 = 2729 m c) Fresnel ellipsoids represent locations where the excessive path length is constant and an integer multiple of half wavelengths For 166 MHz we find a wavelength of 1807 m d) The Fresnel-Kirchoff diffraction parameter can be obtained from the geometric simplification e) The corresponding diffraction gain is -21 dB which can be obtained by computing the Fresnel integrals or read directly from a knife-edge diffraction gain plot as given in the notes It is additionally possible to use the approximation formula given in the lecture handouts f) It is not possible to achieve half-power transmission by adjusting the frequency The half-power point corresponds to -6 dB on the diffraction gain plot or a diffraction parameter of 0 As seen in the formula in part d) the only way to get ν to approach 0 is to increase the transmission wavelength to infinity Alternatively we can dig the mountain until its height matches the line of sight This would require digging out 35653 meters as calculated in part a)
mdash 3 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 2 (16 Points) A three element dipole array is arranged along the y-axis as shown in the figure below
xφ
y
λ
The dipoles conduct current in the +z direction ( ) or the ndashz direction ( ) as indicated and the current amplitudes are equal They are separated by a distance λ
3 Points a) Determine the azimuthal (ie x-y plane) array factor as a function of φ and λ
3 Points b) At what angle in degrees does the first major lobe peak occur
3 Points c) At what angle in degrees does the first null occur
4 Points d) Sketch a polar plot of the antenna power pattern Make sure to include all lobes
(including minor and grating lobes) Clearly indicate the angular orientation of each
lobe
3 Points e) Determine the gain ratio relating the largest lobe and the smallest lobe
G(major)G(minor)
mdash 4 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 2 a) The array factor for the field is given by b) The first peak occurs when the array factor is maximum in magnitude which happens when the cosine term of the array factor is equal to -1 c) The first null occurs when the cosine term equals one half d) The power pattern is proportional to the array factor (and is not influenced by the dipole pattern which is uniform in the xy plane) Because the array factor is descriptive of the electric field the power is proportional to e) The maximum peak occurs at 30 degrees with a normalized amplitude of 9 The minor lobes as seen in the sketch of d) occur at 0 90 180 and 270 degrees and all have an equal normalized amplitude of 1 The gain ratio G(major)G(minor) = 91 = 9
mdash 5 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 3 (12 Points) The electric field of a plane wave propagating in air is given by the phasor
((5 2 ) 4 expx y )E j e e jkz⎡ ⎤= + minus⎣ ⎦ where is the wavenumber and k λ the wavelength The
plane wave is incident upon a RHCP helical antenna of directivity radiation resistance and loss resistance
0 10 dBiD =90 radR = Ω 10 LR = Ω The antenna is perfectly matched and
the frequency of operation is 3 GHzf =
4 Points a) Determine the polarization state of the incident wave (linear circular elliptical
polarization right- or left handed)
3 Points b) If according to Maxwellrsquos equations the magnetic field component of a plane wave is
given by k EHωμtimes
= ω being the angular frequency find the time-averaged power
density of the plane wave
3 Points c) Determine the polarization factor PLF for the described arrangement
2 Points d) Determine the power received by the antenna
mdash 6 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
61 h
γ β
x
Solution 1 The center leg now has a height 871-479 = 392 Angles β and γ are calculated as As α=β+γ we find α = 877deg h can be found by using congruent triangles to find the length of the section of the center leg below the line of sight and subtracting it from the total length of the center leg b) The diffracted path length is The excess path length is then the diffracted path length minus the ground distance or Δ = 958229 ndash 9555 = 2729 m c) Fresnel ellipsoids represent locations where the excessive path length is constant and an integer multiple of half wavelengths For 166 MHz we find a wavelength of 1807 m d) The Fresnel-Kirchoff diffraction parameter can be obtained from the geometric simplification e) The corresponding diffraction gain is -21 dB which can be obtained by computing the Fresnel integrals or read directly from a knife-edge diffraction gain plot as given in the notes It is additionally possible to use the approximation formula given in the lecture handouts f) It is not possible to achieve half-power transmission by adjusting the frequency The half-power point corresponds to -6 dB on the diffraction gain plot or a diffraction parameter of 0 As seen in the formula in part d) the only way to get ν to approach 0 is to increase the transmission wavelength to infinity Alternatively we can dig the mountain until its height matches the line of sight This would require digging out 35653 meters as calculated in part a)
mdash 3 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 2 (16 Points) A three element dipole array is arranged along the y-axis as shown in the figure below
xφ
y
λ
The dipoles conduct current in the +z direction ( ) or the ndashz direction ( ) as indicated and the current amplitudes are equal They are separated by a distance λ
3 Points a) Determine the azimuthal (ie x-y plane) array factor as a function of φ and λ
3 Points b) At what angle in degrees does the first major lobe peak occur
3 Points c) At what angle in degrees does the first null occur
4 Points d) Sketch a polar plot of the antenna power pattern Make sure to include all lobes
(including minor and grating lobes) Clearly indicate the angular orientation of each
lobe
3 Points e) Determine the gain ratio relating the largest lobe and the smallest lobe
G(major)G(minor)
mdash 4 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 2 a) The array factor for the field is given by b) The first peak occurs when the array factor is maximum in magnitude which happens when the cosine term of the array factor is equal to -1 c) The first null occurs when the cosine term equals one half d) The power pattern is proportional to the array factor (and is not influenced by the dipole pattern which is uniform in the xy plane) Because the array factor is descriptive of the electric field the power is proportional to e) The maximum peak occurs at 30 degrees with a normalized amplitude of 9 The minor lobes as seen in the sketch of d) occur at 0 90 180 and 270 degrees and all have an equal normalized amplitude of 1 The gain ratio G(major)G(minor) = 91 = 9
mdash 5 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 3 (12 Points) The electric field of a plane wave propagating in air is given by the phasor
((5 2 ) 4 expx y )E j e e jkz⎡ ⎤= + minus⎣ ⎦ where is the wavenumber and k λ the wavelength The
plane wave is incident upon a RHCP helical antenna of directivity radiation resistance and loss resistance
0 10 dBiD =90 radR = Ω 10 LR = Ω The antenna is perfectly matched and
the frequency of operation is 3 GHzf =
4 Points a) Determine the polarization state of the incident wave (linear circular elliptical
polarization right- or left handed)
3 Points b) If according to Maxwellrsquos equations the magnetic field component of a plane wave is
given by k EHωμtimes
= ω being the angular frequency find the time-averaged power
density of the plane wave
3 Points c) Determine the polarization factor PLF for the described arrangement
2 Points d) Determine the power received by the antenna
mdash 6 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Problem 2 (16 Points) A three element dipole array is arranged along the y-axis as shown in the figure below
xφ
y
λ
The dipoles conduct current in the +z direction ( ) or the ndashz direction ( ) as indicated and the current amplitudes are equal They are separated by a distance λ
3 Points a) Determine the azimuthal (ie x-y plane) array factor as a function of φ and λ
3 Points b) At what angle in degrees does the first major lobe peak occur
3 Points c) At what angle in degrees does the first null occur
4 Points d) Sketch a polar plot of the antenna power pattern Make sure to include all lobes
(including minor and grating lobes) Clearly indicate the angular orientation of each
lobe
3 Points e) Determine the gain ratio relating the largest lobe and the smallest lobe
G(major)G(minor)
mdash 4 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 2 a) The array factor for the field is given by b) The first peak occurs when the array factor is maximum in magnitude which happens when the cosine term of the array factor is equal to -1 c) The first null occurs when the cosine term equals one half d) The power pattern is proportional to the array factor (and is not influenced by the dipole pattern which is uniform in the xy plane) Because the array factor is descriptive of the electric field the power is proportional to e) The maximum peak occurs at 30 degrees with a normalized amplitude of 9 The minor lobes as seen in the sketch of d) occur at 0 90 180 and 270 degrees and all have an equal normalized amplitude of 1 The gain ratio G(major)G(minor) = 91 = 9
mdash 5 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 3 (12 Points) The electric field of a plane wave propagating in air is given by the phasor
((5 2 ) 4 expx y )E j e e jkz⎡ ⎤= + minus⎣ ⎦ where is the wavenumber and k λ the wavelength The
plane wave is incident upon a RHCP helical antenna of directivity radiation resistance and loss resistance
0 10 dBiD =90 radR = Ω 10 LR = Ω The antenna is perfectly matched and
the frequency of operation is 3 GHzf =
4 Points a) Determine the polarization state of the incident wave (linear circular elliptical
polarization right- or left handed)
3 Points b) If according to Maxwellrsquos equations the magnetic field component of a plane wave is
given by k EHωμtimes
= ω being the angular frequency find the time-averaged power
density of the plane wave
3 Points c) Determine the polarization factor PLF for the described arrangement
2 Points d) Determine the power received by the antenna
mdash 6 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Solution 2 a) The array factor for the field is given by b) The first peak occurs when the array factor is maximum in magnitude which happens when the cosine term of the array factor is equal to -1 c) The first null occurs when the cosine term equals one half d) The power pattern is proportional to the array factor (and is not influenced by the dipole pattern which is uniform in the xy plane) Because the array factor is descriptive of the electric field the power is proportional to e) The maximum peak occurs at 30 degrees with a normalized amplitude of 9 The minor lobes as seen in the sketch of d) occur at 0 90 180 and 270 degrees and all have an equal normalized amplitude of 1 The gain ratio G(major)G(minor) = 91 = 9
mdash 5 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 3 (12 Points) The electric field of a plane wave propagating in air is given by the phasor
((5 2 ) 4 expx y )E j e e jkz⎡ ⎤= + minus⎣ ⎦ where is the wavenumber and k λ the wavelength The
plane wave is incident upon a RHCP helical antenna of directivity radiation resistance and loss resistance
0 10 dBiD =90 radR = Ω 10 LR = Ω The antenna is perfectly matched and
the frequency of operation is 3 GHzf =
4 Points a) Determine the polarization state of the incident wave (linear circular elliptical
polarization right- or left handed)
3 Points b) If according to Maxwellrsquos equations the magnetic field component of a plane wave is
given by k EHωμtimes
= ω being the angular frequency find the time-averaged power
density of the plane wave
3 Points c) Determine the polarization factor PLF for the described arrangement
2 Points d) Determine the power received by the antenna
mdash 6 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Problem 3 (12 Points) The electric field of a plane wave propagating in air is given by the phasor
((5 2 ) 4 expx y )E j e e jkz⎡ ⎤= + minus⎣ ⎦ where is the wavenumber and k λ the wavelength The
plane wave is incident upon a RHCP helical antenna of directivity radiation resistance and loss resistance
0 10 dBiD =90 radR = Ω 10 LR = Ω The antenna is perfectly matched and
the frequency of operation is 3 GHzf =
4 Points a) Determine the polarization state of the incident wave (linear circular elliptical
polarization right- or left handed)
3 Points b) If according to Maxwellrsquos equations the magnetic field component of a plane wave is
given by k EHωμtimes
= ω being the angular frequency find the time-averaged power
density of the plane wave
3 Points c) Determine the polarization factor PLF for the described arrangement
2 Points d) Determine the power received by the antenna
mdash 6 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Solution 3 a)
( )
( ) ( )
( ) ( )
( )
218180
(5 2 ) 4 exp
(5 2 )exp 4exp
538 exp 4exp
218538exp 4exp180
x y
x y
jx y
x y
E j e e jkz
j jkz e jkz e
e jkz e jkz e
jkz j e jkz j e
π
π π
⎡ ⎤= + minus =⎣ ⎦
= + minus =
= minus
⎛ ⎞= + + +⎜ ⎟
⎝ ⎠
=
So in the time-domain representation
( )218538cos 4cos180
x yE t kz e t kz eω π ω⎛ ⎞
= + + + + +⎜ ⎟⎝ ⎠
π
0
0
5384
1582218
180
x
y
y xx
y
EE
φ φ φφ
φ
= ⎫⎪= ⎪Δ = minus =⎬= ⎪⎪= ⎭
rarr RH elliptical polarization
b)
( )
( )
( )
0
0
0
0
2 (5 2 ) 4 exp
2 (5 2 ) 4 exp2
(5 2 ) 4 exp
z x y
y x
y x
H e j e e jk
j e e jkzc
j e e jkz
πλωμπλ
λ π μ
εμ
⎡ ⎤= minus times + minus =⎣ ⎦
⎡ ⎤= minus + + =⎣ ⎦
⎡ ⎤= + +⎣ ⎦
z
Therefore the power density is given through the Poynting vector
mdash 7 19 mdash
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
( )
0
0
0
0
0
0
02
0
1 Re21 Re (5 2 ) 4 (5 2 ) 42
1 Re (5 2 )(5 2 ) 162
1 Re 25 10 10 4 162
225 =-005972
x y y x
z
z
z z
S E H
j e e j e e
j j e
j j e
We e m
εμ
εμ
εμ
εμ
= times =
⎡ ⎤ ⎡ ⎤= minus + minus times minus + =⎣ ⎦ ⎣ ⎦
= minus + minus + =
= minus + minus + + =
= minus
c)
(5 2 ) 4 (5 2 ) 4 (5 2 ) 4
(5 2 )(5 2 ) 16 25 10 10 4 16 45
x y x y x yj e e j e e j e e
j j j j
⎡ ⎤ ⎡+ minus = + minus minus minus⎣ ⎦ ⎣
= + minus + = + minus + + =
⎤ =⎦
22
1 (5 2 ) 44512
1 65 2 490 90
x yt
x yr
T
t r
j e e
e je
PLF j j
ρ
ρ
ρ ρ
⎡ ⎤= + minus⎣ ⎦
⎡ ⎤= minus⎣ ⎦
= sdot = + + =1
d)
( )0 10
0
22
0
10 dBi 10log
10
01 m
09
009 m4 4
a
a
radcd
rad L
aem cd
rec em
D D
Dcf
ReR R
A e D
P S A PLF
λ
λπ π
= =
rArr =
= =
= =+
= =
=
0 rArr
mdash 8 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Problem 4 (14 Points) A rectangular aperture mounted on an infinitely conducting ground plane has a length a and width b The aperture is excited with two orthogonal modes first a field xE with uniform distribution (see Figure 41) and second the TE50 mode with electric field distribution
50 05 2 2cos
2 2TE y
a axE e E x
b ba y
π ⎧ ⎫primeminus le le⎪ ⎪⎛ ⎞prime= ⎨ ⎬⎜ ⎟⎝ ⎠ primeminus le le⎪ ⎪⎩ ⎭
2 Points a) Draw schematically the electric field lines for the TE50 mode inside the rectangular
aperture
b) Draw schematically the equivalent problem (with magnetic current sources SM ) for
each mode separately
2 Points
4 Points c) Determine far field of the TE50 mode (E-field only)
4 Points d) Determine the total far field of the antenna assuming both modes are excited with the
same amplitude (E-field only)
2 Points e) What is the polarization of the total wave in the broadside direction (θ = 0deg)
Figure 41 Uniform E-Field distribution in x direction
mdash 9 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Solution 4 a)
b)
mdash 10 19 mdash
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
c) For the TE50 mode consider an equivalent array of 5 apertures in x direction (Each aperture has a TE10 mode and alternating polarity) Etot = Eelement AF For each TE10 mode
( )( ) ( )
( )( ) ( )
1 0 2 2
1 0 2 2
cosexp( ) sin 5sin2 5 2
5 2
cosexp( ) sin 5cos cos2 5 2
5 2where
sin cos2
sin sin2
Xa jkr YE j bkEr Y X
Xa jkr YE j bkEr Y X
aX k
bY k
θ
φ
π φπ π
π θ φπ π
θ φ
θ φ
minus= minus
minus
minus= minus
minus
=
=
1 1 exp( cos ) exp( 2 cos 2 )
5sin2exp( cos ) exp( 2 cos 2 )1sin2
2where cos cos cos sin5
AF jkd j j kd j
jkd j j kd j
akd k d
γ π γ π
ψγ π γ π
ψ
πψ γ π γ φ θλ
= + + + + +
⎛ ⎞⎜ ⎟⎝ ⎠+ minus minus + minus minus =⎛ ⎞⎜ ⎟⎝ ⎠
= + = = =
Note Do not normalize the array factor since the fields must be superposed to the second mode Therefore
( )( ) ( )
( )( ) ( )
1 1 1
0 2 2
1 1 1
0 2 2
5sincosexp( ) sin 25sin12 5 2 sin5 2 2
5sincosexp( ) sin 25cos cos12 5 2 sin5 2 2
t
t
E E AF
Xa jkr Yj bkEr Y X
E E AF
Xa jkr Yj bkEr Y X
θ θ
φ φ
ψπ φ
π π ψ
ψπ θ φ
π π ψ
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
= =
⎛ ⎞⎜ ⎟minus ⎝ ⎠= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠
mdash 11 19 mdash
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
where
sin cos2
sin sin2
cos sin5
aX k
bY k
ak
θ φ
θ φ
ψ φ θ π
=
=
= +
d) For the mode with uniform distribution along the x axis (with constant phase along b)
2
2
0 0 0
2 2
0 2 2
0
exp( )4
exp( )4
For 2 2
2cos exp( sin cos )exp( sin sin )
sin sin cos sin sin2 22cossin cos
2
x s z x y
b
b
jkrE jk LrjkrE jk Lr
E e E M e e E e E
L E jkx jky dx dy
bk kE b
k
θ φ
φ θ
α
φα
π
π
φ θ φ θ
α θ φφ α α θ φ
minus minus
minus= minus
minus= +
= = minus times = minus
prime prime= minus =
⎛ ⎞⎜ ⎟⎝ ⎠= minus
int int φ prime prime
0
2 0
2 0
sin
sin sin2
sin sin2cos
with sin cos sin sin2 2
Then
exp( ) sin sincos2
Similarly
exp( ) sin sincos sin2
bk
X YE bX Y
bX k Y k
jkr X YE jk E br X Y
jkr X YE jk E br X Y
θ
φ
θ φ
θ φ
φ α
α θ φ θ φ
φ απ
θ φ απ
⎛ ⎞⎜ ⎟⎝ ⎠ =
= minus
= =
minus= +
minus= minus
mdash 12 19 mdash
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Finally
( )( ) ( )
( )( ) ( )
1 2
0 2 2
1 2
0 2 2
5sincosexp( ) sin sin25sin cos12 10 sin5 2 2
5sincosexp( ) sin 25cos cos14 10 sin5 2 2
total t
total t
E E E
Xjkr Y Xjk bEr Y XX
E E E
Xjkr Yjk bEr Y X
θ θ θ
φ φ φ
ψπα φ φ
π π ψ
ψπα φ θ
π π ψ
= + =
⎞⎛ ⎛ ⎞⎟⎜ ⎟⎜minus ⎝ ⎠ ⎟⎜= minus +
⎛ ⎞ ⎟⎜ minus ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + =
⎛ ⎛ ⎞⎜ ⎟⎜minus ⎝ ⎠⎜= minus⎛ ⎞minus ⎜ ⎟⎝ ⎠⎝
sin2 XX
⎞⎟⎟+⎟⎜
⎜ ⎟⎠
e) The polarization is linear since the two orthogonal components of the electric field Eθtotal and Eφtotal have the same phase
mdash 13 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Problem 5 (16 Points) A thin monopole of length 0375λ and with a
sinusoidal current distribution is placed above
an infinite perfect electric conductor (PEC)
(Figure 51) The monopole is fed at its lower
end by a 50 Ω transmission line Figure 51 Monopole over infinite PEC
0375 λ
50 Ω
For parts a) and b) of the question use the graphs of Figure 52 Include the graphs with your solution
4 Points a) Determine the radiation resistance and the input resistance of the monopole from the
dipole data shown in Figure 52
3 Points b) Determine the directivity of the monopole
4 Points c) Assuming the monopole is lossless and resonant at the given length find its maximum
effective area in terms of the wavelength
5 Points d) Design a rectangular aperture antenna mounted on an infinite ground plane with TE10-
mode field distribution at its opening The antennarsquos longer side is twice the size of its
shorter side (ie a = 2b) and the maximum effective area is the same as that of the
monopole Express the dimensions of the aperture in terms of the wavelength
Figure 52 Dipole antenna parameters vs length
mdash 14 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Solution 5 a) The monopole and its equivalent dipole with the current distribution are shown in Figure 1
symmetry axis
0375 λ
Iin
075 λ
Figure 1 Monopole over ground and its equivalent dipole
In the graph below we find that the radiation resistance of the 075λ dipole is
(075 ) 180r d r dipoleR R λ= = Ω (1 point)
18
180
075
360
Since
i the monopole has the same field as the dipole in the upper half-plane and no field in
the lower half-plane
ii the field of the dipole is symmetric over the ground plane
mdash 15 19 mdash
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
the monopole will radiate only half the power radiated by dipole Remember the
radiated power is 2 21 sin
2rad SP E r dθ dθ θ φ
η= intint sdot sdot Therefore the radiation resistance of
the monopole is half that of the dipole (since 20
12rad rP I R= sdot sdot ) Thus we can write
(0375 ) 2
90r m r monople r d
r m
R R R
R
λ= =
= Ω (1 point)
Input resistance of the monopole is
2
2
903sin ( ) sin ( )4
180
r min m
monopol
in m
RR
k l
R
π= =sdot
= Ω
Ω
Ω
(2 points)
The input resistance of the monopole can also be found from the graph as follows The input
resistance of the 075λ dipole is
360in dR = (from the graph)
The input resistance of the monopole is half that of the dipole ie
1802
in din m
RR = = Ω
b) Directivity of an antenna is given as max0
4
rad
UDPπ
=
Since 12rad monopole rad dipoleP P= sdot and max maxmonopole dipoleU U= the directivity of the monopole is
twice that of the dipole
From the graph above we can obtain the directivity of the dipole as D0dipole = 18
Therefore the directivity of the monopole is
0 0 36 556m monopoleD D d= = = B
c)The effective area of the monopole is
( )2
2 01
4em m mA Dλπ
= minus Γ sdot sdot (1 point)
mdash 16 19 mdash
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
The reflection coefficient between the feeding transmission line and the monopole is
0
0
0565
in m
in m
R ZR Z
minusΓ =
+
Γ =
Note since the monopole is resonant the complex part of the input impedance is
compensated for
Given all the data the maximum effective area of the monopole is
22
245 01954em mA λ λπ
= sdot = sdot
d) Since effective area of a TE10 distribution aperture on ground plane is
we have
081em aA a= sdot sdotb
2081 0195 2ab and a bλsdot = sdot =
Finally the solution can be written as
07 035a and bλ λasymp asymp
mdash 17 19 mdash
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Problem 6 (12 Points) A patch antenna is built on a substrate with the dielectric constant of 60rε = The frequency
of operation is GHz ( TM mode) and the substrate height is 5 x010 h1 = 0254 cm
a) Determine dimensions W and L of the patch antenna 3 Points
3 Points b) The dielectric constant is now modified to 10rε = What is the new resonant frequency
of the antenna
3 Points c) Modify the antenna geometry for 10rε = while keeping unchanged such that the
resonance frequency remains at 5 GHz
1h
3 Points d) A communication device at 5 GHz needs the patch antenna with the biggest bandwidth
of the two calculated above Which of the two patch antennas has the largest
bandwidth Explain why this is the case Propose a feeding mechanism for maintaining
the bandwidth of the antenna Explain the choice you made
Figure 61
mdash 18 19 mdash
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-
D-ITET Antennas and Propagation September 7 2007
Solution 6 a)
0
12
0
2 160252 1
1 1(1 12 ) 496752 2
( 03)( 0264)0412 10822
( 0258)( 08)
1 2 112862
r r
r rreff
reff
reff
r r
cw cmf
e e hww h
L h mmw h
cL L cmf
ε
ε
εε
ε
minus
= =+
+ minus= + + =
+ +Δ = sdot sdot =
minus +
= minus Δ =
b)
0
1
( 03)( 0264)0412 16952
( 0258)( 08)
102132( 2 )
reff
reff
reff
rreff
w L given
w hL h mm
w h
cf GHzL L
ε
εε
ε
=
+ +Δ = sdot sdot =
minus +
= =+ Δ
c) Same procedure as in a) with
1 0254 5 129979
1
1755526468
r
reff
h cm f GHzw cm
L mmL cm
ε
ε
= =rArr =
=
Δ ==
=
d) 1 1
min ( ) )L r
L
f VSWR BWf Q VSWR
Q antenna in cε
Δ minus=
rArr rArr
sim
mdash 19 19 mdash
- Problem 1 (16 Points)
- Solution 1
- Problem 2 (16 Points)
- Solution 2
- Problem 3 (12 Points)
- Solution 3
- Problem 4 (14 Points)
- Solution 4
- Problem 5 (16 Points)
- Solution 5
- Problem 6 (12 Points)
- Solution 6
-