Final Exam Antenna 20051
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D-ITET
Antennas and Propagation
Student-No.:......................................... Student-No.:................... ............................................ ............................ ...... Name:
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Antennas and Propagation Spring 2005 March 17, 2005, 09:00 am – 12:00 noon Dr. Ch. Fumeaux, Prof. Dr. R. Vahldieck
This exam consists of 6 problems. The total number of pages is 19, including the cover page. You have 3 hours to solve the problems. The maximum possible number of points is 67. Please note: •
This is an open book exam.
•
Attach this page as the front page of your solution booklet.
•
All the calculations should be shown in the solution booklet to justify the solutions.
•
Please, do not use pens with red ink.
•
Do not forget to write your name on each solution sheet.
•
Please, put your student card (LEGI) on the table.
•
Possible further references of general interest will be written on the blackboard during the examination. Problem
Points
1 2 3 4 5 6 Total
— 1 / 19 —
Initials
D-ITET
Antennas and Propagation
March 17, 2005
Problem 1 (10 Points) Assume a receiver is located d = 10 km from a 50 W transmitter. The receiver and transmitter are mounted on 2.5 m and 5 m high posts, respectively. The carrier frequency is 900 MHz. The receiver and transmitter antennas have gains G r = 1 and G t = 2 , respectively. transmitter receiver
5m 2.5 m
ground
d
=
10 km
2 Points
a)
2 Points
b) Find Find additi additiona onall power power loss loss ( Lref ) in dB due to reflec reflectio tions ns from from the ground ground..
3 Points
c)
Find free space loss and and received received power power if reflections from earth are neglected. neglected.
Assume a wall positioned positioned in between the transmitter and and receiver, 8 km away from the transmitter. Calculate the height of the wall so that the power loss due to the knife-edge diffra diffracti ction on is is the the same same as as the the powe powerr loss loss due to refl reflect ection ion from from Earth Earth ( Lref ) in b).
3 Points
d)
What would be the maximum height height of the wall so so that the the power loss due to the diffraction is negligible? Can such a wall be built?
— 2 / 19 —
D-ITET
Antennas and Propagation
Solution 1 a)
f = 900 MHz ⇒ λ =
c = 0.333 m f
Free space loss:
LLOS LLOS
2
0.333 = 1⋅ 2 ⋅ 4π ⋅ 10 ⋅ 102 = 1.4072 ⋅ 10−11 = −108.51 dB dB
( )
λ = GrG t 4πd
(
2
)
Received Power PLOSr = Pt ⋅ LLOS
P LOSr = 7.036 ⋅ 10−10 W = −91.5 d dB BW = −61.5 d dB Bm
or P t = 50 W = 16.989dBW 9dBW
PLOSr = Pt + LLOS = 16.989 − 108.51 = −91.5dBW 5dBW
b) The power loss due to the reflection from the ground is
Lref
2ht hr
2
( d ) ( 2π 2h h =( ⋅ = 0.00222 d ) λ
Lref = 1 + Γ ⋅ exp jk t r
≈ k⋅
2hthr 2 d
)
2
Lref = −26.535dB
The plane earth loss (PEL) is LPEL = LLOS + Lref = −135.045dB
— 3 / 19 —
March 17, 2005
D-ITET
Antennas and Propagation
March 17, 2005
c) Power loss due to the diffraction is the t he same as the power loss due to the reflection, thus Ldiff = Lref = −26.535dB
From the graph knife-edge diffraction vs. parameter υ, we see that
Ldiff = 20 log
υ=h
υ
> 2.4. Thus:
⇒ υ = 4.78 ( 0.225 υ )
2 (d1 + d 2 ) λd1d 2
⇒ h =
υ
2 (d1 + d 2 ) λd1d 2
h = htot
4.78 = 78.1m 2 ( 80 8000 + 2000 ) 0.33 0.3333 ⋅ 8000 8000 ⋅ 2000 2000 = h + h r = 80.5m
d) st
The diffraction loss is negligible if most of the 1 Fresnel zone is clear (uncovered by the wall). More specifically, the Fresnel zone clearance condition has to be satisfied, i.e. υ = −0.8 ⇒ h = −13.063
htot = hr + h = −10.563m Therefore, such wall cannot be built.
— 4 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Problem 2 (13 Points) An infinitesimal dipole of length l is placed at a distance s from a ground plane and at an angle of 45 degrees from the vertical axis, as shown in the figure below. The dipole lies in the yz -plane.
y
45°
45°
normal
x
s
3 Points
a)
Determine the location and direction of the image source, which which can be used to account account for reflections of the ground plane. Your answer should be in a form of a very clear sketch.
6 Points
b)
With the coordinate system system given given in the figure, find the expression expression for the total far field electric field in the 2nd quadrant of the yz -plane (the shaded region).
4 Points
c)
Find the smallest non-zero distance s for which the total field obtained in b) is zero along the normal.
— 5 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Solution 2 a) The image dipole is shown in the figure below. Tangential field at the ground plane is zero. Eq2
Eq1
45
45°
normal
b) The original dipole and the image form an array as shown in the sketch below q y r2 r1 45°
45°
45°
45°-q normal
x
s
y
r1 = r − s ⋅ cos ( 45 − θ ) ⎫ ⎪
⎪ ⎬ r2 = r + s ⋅ cos cos ( 45 − θ )⎪⎪ ⎭ r1 = r2 = r magnitude
z’
phase
(*)
— 6 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
The field produced by the original dipole is kI 0le − jkr 1 E1 = E1θ = j η sin θ ⋅ aθ 4πr 1 The image dipole is directed along the z’ direction and produces field in
ψ
direction, defined
on the graph above. Thus the image dipole field is given as kI 0le − jkr 2 E 2 = E2 ψ = j η sin ψ ⋅ aψ 4πr 2 where
sin ψ = 1 − cos2 ψ = Thus kI 0le − jkr 2 E 2 = E 2ψ = j η 4πr 2
1 − az ' ⋅ ar
2
=
2
1 − ( sin θ sin φ )2
1 − ay ⋅ ar =
1 − ( sin θ si sin φ )2 ⋅ aψ
In the second quadrant of the yz-plane we have
φ = 270°,
Thus: kI 0le − jkr 2 E 2 = E 2 ψ = −j η 1 − ( sin θ )2 ⋅ a θ 4πr 2 kI 0le − jkr 2 cos θ ⋅ a θ E 2 = E 2 ψ = −j η 4πr 2
a ψ = −a θ
The total field is equal to the t he sum of the two fields, i.e.
E tot
⎡ kI 0le− = E1 + E2 = ⎢ j η ⎢⎣ 4πr1
1
kI 0le − sin θ − j η 4πr 2 jkr
2
jk⎤r
cos θ ⎥ ⋅ a θ ⎥⎦
nd Expresing r 1 and r 2 in terms of r (*), we obtain the total field in the 2 quadrant as kI le − jkr ⎡ + cos( 45j−ksθ ) − cos( 45 j− ksθ ) ⎤ ⋅ a s i n − c o s E tot = j η 0 e e θ θ ⎦ θ 4πr ⎣
c) Along the normal, we have θ = 45 Thus, the field can be expressed as kI 0le − jkr s − sin45 ⎡⎣ e + jk− E tot = j η e 4πr kI 0le − jkr 2 sin ( ks ) ⋅ aθ E tot = −j 2η 4πr 2
j⎤ks ⎦ ⋅ a θ
0, 1, 2, ... So E tot = 0 ⇐ ks = n π, n = 0, n = 1 ⇒ smallest non-zero length s is s =
λ
2
— 7 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Problem 3 (12 Points) Given are three radiation patterns that are taken from rectangular horn antennas. The horn antennas are fed by a rectangular waveguide WR90 (dimension a = 22.86 mm and
b = 10.16 mm) operated in the dominant TE10 mode at f = 11 GHz.
E-Plane H-Plane
(1)
2 Points
a)
(2)
(3)
Relate each of the three radiation pattern to a type of rectangular rectangular horn antenna. antenna. Give Give the physical explanation of your choice!
2 Points
b)
Which effect is responsible responsible for large back radiation radiation of the above patterns? What What would you recommend in order to decrease the back-side radiation?
2 Points
c)
Estimate the dimensions a 1 and b1 of a pyramidal horn aperture antenna required to exhibit a gain of G 0 = 17 dBi at f = 11 GHz if a typical aperture efficiency of 50% is assumed. The side length ratio is a1 / b1 = 2 .
2 Points
d)
Determine the maximum power power that can can be received by the antenna antenna designed designed in c) c) if it is illuminated by a plane wave of the frequency f = 11 GHz with an amplitude of E 0 = 30 mV/m.
2 Points
e)
Determine the maximum maximum power that can be received received by the feeding feeding waveguide waveguide (without the horn antenna) if it is illuminated by the same plane wave as in d).
2 Points
f)
Comparing the aperture aperture efficiencies efficiencies of a horn antenna antenna and a waveguide, waveguide, why are waveguides, having the same dimension as the aperture of a horn antenna, not used as radiating elements? Find an explanation.
— 8 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Solution 3 a) Pattern (1)
H-plane rectangular horn antenna
Pattern (2)
pyramidal horn antenna
Pattern (3)
E-plane rectangular horn antenna
The larger the aperture in one direction, the narrower the beam becomes in that direction (until a certain limit). The pattern of the pyramidal horn, that has a flaring in both directions, is a combination of both the E-plane, and the H-plane pattern.
b) The back-radiation is caused by diffraction at the horn aperture edges. Solutions to decrease the back-side radiation are: 1. Corrugated horns, where a similar boundary condition is enforced for both, the E and the H -field, i.e. both fields exhibit a tapered field distribution. Additionally to a reduced back-side radiation, the pattern becomes rotationally symmetric. 2. Aperture matched horns, where a curved surface section is added to the outside of the aperture edges. This decreases diffraction, that occurs at the sharp edges of a the aperture of a regular horn and additionally provides a smooth transition from the horn to the free-space f ree-space impedance.
c) The connection between antenna gain and effective aperture size at a certain frequency is given by 4π G0 = Ae λ If an aperture efficiency of ηap = 0.5 is assumed, the physical aperture size can be determined 4π 4π 1 G 0 = 2 ηap A a 1 ⋅ b1 = ap λ λ2 2 λ2 a1 ⋅ b1 = 2 G 0 4π
— 9 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
The gain is given as G 0 = 17 dBi, which is corresponding to G 0 = 50 and the wavelength at f = 11 GHz is λ = 27.25 mm.
Therefore the physical aperture is 2
a1 ⋅ b1 = 100
27.255 ⋅ 10−3 ) ( 27.2 4π
= 0.006 m2
with the given ratio of a1 / b1 = 2 this gives a 1 =
2 ⋅ 0. 0.006 m = 109.55 mm , and thus
b1 = 54.775mm . At the given frequency this corresponds to a1 ⋅ b1 ≈ 8λ 2 .
— 10 / 19 —
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Antennas and Propagation
March 17, 2005
d) The maximal received power is given by P = Ae ⋅ W , where the effective aperture is
1 ⋅ 0.006m2 and the power density of the incident wave is 2 2 1 E 0 1 900 ⋅ 10−6 W −6 W ≈ = ⋅ W = 1.2 1. 2 10 2 Z w 2 120π m2 m2 1 1 900 ⋅ 10−6 W = 3.6 nW . The maximal received power thus is P horn = 0.006 ⋅ 2 2 120π known Ae =
e) Because of the TE 10 field distribution in the feeding waveguide, the aperture efficiency is determined as ηap =
Pwaveguide
8 π2
= 0.81 and therefore
−6 1 E 0 2 −3 −3 1 900 ⋅ 10 W = ηap ⋅ a ⋅ b = 0.81 ⋅ 22.86 ⋅ 10 ⋅ 10.16 ⋅ 10 2 Z w 2 120π = 0.225nW
f) The larger dimensions (in terms of wavelength) of such a waveguide enable the support of higher order modes in the waveguide. Higher order modes exhibit different group velocities which results in a declination of signal transmission and antenna excitation.
— 11 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Problem 4 (11 Points) You have the choice of two dielectric substrates, one with εr1 = 1.5 , the other with εr2 = 4 . Both have a thickness of h = 1.5 mm. The microstrip antenna should be designed
to radiate at the frequency f r = 2 GHz. W
L
W
L h
w 0
h
er
er
(a)
2 Points
a)
w 1
w 0
(b)
Which dielectric dielectric substrate substrate would you prefer prefer as a dielectric dielectric layer for a microstrip antenna antenna shown in figure (a) and why?
4 Points
b)
Determine the length L and the width W of a linearly polarized patch on the above chosen substrate with the help of the transmission line model.
2 Points
c)
Redesign the patch patch neglecting neglecting fringing fields. Find Find the error of frequency shift that is arising from this neglect.
3 Points
d)
With the matching network (microstrip line 1: width w 1 = 1.62 mm and length λg / 4 at f r = 2 GHz) shown in figure (b), the microstrip patch is matched to the microstrip line 0 with the width w 0 . Give the characteristic impedance of the line 0. Assume there is no mutual coupling between the slots of the microstrip patch and that W λ0 is valid.
— 12 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Solution 4 a) The material with the lower permittivity εr = 1.5 is to choose because of several reasons:
1
P rad Q Q rad = = tot where the radiation Q-factor is 1. Radiation efficiency η = 1 P tot Q rad Q tot proportional to Q rad ∼ εr . The smaller Q rad (and thus εr ), the lower the losses and the higher the radiation r adiation efficiency. 1 1 f = 2. Bandwidth BW , and BW ∼ respectively. The lower εr , the f0 Q tot εr higher the bandwidth of the antenna. 3. Directivity is higher for lower εr , see see graph on slide 8.44 (of lecture notes 2004).
b) The width W can be determined by W =
c 0 2 f r
2 = 67.036 mm. εr + 1
The effective permittivity is
εeff =
εr + 1
2
+
h ⎤ −1 2 = 1.47 ⎢ 1 + 12 ⎥ W ⎦ ⎣
εr − 1 ⎡
2
The length extension caused by fringing effects can be calculated W ( εeff + 0.3 ) + 0 . 26 4 h = 0.891 mm L = h ⋅ 0.412 W ( εeff − 0.258 ) + 0.8 h
(
) )
(
The total length of the patch is therefore c 0 L = − 2L = 59.992 mm 2 f r εeff
c) Without effective permittivity and fringing fields the length of a patch is determined by c 0 λ = = 61.195 mm L* = 2 f r εr 2 In reality, the resonant frequency for this case would be c 0 f r * = = 1.962 GHz. 2 ( L * +2L ) εeff
— 13 / 19 —
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March 17, 2005
d) The input impedance of the microstrip patch can be found by (no mutual coupling) 1 ⎛W ⎞⎟2 1 = 225 Ω with G 1 = ⎜⎜ ⎟ = 2.222 mS. Z in = 90 ⎝ λ0 ⎠ G 1 A quarter-wavelength transformer with the width w 1 is used to match the patch with the impedance Z in to the microstrip line with the impedance Z 0 . The impedance of this transformer has to be Z1 =
Z 0 ⋅ Z in which can be found from slide 8.26
Y + jYC tan(β L) Yin + jY1 tan(β L1) Y2 = Y C 2 , here that is Y0 = Y 1 and β L1 = π / 2 YC + jY2 tan(β L) Y1 + jYin tan(β L1) and thus tan(β L1 ) → ∞ . Hence
Y1 Y 2 = 1 or Z12 = Z 0 ⋅ Z in . Yin Y in 120π = 106.1 Ω W ⎡W ⎤ + 1.393 + 0. 0.667 ln ln + 1.444 ⎥ εeff1 ⎢ h ⎣h ⎦
Y0 = Y 1 Z 1 ≈
with εeff1 =
(
εr + 1
2
+
εr − 1 ⎡
2
)
h ⎤ ⎢ 1 + 12 ⎥ w 1 ⎥⎦ ⎢⎣
−1 2
= 1.32 .
Z 12 = 50 Ω line. Thus, the patch is matched to a Z 0 = Z in
— 14 / 19 —
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March 17, 2005
Problem 5 (11 Points) Design an ordinary end-fire array with its maximum radiation directed towards 180 degrees and half-power beamwidth (HPBW) of 30 degrees. The spacing between the elements is λ / 4 , and the array’s length is much larger than the spacing.
3 Points
a)
Determine number number of of elements elements and progressive phase shift between between the elements (in degrees).
2 Points
b)
Estimate the array’s directivity.
2 Points
c)
Redesign the array in order to increase its directivity for the end-fire radiation. radiation. The number of elements, spacing between them and amplitude uniformity should stay the same.
2 Points
d)
How much can you increase directivity maximally?
2 Points
e)
Could you use the same method method to increase directivity directivity of the original array (designed in a)), if the spacing between the elements was λ / 2 ? Why?
— 15 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Solution 5 a) Ordinary end-fire array has maximum and 180 degrees if 2π λ π β = kd = ⋅ = = 90 2 λ 4 HPBW of an ordinary end fire array is given by 1.391λ HPBW = Θh = 2 ⋅ cos−1 1 − πdN
(
)
Thus, the total number of elements is 1.391λ N = = 52 elements. Θ ⎡ ⎤ h πd ⎢ 1 − cos 2 ⎥
(
⎣
)⎦
b) Directivity of an ordinary end-fire array is given as d D0 = 4N = 52 = 17.16 dB λ
( )
c) To improve the directivity, given everything else staying the same, the progressive phase shift should be calculated from π π π β = kd + = + = 1.63 rad = 93.46 2 52 N The obtained array is Hansen-Woodyard end-fire array with maximum beam at 180 degrees.
d) Directivity of a Hansen-Woodyard array shows improvement of 1.789 times or 2.526 dB over the ordinary end-fire array.
e) No, because for the Hansen-Woodyard array to have the maximum in the desired direction the spacing between the elements have to be around λ / 4 . For the spacing of λ / 2 , the side lobes would have larger maxima than the main lobe.
— 16 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Problem 6 (10 Points) A generator with impedance Z G = ( 50 + j 20 ) Ω and supply power P S = 40 dBm is feeding a transmitting antenna. The transmitting antenna is a lossless, half-wavelength dipole and is oriented according to the figure. An airplane (RCS of σ = 20 m2 , height
h = 3 km, the dimensions of the airplane are much smaller than h ) is illuminated by the antenna under an angle of θt = 30 and scatters the electromagnetic wave. Due to the scattering, the polarization of the incident wave is turned by φs = 30 in the xy -plane. In an angle of θr = 45 , a pyramidal horn antenna (gain G 0 = 23 dBi) is employed as a receiver. The orientation and the polarization of the horn antenna are depicted in the figure. airplane
d t
d r
h qt
30°
qr
45°
z y
transmitter
2 Points
a)
x
®
E
receiver
For which frequency is the power at the receiver higher, for f 1 = 8 GHz or f 2 = 16 GHz if the transmitter in both cases is a half-wavelength dipole? What is the
difference of the received power between both cases in dB? 4 Points
b)
Calculate the power P c that is captured by the airplane for a carrier frequency of f = 8 GHz.
4 Points
c)
Determine the power P r received by the horn antenna.
— 17 / 19 —
D-ITET
Antennas and Propagation
March 17, 2005
Solution 6 a) The received power in a free-space propagation is proportional to P r ∼ λ 2 . Thus, the higher the frequency, the higher are the free-space losses. Doubling the frequency, the received powe powerr dec decre reas ases es by a fac facto torr of of 4 ( 6 dB resp respec ecti tive vely ly). ).
b) The input impedance of a half-wavelength dipole is known as Z in = ( 73 + j 42.5 ) Ω . Thus the
generator is not matched Z − Z G Γ = in = 0.233 . Z in + Z G
to
the
dipole,
resulting
in
a
return
The distance between the transmitter and the airplane is
cos 30 30 = 3000 / dt = h / co
3 m = 3464 m . 2
The directivity (lossless half-wavelength dipole) is given by
I 0 2 3 η 2 sin θ U (θ) sin3 θ 8 π D ( θ ) = 4π . ≈ 4π =4 P rad 2.435 I 0 2 η 2.435 8π
In direction of the airplane θt = 30 this gives D ( θt ) = 4 The power of the generator is
1 1 Re {U G ⋅ I 0∗ } = I 0 2 Re {Z G + Rrad + X A } 2 2 1 = I 0 2 ( RG + Rrad ) 2
PS =
I 0
2
=2
P S RG + Rrad
Pt = Prad =
1 R = 5.93 W. I 0 2 Rrad = PS rad 2 RG + Rrad
For the captured power this means P Pc = σWt = σD ( θt ) t 2 = 161.6 nW. 4πd t
— 18 / 19 —
sin 3 θt = 0.205 . 2.435
loss
of
D-ITET
Antennas and Propagation
March 17, 2005
c) For the received power the polarization has to be taken into account. Lecture notes, slide 2.26 (SS 2004): The PLF can be described using the angle between two 2 2 unit vectors PLF = ρw ⋅ ρa cos 30 = 0.75 . OR: y ’
®
rw 30°
x ’
®
ra
Using the coordinate system shown in the figure above, the polarization vector of the 3 1 ax ' + a y ' and of the receiving antenna ρa = a x ' . Thus the scattered wave is ρw =
2
2 polarization loss factor is PLF = ρw ⋅ ρa
2
= 0.75 . The gain of the antenna is
3d Bi 200 . G 0 = 23dB
The distance between the airplane and the receiver cos 45 45 = 3000 ⋅ 2 m = 4243 m dr = h / co
The received power therefore is determined by P λ2 G0 PLF ⋅ c 2 = 11.95 ⋅ 10−18 W. Pr = Ae ⋅ W c = 4π 4πd r
— 19 / 19 —
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