E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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E&CE 476 Mid-term Examination Date: February 20, 2006 Time: 1 ½ hours Instructor: S. Safavi-Naeini P1. An antenna system is composed of three half-wave (resonant) dipoles, located along x-axis, y-axis and z-axis respectively. All are centered at the origin and fed by three identical 0.1 tje ω [Amp] current sources at a frequency of 1GHz. The input impedances are 73 ohm.
a) Find the total electric field and the radiation pattern in xy-plane ( °= 90θ ). Draw an approximate plot of the pattern as a function of φ in this plane.
b) What are the maximum radiation directions in xy-plane. c) Estimate the gain in xy-plane. (Hint: Find the input power from the knowledge of input impedances)
Solution: a) Electric field due to z-directed dipole in xy-plane:
)()(
: thereforeplane,-on xyˆˆ,2
)2/sin()2/(sin
)2/cos(2
cos
2)(
0000
200
rErE
zr
eIjZr
eIjZrE
z
rjkrjk
rr
r
−=
−==
≈−−
θ
θ θππ
π
ππ
π
Electric fields due to x-directed dipole:
φ
φφ
φπ
πφ
ˆ is plane-in xy dipole
ofdirection therespect to with conical dipole directed-for x becausesinsin
cos2
cos
2)(
200
0
≈−
reIjZ
rErjkr
Electric fields due to y-directed dipole:
.̂- isr unit vecto its and
)2/( is plane-in xy dipole ofdirection therespect to with angle conical dipole, directed-yfor because
coscos
sin2
cos
2)2/sin(
)2/(sin
)2/cos(2
cos
2)( 2
002
0000
φ
φπ
φφ
φπ
πφπ
φπ
φππ
πφ
−
−=−
−
−
−≈−−
reIjZ
reIjZ
rErjkrjkr
z
y
x
E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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Therefore the total field is:
±
+
=
≤≤+≤≤−
±
+−=−
2
200
2
2200
cos
sin2
cos
sin
cos2
cos1
2
2for )( ,0for )(,ˆcoscos
sin2
cossin
sin
cos2
cosˆ
2
0
φ
φπ
φ
φπ
π
πφππφφφφ
φπ
φφ
φπ
π
rIZ
E
zreIjZ
Erjk
r
r
b) To draw a rough sketch of the pattern in xy-plane we note that θEEz = is constant. The ϕE part, given
below:
πφππφφ
φπ
φ
φπ
πφ2for )( ,0for )(,,
cos
sin2
cos
sin
cos2
cos
200 ≤≤+≤≤−
±
=
rIZ
E , is simply the sum or
difference two dipole patterns (red dotted line), one rotated by o90 . This is plotted blow in arbitrary unit.
E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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The plot of total field,2
Er
, ignoring the constant terms, is simply obtained by adding 1 (a circle of radius one in
polar coordinate system) to the previous polar plot squared, which is shown below (arbitrary unit).
The above plots show that the radiation maxima are along oo 315,135=φ . c) Gains along the above directions is:
[ ] [ ]
( ) )5.1(707.0219
12029.121929.1
]733/[29.1
)]4/(733/[58.222
)]4/(/[58.222)4/(
)2/(
:have we/2or 0for
)]4/(/[cos
sin2
cos
sin
cos2
cos1
22)4/(
)2/(
020
20
0
220
2002
200
2
0
2
2
2
200
2
0
2
dBZ
IIZ
rIr
IZrP
rIZ
rP
ZEG
rPr
IZrP
ZEG
inin
inin
−=×
≈×
=××=
=×
=
==
=
−
+
==
ππ
ππ
πππ
πφ
πφ
φπ
φ
φπ
ππ
r
r
E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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P2. Shown below are two half-wave resonant dipoles at a distance of 0.4 λ , radiating at a frequency of 2.4 GHz. Dipole #1 is driven by a current source of amplitude 0.05 [A]. Dipole # 2 is terminated by a variable capacitor, C.
a) Assuming that the self-impedances ( 2211 ZZ = ) are approximately the same as the input impedance of one resonant dipole in free space, namely 73 ohm, write down the impedance matrix relationship between the antennas’ terminal currents and voltages.
b) Find the input impedance at the antenna #1, and the input power in terms of C respectively. c) Find the input (excitation) current at antenna # 2 as a function of C. d) Determine the array factor and antenna system radiation pattern in terms of C. e) Estimate C such that the antenna system has an end-fire pattern (maximum radiation along z). Find
directivity of the antenna array for this case. Solution: a)
][344 where,7373
1221122
21211
Ω−≈+=+=
jZIIZVIZIV
b) , c) From above equation and the V-I relationship for a capacitor:
jXjXZ
IV
ZjX
IZICXjXICjIV
−+≈
−−==⇒
−−=⇒××===−=
731156
7373
7373
104.22,/1,)/(2
12
1
11
1122
9222 πωωω
d)
φθαθπα
απ
θπθ
cossincos,)(/)cos8.0exp(73
3441
sin
cos2
cos
)cos8.0exp(73
344105.0)cosexp(
73105.0
2max
2
2
2
2
12
=−
−−
=
−
−−=
−
−=
CFjjX
jF
jjX
jjkd
jXZ
AF
e) (+10%) For end-fire radiation ( o0max =θ ), C, should maximize the following expression:
( )
dB16.461.279206
}Re{
31.1120}97for {
}Re{
)8.0exp(73
3441
}Re{
)8.0exp(73
3441
42/)}(Re{
)0(42
1
)4/(
)0(}excitationcurrent unity with dipole one frompower radiated{
)4/(
)2/(
pF][7.0]F[107.09773
)30(56)8.0exp(
73344
1
1
2
1
2
0
211
)0(
22
10
2
211
2
22
20
02
2
2
0
2
1222
222
2
===≈=−
−−
=−
−−
=
×=
×==
=×≈⇒≈⇒+
++≈
−−
− −
ZX
Z
jjX
jZ
IZ
jjX
jI
Z
rIZ
AFr
ZZ
rP
AF
rP
ZED
CXX
Xj
jXj
AF
radrad
ππ
ππ
π
πππ
π
44444 844444 76
r
=1I 0.05tje ω C
2I z
E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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P3. A pyramidal horn has a square aperture of dimensions 3 [cm] by 3 [cm]. The horn is connected, through a TE10 rectangular waveguide, to a perfectly impedance-matched 10GHz source with 1 [W] output power.
a) Find the amplitude of electric field at: 1) the center of the horn aperture, and 2) a distance of 1 [km], along the axis (z-axis) from the horn.
b) Determine the far-field pattern in yz-plane ( o90=φ ) Solution: a)
1) 23
3/2003/100
)/(,3
100/,10
32
00
0010222
0102
0 ZZ
ZkYakk w ====−=×= ππ
βπ
πβπ
aperture) ofcenter at the field(]V/m[7.323
4109
[W] 14 0
200
420 =⇒××
×=== EEZEY
abP
-
wrad
2) Along the z-axis ( o0=θ ):
( )
( )( )
( ) ]V/m[71]V/m[101.77.31091000
11032
)0],m[1000(
1)0],m[1000(
1 cos
1 sin
542
2
100002
022
200
200
0
0
µπ
π
θ
πθ
πϕ
πϕ
θθ
ϕ
θ
=×=×××
===
==+===⇒
=
=
−−
=−
−
rE
abEr
kEErE
abEr
ejkE
abEr
ejkE
rrjk
rjk
r
r
b)
( )
2
0
0
2
20
0
00
sin2
sin2
sin)90,(
1
sin2
sin2
sin
0
==⇒
=−
θ
θφθ
πθ
θ
θ bk
bk
Fbk
bk
abEr
ejkE
rjko
P4. A high-gain narrow beam reflector antenna is required for a 30 GHz radio link. The Half-Power Beam-width (HPB) should be less than o5.1 .
a) Assuming a second degree (n = 2) “Parabolic taper on a pedestal” aperture illumination, find an estimate of the edge illumination that results in the minimum reflector diameter, 2a, which is required for such a narrow beam and for a SLL better than -30 dB.
b) Assuming that ae = se = 90%, and e = 96%, find the taper efficiency, te , the total aperture efficiency, ape , and the gain of your design.
Solution: a) The -30 dB SLL falls between the lines corresponding to SSL=-29.5 dB and -31.7 dB in Table 7.1. A simple linear estimate for HP is )2/(207.1 aλ . The corresponding edge illumination becomes: C=-12.45 dB (0.238). The taper efficiency is estimated as 824.0=tε .
b) The total aperture efficiency becomes: 64.096.09.09.0824.0 =×××=ape . The minimum diameter required is obtained from the HP = 026.05.1 =o [rad] and 464.0026.0/01.0207.1026.0/207.12 =×== λa [m] or 46.4 [cm]. The gain is therefore 13600)()/4(64.0 22 =××= aG πλπ or 41.3 [dB].
E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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Appendix Half-wave dipole antenna:
Far-field:
≈−
θ
θπ
πµ
20
00
sin
cos2
cos
2)(
0
rkeI
rArjk
zr , dB)15.2(64.1,/,sin)( 0 === DZEHrAjE z θϕθ θω
r,
Mutual impedance plots: Horn Antenna (Rectangular Aperture)
( )( )
( )( )
−
=
−
=
−
−
20
2
0
0
0
00
20
2
0
0
0
00
cossin
cossin2
cos
sinsin2
sinsin2
sincoscos
cossin
cossin2
cos
sinsin2
sinsin2
sinsin
0
0
ϕθπ
ϕθ
ϕθ
ϕθϕθ
ϕθπ
ϕθ
ϕθ
ϕθϕ
ϕ
θ
ak
ak
bk
bk
abEr
ejkE
ak
ak
bk
bk
abEr
ejkE
rjk
rjk
204
EYab
P wrad = , ab
64
)4/(
)2/(20010
2
0
2
max
λλβπ==
rP
ZED
rad
r
, )/(,/ 00222
010 ZkYak w βπβ =−=
E&CE 476, University of Waterloo S. Safavi-Naeini, 2006
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