antennas and microwaves engineering (650427) · antenna (hertzian dipole) if the center of this...
TRANSCRIPT
Philadelphia University
Faculty of Engineering Communication and Electronics Engineering
Part 6 Dr. Omar R Daoud 1
ANTENNAS and MICROWAVES
ENGINEERING
(650427)
4/15/2018 Dr. Omar R Daoud 2
Antenna Types
Short Antennas If the current distribution of a radiating element is known, it’s possible to
calculate the radiated fields by a direct integration but, the integrals can be
very complex.
For time harmonic fields, integration is performed to find a phasor called
retarded vector magnetic potential, which then followed by simple
differentiation to find the H field.
Electric fields, and analogous term for H field is the vector magnetic
potential A, often used for antenna calculations (we now seek a relation
between vector A and a current source).
The derivations lead to
VE
ddo
dsS dv
R
JA
4
00
The vector magnetic potential
SS 00 AB
The magnetic flux density
SSr 00Re2
1,, *
HEP
The time averaged power radiated
000 /SS BH
SrS 000 HaE
4/15/2018 Dr. Omar R Daoud 3
Antenna Types
Short Antennas
Hertzian Dipole
Suppose that a short line of current,
tIti cos)( 0
It’s placed along the z axis as shown.
z
jkRs
SR
eIaA
4
00
aH sin4
0R
keIj
jkRs
S
aE sin4
00R
keIj
jkRs
S
The time averaged power density at observation point is:
rR
kIr aP
2
22
22200 sin
32,
The term in brackets is the max power
density. The sin2θ term is the normalized
radiation intensity Pn(θ) plotted as :
38sinsinsin 22 dddp
The pattern solid angle
The directivity is 5.14
max
p
D
The total power radiated
2
0
2
2
22
22
0
2
02
max
2 4032
IR
IkrPrP pprad
2280
radR
For Hertzian dipole, where l<<λ, Rrad will be very small and the
antenna will not efficiently radiate power. Larger dipole antennas,
have much higher Rrad and thus more efficient.
4/15/2018 Dr. Omar R Daoud 4
Antenna Types
Short Antennas
Small Loop
The time averaged power density at observation point is:
38sinsinsin 22 dddp
The pattern solid angle
The directivity is 5.14
max
p
D
The total power radiated
aA sin1
4 2
00
jkRsS ejkR
R
SI
2aS
aH
jkRsS e
R
SkI sin
4 0
00
aE
jkRsS e
R
SkI sin4
00
rR
kSIr aP
2
220
2220
20
2
sin32
,
220
2220
20
2
max32 R
kSIP
Since the normalized power function is the same as Hertzian dipole, then
2
2
20
30
3
4
SIPrad
2
2
4320
SRrad
The equations also valid for
multi turn loop, as long as the
loop small compared to
wavelength. For N circular
loop, S=Nπa2 and for square
coil N loops, with each side
length b, S=Nb2
Radiation patterns of loops with various circumferences
4/15/2018 Dr. Omar R Daoud 5
Antenna Types
Short Antennas
Hertzian Dipole
Example
Suppose a Hertzian dipole antenna is 1 cm long and is excited by a 10 mA
amplitude current source at 100 MHz. What is the maximum power density
radiated by this antenna at a 1 km distance? What is the antenna’s radiation
resistance?
Solution:
The max. power radiated is:
2 2 22 2 2
max 2 2 2 2 2 2
2 0.010 0.0101200.052
32 32 3 1000
o oI pWP
r m
l
The antenna radiation resistance : 2 2
2 2 0.0180 80 8.8
3radR m
l
8
6
3 10, 3 .
100 10 1
c x m sc f m
f x s
4/15/2018 Dr. Omar R Daoud 6
Antenna Types
Dipole Antennas A drawback to Hertzian dipole as a practical antenna is its small
radiation resistance. A longer will have higher radiation resistance,
becomes more efficient. It as an L long conductor conveniently
placed along the z axis with current distribution i (z,t).
The time averaged power radiated is:
where, the pattern function is given by:
Therefore, the normalized power function is:
And the max time averaged power density is then :
Assume sinusoidal current distribution on
each arm
For simplicity, assume phase term =0,
and make use of current distribution
term with magnetic field equation for a
Hertzian dipole.
rFr
Ir a P
2
2015
,
2
sin
2coscos
2cos
kLkL
F
max
F
FPn
max2
20
max
15
F
r
IP
4/15/2018 Dr. Omar R Daoud 7
Antenna Types
Dipole Antennas
HPBW means half plane BW or 3-dB BW
As length of antenna increases:
beam becomes narrower,
and directivity increases
3D and 2D amplitude patterns for a thin dipole
of l=1.25λ
4/15/2018 Dr. Omar R Daoud 8
Antenna Types
Dipole Antennas
Half wave Dipole Antennas
It is a smallest resonant dipole antenna.
It has a convenient radiation resistance.
kL/2 = π/2
With the F(θ) is 1, the maximum power density
is:
Therefore, the normalized power density is:
rr
Ir a P
2
2
2
20
sin
cos2
cos15
,
2
2015
,r
Ir
P
2
2
cos
cos2
cos
nP
The current distribution and
normalized radiation pattern for
a half wave dipole antenna.
4/15/2018 Dr. Omar R Daoud 9
Antenna Types
Dipole Antennas Half wave Dipole Antennas
The pattern solid angle is
with L = λ/2, then the directivity is:
The radiation resistance is given by:
Its radiation resistance much higher than of Hertzian
dipole
Radiates more efficiently.
Easier to construct an impedance matching network for this
antenna impedance.
It contains a reactive components, Xant, where for a λ/2 dipole
antenna it is equal to 42.5Ω . Therefore, total impedance by
neglecting Rdiss.
The current distribution and
normalized radiation pattern for
a half wave dipole antenna. 658.7 p
640.14
max
p
D
Its directivity is slightly higher than
the directivity of Hertzian dipole.
pradrad PrRIP max22
02
1 2.73
30pradR
5.422.73 jZant
For impedance matching, need to make reactance zero (in resonant
condition). So, it can be achieved by making the antenna slightly
shorter (reduced in length until reactance vanishes).
4/15/2018 Dr. Omar R Daoud 10
Antenna Types
Dipole Antennas Half wave Dipole Antennas
Example
Find the efficiency and maximum power gain of a λ/2 dipole antenna constructed with AWG#20
(0.406 mm radius) copper wire operating at 1.0 GHz. Compare your result with a 3 mm length dipole
antenna (Hertzian Dipole) if the center of this antenna is driven with a 1.0 GHz sinusoidal current.
Solution
first find the skin depth of copper at 1.0 GHz,
The wire area over which current is conducted by:
m
fcu
6
7790
1009.2108.5104101
11
It is much smaller than the
wire radius
291033.52 maS cu
At 1 GHz, the wavelength is
0.3m and the λ/2 is 0.15m long
4/15/2018 Dr. Omar R Daoud 11
Antenna Types
Dipole Antennas Half wave Dipole Antennas
Example
Find the efficiency and maximum power gain of a λ/2 dipole antenna constructed with AWG#20
(0.406 mm radius) copper wire operating at 1.0 GHz. Compare your result with a 3 mm length dipole
antenna (Hertzian Dipole) if the center of this antenna is driven with a 1.0 GHz sinusoidal current.
Solution
The ohmic resistance is then:
for Hertzian Dipole, the ohmic resistance of the small dipole is :
485.01
SRdiss
the radiation resistance for
half wave dipole is 73.2Ω,
99.0485.02.73
2.73
e 63.1640.199.0maxmax eDG
mS
Rdiss 7.91
mm
mRrad 79
3.0
10380
23
2 89.07.979
79
mm
me
34.15.189.0maxmax eDG
Thus, the half wave dipole is clearly more efficient
with a higher gain than the short dipole.
4/15/2018 Dr. Omar R Daoud 12
Antenna Types
Dipole Antennas Dipole Antennas
Example
A dipole of the length 2l = 3 cm and
diameter d = 2 mm is made of copper wire
(s = 5.7 107 S/m) for mobile
communications. If the operational
frequency is 1 GHz, Then
calculate its input impedance,
radiation resistance and radiation
efficiency;
if this antenna is also used as a
field probe at 100 MHz for EMC
applications, find its radiation
efficiency again, and express it in
dB.
Solution
Since the frequency is 1 GHz, the wavelength λ = 30 cm, 2l /λ =
0.1 and βl = 0.1π.
The Skin depth
It is much smaller than the radius of the wire. Thus, we need to
compute the loss resistance of the dipole:
Since Za has not taken the loss resistance into account, the more
accurate input impedance is
And then, the radiation efficiency is
𝑍𝑎 ≈ 20 𝛽𝑙 2 −𝑗120(ln(
2𝑙𝑑− 1)
(𝛽𝑙)= 1.93 − 𝑗652Ω
𝛿 ≈1
𝜋𝑓𝜇𝜎 =
1
𝜋×109×4𝜋×10−7×5.7×107= 2.1 × 10−6m
𝑅𝐿 ≈2𝑙 𝑓𝜇
𝑑 𝜋𝜎= 0.04Ω
𝑍𝑎 = 1.97 − 𝑗652Ω
𝑒 =1.93
1.93 + 0.04= 97.97%
4/15/2018 Dr. Omar R Daoud 13
Antenna Types
Dipole Antennas Dipole Antennas
Example
A dipole of the length 2l = 3 cm and
diameter d = 2 mm is made of copper wire
(s = 5.7 107 S/m) for mobile
communications. If the operational
frequency is 1 GHz, Then
calculate its input impedance,
radiation resistance and radiation
efficiency;
if this antenna is also used as a
field probe at 100 MHz for EMC
applications, find its radiation
efficiency again, and express it in
dB.
Solution
The frequency is now 100 MHz and the wavelength λ = 300 cm,
thus 2l/λ = 0.01 and βl = 0.01π. This is an electrically small
antenna; we can use the same approach as in previous to
obtain:
𝑍𝑎 ≈ 20 𝛽𝑙 2 −𝑗120(ln
2𝑙𝑑− 1
(𝛽𝑙)= 0.0197 − 𝑗6524.3Ω
𝛿 ≈1
𝜋𝑓𝜇𝜎 = 6.67 × 10−6m
𝑅𝐿 ≈2𝑙 𝑓𝜇
𝑑 𝜋𝜎= 0.0126Ω
4/15/2018 Dr. Omar R Daoud 14
Antenna Types
Monopole Antennas A monopole antenna is excited by a current source at
its base.
By image theory, the current in the image will be
the same with the current in actual monopole. The
pair of monopole resembles a dipole antenna.
A monopole antenna placed over a conductive plane
and half the length of a corresponding dipole
antenna will have identical field patterns in the
upper half plane.
Consider the construction of half wave
dipole for an AM radio station
broadcasting at 1 MHz. At this f, the
wavelength is 300m long and the half
wave dipole antenna must be 150m tall.
We can cut this in half, by employing
image theory to build a quarter wave monopole antenna that is only 75m
tall!!
For the upper half plane (00<
θ<900), the time averaged power,
max power density and normalized
power density for the quarter wave
monopole is the same with half
wave dipole. But the pattern solid
angle is different.
dPnp 28.34
829.3 max
p
p D
6.3630
pradR
25.216.36 jZant Normalized power density is
zero for (900< θ<1800)
4/15/2018 Dr. Omar R Daoud 15
Antenna Types-Summary 13
4/15/2018 Dr. Omar R Daoud 16
Antenna Types
Antenna Arrays The motivation is to
achieve desired high gain or radiation pattern,
the ability to provide an electrically scanned beam.
It consists of more than one antenna element
They are strategically placed in space to form an array with desired characteristics which are achieved by varying the feed (amplitude and phase) and relative position of each radiating element;
The main drawbacks are
the complexity of the feeding network required and
the bandwidth limitation (mainly due to the feeding
network)
where An is the amplitude, n is the relative
phase, Ee is the radiated field of the antenna
element, and AF is called array factor.
Thus the radiation pattern of an array is the product of the pattern of individual element antenna with the (isotropic source) array pattern.
4/15/2018 Dr. Omar R Daoud 17
Antenna Types
Antenna Arrays A properly spaced collection of antennas, can have
significant variation in φ leading to dramatic
improvements in directivity. It can be designed to give a particular shape of
radiating pattern.
Control of the phase and current driving each array
element along with spacing of array elements can
provide beam steering capability.
For simplification:
All antenna elements are identical
The current amplitude is the same feeding each element.
The radiation pattern lies only in xy plane, θ=π/2
The radiation pattern then can be controlled by: controlling the spacing between elements or
controlling the phase of current driving for each element
consider a pair of dipole antennas driven
in phase current source and separated by
λ/2 on the x axis.
Assume each antenna radiates
independently, at far field point P, the
fields from 2 antennas will be 180 out-of-
phase, owing to extra λ/2 distance travel
by the wave from the farthest antenna
fields cancel in this direction.
At point Q, the fields in phase and adds.
The E field is then twice from single
dipole, fourfold increase in power
broadside array max radiation is
directed broadside to axis of elements.
4/15/2018 Dr. Omar R Daoud 18
Antenna Types
Antenna Arrays A properly spaced collection of antennas, can have
significant variation in φ leading to dramatic
improvements in directivity. It can be designed to give a particular shape of
radiating pattern.
Control of the phase and current driving each array
element along with spacing of array elements can
provide beam steering capability.
For simplification:
All antenna elements are identical
The current amplitude is the same feeding each element.
The radiation pattern lies only in xy plane, θ=π/2
The radiation pattern then can be controlled by: controlling the spacing between elements or
controlling the phase of current driving for each element
Modify with driving the pair of dipoles
with current sources 180 out of phase.
Then along x axis will be in phase and
along y axis will be out of phase, as
shown by the resulting beam pattern
endfire array max radiation is
directed at the ends of axis containing
array elements.
4/15/2018 Dr. Omar R Daoud 19
Antenna Types
Antenna Arrays Pair of Hertzian Dipoles
The find radiated power
Consider a pair of z oriented Hertzian
dipole, with distance d, where the total
field is the vector sum of the fields for
both dipoles and the magnitude of
currents the same but a phase shift
between them.
aaEEE2
20
1
1020100
44
21
R
keIj
R
keIj
jkRs
jkRs
SStotS
jss eIIII 0201
Assumption,
2121 RrR
cos2
cos2
21d
rRd
rR
r
rtotSSS
dk
R
kI
Er
a
aHEP
2cos
2cos4
32
2
1Re
2
1,
2,
2
22
22200
200
*
rarrayunit FFr a P
,
2,
Unit factor, Funit is the max time
averaged power density for an
individual antenna element at
θ=π/2
22
22200
32 R
kIFunit
The array factor, Farray is
2cos4 2
arrayF coskd
4/15/2018 Dr. Omar R Daoud 20
Antenna Types
Antenna Arrays Pair of Hertzian Dipoles
Example
The λ/2 long antennas are driven in phase and are λ/2 apart. Find:
The far field radiation pattern for a pair of half wave dipole shown.
The maximum power density 1 km away from the array if each antenna is driven by a 1mA
amplitude current source at 100 MHz.
Solution:
At 100 MHz, λ = 3m, so that 1 km away is definitely in far field. For a half wave dipole.
At θ = π/2
2
sin
2coscos
2cos
kLkL
F
rFr
Ir a P
2
2015
,
rrr
IrPr a aP
2
2
2
20
sin
cos2
cos15
,,2
2015
r
IFunit
The array factor, with d = λ/2 and =0 (due to
antennas are driven at the same phase) :
cos
2cos4 2
arrayF
cos
2cos
60,
2, 2
2
20
r
IFFr rarrayunit a P
4/15/2018 Dr. Omar R Daoud 21
Antenna Types
Antenna Arrays Pair of Hertzian Dipoles
Example
The λ/2 long antennas are driven in phase and are λ/2 apart. Find:
The far field radiation pattern for a pair of half wave dipole shown.
The maximum power density 1 km away from the array if each antenna is driven by a 1mA
amplitude current source at 100 MHz.
Solution: The normalized power function is:
This can be plotted as :
2
sin
2coscos
2cos
kLkL
F
rFr
Ir a P
2
2015
,
cos2
cos
,2
,
,2
,
,2
2
max
rP
rP
nP
The maximum radiated power density at 1000m is :
22
23
2
20
max 191000
106060
m
pW
r
IP
4/15/2018 Dr. Omar R Daoud 22
Antenna Types
Antenna Arrays N-Element Linear Arrays
The procedure of two-element array can be
extended for an arbitrary number of array
elements, by simplifying assumptions: The array is linear (antenna elements are evenly
spaced, d along a line).
The array is uniform (each antenna element driven
by same magnitude current source, constant phase
difference, between adjacent elements).
10
2030201 ,....,, Nj
sNj
sj
ss eIIeIIeIIII
rNjjj
jkR
totS eeeR
keIj aE
12000 ...1
4
coskd
2sin
2sin
2
2 N
Farray
2
maxNFarray
The normalized power pattern for these elements is :
2sin
2sin
1
2
2
2max
N
NF
FP
array
arrayn
4/15/2018 Dr. Omar R Daoud 23
Antenna Types
Antenna Arrays N-Element Linear Arrays
Example
Five antenna elements spaced λ/4 apart with progressive phase steps 300. The antennas
are assumed to be linear array of z oriented dipoles on the x axis. Find:
The normalized radiation pattern in xy plane
The plot of the radiation pattern.
Solution:
To find the array factor, first need to find psi, Ψ:
Inserting this ratio to array factor,
6cos
218030cos
4
2
cos
kd
12cos
4sin
12
5cos
4
5sin
2sin
2sin
2
2
2
2
N
Farray 252
max NFarray
4/15/2018 Dr. Omar R Daoud 24
Antenna Types
Antenna Arrays N-Element Linear Arrays
Example
Five antenna elements spaced λ/4 apart with progressive phase steps 300. The antennas
are assumed to be linear array of z oriented dipoles on the x axis. Find:
The normalized radiation pattern in xy plane
The plot of the radiation pattern.
Solution:
The normalized radiation pattern is :
12cos
4sin
12
5cos
4
5sin
25
1
2
2
nP
4/15/2018 Dr. Omar R Daoud 25
Antenna Types
Helical Antennas Modes of Operation
Normal (Broadside) (the maximum radiation of the
array is directed normal to its axis ( = 00))
Axial (End-fire) – Most practical (if the maximum
radiation is directed along the axis of the array ( =
900)) Circular polarization can be achieved over a wider
bandwidth (usually 2:1)
More efficient
Important Parameters
C
S
D
S 11 tantan
α = 0o (flat loop)
α = 90o (linear wire)
220 CSL = single turn
220 CSNNLLn
D = diameter of helix
S = spacing between turns
C = circumference of helix
L = length of one turn
AR = axial ratio
= pitch angle
4/15/2018 Dr. Omar R Daoud 26
Antenna Types
Helical Antennas Modes of Operation
Normal (Broadside) (NL0 << λ)
Dipole:
Loop:
Δφ = j = 90o
sin4
0
r
SeIkjE
rjko
o
sin
4
2 0
22
r
eIDkE
rjko
o
20
2
24
D
S
Dk
S
E
EAR
o
For this special case,
1
22
0 D
SAR
SCD 02
SC 02 000222
tan
DS
S
S
D
S
The radiated field is circularly polarized in all
directions other than θ = 00
D = diameter of helix
S = spacing between turns
C = circumference of helix
L = length of one turn
AR = axial ratio
= pitch angle
4/15/2018 Dr. Omar R Daoud 27
Antenna Types
Helical Antennas Modes of Operation
Axial (End-fire) – Most practical
oo 1412 00
3
4
4
3 C
3N
(C ≈ λ0 near optimum)
0
140
CR
Accuracy (± 20%) NSC
HPBW2
3
052
NSCFNBW
23
0115(deg)
3
0
2
15
SCNDo
N
NAR
2
12
D = diameter of helix
S = spacing between turns
C = circumference of helix
L = length of one turn
AR = axial ratio
= pitch angle
HPBW = half power beamwidth
FNBW = first null beamwidth
Do = Directivity
R = Radiation Resistance
4/15/2018 Dr. Omar R Daoud 28
Antenna Types
Helical Antennas Modes of Operation
Axial (End-fire) – Most practical
Example:
Design a circularly polarised helix antenna of an end-fire radiation pattern with a
directivity of 13 dBi. Find out its radiation resistance, HPBW, AR and radiation pattern.
4/15/2018 Dr. Omar R Daoud 29
Antenna Types
Helical Antennas
4/15/2018 Dr. Omar R Daoud 30
Antenna Types
Helical Antennas Feed Design
The nominal impedances of ordinary helices is 100-200 Ω. However, for many
practical TLs, it is desired to make it 50 Ω, and can be accomplished in many
ways.
One way is to properly design the first ¼ turn of the helix next to the feed.
This is done by flattening the wire in the form of a strip width, w, and nearly
touching the ground plane which is covered by a dielectric slab of height (h):
The helix transitions from the strip to the regular wire gradually during the ¼ to
½ turns.
2377
00
Z
wh
where
w – width of the strip starting at feed
εr – dielectric constant of the dielectric slab
Z0 – characteristic impedance of the input TL
4/15/2018 Dr. Omar R Daoud 31
Antenna Types
Yagi Uda Array Antennas The driven element (feeder) is the very heart of the
antenna. It determines the polarisation and centre
frequency. For a dipole, the recommended length is
about 0.47 to ensuring a good input impedance to a
50 Ω feed line.
The reflector is longer than the feeder to force the
radiated energy towards the front. The optimum
spacing between the reflector and the feeder is
between 0.15 to 0.25 wavelengths.
The directors are usually 10 to 20% shorter than the
feeder and appear to direct the radiation towards the
front. The director to director spacing is typically 0.25
to 0.35 wavelengths,
The number of directors determines the maximum
achievable directivity and gain.
Typical Sizes
Director lengths: (0.4 – 0.45) λ
Feeder length: (0.47 – 0.49) λ
Reflector length: (0.5 – 0.525) λ
Reflector – feeder spacing:
(0.2 – 0.25) λ
Director spacing: (0.3 – 0.4) λ
4/15/2018 Dr. Omar R Daoud 32
Antenna Types
Yagi Uda Array Antennas Applications
Amateur radio
TV antenna (usually single or a few channels)
Frequency ranges HF (3-30 MHz)
VHF (30-300 MHz)
UHF (300-3000 MHz)
Design sizes and Directivities
D0 (λ/2) = 1.67 = 2.15 dB 0.001 ≤ d/λ0 ≤ 0.04
0.002 ≤ D/λ0 ≤ 0.04
Typical Sizes
Director lengths: (0.4 – 0.45) λ
Feeder length: (0.47 – 0.49) λ
Reflector length: (0.5 – 0.525) λ
Reflector – feeder spacing:
(0.2 – 0.25) λ
Director spacing: (0.3 – 0.4) λ
4/15/2018 Dr. Omar R Daoud 33
Antenna Types
Yagi Uda Array Antennas Design Procedure
Specify: Center frequency (fo)
Directivity
do/λ (diameter of parasitic elements)
D/ λ (diameter of boom)
Find: Lengths of directors and reflector(s)
Spacing of directors and reflector(s)
Typical Sizes
Director lengths: (0.4 – 0.45) λ
Feeder length: (0.47 – 0.49) λ
Reflector length: (0.5 – 0.525) λ
Reflector – feeder spacing:
(0.2 – 0.25) λ
Director spacing: (0.3 – 0.4) λ
4/15/2018 Dr. Omar R Daoud 34
Antenna Types
Yagi Uda Array Antennas Example
Given: Center frequency (fo) = 50.1 MHz,
Directivity (relative to λ/2) = 9.2 dB, d =
2.54 cm, D = 5.1 cm. Using these
parameters, design a Yagi Uda Array
antenna by finding the element spacing,
lengths and total array length.
Solution
λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ
= 8.52 x 10-3
Step 1: Find L and N from Table 10.6 L = 0.8 λ (from the given
directivity of 9.2 dB)
N = 5 (3 directors, 1 reflector, 1 feeder)
4/15/2018 Dr. Omar R Daoud 35
Antenna Types
Yagi Uda Array Antennas Example
Given: Center frequency (fo) = 50.1 MHz,
Directivity (relative to λ/2) = 9.2 dB, d =
2.54 cm, D = 5.1 cm. Using these
parameters, design a Yagi Uda Array
antenna by finding the element spacing,
lengths and total array length.
Solution
λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ
= 8.52 x 10-3
Step 1:
For d/λ = 0.0085:
424.0
428.0
0482.0
''4
''5
''3
''1
l
ll
l
Marked by dot ()
Step 2: From Fig., draw vertical line
through d/λ = 0.00424
442.0
0485.0
''5
''3
''1
ll
lMarked in the Fig. by
(x)
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Antenna Types
Yagi Uda Array Antennas Example
Given: Center frequency (fo) = 50.1 MHz,
Directivity (relative to λ/2) = 9.2 dB, d =
2.54 cm, D = 5.1 cm. Using these
parameters, design a Yagi Uda Array
antenna by finding the element spacing,
lengths and total array length.
Solution
λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ
= 8.52 x 10-3
Step 3:
Measure Δl between l3’’, l5’’ and l4’’. Transpose
to l3’, l5’ to find l4’. Read l4’.
438.0'4 l
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Antenna Types
Yagi Uda Array Antennas Example
Given: Center frequency (fo) = 50.1 MHz,
Directivity (relative to λ/2) = 9.2 dB, d =
2.54 cm, D = 5.1 cm. Using these
parameters, design a Yagi Uda Array
antenna by finding the element spacing,
lengths and total array length.
Solution
λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ
= 8.52 x 10-3
Step 4:
From the Fig., for D/λ = 0.00852
Δl = 0.005 λ
443.0005.0438.0005.0
447.0005.0442.0005.0
490.0005.0485.0005.0
'44
'353
'11
ll
lll
ll
Therefore,
Total array length: 0.8 λ
The spacing between directors: 0.2 λ
The reflector spacing: 0.2 λ
The actual elements length:
L3 = L5 : 0.447λ
L4 : 0.443λ
L1 : 0.490λ
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Antenna Types
Log-Periodic Antennas The antenna is divided into the so called active region
and inactive regions.
The role of a specific dipole element is linked to the
operating frequency: if its length, L, is around half of the
wavelength, it is an active dipole and within the active
region; Otherwise it is in an inactive region and acts as a
director or reflector as in Yagi-Uda antenna
The driven element shifts with the frequency – this is why
this antenna can offer a much wider bandwidth than the
Yagi-Uda. A travelling wave can also be formed in the
antenna.
The highest frequency is basically determined by the
shortest dipole length while the lowest frequency is
determined by the longest dipole length (L1).
Other parameters such as the directivity or the length
of the antenna) are required to produce an optimised
design.
In practice, the most likely scenario is that the
frequency range is given from fmin to fmax, the
following equations may be employed for
design
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Antenna Types
Log-Periodic Antennas Example Design a log-periodic dipole antenna to cover all UHF
TV channels, which is from 470 MHz for channel 14 to
890 MHz for channel 83. Each channel has a
bandwidth of 6 MHz. The desired directivity is 8 dBi
Solution The given three parameters are: fmin = 470MHz, fmax = 890MHz, and
D = 8 dBi.
From the table, we can see that, for the optimum design,
the scaling factor τ = 0.865,
the spacing factor σ = 0.157, and
the apex angle α = 12.13.
That means at least six elements are required. To be on the safe side we
should use seven or even eight elements to be sure the desired directivity
will be achieved.
If N = 8, we can afford to start from a lower
frequency, say 400 MHz, thus
Ln = the length of element n, and n = 1, 2, . . , N;
sn = the spacing between elements n and (n + 1);
dn = the diameter of element n;
gn = the gap between the poles of element n.
Zo= Characteristic Impedance
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Antenna Types
Log-Periodic Antennas Example Design a log-periodic dipole antenna to cover all UHF
TV channels, which is from 470 MHz for channel 14 to
890 MHz for channel 83. Each channel has a
bandwidth of 6 MHz. The desired directivity is 8 dBi
Solution The given three parameters are: fmin = 470MHz, fmax = 890MHz, and
D = 8 dBi.
The spacing between the elements is as follows
The total length of the antennas is
Ln = the length of element n, and n = 1, 2, . . , N;
sn = the spacing between elements n and (n + 1);
dn = the diameter of element n;
gn = the gap between the poles of element n.
Zo= Characteristic Impedance
Aperture-Type Antennas They are often used for higher frequency applications (> 1GHz)
than wire-type antennas.
The relation between the aperture E field to the radiated field is
as follows:
Then , the directivity will be
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Antenna Types
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Antenna Types
Aperture-Type Antennas Example An open waveguide aperture of dimensions a long x
and b along y located in the z = 0 plane. The field in
the aperture is TE10 mode and given by
Find
the radiated far field
the directivity.
Solution The far field can be obtained using
In the φ = 0 plane (H-plane), i.e. the xz plane, we
have kx = β sin θ, ky = 0, and
In the φ = π/2 plane (the E-plane), i.e. the yz plane, we have kx = 0, ky = β sin θ, and
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Antenna Types
Aperture-Type Antennas Example An open waveguide aperture of dimensions a long x
and b along y located in the z = 0 plane. The field in
the aperture is TE10 mode and given by
Find
the radiated far field
the directivity.
Solution
Typical radiation patterns of an open waveguide in
(a) the H-plane; (b) the E-plane
Aperture-Type Antennas Horn Antennas
Horn antennas are the simplest and one of the
most widely used microwave antennas – the
antenna is nicely integrated with the feed line
(waveguide) and the performance can be easily
controlled. They are mainly used for standard antenna gain
and field measurements, feed element for reflector
antennas, and microwave communications.
Design In the H-Plane, the dimensions are linked by
In the E-Plane, the dimensions are linked by
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Antenna Types
To make this horn, we must have
The directivity:
The design equation
For the optimum design, use a first guess approximation
Aperture-Type Antennas Horn Antennas
Example:
Design a standard gain horn with a directivity of 20 dBi at
10 GHz. WR-90 waveguide will be used to feed the horn.
Solution:
The directivity is D = 20 dBi = 100, wavelength λ = 30 mm and the
dimensions of the waveguide are a = 22.86 mm and b = 10.16 mm.
Step 1: Compute the dimension A from the design Equation
• The parameters for the optimum horns are:
• The design equation
• Plot the function
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Antenna Types
Function Y(A) vs. A
around A = 135 (mm). As seen in the Figure, the
actual solution is readily found as
Aperture-Type Antennas Horn Antennas
Example:
Design a standard gain horn with a directivity of 20 dBi at
10 GHz. WR-90 waveguide will be used to feed the horn.
Solution: Step 2: Find B
B = 104.82 mm
Step 3: Find the remaining dimensions
R1 = 199.41mm;
R2 =183.14 mm;
lH = 210.36 mm and RH = 165.38 mm;
lE = 190.49 mm and RE = 165.38 mm.
The half-power beamwidths in the two principal planes for this
antenna are
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Antenna Types
Step 4: Check if RH and RE are the same. If not,
it means that the solution of A in Step 1 is not accurate enough.
From the results in Step 3, we can see that RH and RE are identical, thus the design is very good.
However, if we used the guessing value A = 0.45λ √D = 135 (mm) as the solution, it would give RH =
168.21 mm and RE = 162.73 mm. They are
obviously different, which means that the design
needs to be revised.
Aperture-Type Antennas Reflector Antennas
Reflector antennas can offer much higher gains
than horn antennas and are easy to design and
construct.
The most widely used antennas for high frequency
and high gain applications in radio astronomy,
radar, microwave and millimetre wave
communications, and satellite tracking and
communications.
The most popular shape is the paraboloid –
because of its excellent ability to produce a pencil
beam (high gain) with low sidelobes and good
cross-polarisation characteristics
The reflector design problem consists primarily of matching the feed antenna pattern to the reflector. The usual goal is to have the feed pattern about 10 dB down in the direction of the rim, that is the edge taper = (the field at the edge)/(the field at the
centre) ≈10 dB.
4/15/2018 Dr. Omar R Daoud 47
Antenna Types Paraboloidal and Cassegrain reflector antennas
Directivity:
Half-power beamwidth
Let us assume that the diameter of the reflector is 2a and the focal length is F. Any point on this paraboloid
must satisfy the following condition:
where the subtended/angular aperture angle θ0 (also
known as the edge angle) is determined by the
reflector diameter and the focal length:
If g(θ) is the power radiation pattern of the feed located at
the focus – it is circularly symmetrical (not a function of φ).
The aperture efficiency is
Aperture-Type Antennas Reflector Antennas
The aperture efficiency is generally viewed as
the product of the
Spillover efficiency: the fraction of the total
power intercepted and collimated by the reflector;
it reduces the gain and increases the side-lobe
levels.
Taper efficiency: the uniformity of the amplitude
distribution of the feed pattern over the surface of
the reflector.
Phase efficiency: the phase uniformity of the field
over the aperture plane; it affects the gain and side
lobes.
Polarization efficiency: the polarization uniformity
of the field over the aperture plane.
Blockage efficiency: by the feed; it reduces gain
and increases side-lobe levels. The support
structure can also contribute to the blockage.
Random error efficiency: over the reflector
surface.
4/15/2018 Dr. Omar R Daoud 48
Antenna Types Paraboloidal and Cassegrain reflector antennas
Directivity:
Half-power beamwidth
Let us assume that the diameter of the reflector is 2a and the focal length is F. Any point on this paraboloid
must satisfy the following condition:
where the subtended/angular aperture angle θ0 (also
known as the edge angle) is determined by the
reflector diameter and the focal length:
If g(θ) is the power radiation pattern of the feed located at
the focus – it is circularly symmetrical (not a function of φ).
The aperture efficiency is
Aperture-Type Antennas Reflector Antennas
Example:
A circular parabolic reflector has F/2a = 0.5. The field
pattern of the feed antenna is E(θ) = cos θ, θ < π/2. Find the
edge taper, spillover efficiency and aperture efficiency.
Solution:
The edge angle is
The edge taper is
, but
The Spillover efficiency is
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Antenna Types
the spillover loss is about −1.08 dB.
The aperture efficiency
Antenna Equivalent circuit
Slots Antennas They are very low-profile and can be
conformed to basically any configuration,
thus they have found many applications, for
example, on aircraft and missiles.
The radiated field by the slot is the same as
the field radiated by its equivalent surface
electric current and magnetic current which
were given by
where E and H are the electric and magnetic fields within
the slot and is the unit vector normal to the slot surface
S.
For a half-wavelength slot, its equivalent
electric surface current JS = 0, the remaining
source at the slot is its equivalent magnetic
current MS (it would be 2MS if the conducting
ground plane were removed using the imaging
theory).
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Antenna Types
Slot waveguide antenna array: widely used for radar
n
Antenna Equivalent circuit
Slots Antennas From Babinet’s Principle, the input
impedance of the slot antenna is given as
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Antenna Types
Slot waveguide antenna array: widely used for radar
Microstrip/Patch Antennas Ease of construction and integration, relatively low
cost, compact low profile configuration and good
flexibility
Typical applications for 1 - 20 GHz
To be a resonant antenna, the length L should be
around half of the wavelength (the antenna can be
considered as a l/2 transmission line resonant cavity
with two open ends where the fringing fields from the
patch to the ground are exposed to the upper half
space (z > 0) and are responsible for the radiation.)
The radiation mechanism is the same as the slot
line (there are two radiating slots on a patch antenna).
As a resonant cavity, there are many possible
modes (as waveguides), thus a patch antenna is
multi-mode and may have many resonant frequencies
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Antenna Types
Microstrip/Patch Antennas
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Antenna Types
Directivity
Input impedance
Bandwidth
Optimised width:
Resonant freq.:
Length:
Microstrip/Patch Antennas Example
RT/Duroid 5880 substrate ( and d = 1.588 mm) is to be used to make a
resonant rectangular patch antenna of linear polarisation;
Design such an antenna to work at 2.45 GHz for Bluetooth
applications;
Estimate its directivity;
If it is to be connected to a 50 ohms microstrip using the same
PCB board, design the feed to this antenna;
Find the fractional bandwidth for VSWR < 2.
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Antenna Types
Microstrip/Patch Antennas Example
RT/Duroid 5880 substrate ( and d = 1.588 mm) is to be used to make a
resonant rectangular patch antenna of linear polarisation;
Design such an antenna to work at 2.45 GHz for Bluetooth
applications;
Solution:
Step #1: Find W Step #2: Calculate the effective permittivity
Step #3: Compute the extension of the length Step #4: Determine the length L
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Antenna Types
Microstrip/Patch Antennas Example
RT/Duroid 5880 substrate ( and d = 1.588 mm) is to be used to make a
resonant rectangular patch antenna of linear polarisation;
Estimate its directivity;
Since the wavelength at 2.45 GHz is 122.45 mm > W, from
It is about 6.6 or 8.2 dBi.
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Antenna Types
Microstrip/Patch Antennas Example
RT/Duroid 5880 substrate ( and d = 1.588 mm) is to be used to make a
resonant rectangular patch antenna of linear polarisation;
If it is to be connected to a 50 ohms microstrip using the same
PCB board, design the feed to this antenna;
Step #1: Input Impedance
Step #2: Characteristics Impedance
Step #3: Transition line width
Step #4: Transition line length
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Antenna Types
Does not match well
with a 50 standard
microstrip and therefore
a quarter-wavelength
transformer is used to
connect them
Microstrip/Patch Antennas Example
RT/Duroid 5880 substrate ( and d = 1.588 mm) is to be used to make a
resonant rectangular patch antenna of linear polarisation;
If it is to be connected to a 50 ohms microstrip using the same
PCB board, design the feed to this antenna; Step #5: The microstrip width
Find the fractional bandwidth for VSWR < 2.
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Antenna Types