spring 2006 dr. stuart longcourses.egr.uh.edu/ece/ece5318/longpwrptsch6-0221a71.pdf · spring 2006...
TRANSCRIPT
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Chapter 6Chapter 6Part 1Part 1
Array BasicsArray Basics
ECE 5318/6352ECE 5318/6352Antenna EngineeringAntenna Engineering
Spring 2006Spring 2006Dr. Stuart LongDr. Stuart Long
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Array RationaleArray Rationale
Single elementsSingle elements
Usually broad beamwidthRelatively low directivity
Often system requirements demand higher directivitiesOften system requirements demand higher directivities
Can increase electrical sizeor
form an “assembly” of elements in an electrical and geometrical
configuration called an
ARRAYARRAY
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Assumptions for our coverageAssumptions for our coverage
1. All elements are identical2. No coupling between elements3. Total field is vector sum of individual
radiation patterns
Use array to control overall pattern shape by:Use array to control overall pattern shape by:
1. Geometrical configuration of overall array(linear, circular, rectangular, spherical, …)
2. Relative displacement between radiators3. Amplitude of excitement of each element4. Excitation phase of each element5. Individual element pattern
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Two element arrayTwo element array
simplest casesimplest case
ψ
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Two identical horizontal sources along the z axis, a distance “d ” apart, constant amplitudes; upper element leads lower one in phase by amount “β “
look at [ y - z ] plane (φ = 90°)only there
sin ψ = cos θ
Similar case of horizontal dipole above a ground plane, except 2nd. source not necessarily 180° out of phase – but similar analysis
Two element array Two element array (cont)(cont)
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Total FieldTotal Field
Two element array Two element array (cont)(cont)
22
)2
(
2 cos4
2
θπ
ηβ
reIkjE
rkj
o
−−
≅
2 cos2dr r θ≅ +
21 EEE +≅t
11
)2
(
1 cos4
1
θπ
ηβ
reIkjE
rkj
o
−−
≅
θcos21drr −≅ [6-2]
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Total FieldTotal Field
[6-3]
Two element array Two element array (cont)(cont)
⎥⎥⎦
⎤
⎢⎢⎣
⎡+≅
⎟⎠⎞
⎜⎝⎛ +
−⎟⎠⎞
⎜⎝⎛ +−
2cos
2cos
cos4
ˆβθβθ
θ θπ
η kdjkdjjkro
t eereIkjaE
1 22 2
1 21 2
ˆ cos cos4
j k r j k r
ot
k I e ejr r
β β
θη θ θ
π
⎛ ⎞ ⎛ ⎞− − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎡ ⎤⎢ ⎥≅ +⎢ ⎥⎢ ⎥⎣ ⎦
E a
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[6-3]
Element Factor Array Factor
[AF][Ee]
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +
≅−
2coscos2cos
4ˆ βθθ
πη
θkd
reIkjE
rkjo
t a
Two element array Two element array (cont)(cont)
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θπ
η cos4
][EereIkj
rkjo
−
≅
⎟⎠⎞
⎜⎝⎛ +
≅2
coscos2[AF] βθkd
as function of kd and β⎟⎠⎞
⎜⎝⎛ +
≅2
coscos(AF) βθkdn
[6-4]
Two element array Two element array (cont)(cont)
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ExampleExample
4λ
=d
242 πλλπ
==dk
( )AF cos cos4n
π θ⎡ ⎤= ⎢ ⎥⎣ ⎦
(a) 0=β
( ) θcosEe =n
Two element array Two element array (cont)(cont)
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ExampleExample
4λ
=d
242 πλλπ
==dk
( ) ( )AF cos cos 14n
π θ⎡ ⎤= +⎢ ⎥⎣ ⎦
(b) 2πβ =
( ) θcosE e =n
Two element array Two element array (cont)(cont)
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ExampleExample
( ) ( )AF cos cos 14n
π θ⎡ ⎤= −⎢ ⎥⎣ ⎦
(c)2πβ −=
( ) θcosEe =n
Two element array Two element array (cont)(cont)
4λ
=d
242 πλλπ
==dk
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General example for TwoGeneral example for Two--Element ArrayElement Array
( ) ⎟⎠⎞
⎜⎝⎛ +
=2
coscosAF βθkdn
nulls whennulls when
( ) πβθ ⎥⎦⎤
⎢⎣⎡ +
±=+2
12cos21 nkd n
0,1, 2, 3,n =
[ ]⎭⎬⎫
⎩⎨⎧
+±−= − πβπλθ )12(
2cos1 n
dn
for first null due to [AF] to exist
2λ
≥d
oror
for not phase shift
( )0=β
Defining Nulls
02
coscos =⎟⎠⎞
⎜⎝⎛ + βθkd
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General example for TwoGeneral example for Two--Element ArrayElement Array
Defining Nulls
( ) nn θcosEe =
null at null at θθ nn = = 9090°°
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Chapter 6Chapter 6Part 2Part 2
NN--Element Uniform Element Uniform Linear ArrayLinear Array
ECE 5318/6352ECE 5318/6352Antenna EngineeringAntenna Engineering
Spring 2006Spring 2006Dr. Stuart LongDr. Stuart Long
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NN--element uniformelement uniformlinear arraylinear array
βjnn eII 1−=
constant amplitudeconstant amplitudeconstant phase shiftconstant phase shiftelements along a lineelements along a line
(e.g. # 3 leads # 2 by β radians)
for convenience assume isotropic sources
AssumptionsAssumptions
[6-6]
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[ ] ( )1
1 cos AF
Nj n
nkd e ψψ θ β −
=
= + ⇒ = ∑let
( ) ( ) ( )βθβθβθ +−+++++ ++++= cos)1(cos2cos1AF kdNjkdjkdj eee
[6-6]
NN--element uniform linear array element uniform linear array (cont)(cont)
( )[ ]∑−
+−+=N
n
kdnje1
cos1AF βθ
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11[AF]
−−
= ψ
ψ
j
jN
ee
ψψψψ jNjjj eeee +++= 2[AF]
[6-7]
NN--element uniform linear array element uniform linear array (cont)(cont)
Multiply RHS by and subtract from [AF]e j ψ
( ) ψψ jNj ee −=− 11[AF]
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ExampleExample
Look at N = 5
⇒
ψψ 41AF jj ee ++++=
ψψψψψ 542[AF] jjjjj eeeee ++++=
( ) ψψ 511[AF] jj ee −=−
11[AF]
5
−−
= ψ
ψ
j
j
ee
NN--element uniform linear array element uniform linear array (cont)(cont)
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NN--element uniform linear array element uniform linear array (cont)(cont)
this term due to asymmetric locationof elements; shifting to center or taking magnitude eliminates it
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−=
−−
=−
−⎥⎦⎤
⎢⎣⎡ −
22
222
1
11[AF] ψψ
ψψψ
ψ
ψ
jj
NjNjNj
j
jN
ee
eeeee
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎦⎤
⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
=⎥⎦⎤
⎢⎣⎡ −
2sin
2sin
[AF] 21
ψ
ψψ
N
eNj
[6-10]
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NN--element uniform linear array element uniform linear array (cont)(cont)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎦⎤
⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
=
2sin
2sin
[AF]ψ
ψN
or ‘Normalized’
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎦⎤
⎢⎣⎡
⎥⎦⎤
⎢⎣⎡
=
2sin
2sin
1[AF]ψ
ψN
Nn
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NN--element uniform linear array element uniform linear array (cont)(cont)
as ψ → 0
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡⎥⎦⎤
⎢⎣⎡
≅
2
2sin
[AF] ψ
ψN
or⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡⎥⎦⎤
⎢⎣⎡
=
2
2sin
1[AF] ψ
ψN
Nn
N→[AF] 1[AF] →nor
[Note: for small arguments ; thus,
for small x (but not necessarily small)]
xxx
=→
sinlim0
ψN
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NullsNullssin x
x
.707
.63
1.0
1.39 π.217
π2
2π
3π2
x
nulls occur when
(where )πψ nN±=
2
xx
nsin[AF] ≅
02
sin =⎟⎠⎞
⎜⎝⎛ ψN
( )cos2 nN kd nθ β π⇒ + = ±
but not n = 0,3,2,1=n
⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡ ±−= − πβ
πλθ
Nn
dn2
2cos 1
[6-11]
NN--element uniform linear array element uniform linear array (cont)(cont)
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No matter how small ψ, always have max at ψ = 0 , (m = 0)
Absolute maximaAbsolute maxima
[ ]⎭⎬⎫
⎩⎨⎧
±−= − πβπλθ m
dm 22
cos 1
m = 0,1,2,3,…Absolute max. when πψ m±=2
⎥⎦
⎤⎢⎣
⎡= −
dm πλβθ
2cos 1
[6-12]
[6-13]
NN--element uniform linear array element uniform linear array (cont)(cont)
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Half Power PointHalf Power Point
Half Power Point ⇒⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡ ±−= −
Ndh782.2
2cos 1 β
πλθ
for small ψ : (see appendix)x
xn
sin[AF] ≅
3 dB point ⇒ when 391.12
±=ψN707.0sin
→x
x
[6-14]
NN--element uniform linear array element uniform linear array (cont)(cont)
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Secondary maxima ⇒ approx. when is max = ± 1(minor lobes)
sin N2
ψ
Secondary maximaSecondary maxima
( )2
122
πψ+±= sN
S = 1,2,3,…1 2 1cos2s
sd N
λθ β ππ
− ⎧ ⎫⎡ + ⎤⎡ ⎤= − ±⎨ ⎬⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦⎩ ⎭ [6-15]
NN--element uniform linear array element uniform linear array (cont)(cont)
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First minor lobe maximaFirst minor lobe maxima
⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡ ±−= −
Ndsπβ
πλθ 3
2cos 1
23
2πψ ±=
NMaxima value of first minor lobe occurs
at NOT at 2π
[6-16]
NN--element uniform linear array element uniform linear array (cont)(cont)
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At this point the magnitude of the AF
(note this side lobe is independent of N )
First minor lobe maximaFirst minor lobe maxima
212.0
231
2
2sin
[AF]
23
≅=⎟⎠⎞
⎜⎝⎛
≅
==
πψ
ψ
πθθ s
N
N
n
[ ] ( ) [ ]dB5.13212.0log20dB −=⇒in
[6-17]
NN--element uniform linear array element uniform linear array (cont)(cont)
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Chapter 6Chapter 6Part 3Part 3
Broadside ArraysBroadside Arrays
ECE 5318/6352ECE 5318/6352Antenna EngineeringAntenna Engineering
Spring 2006Spring 2006Dr. Stuart LongDr. Stuart Long
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BROADSIDE arraysBROADSIDE arrays
maximum radiation in direction normal to line of array maximum radiation in direction normal to line of array ((θθ = 90= 90°°))
⇒ Broadside radiation pattern when all elements are in phase( independent of separation dist “d” )
ββ = 0= 0
00cos90 =⇒=+⇒= ββθθ kdfor
0cos0 =+⇒= βθψ kd whenMax.[6-18]
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BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
NullsNulls
MaximaMaxima
Half Power PointsHalf Power Points
Side Lobe MaximaSide Lobe Maxima
⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡ +
±≅
⎥⎦
⎤⎢⎣
⎡±≅
⎥⎦
⎤⎢⎣
⎡±=
⎥⎦
⎤⎢⎣
⎡±=
−
−
−
−
Ns
d
dN
dm
dNn
s
h
m
n
122
cos
391.1cos
cos
cos
1
1
1
1
λθ
πλθ
λθ
λθ ,3,2,,3,2,1
NNNnn
≠=
,3,2,1=s
,3,2,1,0=m
assuming 1<<λ
π d
assuming 1<<λ
π d
[Table 6.1]
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ββ = 0= 0
ΘS
NOTE FIRST SIDE LOBE BEAMWIDTHNOTE FIRST SIDE LOBE BEAMWIDTH
(not the width of any beam)
First Null BeamwidthFirst Null Beamwidth
Half Power BeamwidthHalf Power Beamwidth⎭⎬⎫
⎩⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−=Θ −
dNh πλπ 391.1cos
22 1
⎭⎬⎫
⎩⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−=Θ −
dNnλπ 1cos
22
[Table 6.2]
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
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Example Example
NullsNulls
Absolute MaximaAbsolute Maxima
0;4
;10 === βλdN
3754cos
6652cos
52cos
1
1
1
=⎥⎦⎤
⎢⎣⎡±=
=⎥⎦⎤
⎢⎣⎡±=
⎥⎦⎤
⎢⎣⎡±=
−
−
−
n
n
nn
θ
θ
θ ,3,2,,3,2,1
NNNnn
≠=
⇒==5
2
410
nndN
nλλλ
,3,2,1,0=m⇒= md
m 4λ [ ] ( ) 900cos4cos 11 ==±= −− mmθ
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
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34
Half Power BeamwidthHalf Power Beamwidth
Example Example
( ) 4.208.79902 =−=Θh
Half Power PointsHalf Power Points
⇒⋅
⋅=
52391.1391.1
ππλdN
8.795
2391.1cos1 =⎥⎦⎤
⎢⎣⎡
⋅⋅
±= −
πθh
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
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Side Lobes MaximaSide Lobes Maxima
01010cos
53106cos
10122cos
1
1
1
=⎥⎦⎤
⎢⎣⎡±=
=⎥⎦⎤
⎢⎣⎡±=
⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡ +
±=
−
−
−
s
s
ss
θ
θ
θ⇒⎥⎦⎤
⎢⎣⎡ +
=⎥⎦⎤
⎢⎣⎡ +
10122
1012
2ss
dλ ,3,2,1=s
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
Example
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at θ = 53°
Array Factor MagnitudeArray Factor Magnitude
Example Example
10 2 5sin sin cos sin cos12 2 4 2[AF] 2.19
2 0.45sin sin cos sin cos2 4 2
n
N π λ πψ θ θλ
ψ π λ πθ θλ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠≅ = = = =⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
[ ] ( ) [ ]1AF 2.19 0.219 20 log 0.219 13.2 dB10n = ⋅ = ⇒ = −
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
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Array Factor MagnitudeArray Factor Magnitude
at θ = 0°
Example Example
5sin12[AF] 1.414
0.707sin4
π
π
⎛ ⎞⎜ ⎟⎝ ⎠≅ = =⎛ ⎞⎜ ⎟⎝ ⎠
[ ] ( ) [ ]1AF 1.414 0.1414 20 log 0.1414 17.0 dB10n = ⋅ = ⇒ = −
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
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NullsNulls
20.4°
θ 53°
37°
79.8°
66°
MaximaMaxima
0.14140°
0.21953°
1.090°
[AF]nθm
Half PowerHalf Power
Θh = 2(90-79.8) = 20.4°θ h = 79.8°
20.4°
Summary Summary
BROADSIDE arrays BROADSIDE arrays (cont)(cont)
ββ = 0= 0
at θ n = 66° & 37°
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39
Chapter 6Chapter 6Part 4Part 4
Ordinary Ordinary Endfire ArrayEndfire Array
ECE 5318/6352ECE 5318/6352Antenna EngineeringAntenna Engineering
Spring 2006Spring 2006Dr. Stuart LongDr. Stuart Long
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ORDINARY ENDFIRE ARRAYSORDINARY ENDFIRE ARRAYSMaximum radiation in direction along line of array Maximum radiation in direction along line of array
ββ == ±± kdkd
for max. in for max. in θθ = 0= 0°°⇒⇒
kd cos θ + β = 0 ⇒
ψ = 0 for θ = 0° ⇒
β = - kd
for max. in for max. in θθ = 180= 180°° ⇒⇒
ψ = 0 for θ = 180° ⇒
kd cos θ + β = 0 ⇒
β = + kd
[6-20]
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41
ORDINARY ENDFIRE arrays ORDINARY ENDFIRE arrays (cont)(cont)
NullsNulls
MaximaMaxima
Half Power PointsHalf Power Points
Side Lobe MaximaSide Lobe Maxima
ββ == ±± kdkd
⎭⎬⎫
⎩⎨⎧
⎥⎦⎤
⎢⎣⎡ +
−≅
⎥⎦
⎤⎢⎣
⎡−≅
⎥⎦
⎤⎢⎣
⎡−=
⎥⎦
⎤⎢⎣
⎡−=
−
−
−
−
Ns
d
dN
dm
dNn
s
h
m
n
122
1cos
391.11cos
1cos
1cos
1
1
1
1
λθ
πλθ
λθ
λθ,3,2,
,3,2,1NNNn
n≠=
,3,2,1=s
,3,2,1,0=m
assuming 1<<λ
π d
π dλ
assuming << 1
[Table 6.3]
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First Null BeamwidthFirst Null Beamwidth
Half Power BeamwidthHalf Power Beamwidth
ββ == ±± kdkd
ExampleExample
kddN −=== βλ ;4
;10
⎭⎬⎫
⎩⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−≅Θ −
dNh πλ391.11cos2 1
⎭⎬⎫
⎩⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−≅Θ −
dNnλ1cos2 1
hh θ2=ΘHPBW HPBW ⇒⇒
[Table 6.4]
ORDINARY ENDFIRE arrays ORDINARY ENDFIRE arrays (cont)(cont)
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43
Chapter 6Chapter 6Part 5Part 5
Scanning Scanning Phased ArrayPhased Array
ECE 5318/6352ECE 5318/6352Antenna EngineeringAntenna Engineering
Spring 2006Spring 2006Dr. Stuart LongDr. Stuart Long
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SCANNING PHASED ArraysSCANNING PHASED Arrays
Can Can ““scanscan”” beam between endfire and broadsidebeam between endfire and broadsideby allowing phase shift by allowing phase shift ββ to beto be
-- kd kd << ββ <<kdkd
0 ≤ β ≤ kd -kd ≤ β ≤ 0or
θθo o
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for for maximiummaximium radiation at radiation at θθ == θθoo ⇒⇒
-- kd kd << ββ << kdkd
kd cos θo + β = 0 ⇒
ψ = 0 for θ = θo ⇒
β = - kd cos θo [6-21] θθo o
PHASED Arrays PHASED Arrays (cont)(cont)
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46
PHASED Arrays PHASED Arrays (cont)(cont)
-- kd kd << ββ <<kdkd
(use general expressions for nulls, max, sidelobes)
For HPBW the main beam is no longer symmetric about θ = 90° or θ = 0° ;
subtracting the two θ h (+ and – sign)
⎟⎟⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=Θ −−
dkNdkN ooh782.2coscos782.2coscos 11 θθ
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47
PHASED Arrays PHASED Arrays (cont)(cont)
-- kd kd << ββ <<kdkd
L••
•
••
•
L = (N-1)d
L + d = Ndcan also write Θh as
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+−⎟⎟⎠
⎞⎜⎜⎝
⎛+
−=Θ −−
dLdL oohλθλθ 443.0coscos443.0coscos 11
[6-22]
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Example Example
60;4
;10 === odN θλ
785.04
60cos2
60cos4
2cos −=−=−=−=−=ππλ
λπθβ okd
7.234.472.715.2
443.021cos
5.2443.0
21cos 11 =−=⎟
⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ −=Θ −−
h
PHASED Arrays PHASED Arrays (cont)(cont)
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Example Example
-12.6°
+11.2°
71.2°
47.4°
NOTE: MAIN BEAM IS NOTNOTE: MAIN BEAM IS NOTSYMMETRIC ABOUT ITSSYMMETRIC ABOUT ITSMAX ATMAX AT θθ = 60= 60°°
ALSO NOTE ROTATIONALALSO NOTE ROTATIONALSYMMETRY GIVES 3SYMMETRY GIVES 3--DDPATTERN OF CONICAL BEAMPATTERN OF CONICAL BEAM
PHASED Arrays PHASED Arrays (cont)(cont)
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50
Chapter 6Chapter 6Part 5Part 5
DirectivityDirectivity
ECE 5318/6352ECE 5318/6352Antenna EngineeringAntenna Engineering
Spring 2006Spring 2006Dr. Stuart LongDr. Stuart Long
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51
DIRECTIVITYDIRECTIVITY““the ratio of the radiation intensity in a given direction from tthe ratio of the radiation intensity in a given direction from the he antenna to the radiation intensity averaged over all directionsantenna to the radiation intensity averaged over all directions””
radoo P
UUUD max4π==
Fig. 2.13 3-D directivity pattern of a λ/2 dipole
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For BROADSIDE For BROADSIDE and small spacingand small spacing
( ( d d << << λλ ))
For ORDINARY ENDFIRE For ORDINARY ENDFIRE and small spacingand small spacing
( ( d d << << λλ ))ψ
ψ
2
2sin
[AF] N
N
n
⎟⎠⎞
⎜⎝⎛
≅
ZZ
n
sin[AF] ≅
θ
θ
cos2
cos2
sin[AF]
kdN
kdN
n
⎟⎠⎞
⎜⎝⎛
≅
θcos2
kdNZ =where
1cos2
1cos2
sin[AF]
−
⎟⎠⎞
⎜⎝⎛ −
≅θ
θ
kdN
kdN
n
1cos2
−= θkdNZwhere
[6-38] [6-45]
DIRECTIVITY DIRECTIVITY (cont)(cont)
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53
DIRECTIVITY DIRECTIVITY (cont)(cont)
For BROADSIDE For BROADSIDE and small spacingand small spacing
( ( d d << << λλ ))
For ORDINARY ENDFIRE For ORDINARY ENDFIRE and small spacingand small spacing
( ( d d << << λλ ))( )[ ]
22 sinAF)( ⎟
⎠⎞
⎜⎝⎛==
ZZU nθ
90at1max == θU 180or0at1max == θU
rado P
UD max4π=
→2
dkNwith approx. that large; can calculate Prad
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∫∫∫∫∫∫ ⎟⎠⎞
⎜⎝⎛==Ω=
Ω
ππππθθφθθφ
0
22
00
2
0sinsinsin d
ZZddUddUPrad
θθππ
dZ
Z sinsin22
0∫ ⎟⎠⎞
⎜⎝⎛=
[2-13], [6-39]
DIRECTIVITY DIRECTIVITY (cont)(cont)
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For BROADSIDE For BROADSIDE and small spacingand small spacing
( ( d d << << λλ ))
For ORDINARY ENDFIRE For ORDINARY ENDFIRE and small spacingand small spacing
( ( d d << << λλ ))
θcos2
kdNZ =
θθ dkdNZd sin2
−=
ZdZ
ZdkNP
Nkd
Nkdrad ∫ ⎟⎠⎞
⎜⎝⎛−=
2/
2/
2sin
2
2π
)1(cos2
−= θkdNZ
θθ dkdNZd sin2
−=
ZdZ
ZdkNP
Nkd
rad ∫ ⎟⎠⎞
⎜⎝⎛−=
0
2sin
2
2π
[6-40] [6-47]
DIRECTIVITY DIRECTIVITY (cont)(cont)
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with approx. that Ν k d
2→ large
For BROADSIDE For BROADSIDE and small spacingand small spacing
( ( d d << << λλ ))
For ORDINARY ENDFIRE For ORDINARY ENDFIRE and small spacingand small spacing
( ( d d << << λλ ))
ZdZ
ZdkN
Prad ∫∞+
∞⎟⎠⎞
⎜⎝⎛=
sin4π
[6-42]
ππdkN
Prad4
=
λπ dN
PUD
rado
24 max ==
ZdZ
ZdkN
Prad ∫∞+
⎟⎠⎞
⎜⎝⎛=
0
2sin4π
[6-49]
ππ 24dkN
Prad =
λπ dN
PUD
rado
44 max ==
DIRECTIVITY DIRECTIVITY (cont)(cont)
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57
For BROADSIDE For BROADSIDE and small spacingand small spacing
( ( d d << << λλ ))
For ORDINARY ENDFIRE For ORDINARY ENDFIRE and small spacingand small spacing
( ( d d << << λλ ))
for large array
[6-42]λ
π dNP
UDrad
o24 max ==
NdLdL =⇒>>
λLDo 2=
for large array
[6-49]λ
π dNP
UDrad
o44 max ==
NdLdL =⇒>>
λLDo 4=
[6-44] [6-49]
DIRECTIVITY DIRECTIVITY (cont)(cont)
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For BROADSIDE For BROADSIDE and small spacingand small spacing
( ( d d << << λλ ))
For ORDINARY ENDFIRE For ORDINARY ENDFIRE and small spacingand small spacing
( ( d d << << λλ ))
ExampleExampleExampleExample
ExampleExample Note:Note:
The directivity for the ENDFIRE case is exactly twice that for the broadside case since it is unidirectional instead of bidirectional
L2 4.5oDλ
=
2 5.0oNdDλ
=
4;10 λ
== dN
2;20 λ
== dN
19 or 20oD
5HPBW =
3.20Pozar =⇒ oD
2;20 λ
== dN
40or38=oD
32HPBW =
40Pozar =⇒ oD
DIRECTIVITY DIRECTIVITY (cont)(cont)
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DIRECTIVITY DIRECTIVITY (cont)(cont)
Fig. 6.12 HalfFig. 6.12 Half--Power Beamwidth for Broadside, OrdinaryPower Beamwidth for Broadside, OrdinaryEndEnd-- Fire, and Scanning Uniform Linear ArraysFire, and Scanning Uniform Linear Arrays
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GRATING LOBESGRATING LOBES
When visible range (VR) includes When visible range (VR) includes ψψ= = --22ππ oror ψψ = 2= 2ππ, a second major , a second major lobe of same magnitude N is produced lobe of same magnitude N is produced
-- This is called a This is called a ““grating lobegrating lobe”” --
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⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
2sin
2sin
)(ψ
ψ
ψ
N
AF
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GRATING LOBES GRATING LOBES (cont)(cont)
π≥kd
⎟⎠⎞
⎜⎝⎛ −≤
Nkd ππ 22
⎟⎠⎞
⎜⎝⎛ −≤
Nkd
2ππ
π2≥kd
⎟⎠⎞
⎜⎝⎛ −≤
Nkd ππ2 ⎟
⎠⎞
⎜⎝⎛ −−≤ βππ
Nkd 2
( )βπ −≥ 2kd
Broadside Broadside Other phases Other phases
For no part of grating lobe
For complete grating lobe
But to keep second lobe from being grater than side lobe need
Endfire Endfire
⎟⎠⎞
⎜⎝⎛ −≤
Nkd ππ ⎟
⎠⎞
⎜⎝⎛ −−≤ βππ
Nkd 22
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GRAPHICAL SOLUTION FOR ARRAYS GRAPHICAL SOLUTION FOR ARRAYS
ζζ )(sin)( =nAF
ExampleExample
Will see several functions like
)cos()( δγζ += cff
kdN2
=c)cos(2
βθζ += kdN βδ2N
=where
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Often easier to plot in rectangular form and then convert to polar plot for actual function of real angle θ
)(ζf
GRAPHICAL SOLUTION FOR ARRAYSGRAPHICAL SOLUTION FOR ARRAYS(cont)(cont)
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1.First, draw rectangular plot of
GRAPHICAL SOLUTION FOR ARRAYSGRAPHICAL SOLUTION FOR ARRAYS(cont)(cont)
)(ψAF ψvs.
2.Then, draw a semicircle of radius underneath,
which is offset by an amount
kdβ
3.Draw vertical lines to intersect semicircle
5.Mark corresponding magnitudes on radial lines
4.Draw radial lines to points of intersection
6.Connect points
ProcedureProcedure
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ExampleExample
GRAPHICAL SOLUTION FOR ARRAYSGRAPHICAL SOLUTION FOR ARRAYS(cont)(cont)
5–element Uniform Linear array
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
2sin
2sin
)(ψ
ψ
ψ
N
AF
for N = 5
kd = 1.2π
β = - 0.2π
βθψ += coskd
25
n πψ = ±
Prin. Max. at ψ = 0 , mag. = 5
First null at ψ = ± 0.4π
Nulls at
(2n 1)5
πψ += ±Secondary Max. at
23.1)3.0(sin
1==
π1st. Side Lobe mag.
(±0.8π, ±1.2π, ±1.6π)
(±0.6π, ±π, ±1.4π)
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GRAPHICAL SOLUTION FOR ARRAYSGRAPHICAL SOLUTION FOR ARRAYS(cont)(cont)
(First, draw rectangular plot of ))(ψAF ψvs.
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1.23
Semicircular radius kd = 1.2πOffset by β = -0.2π
Rectangular Rectangular
to Polarto Polar
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Rectangular Rectangular
to Polarto Polar
Fig. 6.16 Rectangular to polar plot graphical solution
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Broadside ⇒ β = 0 (no offset)Max. always along θ = 90º
Smaller kd gives broader beamwidth
If kd approaches 2πadditional main lobescan appear – called
“GRATING LOBES”
VISIBLE RANGE:values of ψ whichcorrespond to real
angles θ
NOTESNOTES
Larger kd gives narrower beamwidth
–kd ≤ ψ ≤ kd
GRAPHICAL REPRESENTATION GRAPHICAL REPRESENTATION FOR BROADSIDEFOR BROADSIDE
kdkd
kd
θ
θ
θ
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GRAPHICAL GRAPHICAL REPRESENTATION REPRESENTATION
FOR ENDFIREFOR ENDFIRE
kd ≤ (2π - 2π/N)
Example for no part of grating lobe
β = - kd
β = kd
kd
kd