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1
Chapter 4 : Linear Wire Antenna
• Infinitesimal Dipole
• Small Dipole
• Finite Length Dipole
• Half-Wavelength Dipole
• Linear Elements near or on Infinite Perfect
Conductors
2
Infinitesimal Dipole• Length l << λλλλ
• Used to represent capacitor-plate (top-hat-loaded) antennas
• Capacitive loading to maintain the uniform current
3
Radiated Field
• The current on the infinitesimal dipole is
assumed to be constant, i.e.,
• Recall that
• Since the dipole is infinitesimal, the following approximations hold:
0ˆ)'( Izz =I
')',','(4
),,( dlR
ezyxzyx
jkR
C
−
∫= JAπµ
0''' === zyx
constant|'| ==−= rrrRrr
'' dzdl =
4
Radiated Field (2)
jkrl
l
jkre
r
lIzdze
r
Izzyx
−
−
− == ∫ πµ
πµ
4ˆ'
4ˆ),,( 0
2/
2/
0A
0;sin;cos =−== φθ θθ AAAAA zzr
,)(ˆ
)(sin
1ˆ)sin(
sin
ˆ
∂∂
−∂∂
+
∂∂
−∂∂
+
∂∂
−∂∂
=×∇
θφ
φθθ
φθ
θθ
θ
φθ
φ
r
r
ArA
rr
rAr
A
r
AA
r
rA
Thus
Since A has only z component,
and
∂∂
−∂∂
=θµ
φ θrA
rArr
)(1ˆH
The magnetic field becomes:
5
Radiated Field (3)
jkr
r
ejkrr
lkIjH
HH
−
+=
==
11
4
sin
0
0
πθ
φ
θ
0
)(
111
4
sin
11
2
cos
2
0
2
0
=
−+=
+=
−
−
φ
θ πθ
η
πθ
η
E
ekrjkrr
lkIjE
ejkrr
lIE
jkr
jkr
r
Therefore,
Likewise, the electric field can be found to be
6
Power Density and
Radiation Resistance
• Poynting vector
)ˆˆ(2
1
)ˆ()ˆˆ(2
1)(
2
1
**
**
φφθ
φθ
θ
φθ
HEHEr
HEEr
r
r
−=
×+=×= HEW
+=
−=
232
2
0
32
22
0
)(
11
16
sincos||
)(
11
sin
8
krr
lIkjW
krj
r
lIWr
πθθ
η
θλ
η
θ
−==
⋅+=⋅=
∫ ∫
∫ ∫∫∫
3
2
02
0 0
2
2
0 0
2
)(
11
3sin
sinˆ)ˆˆ(
krj
lIddrW
ddrrWWrdP
r
r
S
λπ
ηφθθ
φθθθ
π π
π π
θsW
Outgoing Power
Components of
Poynting Vector
7
Power Density and
Radiation Resistance (2)
3
2
0
)(
1
3)
~~(2
kr
lIWW em λ
πηω −=−
)~~
(2 emrad WWjPP −+= ω
direction radialin All power; (reactive)
imaginary average-time)~~
(2
energy electric average-time~
energy magnetic average-time~
power radiated average-time
power
=−
=
=
=
=
em
e
m
rad
WW
W
W
P
P
ω
2
2
2
803
2
=
=λ
πλ
πη
llRr
Radiation Resistance
Complex Power
rrad RIlI
P2
0
2
0 ||2
1
3==
λπ
η
Reactive Power
8
Far Field (kr >> 1)• For kr >> 1, the fields can be approximated as
1
0
4
sin
4
sin
0
0
>>
===≅
≅
≅
−
−
kr
HHEE
er
lkIjH
er
lkIjE
rr
jkr
jkr
θφ
φ
θ
πθ
πθ
η
ηφ
θ ==H
EZw
space)-freefor (120 impedance intrinsic
impedance wave
Ω=
=
πη
Zw
Ratio of E and H:
TEM Wave
9
Directivity
2
22
02* sin
42ˆ||
2
1ˆ)Re(
2
1
r
lkIrErav
θπ
ηη θ ==×= HEW
22
2
2
02 |),,(|2
sin42
φθη
θπ
ηθ rE
rlkIWrU av ===
2
0max
42 πη lkI
U =
2
34 max
0 ==radP
UD π
πλ
πλ
8
3
4
2
0
2
== DAem
Time-average power density:
Radiation intensity:
Maximum radiation intensity:
Maximum directivity:
Maximum effective area:
10
Small Dipole
• Length λ/50 <λ/50 <λ/50 <λ/50 < l < λ/10λ/10λ/10λ/10
11
Radiated Field• The current on the small dipole is assumed to
be a triangular function, i.e.,
≤≤−+
≤≤−=
0'2/),'2
1(ˆ
2/'0),'2
1(ˆ
)'(
0
0
zlzl
Iz
lzzl
Iz
zI
−+
+=
∫
∫−
−
−
2/
00
0
2/0
')'2
1(
')'2
1(4
ˆ),,(
ljkR
l
jkR
dzR
ez
lI
dzR
ez
lIzzyx
πµ
A
where I0 is a constant. Vector potential becomes:
Approximating R ~ r yields the maximum phase error
kl/2 = π/10 for l=λ/10.
12
Radiated Field (2)
1
0
8
sin
8
sin
0
0
>>
===≅
≅
≅
−
−
kr
HHEE
er
lkIjH
er
lkIjE
rr
jkr
jkr
θφ
φ
θ
πθ
πθ
η
jkr
z er
lIzAzzyx
−==π
µ42
1ˆˆ),,( 0A
Using R~r:
The far-field can be given by
which is one-half of that for the infinitesimal dipole.
2
2
2
0
20||
2
==λ
πl
I
PR rad
rRadiation resistance:
13
Field Separation• For a very thin dipole, x’=y’=0, thus
)'cos'2()''2()(
)'()'()'()'(
222222
222222
zrzrzzzzyx
zzyxzzyyxxR
+−+=+−+++=
−++=−+−+−=
θ
θcos; where 2222rzzyxr =++=
14
15
Field Separation (2)
then,'cos'2
Let2
2
r
zrzx
+−=
θ
L+
+
+−= θθθθ 2
3
2
22
sincos2
'1sin
2
'1cos'
z
r
z
rzrR
Recall also the Taylor expansion:
L+++= 2
!2
)0('')0(')0()( x
fxffxf
L1682
1)1(32
2/1 xxxx +−+=+
which yields the same result:
)'cos'2(1
1 2
2zrz
rrR +−+= θ
Recall that
16
Far Field
θcos'zrR −=
2when
2
'sin
2
'1 2
max
22 π
θθ ==
r
zz
r
82
'2 π≤
r
zk
By retaining only the first two terms, i.e.,
λ
22lr ≥
λ
22when
Dr ≥
termphasefor 'ˆcos'
termamplitudefor
rrrrrR
rRr⋅−=−≈
≈
θ
The most significant neglected term has the maximum value
A maximum total phase error of π/8 is acceptable, thus
Far-field approximation
2/'2/ lzl ≤≤−
17
Radiating Near Field
+−= θθ 2
2
sin2
'1cos'
z
rzrR
[ ] 0cos2sinsin2
'sincos
2
'1 22
2
32
3
2=+−=
∂∂
θθθθθθ r
zz
r
[ ] 0cos2sin1
22 =+− =θθθθ
By retaining only the first three terms, i.e.,
)2(tan 1
1 ±= −θ
The most significant neglected term is the fourth term. In
order to find its maximum value, one can differentiate the fourth term with respect to θ, and the result is set to 0, i.e.,
yields
18
Radiating Near Field (2)
83123
2
3
1
8sincos
2
'2
3
2
3
tan
2/'
2
2
3
1
πλ
πλπ
θθθθ
≤
=
=
−== r
l
r
l
r
kz
lz
≥
λ
32
33
2 lr
λ
3
62.0l
r ≥
which reduces to
or
If the maximum total phase error is allowed to be π/8,
Near-field regionλλ
32
62.02 D
rD
≥≥
19
Finite Length Dipole
• Length > λλλλ/10
• Current on the finite length dipole assuming
the wire is very thin
≤≤−
+
≤≤
−
=
0'2/,)'2
(sinˆ
2/'0,)'2
(sinˆ
)'(
0
0
zlzl
kIz
lzzl
kIz
zeI
where I0 is a constant. This distribution assumes
that the antenna is center-fed and the current
vanishes at the end points.
20
Radiated Field
0
'4
sin)'(
'4
sin)'(
===≅
≅
≅
−
−
θφ
φ
θ
πθ
πθ
η
dHdHdEdE
dzeR
zkIjdH
dzeR
zkIjdE
rr
jkRe
jkRe
The electric and magnetic field components in
the far field for the infinitesimal dipole dz’ are
given by
Using the far-field approximation yields
'4
sin)'( cos'dzee
r
zkIjdE
jkzjkre θθ π
θη −≅
21
Radiated Field (2)
444 3444 2144 344 21factor space
2/
2/
cos'
factorelement
2/
2/')'(
4
sin
== ∫∫ −
−
−
l
l
jkz
e
jkrl
ldzezI
r
kejdEE
θθθ π
θη
factor) (spacefactor)(element field total ×=
The total electric field can be obtained by
summing up contributions from all infinitesimal
dipoles, i.e.,
Thus, the electric field of the finite length dipole
can be given by
−+
+=
∫
∫−−
2/
0
cos'
0
2/
cos'0
')'2
(sin
')'2
(sin4
sin
ljkz
l
jkzjkr
dzezl
k
dzezl
kr
ekIjE
θ
θθ π
θη
22
Radiated Field (3)
−
≅−
θ
θ
πηθ
sin
2coscos
2cos
2
0
klkl
r
eIjE
jkr
)]cos()sin([)sin(22
γββγβαβα
γβα
α +−++
=+∫ xxe
dxxex
x
Using
Likewise, the magnetic field can be given by
yields
−
=≅−
θ
θ
πηθ
φsin
2coscos
2cos
2
0
klkl
r
eIj
EH
jkr
23
Current Distributions and
Radiation Pattern
24
Radiation Pattern for l=1.25λλλλ
25
Power Density and
Radiation Intensity
2
22
2
02
**
sin
2coscos
2cos
8
||ˆ||
2
1ˆ
)ˆˆRe(2
1)Re(
2
1
−
==
×=×=
θ
θ
πη
η
φθ
θ
φθ
klkl
r
IrEr
HEav HEW
Time-average power density:
Radiation intensity:2
2
2
02
sin
2coscos
2cos
8
||
−
==θ
θ
πη
klkl
IWrU av
26
Radiated Power
θθ
θ
πη
φθθ
π
π π
d
klkl
I
ddUUdPrad
∫
∫ ∫∫∫
−
=
=Ω=Ω
0
2
2
0
2
0 0
sin
2coscos
2cos
4
||
sin
−+++
− +−+=
)](2)2()2/ln()[cos(2
1
)](2)2()[sin(2
1)()ln(
4
|| 2
0
klCklCklCkl
klSklSklklCklCI
P
ii
iiirad πη
constant) s(Euler' 1566495772.0=C
Radiated power can be obtained by
which can be given by
where
(1)
27
Radiated Power (2)
∫∫ ∞
∞=−=
x
xi dy
y
ydy
y
yxC
coscos)(
∫=x
i dyy
yxS
0
sin)(
∫
−=
−+=
x
in
iin
dyy
yxC
xCxCxC
0
cos1)( where
)()ln()(
Ci(x) is related to Cin(x) by
Cosine integral
Sine integral
0 5 10 15 20 25 30 35 40 45 50−5
−4
−3
−2
−1
0
1
2
3
4
5
Ci(x)
Si(x)
Cin
(x)
28
Radiation Resistance and
Input Resistance
rinin R
IR
I
2
||
2
|| 2
0
2
=
−+++
− +−+==
)](2)2()2/ln()[cos(2
1
)](2)2()[sin(2
1)()ln(
2||
22
0
klCklCklCkl
klSklSklklCklCI
PR
ii
iiirad
r πη
r
in
in RI
IR
2
0
=
Radiation resistance becomes
Input resistance can be given by
Since
For a dipole of length l,
=2
sin0
klII in
=
2sin 2 kl
RR r
in
Input Resistance
assuming loss-less
29
Directivity
Q
F
P
UD
rad
maxmax0
)(24
θπ ==
−+++
− +−+=
)](2)2()2/ln()[cos(2
1
)](2)2()[sin(2
1)()ln(
klCklCklCkl
klSklSklklCklCQ
ii
iii
Directivity is given by
and
where 2
sin
2coscos
2cos
)(
−
=θ
θθ
klkl
F
30
Radiation resistance, input
resistance and directivity
31
Half-wavelength Dipole
≅−
θ
θπ
πηθ
sin
cos2
cos
2
0
r
eIjE
jkr
=≅−
θ
θπ
πηθ
φsin
cos2
cos
2
0
r
eIj
EH
jkr
θπ
ηθ
θπ
πη 3
22
2
0
2
22
2
0 sin8
||
sin
cos2
cos
8
||
r
I
r
IWav ≅
=
Let l = λ/2, then
θπ
ηθ
θπ
πη 3
2
2
0
2
2
2
02 sin8
||
sin
cos2
cos
8
|| IIWrU av ≅
==
Power density
Radiation Intensity
32
Half-wavelength Dipole (2)
435.2)2()2ln()2( ≅−+= πππ iin CCC
5.4273 jZ in +=
Radiated Power
Radiation Resistance
)2(8
||cos1
4
||
sin
cos2
cos
4
||
2
02
0
2
0
0
2
2
0
ππ
ηπ
η
θθ
θπ
πη
π
π
in
rad
CI
dyy
yI
dI
P
∫
∫
=
−=
=
643.1435.2
44 max
0 ≅==radP
UD π
73)2(4||
22
0
≅== ππη
inrad
r CI
PR
Input Impedance
Directivity
where
(2)
33
Wire antennas near or on infinite
perfect conductor
34
Image Theory
PMCon tanˆ0 0H =×→= nH
NOTE: The fields
obtained are valid
only in the top
half-plane.
PECon tanˆ0 0E =×→= nE
35
Vertical Electric Dipole
1
1
0 sin4
1
θπ
ηθr
lekIjE
jkrd
−
=
2
2
0
2
2
0
sin4
sin4
2
2
θπ
η
θπ
ηθ
r
lekIj
r
lekIjRE
jkr
jkr
v
r
−
−
=
=Direct Component
Reflected Component
36
Vertical Electric Dipole (2)
rrr ≅≅ 21
θθ cos]cos2[ 222
1 hrrhhrr −≅−+=
In far field:
θθ cos]cos2[ 222
2 hrrhhrr +≅++=Phase term
Amplitude term
Parallel ray
approximation
37
Vertical Electric Dipole (3)
00
0)]coscos(2[sin4
0
<=
≥≅−
z
zkhr
lekIjE
jkr
θθπ
ηθTotal Electric Field
12
lobes ofnumber +≅λh
38
Vertical Electric Dipole (4)
)cos(cossin2
ˆ||2
1ˆ)Re(
2
1 22
2
0
2
2* θθλ
ηη θ kh
lI
rrErav ==×= HEW
)cos(cossin2
||2
22
2
022
θθλ
ηη θ kh
lIE
rU ==
2
0max
2 λη lI
U =
+−=
== ∫∫ ∫
32
2
0
2/
0
2
0
2/
0
)2(
)2sin(
)2(
)2cos(
3
1
sin2sin
kh
kh
kh
khlI
dUddUPrad
λπη
θθπφθθππ π
Power Density
Radiation Intensity
Maximum Radiation Intensity
Radiated Power
(3)
39
Vertical Electric Dipole (5)1
32
max0
)2(
)2sin(
)2(
)2cos(
3
124
−
+−==
kh
kh
kh
kh
P
UD
rad
π
+−
==32
2
2
0 )2(
)2sin(
)2(
)2cos(
3
12
||
2
kh
kh
kh
khl
I
PR rad
r λπη
Directivity
Radiation Resistance
40
λλλλ/4 Monopole
25.215.36
)5.4273(2
1
(dipole)2
1
j
j
ZZ inin
+=
+=
=
41
Fields due to y-directed dipole
r
leIy
jkr
πµ
4ˆ 0
−
=A
φθπ
µφθθ sincos
4sincos 0
r
leIAA
jkr
y
−
==
φπ
µφφ cos
4cos 0
r
leIAA
jkr
y
−
==
φπ
µωω
φθπ
µωω
φφ
θθ
cos4
sincos4
0
0
r
leIjAjE
r
leIjAjE
jkr
jkr
−
−
−=−≅
−=−≅
Vector Potential A
Recall that )()( Aj
Ajrrrr
⋅∇∇−−=ωµε
ωE
Far-field Electric Field
42
Fields due to y-directed dipole (2)
ψψζ
ψζψζψψψζψζψ
cosˆsinˆˆ
sinˆsincosˆcoscosˆˆ
cosˆsinsinˆcossinˆˆ
xz
yxz
yxzr
+−=
−+=
++=
ψπ
µωψ sin
4
0
r
leIjE
jkr−
≅
then
Far-field Electric Field
Introduce a “new” spherical coordinate system (r,ψ,ζ) such that
ψπη
ωµζ sin
4
0
r
leIjH
jkr−
≅Far-field Magnetic
Field
(This can be obtained by letting (x,y,z)->(z,x,y) and (ψ,ζ)->(θ,φ))
43
Horizontal Electric Dipole
ψπ
ηψ sin4 1
01
r
lekIjE
jkrd
−
=ψ
πη
ψπ
ηψ
sin4
sin4
2
0
2
0
2
2
r
lekIj
r
lekIjRE
jkr
jkr
h
r
−
−
−=
=Direct Component
Reflected Component
44
Horizontal Electric Dipole (2)
rrr ≅≅ 21
θθ cos]cos2[ 222
1 hrrhhrr −≅−+=θθ cos]cos2[ 222
2 hrrhhrr +≅++=
In far field:
Phase term
Amplitude term
φθψ sinsinˆˆcos =⋅= ry
45
Horizontal Electric Dipole (3)
)]cossin(2[sinsin14
220 θφθπ
ηψ khjr
lekIjE
jkr
−≅−
Total Electric Field
λh2
lobes ofnumber ≅
46
Horizontal Electric Dipole (4)
)cos(sin)sinsin1(2
ˆ 222
2
0
2θφθ
λη
khlI
rrav −=W
)cos(cos)sinsin1(2
||2
222
2
022
θφθλ
ηη θ kh
lIE
rU −==
+−−=
== ∫∫ ∫
32
2
0
2/
0
2
0
2/
0
)2(
)2sin(
)2(
)2cos(
2
)2sin(
3
2
2
sin2sin
kh
kh
kh
kh
kh
khlI
dUddUPrad
λπ
η
θθπφθθππ π
Power Density
Radiation Intensity
Radiated Power
+−−
==32
2
2
0 )2(
)2sin(
)2(
)2cos(
2
)2sin(
3
2
||
2
kh
kh
kh
kh
kh
khl
I
PR rad
r λπη
Radiation Resistance
47
Horizontal Electric Dipole (5)
>
≤
=
2/)(
4
2/)(
)(sin4 2
0
π
π
khkhR
khkhR
kh
D
223220
15
322
15
8
3
2
3
2
=
+−
=→
λλπ
ηλπ
λπη
hlhlR
kh
r
Directivity
For small kh
Maximum Radiation Intensity
==>
=≤
=
1)cossin( and 0,2/2
0,2/)(sin2
max
2
0
2
2
0
max
θφπλ
η
θπλ
η
khkhlI
khkhlI
U
48
Horizontal Electric Dipole (6)
+−−=
32 )2(
)2sin(
)2(
)2cos(
2
)2sin(
3
2)(
kh
kh
kh
kh
kh
khkhR
21
220
0
)sin(5.7)(
15
8
3
2
3
2)(sin4
=
+−−→
=kh
khkhkhD
khFor small kh
where
top related