Infinitesimal Dipole. Outline Maxwell’s equations – Wave equations for A and for Power:...
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Transcript of Infinitesimal Dipole. Outline Maxwell’s equations – Wave equations for A and for Power:...
Infinitesimal Dipole
Outline
• Maxwell’s equations– Wave equations for A and for
• Power: Poynting Vector• Dipole antenna
Maxwell Equations
Jt
DH
vD
0 B
t
BE
Ampere:
Faraday:
Gauss:
Constitution Relation
EJt
J
HB
ED
Vector Magnetic PotentialA
HAB • Applying in Faraday’s Law:
0
t
AE
0
: vectorial
Identidad
t
AE
At
At
E
is the Electric Scalar Potential
Ampere’s Law:J
t
EB
Jt
DH
2
221
t
A
tJAA
t
A
tJAA
21
AAA 2
Jt
AA
tA
2
22
Lorentz’ condition
tA
• Assuming
• The wave equation
Jt
AA
tA
2
22
Jt
AA
2
22
Gauss’ Law
vD
vE
v
t
A
2
v
t
2
22
v
t
A
v
tt
2
Wave equation
• For sinusoidal fields (harmonics):
whereJAkA
JAjA
Jt
AA
22
22
2
22
)(
22 k
Outline
• Maxwell’s equations– Wave equations for A and for
• Power: Poynting Vector• Dipole antenna
Poynting Vector *
2
1HES
*Re2
1HESave
yx EyExE ˆˆ
yx HyHxH ˆˆ
)(2
)(1
ztjy
ztjx
eEE
eEE
)(2
)(1
ztj
y
ztjx
eHH
eHH
*ˆˆˆˆRe2
1yxyxave HyHxEyExS
Average Poynting Vector S yxyxave HyHxEyExS ** ˆˆˆˆRe
2
1
zHEHES xyyxave ˆRe2
1 **
zeHEHES zave ˆcos
2
1 2*12
*21
zeZ
E
Z
ES z
ooave ˆcos
2
1 2
2
2
2
1
zEER
So
ave ˆ2
1 2
2
2
1
In free space:
Outline
• Maxwell’s equations– Wave equations for A and for
• Power: Poynting Vector• Infinitesimal Dipole antenna
Find A from Dipole with current J
• Line charge w/uniform charge density, L
Jt
AA
2
22
x
z
r
r
0
Jz
)()( yxIoo
Az zyxJAkA zzz ,,22
Assume the simplest solution Az(r):
To find….
a
sin
1aa
AA
r
AA r
F
r
F
rr
Fr
rF r
sin
1sin
sin
11 2
2
zz rAdr
d
rrA
2
22 1
)(
AA 2
Assume the simplest solution Az(r):
Homogenous Equation (J=0)
01 2
2
2
zz AkrAdr
d
r
022
2
zz rAkrAdr
d
jkrjkrz ececrA 21
Which has general solution of:
Apply B.C.• If radiated wave travels outwards from the source:
• To find C2, let’s examine what happens near the source. (in that case k tends to 0)
• So the wave equation reduces to
r
ecrA
jkr
z
2)(
r
crAz
2)(
jkrz ecrrA 2)(
)()(2 yxIAA oozz
Now we integrate the volume around the dipole:
• And using the Divergence Theorem
zIr
Arddr
r
Aoo
zz
22 4sin
zIdSA ooz
ddrdS sin2
dxdydzyxIdvA ooz )()(
Comparing both, we get:
42
zIc oo
24 r
zI
dr
dA ooz
jkrooz e
r
zIrA
4)(
22
r
c
dr
dAz
Now from A we can find E & H
• Using the Victoria IDENTITY:
• And
• Substitute:
zrzzz aarAr
aAH ˆˆ)(1
ˆ1
zzzzzz aAaAaA ˆˆˆ
zzaAAH ˆ11
)( GfGffG
sinˆˆˆ aaa zr
ˆsinˆ1
Har
AH z
ˆsinˆ1
Har
AH z
sin
1
4 2jkro e
rr
jklIH
jkrooz e
r
zIrA
4)(
The magnetic filed intensity from the dipole is:
Now the E field:
t
EH
Jt
DH
rr ArA
rrrA
r
A
r
AA
rrA
1ˆsin
1ˆsin
sin
1ˆ
ˆˆ
ˆsin
sin
1ˆ
ErEj
rHrr
HrrH
r
HEj
The electric field from infinitesimal dipole:
cossin21
4sin
1
sinsin
1
2jkro
r
err
jklI
r
Hr
Ej
cos1/
2 32jkro
r erjr
lIE
sin1
4
1
22 jkro e
rr
jkk
r
ljI
rHrr
Ej
sin
1/
4 32jkro e
rjrr
jlIE
General @Far field r> 2D2/
sin
1
4 2jkro e
rr
jklIH
cos1/
2 32jkro
r erjr
lIE
sin
1/
4 32jkro e
rjrr
jlIE
0rE
sin
4jkro e
r
jlIE
sin
4jkro e
r
jklIH
Note that the ratio of E/H is the intrinsic impedance of the medium.
Power :Hertzian Dipole
ˆˆ
2
12
1
**
*
HErHE
HES
r
ddrHErdASAdSP sin2
1ˆ 2*
2
0 0
2322
2
211/
sin32 rr
jk
rjrr
jlIS or
3
222
21
1sin8 krr
lIS or
sin1
4sin
1/
42
12
*
32jkrojkro
r err
jklIe
rjrr
jlIS
dr
krr
lIdP o sin
11sin
82
32
22
2
3
2
13 kr
jlIP o
Radiation Resistance
reactivarado PP
kr
jlIP
3
2
13
radoo
rad RIlI
P2
2
2
1
3
3
2
13
2
kr
jlZ
2
280
lRrad