Review of wave equations

Non-homogeneous helmholt’s wave equation for retarded

electric scalar potential
v
 V k V 
2 2


Non-homogeneous Helmholt’s wave equation for retarded
magnetic vector potential

 A  k A   J
2 2

Solution of V and A
~
~ Q ~
V (r )  e  j r
~  J  jr
4r A(r )   e dv
4 r
~ 1 ~  jr
V (r )   e dv
4 r
 Hertzian Dipole Antenna

A very small filament of current carrying element which produce fields

in a dielectric medium. ~ dq~
Current in phasor form I   jq~
dt
 q~ P
Replacing Ĵdv by Ĩdzâz we can express the magnetic
~  r
l I vector potential as
~
~  I  j r
 q~ Az   e dz (1)
4 c r
If over the short dipole the current is same and r>>l , we can write the
eq.1 as ~  ~  jr
Az  I le
In time domain form 4r
~ 
Az  I 0l cost  r  Where I(t)=I0cost
4r
Wave propagating away from the dipole in r-direction with a phase
constant , phase velocity of up=/

à in spherical co-ordinate system

âz sinâ
Ã=âzAz

cosâr
 ~  jr  
 I le cosar  sin  a 
4r H    A
~ 1

 1  aˆ r aˆ r aˆ r sin 
H   A
 
 A 
1   
r sin  r
2
 
Ar rA (r sin  ) A
~
~ j I l  1 
H 1   sin e aˆ
 j r

4r  j r 
To find electric field
~ 1 ~
E  H aˆ r aˆ r aˆ r sin 
j
 1   
 H  2
r sin  r  
Hr rH  (r sin  ) H 

~ ~
~ I l  1  jI l  1 1 
E  1   cos e  j r
a
ˆ   1    sin e  j r
aˆ
2r  j r  4r  j r  r 
2 r 2 2

 is the intrinsic impedance of the dielectric medium.

Near zone fields

r<<1, thus the term 12 , 13 are predominant and e-jr 1

r r

Therefore
~
~ Il
H sin aˆ
4r 2

~
~ I l  1 
E 1  2 cosaˆ r  sin aˆ 
4r 
2
j r 

Replacing Ĩ=jq and / by 1/, and (1+1/jr )can be

approximated as 1/ jr the E field expression can be written as

~ q~l  2 cos sin  

E  aˆ r  3 aˆ 
4  r 3
r 
 1/r2 is referred as induction term.
Power density in near zone field

 1  *
S  EH
2
2 2
Il 
 -j sin 2
 a
32 r 
2 5 r

Purely reactive in other words the average power in the near zone
field is zero. –j indicates that the near zone region behaves like a
capacitor.
 Far Radiation fields

P is far away , βr>>1, 1/r term will be dominant,

~
~ jI  l  j r
H sin e aˆ
4r
~
~ j I l  j r
E sin e aˆ
4r
Far field propagates in the radial direction , transverse component ,
E and H are perpendicular to each other.
E
 377
H
Normalized radiation pattern

E  ,  
E  ,  n   sin 
E  ,  max
=0

=900

=1800
 1
S  EH 
2
*

2

1 2
E â r 
2 2
I l
   sin  â r
2 2

32 r
2 2

Real power and is directly radiated radially outward. It represent

average power/area.

Normalized power pattern f  ,    sin 

2

=0

=900
Pn
=1800
 
Total power passing through Prad   s .ds
I l sin  2 2
 
2 2 2
Prad    r sin dd
32 r
2 2

I l   3
2 2 2 2
Prad   sin d  d
32 0
2
0
 2
Prad   l I
2 2 2   120  

12 
2 2
 l 2 2 l 
Prad     I  40   I 2
3    
 Radiation resistance
Far region can be simulated by a resistance called radiation resistance
1 2
Prad  I Rrad
2
2
l 2 1 2
2
40   I  I Rrad
  2
2
l 2
Rrad  80  
 
 Directivity gain and Directivity
Directive gain as the ratio of power density radiated by the dipole to
the average power density
2
4r s
G  1.5Sin 2
Prad
When G is maximum is called directivity. D=1.5 =1.76dB
An electric dipole of length 50 cm is situated in free space
Imax=25 A,f=10 MHz. (i) Find E and H in the far zone. (ii) Pavg
and (ii) Rr
=2f=6.283x107, β=0.209, I=25 A, L=0.5 m

~ j 0.208  j 0.209 r
H sin e aˆ
r
~ j 78.416  j 0.209 r
E sin e aˆ
r
2 2
I l
   sin  â r
2 2

32 r
2 2
Average power density in the radial direction is

8.15 2
S r  2 sin 
r
Total power crossing a spherical surface at r
2
2
l 2 2
Prad  40   I  68.25 W/m
 
Radiation resistance
2
l 2
Rrad  80    0.22
 
 A half wave dipole antenna
R P
/4
r Assuming current distribution on the antenna is
Iˆ  I 0 cos z

-/4
At a point P(r,,) from the antenna , the magnetic vector potential is

~  4 cos  z
 j R
A( z )  I0  e dz
4  R

4
P is far away from antenna R=r-zcos
j z  j z
e e
cos z 
2
P is far away from antenna, R=r-zcos to account for the phase
shift and Rr for the distance in the denominator


j z  j z
~  4 e  e  j ( r  z cos )
A( z )  I0  e dz
4  2r

4




8r
4
 
I 0e  jr  e jz  e  jz e jz cos dz


4
 After integration we will get
  
 cos cos  
~   j r  2 
Az  I 0e  
2r  sin 
2

 
Magnetic field intensity in far field region
  
 cos cos  
~ j ~ j  j r  2  aˆ
H sin Az aˆ  I 0e   
 2r  sin  
 
 The corresponding electric field intensity is
  
 cos cos  
~ j  j r  2  aˆ
E I 0 e   
2r  sin  
 
Average power per unit area radiated by the half-wave antenna is

1
S  EH 
2
*
 1 2
2
E â r 
 2 
 cos  cos  
1 2  2  aˆ
 2 2 I 0  2  r
8 r  sin  
 
Total power radiated by the half wave dipole antenna

 2 
2  cos  cos  
I 0   2  2
Prad  2 0  2  d 0 d
8  sin  
 

After numerical integration , the value we obtained 1.219

1.219 2
Prad  I 0
4
Radiation resistance of the half wave dipole antenna
1 2
Prad  I 0 Rrad
2
1 2 1.219 2
I 0 Rrad  I 0
2 4
1.219
Rrad    73.14
2
High value of radiation resistance makes a half-wave
dipole antenna very effective for radiating considerable
amounts of power.
P2. The amplitude of the electric field intensity broadside to a halfwave
dipole antenna at a distance of 15 km is 0.1 V/m in free space. If the
operating frequency is 100 MHz, determine the length of the antenna and
the total power that it radiates. Also writes the general expression for the
electric and magnetic field intensities in the time domain.
Hence the length of
f=100 MHz, =2f=6.28319x108
rad/s the half wave dipole is
β=/c=2.094, =3m 1.5 m.
=900 and r=15 km, the maximum amplitude of the current is
2r
I0  E  25 A
0
Radiation resistance of the halfwave dipole antenna is 73.14 , thus
the power radiated by the antenna is
1 2
Prad  I 0 Rrad  22.86kW
2
Prob-1

Prob-2