Explanation of the infinitesimal dipole element used in wire antennas, calculation of radiated fields, radiation patterns, directivity, and far-field patterns. Includes illustrations and equations for a comprehensive understanding.

• Antenna Patterns
• Infinitesimal Dipole
• Directivity
• Radiation Patterns
• Electromagnetics

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Understanding Antenna Patterns and Directivity of Infinitesimal Dipole Elements

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1. ECE 3317 Applied Electromagnetic Waves Prof. David R. Jackson Fall 2023 Notes 22 Antenna Patterns z 30 30 z = 0 60 60 I 0 dB -10 dB -20 dB m l y -30 dB x 120 120 150 150 1

2. Infinitesimal Dipole The infinitesimal dipole current element is shown below. z I l y x The dipole moment (amplitude) is defined as Il. The infinitesimal dipole is the foundation for many practical wire antennas. From Maxwell s equations we can calculate the fields radiated by this source (e.g., see Chapter 14 of the Hayt and Buck textbook). 2

3. Infinitesimal Dipole (cont.) The exact fields of the infinitesimal dipole in spherical coordinates are 1 r 1 Il = + jk r 1 cos E e 0 0 r 2 2 jk r 0 1 r 1 1 Il = + + jk r ( ) 1 sin E j e 0 0 2 4 ( ) jk r jk r 0 0 1 r 1 Il = + jk r ( ) 1 sin H jk e 0 0 4 jk r 0 3

4. Infinitesimal Dipole (cont.) In the far field (r ) we have: jk r Il e 0 ( ) , , ( ) sin E r j 0 4 r jk r Il e 0 ( ) , , ( ) sin H r jk 0 4 r Hence, we can identify Il Recall: ( ) = FF , ( )sin E j 0 4 jk r e 0 ( ) ( ) , , = FF , E r E r Il ( ) = FF , ( )sin H jk 0 4 4

5. Infinitesimal Dipole (cont.) The radiation pattern is shown below. Il = FF ( )sin E j 0 4 z ( ) FF , E ( ) 0 = dB , 20log 30 30 ( ) 10 FF , E 45o 0 m 60 60 ( ) ( ) = dB , 20log sin 10 0 dB -9 -6 -3 ( ) ( = o sin 45 1/ 2 ) 3 [dB] = 20log 1/ 2 10 120 120 HPBW = 90o 150 150 5

6. Infinitesimal Dipole (cont.) The directivity of the infinitesimal dipole is now calculated 2 ( ) ( ) ( ) r FF , / 2 E , S ( ) 0 = = r , D ( ) ) ( 2 2 / 4 1 P 2 ( ) ( ) FF 2 , / 2 sin E r d d rad 0 2 4 r 0 0 2 ( ) FF 4 , E = 2 2 ( ) FF , sin E d d Recall: 0 0 Il = FF ( )sin E j 0 4 Hence 4 sin 2 ( ) = , D 2 2 sin sin d d 0 0 6

7. Infinitesimal Dipole (cont.) Evaluating the integrals, we have: 4 sin 2 ( ) = , D 2 2 sin sin d d 0 0 Hence, we have 2 4 sin = 3 2 ( ) D = 2 , sin ( ) 2 2 sin sin d 0 2 2sin = 3 sin d 0 2sin 4/3 2 = 7

8. Infinitesimal Dipole (cont.) Il 3 2 ( ) = D = FF 2 ( )sin E j , sin 0 4 z 30 30 ( ( ) D = = , 1.0 ) o 54.7 60 60 0 dB -9 -6 -3 ( ) 90 D = ( = , 1.5 ) o 120 120 150 150 The far-field pattern is shown, with the directivity labeled at two points. 8

9. Wire Antenna A center-fed wire antenna is shown below. z h ( ) I z Feed y ( ) 0 I = 2 l h = x (0) I I 0 h A good approximation to the current is: I ( ) = ( ) I z sin k h z 0 k h ( ) 0 sin 0 9

10. Wire Antenna (cont.) A sketch of the current is shown below for two cases. I ( ) = ( ) I z sin k h z 0 k h ( ) 0 sin 0 h h = = 2 l h 2 l h ( ) 0 I ( ) 0 I h h Resonant dipole (l = 0 / 2, k0h = /2) Short dipole (l << 0) Use ( ) sin x x z h z h ( ) = = ( ) I z cos cos I k z I ( ) I z 1 I 0 0 0 0 2 10

11. Wire Antenna (cont.) Short Dipole The average value of the current is I0 / 2. 1 2 ( ) Il ( ) = h I l 0 eff = 2 l h Il = FF ( )sin E j 0 ( ) 0 4 I Infinitesimal dipole: Il = FF ( )sin H jk 0 4 h ( ) 4 Il Il = FF ( )sin eff E j 0 Short dipole (l << 0 / 2) Short dipole: ( ) 4 z h = FF ( )sin eff H jk ( ) I z 1 I 0 0 11

12. Wire Antenna (cont.) For an arbitrary length dipole wire antenna, we need to consider the radiation by each differential piece of the current. z Far-field observation point ( r = ) , , x y z h ( ) I z R 1 2 ( ) 2 dz r = + + 2 2 R x y z z z Feed y Infinitesimal dipole: jk r e Il 0 = ( )sin E j x 0 4 r h h jk R 1( 4 e 0 ( ) I z dz = )sin E j Wire antenna: 0 R h 12

13. Wire Antenna (cont.) z Far-field observation point ( ) = , , x y z r h ( ) I z R dz r z Feed y ( ) 2 = + + 2 2 R x y z z ( ) ( + ) ( ) = + + 2 2 2 2 2 x y z z zz x = + 2 2 2 r z zz h 2 z r zz r = + 1 2 r 2 2 z r z r z r = + 1 2 r 13

14. Wire Antenna (cont.) z Far-field observation point ( ) h = , , x y z r ( ) I z R z dz r Feed y 2 z r z r z r = cos / z r = + 1 2 R r x h 2 z r z r z r z r ( ) ( ) ( ) = + 1 2 cos = 1 2 cos 1 cos cos R r r r r z x + + 1 1 , 1 Note: x x 2 14

15. Wire Antenna (cont.) z Far-field observation point ( ) h = , , x y z r ( ) I z R z dz r Feed y = cos / z r x cos R r z h It can be shown that this approximation is accurate when 2 2D ( ) r = 2 D h 0 15

16. Wire Antenna (cont.) z Far-field observation point ( ) h = , , x y z r ( ) I z R z h jk R 1( 4 e 0 ( ) I z dz dz r = )sin E j 0 R Feed h y = cos / z r Hence, we have: ( ) x r z h cos jk 1( 4 e r 0 ( ) I z dz h )sin E j 0 cos z h + h cos jk r jk z 1( 4 e e 0 0 ( ) I z dz = )sin j ( ) 0 1 / cos r z r h h j k r 1( 4 e 0 ( ) I z dz + co s jk z )sin j e 0 0 r h 16

17. Wire Antenna (cont.) We define the space factor of the wire antenna: h ( ) ( ) I z 0 cos jk z + SF e dz h We then have the following result for the far-field pattern of the wire antenna: jk r 1( 4 e 0 ( ) )sin SF E j 0 r Note: The term in front of the array factor in the above equation is the far-field pattern of the unit-amplitude infinitesimal dipole. 17

18. Wire Antenna (cont.) Using our assumed approximate current function, we have: I ( ) = ( ) I z sin k h z 0 k h ( ) 0 sin 0 Hence h I ( ) ( ) 0 cos jk z + = SF sin 0 k h k h z e dz ( ) 0 sin 0 h The result is (derivation omitted): cos( cos ) cos( sin k ) I k h k h ( ) = SF 2 0 k h 0 0 ( ) 2 sin 0 0 18

19. Wire Antenna (cont.) In summary, we have: jk r 1( 4 e 0 ( ) )sin SF E j 0 r cos( cos ) cos( sin k ) I k h k h ( ) = SF 2 0 k h 0 0 ( ) 2 sin 0 0 Thus, we have: jk r 1 cos( cos ) k h cos( ) e I k h k h 0 ( ) E j h 0 k h 0 0 ( ) 0 2 sin sin r 0 0 19

20. Wire Antenna (cont.) For a resonant half-wave dipole antenna: = = / 4 h k h 0 / 2 0 cos cos jk r 1 e 2 0 ( ) I ( ) E j h ( ) 0 0 2 / 2 sin r or cos cos jk r e j 2 0 ( ) E I h 0 0 2 sin r 20

21. Wire Antenna (cont.) cos cos jk r = = e j / 4 h k h 2 0 ( ) 0 E I h 0 0 2 sin r / 2 0 The directivity is: 2 2 ( ) ( ) FF FF 4 , 4 , E E ( ) = = , D 2 2 2 ( ) ( ) FF FF , sin 2 , sin E d d E d d 0 0 0 The result (from numerical calculations) is: ( ) D = /2, 1.643 21

22. Wire Antenna (cont.) Results = 2 l h 22

23. Wire Antenna (cont.) Radiated Power: 2 1 2 ( ) = FF , sin P E d d rad 2 0 0 0 2 2 cos( cos ) cos( sin k h ) 1 1( 2 I k h k h = ) sin 0 k h 0 0 j h d d ( ) 0 2 sin 0 0 0 0 0 = = = 0 0 0 Simplify using 0 k 0 0 0 0 2 2 1 1 cos( cos ) cos( sin ) I k h k h = j ( ) sin P d d 0 k h 0 0 ( ) 0 rad 2 2 sin 0 0 0 0 23

24. Wire Antenna (cont.) Performing the integral gives us a factor of 2 : 2 1 1 cos( cos ) cos( sin ) I k h k h ( ) = j 2 ( ) sin P d 0 k h 0 0 ( ) 0 rad 2 2 sin 0 0 0 After simplifying, the result is then 2 2 cos( cos ) cos( sin ) I k h k h = sin P d 0 0 k h 0 0 ( ) rad 4 sin 0 0 24

25. Wire Antenna (cont.) 1 2 2 = P R I The radiation resistance is defined from 0 rad rad z 2 P I = R rad h rad 2 = (0) I I 0 0 ( ) I z Circuit Model Feed 0I y ( ) 0 I Z Z = 2 0 l h in = + Z R jX x in in in h = R R in rad For a resonant antenna (l 0 / 2), Xin = 0. 25

26. Wire Antenna (cont.) The radiation resistance is now evaluated. 2 P I = R rad rad 2 0 Using the previous formula for Prad, we have: 2 2 cos( cos ) sin cos( ) k h k h 1 k h = 0 0 R d 0 rad 2 sin( ) 0 0 73 R Resonant 0 /2 Dipole: = = , h k h 0 rad 0 4 2 26

27. Wire Antenna (cont.) The result can be extended to the case of a monopole antenna ( ) I z 0/4 h h Feeding coax 1 2 = monopole in dipole in Z Z (see the next slide) 36.5 R rad 27

28. Wire Antenna (cont.) 1 2 = monopole in dipole in Z Z This can be justified as shown. monopole V 0I dipole V +- 0I + Monopole - Virtual ground E = 1 2 zE = monopole dipole z V V = monopole dipole I I 1 2 Dipole = monopole in dipole in Z Z 28