Prof. David R. Jackson Dept. of ECE Fall 2013 Notes 22 ECE 6340 Intermediate EM Waves 1

Radiation Infinitesimal Dipole Current model: y x l z I y x z 2

Radiation (cont.) Define Then As Letting or 3

Maxwell’s Equations: From (3): (5) Note:  c accounts for conductivity. Radiation (cont.) Note: The Harrington book uses 4

From (1) and (5): or Hence (6) Radiation (cont.) This is the “mixed potential” form for E. 5 or

Substitute (5) and (6) into (2): Use the vector Laplacian identity: The vector Laplacian has his nice property in rectangular coordinates: Radiation (cont.) 6

Hence or Then we have Lorenz Gauge: Choose Radiation (cont.) Note: Lorenz, not Lorentz (as in Lorentz force law)! 7

Take the z component: Hence we have Note: so Radiation (cont.) 8

Spherical shell model of 3D delta function: Radiation (cont.) y x a z SaSa The A z field should be symmetric (a function of r only) since the source is. In spherical coordinates: 9

Assume For so that Next, let we have that Radiation (cont.) 10

We choose exp (-jkr) for r > a to satisfy the radiation condition at infinity. Solution: Radiation (cont.) We choose sin(kr) for r < a to ensure that the potential is finite at the origin. 11

Radiation (cont.) Hence The R function satisfies 12

Also, from the differential equation we have Radiation (cont.) We require 13 (BC #1) (BC #2)

Hence Radiation (cont.) Recall 14 (BC #2) (BC #1)

Substituting Ae -jka from the first equation into the second, we have Radiation (cont.) Hence we have From the first equation we then have 15

Letting a  0, we have Radiation (cont.) Hence 16 so

Final result: Radiation (cont.) y l z I x Note: It is the moment Il that is important. 17

Extensions (a) Dipole not at the origin y x z 18

Extensions (cont.) (b) Dipole not in the z direction x y z 19

(c) Volume current y x z Extensions (cont.) dl dS 20 Hence Consider the z component of the current: The same result is obtained for the current components in the x and y directions, so this is a general result.

(d) Volume, surface, and filamental currents Extensions (cont.) 21 Note: The current I i is the current that flows in the direction of the unit vector.

Extensions (cont.) 22 Note: The current I i is the current that flows in the direction of the contour. Filament of current: C A B I

Fields To find the fields: Note: Ampere’s law (the equation above) for finding the electric field is an alternative to using the following equation: 23

Fields (cont.) Results for infinitesimal, z -directed dipole at the origin: with 24

Note on Dissipated Power Note: If the medium is lossy, there will be an infinite amount of power dissipated by the infinitesimal dipole. Note: The N superscript denotes that the  dependence is suppressed in the fields. (power dissipation inside of a small spherical region V  of radius  ) 25

Far Field Far-field: Note that The far-zone field acts like a homogeneous plane wave. 26

Radiated Power (lossless case) The radius of the sphere is chosen as infinite in order to simplify the fields. 27 We assume lossless media here.

Radiated Power (cont.) or Hence Integral result: 28

Use Simplify using Radiated Power (cont.) Then or 29

Note: The dipole radiation becomes significant when the dipole length is significant relative to a wavelength. Radiated Power (cont.) 30

Far-Field Radiation From Arbitrary Current y x z 31

Far- Field (cont.) Use Taylor series expansion: x 32

As Far- Field (cont.) 33

Physical interpretation: Far- Field (cont.) Far-field point (at infinity) y x z 34

Hence Define: Far- Field (cont.) (This is the k vector for a plane wave propagating towards the observation point.) and 35

The we have where Far- Field (cont.) and 36 with

Fourier Transform Interpretation Then Denote The array factor is the 3D Fourier transform of the current. 37

Magnetic Field We next calculate the fields far away from the source, starting with the magnetic field. 38 In the far field the spherical wave acts as a plane wave, and hence Recall Hence We then have Note: Please see the Appendix for an alternative calculation of the far field.

The far-zone electric field is then given by Electric Field We start with 39

Electric Field (cont.) We next simplify the expression for the far-zone electric field. 40 Recall that Hence

Hence, we have Electric Field (cont.) Note: The exact field is 41 The exact electric field has all three components (r, ,  ) in general, but in the far field there are only ( ,  ) components.

Relation Between E and H Hence or Start with 42 Recall: Also

Summary of Far Field 43 (keep only  and  components of A )

Poynting Vector 44

Assuming a lossless media, Poynting Vector (cont.) In a lossless medium, the Poynting vector is purely real (no imaginary power flow in the far field). 45

Far-Field Criterion Antenna (in free-space) D : Maximum diameter How large does r have to be to ensure an accurate far-field approximation? z The antenna is enclosed by a circumscribing sphere of diameter D. y x 46

Far-Field Criterion (cont.) Let  = maximum phase error in the exponential term inside the integrand. Hence Neglect Error 47

Set Then Far-Field Criterion (cont.) 48

Hence, we have the restriction Far-Field Criterion (cont.) for 49 “Fraunhofer criterion”

Note on choice of origin: Far-Field Criterion (cont.) The diameter D can be minimized by a judicious choice of the origin. This corresponds to selecting the best possible mounting point when mounting an antenna on a rotating measurement platform. 50 Mounted at center Mounted at edge D = diameter of circumscribing sphere D a = diameter of antenna

Far-Field Criterion (cont.) 51 It is best to mount the antenna at the center of the rotating platform. Platform (ground plane) Antenna Rotating pedestal

Wire Antenna Assume y +h z I (z') x -h +h -h Wire antenna in free space 52

Wire Antenna (cont.) Hence 53

Wire Antenna (cont.) Recall that For the wire antenna we have 54 The result is

Hence or Simplify using Wire Antenna (cont.) 55

We then have Wire Antenna (cont.) Note: The pattern goes to zero at  = 0, . (You can verify this by using L’Hôpital’s rule.) 56

Wire Antenna: Radiated Power 57

Radiated Power (cont.) or 58

Input Resistance 59 Z in R in jX in I (0)

 Dipole: Input Resistance (cont.) (resonant) 60

61Example A small loop antenna x y z I a The current is approximately uniform) Assume  = 0: By symmetry: The x component of the array factor cancels for points  and - .

62 Example (cont.) Hence Identity: We then have

63 Approximation of Bessel function: The result is Hence Example (cont.)

Appendix 64 In this appendix we derive the far field expression for the magnetic and electric field from a radiating current source by using the curl operations in spherical coordinates, instead of using the plane-wave approximation for the del operator.

Hence Therefore Magnetic Field (cont.) Note that a is not a function of r, so these derivative terms are O (1). 65

Therefore Magnetic Field (cont.) Hence 66

We then have Magnetic Field (cont.) or Note: The t subscript can be added without changing the result. The t subscript means transverse to r. That is, keep only the  and  components. 67

The far-zone electric field is then given by Electric Field We start with 68

(This follows from the same reasoning as before.) Hence Electric Field (cont.) 69

Electric Field (cont.) We next simplify the expression for the far-zone electric field. 70 Recall that Hence