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

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

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Radiation (cont.) Define Then As Letting or 3

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

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

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

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

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

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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

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

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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

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Radiation (cont.) Hence The R function satisfies 12

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

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

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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

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

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

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

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

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(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.

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(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.

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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

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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

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

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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

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

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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.

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

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

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

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Far-Field Radiation From Arbitrary Current y x z 31

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

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As Far- Field (cont.) 33

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

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

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

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

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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.

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

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

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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.

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

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

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Poynting Vector 44

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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

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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

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

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Set Then Far-Field Criterion (cont.) 48

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

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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

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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

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

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Wire Antenna (cont.) Hence 53

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

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Hence or Simplify using Wire Antenna (cont.) 55

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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

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Wire Antenna: Radiated Power 57

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Radiated Power (cont.) or 58

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

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

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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 - .

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

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

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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.

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

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Therefore Magnetic Field (cont.) Hence 66

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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

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

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

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

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