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Prof. David R. Jackson Dept. of ECE Fall 2013 Notes 26 ECE 6340 Intermediate EM Waves 1

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Equivalence Principle Basic idea: We can replace the actual sources in a region by equivalent sources at the boundary of a closed surface. Keep original fields E, H outside S. Put zero fields (and no sources) inside S. E, HE, H a b ANT S E, HE, H 2

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Equivalence Principle (cont.) Note: ( E a, H a ) and ( E b, H b ) both satisfy Maxwell’s equations. E b = 0 H b = 0 No sources E a = E H a = H 3

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The B.C.’s on S are violated. Introduce equivalent sources on the boundary to make B.C.’s valid. a b Equivalence Principle (cont.) 4 Zero fields Original fields

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Outside S, these sources radiate the same fields as the original antenna, and produce zero fields inside S. This is justified by the uniqueness theorem: Hence Equivalent sources: Zero fields Zero sources S Equivalence Principle (cont.) Maxwell's equations are satisfied along with boundary conditions at the interface. 5

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Note about materials: If there are zero fields throughout a region, it doesn’t matter what material is placed there (or removed). Equivalence Principle (cont.) This object can be added Zero fields (E, H) S 6

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Scattering by a PEC (E, H) Source PEC S 7

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Scattering by a PEC (cont.) S Source (E, H) (0,0)(0,0) Equivalent Sources: (E, H) Source PEC S We put zero fields and sources inside of S, and remove the PEC object. 8

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Conclusion: The conductor can be removed. Equivalent problem: Source (E, H) (0,0)(0,0) S Source PEC S Original problem: Scattering by a PEC (cont.) 9

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Source (E, H) (0,0)(0,0) This integral equation has to be solved numerically. Integral equation for the unknown current so “Electric Field Integral Equation” (EFIE) Note: The bracket notation means “field due to a source.” Scattering by a PEC (cont.) 10

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Scattering by Dielectric Body (E, H) Source (E, H) Note: The body is assumed to be homogeneous. 11

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Exterior Equivalence Replace body by free space (The material doesn’t matter in the zero-field region.) S Source E a = E H a = H E b = 0 H b = 0 S Source (E, H) (0,0)(0,0) 12

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Summary for Exterior S Original problem: Free-space problem: S (0,0)(0,0) (E, H) 13

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Interior Equivalence S No sources or fields outside S H a, E a E b = 0 H b = 0 Place dielectric material in the "dead region" (region b ). E a = E H a = H 14

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Interior Equivalence (cont.) S (0,0)(0,0) E, H When we calculate the fields from these currents, we let them radiate in an infinite dielectric medium. 15

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Summary for Interior S Original problem: S (0,0)(0,0) E, H Homogeneous- medium problem: 16

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Integral Equation S Boundary conditions: Hence: The “–” means calculate the fields just inside the surface, assuming an infinite dielectric region. Note: The “+” means calculate the fields just outside the surface, radiated by the sources in free space. “PMCHWT" Integral Equation* * Poggio-Miller-Chang-Harrington-Wu-Tsai 17 Recall:

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Fields in a Half Space Equivalent sources: (0,0)(0,0) (E, H) Sources, structures, etc. z (E, H) Region of interest ( z > 0) Equivalence surface S (closed at infinity in the z < 0 region) 18

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Hence, we have: Put PEC in zero-field region: The electric surface current on the PEC does not radiate. Fields in a Half Space (cont.) Note: The fields are only correct for z > z (E, H) PEC (E, H) PEC z

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Note: The fields are still correct for z > 0. Now use image theory: Fields in a Half Space (cont.) This is useful whenever the electric field on the z = 0 plane is known. 20 Incorrect fields (E, H) Correct fields z

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Fields in a Half Space: Summary Sources + objects z (E, H) Region of interest ( z > 0) 21 Incorrect fields (E, H) Correct fields z

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Alternative (better when H is known on the interface): M s e does not radiate on PMC, and is therefore not included. Image theory: Incorrect fields (E, H) Correct fields Fields in a Half Space (cont.) 22 PMC z

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Example: Radiation from Waveguide z b y b a x y 23

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Example (cont.) Step #1 Step #2 PEC z z 24 The feeding waveguide was removed from the dead region that was created, and then a continuous PEC plane was introduced. Image theory is applied to remove the ground plane and double the magnetic surface current. Apply equivalence principle Apply image theory

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x a b y z Example (cont.) 25 A three-dimensional view

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a b x y z Solve for the far field of this problem first. (This is the “A” problem in the notation of the duality notes.) Then use: Example (cont.) Use the theory of Notes 22 to find the far field from this rectangular strip of electric surface current. 26 Step #3 Apply duality

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A radiating electric current can be replaced by a magnetic current, and vice versa. Volume Equivalence Principle 27 y x z y x z P. E. Mayes, “The equivalence of electric and magnetic sources,” IEEE Trans. Antennas Propag., vol. 6, pp. 295–29, We wish to have the same set of radiated fields.

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Volume Equivalence Principle (Cont.) 28 Set 1 (electric current source): Hence Therefore, we have

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Volume Equivalence Principle (Cont.) 29 Set 2 (magnetic current source): Hence Therefore, we have

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Volume Equivalence Principle (Cont.) 30 Set Compare: Hence

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Volume Equivalence Principle (Cont.) 31 Hence, the two electric fields are equal everywhere, but the magnetic fields are only the same outside the source region. Next, examine the difference in the two Faraday laws: so This gives us

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Summary 32 y x z y x z Volume Equivalence Principle (Cont.)

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Apply duality on the two curl equations to get two new equations: 33 y x z y x z Volume Equivalence Principle (Cont.)

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34 Similarly, from duality we have Volume Equivalence Principle (Cont.)

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35 y x z y x z Summary Volume Equivalence Principle (Cont.)

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