ECE Spring Prof. David R. Jackson ECE Dept. Notes 1

Transcription

1 ECE 6341 Spring 16 Prof. David R. Jackson ECE Dept. Notes 1 1

2 Fields in a Source-Free Region Sources Source-free homogeneous region ( ε, µ ) ( EH, ) Note: For a lossy region, we replace ε ε c ( / ) εc ε j σ ω ε c jε c ε c εε rc εε ( 1 j tanδ) ( 1 j tanδ) rc E jωµ H H jωε E E H

3 Fields in a Source-Free Region (cont.) (a) (b) E H H 1 µ A E 1 ε F Ampere s law: Faraday s law: E 1 1 jωε µ A H 1 1 jωµ ε F The field can be represented using either A or F 3

4 Fields in a Source-Free Region (cont.) Case (a) Assume we use A: 1 H A µ E jωµ H E jω A ( ) ( E jω A) + E+ jω A Φ Hence E Φ jω A This is the mixed potential form for the electric field. 4

5 Fields in a Source-Free Region (cont.) Next, use H jωε E 1 A jωε Φ jω A µ ( ) ( ) ( ) ( ) A k A j ωεµ Φ A A k A j ωεµ Φ Recall : A ( A) ( A) 5

6 Fields in a Source-Free Region (cont.) ( ) A A k A jωεµ Φ Choose A jωεµ Φ (Loren Gauge) Then A+ k A This is the vector Helmholt equation for the magnetic vector potential. 6

7 Fields in a Source-Free Region (cont.) Case (b) Assume we use F: 1 E F ε H Invoking duality: Ψ jω F Ψmagnetic scalar potential Choose F jωεµ Ψ We then have: F + k F 7

8 Fields in a Source-Free Region (cont.) To be even more general, let a E E + E a H H + H f f where (E a, H a ) and (E f, H f ) each satisfy Maxwell s equations. (This is an arbitrary partition.) The representation is not unique, since there are many ways to split the field. For example, we could use E E a f.1e.9e 8

9 Fields in a Source-Free Region (cont.) We construct the vector potentials so that A ( E, H ) F ( E, H ) a f a f We then have 1 1 E ( A) F jωµε ε 1 1 H A+ F µ jωµε ( ) 9

10 Fields in a Source-Free Region (cont.) Depending on the type of source (outside the source-free region) that is producing the field within the source-free region, it may be more convenient to represent the field using only A or only F. EXAMPLES Electric dipole: choose only A (A is in the same direction as J) Magnetic dipole: choose only F (F is in the same direction as M) This solution was discussed in ECE 634. In principle, one could represent the field of the electric dipole using the F vector potential, but it would be much more difficult than with the A vector potential! 1

11 TE /TM Theorem In a source-free homogeneous region, we can always represent the field using the following form: A F A ˆ ( x, y, ) F ˆ ( x, y, ) ẑ is an arbitrary fixed direction (called here the pilot vector direction). (A proof of this theorem is given later.) This theorem gives us a systematic way to represent the fields in a source-free region, involving only two scalar field components. 11

12 TE /TM Theorem (cont.) Note: A simpler solution often results if we choose the best direction for the pilot vector. EXAMPLE An infinitesimal unit-amplitude electric dipole pointing in the direction: A A ˆ ( x, y, ) The solution is A( xy,, ) e µ 4 π jkr r We only need A and not F. If we had picked the pilot vector direction to be something different, such as x, we would need BOTH A x and F x. 1

13 TE /TM Theorem (cont.) Consider, for example, H a TE / TM property of Fields 1 µ ( A ˆ ) A F 1 a ( A ˆ ˆ A) H µ A ˆ ( x, y, ) F ˆ ( x, y, ) a Hence, H Similarly, E f (A TM ) (F TE ) 13

14 TE /TM Theorem (cont.) The fields are found as follows: ψ A E 1 + jωµε k ψ H x 1 ψ µ y E x 1 ψ jωµε x H y 1 ψ µ x E y 1 ψ jωµε y H Note: There is a factor µ difference with the Harrington text. 14

15 TE /TM Theorem (cont.) ψ F H 1 + jωµε k ψ E x 1 ψ ε y H x 1 ψ jωµε x E y 1 ψ ε x H y 1 ψ jωµε y E Note: There is a factor ε difference with the Harrington text. 15

16 TE /TM Theorem (cont.) From the vector Helmholt equation we have: A+ k A Taking the component gives: ( ) ( ) ˆ A + ˆ k A or Similarly, A + k A F+ kf Hence, any EM problem in a source-free region reduces to solving the scalar Helmholt equation: ψ + k ψ 16

17 Summary of TE /TM Result Source-free homogeneous region Sources ( ε, µ ) ( EH, ) A F Arbitrary pilot direction A F A ˆ ( x, y, ) F ˆ ( x, y, ) ψ + k ψ ψ A or F 17

18 Proof of TE /TM Theorem ( ε, µ ) ( EH, ) S ˆn Source-free region Apply equivalence principle: Keep original material and fields inside S. Put ero fields outside S. Put (ε, µ) outside S. 18

19 Proof of TE /TM Theorem ( ε, µ ) ( EH, ) S e J s ( ε, µ ) ˆn e M s Infinite homogenous space e s e s ( ˆ) ( ˆ) J n H ( ) M n E 19

20 Proof of TE /TM Theorem (cont.) Consider the component of the currents: J M e e A F (from 634) Also, from a horiontal dipole source, we can show that: e xy, + J A F e xy, + M A F This will be established later in the class notes and the homework by solving the problem of a horiontal (x or y directed) dipole source using A and F. Hence, the fields in the source-free region due to all of the equivalent currents may be represented with A and F.

21 Non-Uniqueness of Potentials A and F are not unique. To illustrate, consider: A F ce 1 jk, k ω µε TM field ( ) 1 + jωµε E k A E E x y 1 A jωµε x 1 A jωµε y This set of potentials produces a null field! 1 H E jωµ Hence, adding this set of potentials to a solution does not change the fields. 1