Solving Problems with Line Sources: Addition Theorem Insights
The addition theorem allows us to shift the origin when dealing with line sources outside a cylinder, providing a useful tool for solving such problems. Explore the detailed solutions and derivations presented in the notes by Prof. David R. Jackson for ECE 6341 in Spring 2016.
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ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 14 1
Addition Theorem Allows us to shift the origin for line sources. This is useful for solving problems with a line source outside of a cylinder. y , ) ( 0I x = zA Denote 2
Addition Theorem (cont.) I ( )( 0 ) 2 = 0 H kR = R 4 j ( ) = , x y y ( ) = , x y R x This is the exact solution, but it is not in a useful form. 3
Addition Theorem (cont.) Assume for simplicity 0 = Tube of current model y y 2 2 1 1 ( ,0) 0I ( 0I ,0) x x Surface-current: ( ) = ( ) J A 1 sz 4
Addition Theorem (cont.) + ( ) d = Determine A1: J I 0 sz ( ) = ( ) J A 1 sz = A I so 1 0 I ( ) = ( ) J 0 Hence sz Using the tube of current concept, + = ( ) jn J c e sz n = n 5
Addition Theorem (cont.) 2 1 = ( ) jn c J e d n sz 2 0 2 1 I = jn ( ) 0 e d 2 0 I 1 1 I = 0 = c 0 2 n 2 y The tube of current representation 2 1 x 6
Addition Theorem (cont.) + = jn ( ) a e J k 1 n n = + n = ( ) 2 n jn ( ) b e H k 2 n = n From previous tube of current derivation: ( ) a ( ) = (2) a j c H k n n n 1 I 2 = c 0 n 2 ( ) = b j c J k n n n 2 7
Addition Theorem (cont.) Therefore, we have I ( ) = (2) 0 a j H k n n 4 I ( ) = 0 b j J k n n 4 Hence + I ( ) ( ) = (2) n jn 0 j H k e J k 1 n 4 = + n I ( ) = (2) n jn ( ) 0 j J k e H k 2 n 4 = n 8
Addition Theorem (cont.) For replace 0 + I ( ) ( ) ( ) jn = (2) n 0 j H k e J k 1 n 4 = + n I ( ) ( ) jn = (2) n ( ) 0 j J k e H k 2 n 4 = n y y 0I 2 2 ( ,0) , ) ( 1 1 x x 0I 9
Addition Theorem (cont.) Comparing with the original expression for the potential of the line current, we have + ( ) ( ) (2) n ( ) jn H k J k e n = + n = (2) 0 ( ) H k ( ) ( ) (2) n ( ) jn J k H k e n = n This is the Graf addition theorem. It also holds if k is replaced with k . Johann Heinrich Graf (1852-1918) 10
Addition Theorem (cont.) + ( ) ( ) (2) n ( ) jn H k J k e n = + n = (2) 0 ( ) H k ( ) ( ) (2) n ( ) jn J k H k e n = n y ( , ) 0I x 11
Scattering From a Line Current z a 0I y , ) ( x = zA TMz: 12
Scattering From a Line Current (cont.) + I ( ) ( ) = (2) n ( ) i jn H k J k e 0 n 4 j = n (This is only valid for ) Assume + I ( ) = (2) n ( ) s jn a H k e 0 n 4 j = n a (This is only valid for ) 13
Scattering From a Line Current (cont.) Boundary Conditions ( = a): ( , , ) a = ( , , ) a s i z z Hence = (2) n (2) n ( ) ( ) ( ) a H ka H k J ka n n or ( ) 2 n ( ) H H k ka = ( ) a J ka ( ) 2 n n n ( ) 14
Scattering From a Line Current (cont.) Final result: ( ) 2 n + ( ) ) I H H k ( )( n ) 2 = ( ) s jn ( ) J ka H k e 0 n (2) 4 ( j ka = n n z a 0I Scattered field y , ) ( x 15
