Electromagnetic Scattering by Wedge and Line Source Notes
In these notes from ECE 6341, Spring 2016, Prof. David R. Jackson covers the topic of scattering by a wedge and line source. The discussion includes boundary conditions, Bessel functions, integration techniques, and more, providing a thorough exploration of the electromagnetic phenomena involved.
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ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 18 1
Scattering by Wedge y Line source 0I = ( , ) x Note: = 2 We will generalize to allowing for kz at the end. zA ( , ) Assume TMz: = = 0 , 2 zA at Boundary conditions: 2
Scattering by Wedge (cont.) Let ( ) ( ) ( ) = ( ) = + sin cos h A B (B.C. at = ) ( ) ) ( ) = sin 2 0 =2 - : ( = 2 2 n so n = = ( ) n 2 3
Scattering by Wedge (cont.) ( ) = )sin ( A B k z n n n Bessel function of order n = = 0 0 ( ) n Note: trivial solution n 0 0 n Note: n ( ) 0 J k as n 1 ( x negativeinteger ( ) ~ x J since ( ) + 2 1) 4
Scattering by Wedge (cont.) Hence n = = ( ) n 2 n = 1,2,3 = , . etc ( ) 1 2 5
Scattering by Wedge (cont.) n ( ) = n sin ( ) A a J k 1 z n n = 1 For assume n 0 to match with the interior form. ( ) = (2) sin ( ) A b H k 1 z n n n = 1 n 6
Scattering by Wedge (cont.) B.C. s = = E E 1 2 z z I ( ) = = ( ) H H J 0 2 1 sz where y 1 zA = H 2 0I = ( , ) x = 2 7
Scattering by Wedge (cont.) Hence we have = = A A 1 2 z z 1 1 A A I = ( ) 2 1 0 z z 8
Scattering by Wedge (cont.) y 0I First B.C. : = ( , ) = A A x 1 2 z z = 2 ( ) 2 ) sin ) sin = ( ( ) ( ( ) a J k b H k n n n n n n = = 1 1 n n ,2 Multiply by and integrate over sin ( ) m 9
Scattering by Wedge (cont.) 0, 12 2 m n 2 = sin ( )sin ( ) d ( ) n m = 2 , m n = ( ) x Note: To evaluate this integral, use ( ) 2 = Recall: n ( ) n 2 ( ) 2 2 ( ) ( ) = sin ( )sin ( ) sin sin d nx mx dx n m 0 ( ) 2 = ( ) ( ) a J k b H k Hence m m m m 10
Scattering by Wedge (cont.) Second B.C. : 1( ) ( )( n ) ( ) 2 sin ( ) k b H k a J k n n n n = 1 n I = ( ) 0 ,2 Multiply by and integrate over sin ( ) m 1 1 2 ( )( m ) ( ) 2 2 ) ( ) k (2 b H k a J k m m n I = sin ( ) 0 m 11
Scattering by Wedge (cont.) Solution: 1 I ( ) ( )( n ) 2 = sin a H k 0 ( ) n n k DEN ( ) J k 1 ( ) ( )( n ) 2 = sin b H k 0 n ( ) ( )( n ) n n 2 k DEN H k where ( )( n ) ( ) ( )( n ) ( ) 2 2 = DEN H k J k H k J k n n 2 k = j (Wronskian Identity) 12
Scattering by Wedge (cont.) Generalization: To generalize the solution for arbitrary , we simply multiply the entire solution by exp( ) z jk z zk k k and then make the substitution The solution is then valid for a line source of the form: I e = jk z ( ) I z z 0 13
Edge Behavior ( ) = sin ( ) A a J k 1 z n n n = 1 n As , keep term, since 0 = 1 n 1 ( ( ) ~ J x x + 2 1) Hence ( ) jk z ~ sin ( ) zA a J k e z 1 1 1 so = zA 1 ( ) 1 2 14
Edge Behavior (cont.) Therefore we have: 2 k = E A 1 z z j 2 1 A = 1 ( 0) E k z z 1 z j 2 1 1 A = 1 ( 0) E k z z 1 z j Note: kz= 0 corresponds to a uniform line current, where there is no charge density (and hence no normal electric field). 15
Edge Behavior (cont.) 0 0 z E y as 0I = ( , ) -1< 0 E E if ( , ) 1 1 x 1 = 2 1 ( ) 2 if ( , ) E E Hence ( ) 2 1 2 2 ( , ) E E Therefore (convex corner) if 2 16
Knife Edge Recall: = 11 , E E ( ) 1 2 1 2 1 2 = = = 0 1 1 1 1 so E 17
Knife Edge (cont.) y Parallel Current ( ) x J sz x + = = 0 , : x At = = J H H sz x 1 A = z 1 ( ) ( ) 1 1 a jk z cos J k e z 1 1 18
Knife Edge (cont.) so 1 1 = 1/2 J 1 sz or 1/2 J sz or 1 J sz x 19
Strip in Free Space y Current on Strip J sz x w From conformal mapping: / I = 0 J sz 2 Maxwell function w 2 x 2 20
Knife Edge (cont.) Perpendicular Current y ( ) x J sx x + = = 0 , J H At sx z Note: To have this component, we must use a TEz solution (e.g., using a magnetic current source). J x If we did the TEz solution, the result would show that sx 21
Microstrip line y Longitudinal Total Current Density on a Strip Transverse x Note: w The current has both components, due to the fact that the mode is not exactly TEM (due to the substrate). = j J s s J = sx x jk J j z sz s The longitudinal current and the charge density are even functions, while the transverse current is an odd function. 22
Microstrip line (cont.) y longitudinal transverse x w Fourier-Maxwell Basis Function Expansion: 1 2 1 M m x w ( ) = jk z , cos J x z e a z sz m 2 w = 0 m 2 x 2 ( ) 2 2 1 n x N w ( ) = jk z 2 , sin J x z e x b z sx n 2 w = 1 n 23
Microstrip line (cont.) y longitudinal transverse x w Chebyshev-Maxwell Basis Function Expansion: ( ) + 2 1 1 1 M 2 w x ( ) = 0 m jk z , J x z e a T z 2 sz m m w 2 w = 0 m 2 x 2 2 2 w 4 w N w x j ( ) = 2 jk z , J x z e x b U z 2 1 sx n n 2 = 1 n 24
Meixner* Edge Condition U E This condition must be satisfied at all edges. Mathematically, imposing this condition in the solution of a problem is necessary to ensure a unique solution. C. J. Bouwkamp. A note on singularities occurring at sharp edges in electromagnetic diffraction theory, Physica (Utrecht), vol. 12, pp. 467-474. Oct., 1946. *J. Meixner, Dle kantenbedingung in der theorie du beugung electromagnetischer wellen an vollkommen leitenden ebenen schirm, Ann. Phys., vol. 6, pp 1-9, 1949. 25
Meixner Edge Condition (cont.) Meixner condition: y U E Let s verify this for the wedge: V x a 1 4 2 = U E dV E V 1 E 1 1 4 = d d dz 2( 1) 1 1 E 1 V 26
Meixner Edge Condition (cont.) We require that a 2( 1) 0 d as 1 a or 2 1 d 1 1 a or 2 1 2 1 or = 2 0 as 1 2 0 This will be satisfied since = Recall: ( ) 1 1 2 27
