Electromagnetic scattering. Graduate Course Electrical Engineering (Communications) 1 st Semester, Sharif University of Technology

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1 Electroagnetic scattering Graduate Course Electrical Engineering (Counications) 1 st Seester, Sharif University of Technology

2 Contents of lecture 5 Contents of lecture 5: Scattering fro a conductive wedge The scalar wave equation The vector solutions Introducing a source Solving the equations The scattering proble Special cases Scattering fro a conductive wedge 2

3 Introduction An exaple of another canonical, exactly solvable scattering proble is that of an infinite, perfectly conducting wedge The wedge is assued to have an angle of Note that the wedge is infinitely extended and infinitely long Also, an iportant liiting case is when 0. (Why?) k i Scattering fro a conductive wedge 3

4 Scalar equation This tie the syste has no rotational syetry Nonetheless, the natural waves of the syste can still be found by solution of the wave equation in cylindrical coordinates As before, first consider the scalar wave equation k y k In cylindrical coordinates x Scattering fro a conductive wedge 4

5 Scalar equation The solution should be found for the angles 0 2 Let us consider solutions of the type sin, k (,, ) f, ( )exp k jk cos Where the radial function f satisfies f, k k f f, ( ) AJ k BY k k, k 0 k k k 2 2 Scattering fro a conductive wedge 5

6 Vector solution For the oent we do not ipose any boundary conditions Solutions of the vector wave equation can once again be found: M 1, k ˆ, k ˆ ˆ, k, k k k k N 1 jk ˆ ˆ k, k, k 2, k jk 2 k, k ˆ These vectors can be used to construct the electric field in the region outside the wedge Scattering fro a conductive wedge 6

7 Vector solution But, which kind of function should we choose to satisfy the boundary conditions? For the M vector to be a legitiate electric field its radial coponent should vanish on the wedge surfaces for all, k y 0 for 0,2 jk, f, ( )exp cos k k x Scattering fro a conductive wedge 7

8 Vector solution What about the N vector? Now we should deand for all, k, k 0 for 0, 2 y jk, f, ( )exp sin k k 2 0 x 2 Scattering fro a conductive wedge 8

9 Vector solution solution (horiontal polariation): E, k ˆ, k M, k k k ˆ case 1 j H E N j, k j jk 2 k, k jk, k ˆ 2 ˆ k, k ˆ No electric field along, but agnetic field has a -coponent Scattering fro a conductive wedge 9

10 Vector solution solution (vertical polariation) 1, k jk, k ˆ 2 E N ˆ ˆ, k jk 2 k, k k 1 j H E M j j, k ˆ, k ˆ k k, k case No agnetic field along, but electric field has a coponent Scattering fro a conductive wedge 10

11 Vector solution Here, unlike the proble of a conducting cylinder, the solution extends to =0. To ensure the finiteness of the electroagnetic energy (not necessarily field) near the origin we should have f, ( ) J k k y, J ( k )exp jk cos k, J ( k )exp jk sin k 0 x Scattering fro a conductive wedge 11

12 Vector solution The total electric field which satisfies boundary conditions at surfaces of conducting wedge y and ensures correct behavior k i as 0 Is a superposition of these and vector solutions x Reeber: this representation is for the total field; for the scattering proble we cannot separate the incident and scattered waves in this representation Scattering fro a conductive wedge 12

13 Vector solutions in the scattering proble k i, If the incident wave has a certain, then we know fro the unifority of the syste in the direction that the total solution consists of waves with k k k k k k 2 2 i, i, i, Then, for exaple for the case, the total field can look like E, k i,, k ˆ i, a k k ˆ i, J ( k )cos exp jk, k i, i, Scattering fro a conductive wedge 13

14 Vector solutions in the scattering proble Or: E k i, a J ki, k ( )cos J ( ki, ) sin k ˆ ˆ exp jki, Note: electric field is not necessarily finite as 0, it is electric energy density which should reain integrable Scattering fro a conductive wedge 14

15 Vector solutions in the scattering proble For case or vertical polariation E a jk k k 1, ki, jk, ki, ˆ 2 ˆ 2 i,, k i, ˆ i, J ( k )sin exp jk, k i, i, E ki, ki, jki, a j J( k 2 i, )sin J ( k 2 i, )cos k k k 2 i, 2 k ˆ J ( ki, )sin exp jki, ˆ ˆ Scattering fro a conductive wedge 15

16 Vector solutions in the scattering proble So far, it is fine, but unfortunately it is not so easy to find the constant coefficients a to describe the scattering solution for a particular incident wave propagating in a certain direction For a cylinder we could separate the incident and scattered field fro the beginning and then apply the boundary condition on the cylinder surface Also, it was essential to expand the incident plane wave in Bessel functions in order to be able to solve the proble. This is not possible here because of the non-integer Bessel functions used. Scattering fro a conductive wedge 16

17 Introducing a source The traditional way to treat this proble is by solving the proble in y Current line source presence of the sources Once the proble is solved, the 0 sources is oved to a point infinitely far away. The field generated by such a source is a far one field and 0 x has a plane-wave character. Therefore by taking this liit of the full solution the proble is solved. Scattering fro a conductive wedge 17

18 Introducing a source Let us consider this proble in the sipler case where k = 0. So, we are interested in waves which are unifor along. Fro the field expansion discussed, we expect the total electric field vector to look like ( ) a J ( k )cos J k E sin k ˆ ˆ E ˆ a J ( k)sin Here we used ki, 0 ki, k Scattering fro a conductive wedge 18

19 Introducing a source Let us return to Maxwell equations, with electric and agnetic line current sources along, and unifor along 0 y 0 E j H M H je J i i Magnetic current x i i i i J J ˆ M M ˆ Scattering fro a conductive wedge 19

20 Introducing a source With Maxwell equations in cylindrical coordinates, and fields unifor in -direction: E j H H j E E j H H j E E i E j H M H i H je J Scattering fro a conductive wedge 20

21 Introducing a source Two separate sets for horiontal and vertical polariations Vertical electric field Horiontal electric field E j H H j E E j H H j E H i H je J E i E j H M Horiontal agnetic field Vertical agnetic field Scattering fro a conductive wedge 21

22 Introducing a source 1 st set leads to: E E 2 2 i k E 2 2 jj 2 nd set gives: H H 2 2 i k H 2 2 jm Left hand side sae as the scalar wave equation in cylindrical coordinates for unifor fields along Scattering fro a conductive wedge 22

23 Introducing a source We would like to solve these equations in presence of the conducting wedge, for infinitely thin line sources I i 0 J y M i 0 M Boundary conditions: 0 0 x E E 0 for 0 E E 0 for 2 E E Scattering fro a conductive wedge 23

24 Solving the equations () Consider the case first. Solution in the region 0 which 0, 0 y satisfies boundary conditions at the wedge surfaces, and shows the right behavior at the origin is 0 x, ( )sin E a J k 2 Scattering fro a conductive wedge 24

25 Solving the equations () y Solution in the region which satisfies the wedge boundary conditions, and the radiation condition at the infinity is 0 0, 0 x (2), ( )sin E b H k 0 2 Scattering fro a conductive wedge 25

26 Solving the equations () y Next we have to atch the solutions at the source by deanding that 0 0, 0 0 x E 0, E 0, E j 0 E I Scattering fro a conductive wedge 26

27 Solving the equations () Resulting equations ( is shown explicitly) (2) b H k 0 a J k ( )sin ( )sin (2) b H k 0 a J k ( )sin ( )sin 2 j I 0 0 k0 Scattering fro a conductive wedge 27

28 Solving the equations () (2) b H k0 a J k 0 ( ) ( ) 2 j I b H ( k ) a J ( k ) sin (2) k0 a I0 2 j H k 2 0 (2) W ( k ) 0 sin 0 W J ( k ) H ( k ) J ( k ) H ( k ) (2) (2) j j J ( k0) Y ( k 0) J ( k 0) Y ( k 0) k 0 Scattering fro a conductive wedge 28

29 Solving the equations () Solution for 0 a I H ( k )sin 2 (2) E I H k J k (2), ( )sin ( )sin Scattering fro a conductive wedge 29

30 Plane wave solution () What if we had no wedge? Then the equation to be solved for this case is the sae, but has to be solved for a source in free space E E j I k E The solution is I E H0 kr 4 0 (2) 2 2, R 0 20 cos 0 Scattering fro a conductive wedge 30

31 Plane wave solution () Now, if we ove the source to a point far away in the free space proble then I I 2 E H0 kr jkr j 4 4 kr (2), 0 0 exp / 4 R 2 cos cos R y, exp cos E A jk 0 0 I 2 A exp jk j / k0 0 0 x Scattering fro a conductive wedge 31

32 Plane wave solution () Apart fro its aplitude this is the field of a plane wave propagating towards the origin along the angle 0 A0 exp jk cos 0 A0 exp jki r ki k cos 0, k sin 0, 0 ki r cos, sin, 0 0 Scattering fro a conductive wedge 32

33 Plane wave solution () So, in the original proble, if we take the sae liit, while keeping the aplitude constant, the resulting solution will be that of the total field for the incident plane wave along 0 0 Note: keeping aplitude constant eans that we k s should put 4 k I exp jk j / 4 A Scattering fro a conductive wedge 33

34 Plane wave solution () Result: 4 E A j J k , sin ( )sin This way we have found the coefficients for the total k i electric field when the incident wave is a plane wave with ki k cos 0, k sin 0, Scattering fro a conductive wedge 34

35 Plane wave solution () The corresponding agnetic field is A k 4 H j 0 J k j 2 0, sin ( )sin 1 A 4 H j 0 J k j 2 0, sin ( )cos 1 Scattering fro a conductive wedge 35

36 scattering In a siilar way, the scattering proble can be solved Now we solve the equation for the agnetic field 2 H H 2 2 M i 0 M i k H jm With the boundary condition H 0 for 0, 2 Scattering fro a conductive wedge 36

37 scattering The overall solution can be written as, ( )cos H a J k 0 (2), ( )cos H b H k At the line source, they should be atched by H 0, H 0, H j 0 H M Scattering fro a conductive wedge 37

38 scattering Due to the analogy with the case, we conclude that if the incident agnetic field is 0 0 H A j i exp 0 ki r k i Then the total agnetic field is given by 2 H A j J k , cos ( )cos 2 1 Scattering fro a conductive wedge 38

39 scattering i E But what about the electric field? The (horiontal) incident field is k i E i 0 1 j H ˆ jk rˆ k A exp i i i 0 0 The total electric field is E 1 H H ˆ ˆ ρ j Scattering fro a conductive wedge 39

40 scattering E 2, j A j cos J ( k) sin ( )cos k ρˆ J k ˆ Copare with the general expression found before Scattering fro a conductive wedge 40

41 Special cases Let us consider soe specific y cases. First when=. k i Corresponds to reflection of a plane incident wave by a flat, infinite perfect conductor 0 x Solution for vertical fields () 2, 4 sin ( )sin E A j J k Scattering fro a conductive wedge 41

42 Special cases, 4 sin ( )sin E A j J k We can use the relationship To arrive at exp j cos J0( ) 2 j cos J( ) 1, exp cos exp cos E A0 jk 0 A0 jk 0 Incident wave A exp 0 jki r Scattering fro a conductive wedge 42

43 Special cases The reaining (scattered) ter corresponds to: A0 exp jk cos 0 A0 exp jks r k s k cos 0, k sin 0, 0 y This is what one would have expected. k i k s The case is siilar except 0 for the sign of the reflected (agnetic) field x Scattering fro a conductive wedge 43

44 Special cases Another iportant case is the liit 0 k i This is the liit of a half-infinite plane which is quite iportant 0 0 Total fields:, 2 / 2 sin / 2 ( )sin / 2 E A j J k 0 0 / / 2 0, 2 / 2 cos / 2 ( )cos / 2 H A j J k Scattering fro a conductive wedge 44

45 Nuerical exaples We plot the scattered field as k i function of the observation angle (at a certain distance) for a certain incoing wave 0 0 scattered field: 4 E A j J k , sin ( )sin,, exp k r E, A exp jk cos E E A j s, 0 i 0 0 Scattered field Scattering fro a conductive wedge 45

46 Nuerical exaples () k i 45, k Scattering fro a conductive wedge 46

47 Nuerical exaples () k i 45, k Scattering fro a conductive wedge 47

48 Nuerical exaples () k i 0, k Scattering fro a conductive wedge 48

49 Nuerical exaples () For scattering, we plot the scattered agnetic field as function of the observation angle (at a certain distance) for a certain incoing wave scattered field: 2 H A j J k , cos ( )cos,, exp k r H, A exp jk cos H H A j s, 0 i 0 0 Scattering fro a conductive wedge 49

50 Nuerical exaples () k i 0, k Scattering fro a conductive wedge 50

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