Higher Order Modelling for Computational Electromagnetics - PowerPoint PPT Presentation

Integral equations for EM applications Integral equations for EM applications, and their kernels ζ g ( r )- ∫ K ( r , r ’) g ( r ’) d r ’= f ( r ) ∫ f ( r ) is known and g ( r ) is unknown ( ) g ( ) ζ =0 → eq. of the first kind (EFIE for PEC); ζ≠ 0 → eq. of the second kind (HFIE). ζ≠ 0 → eq of the second kind (HFIE) The kernel must be singular to get well-posed first kind eqs . Very Luckily , in EM, one usually has singular kernels, at least of logarithmic kind for 2D problems or like 1/r for 3D problems bl lik 1/ f 3D bl 20

Representations of Fields in Terms of Potentials (Frequency-Domain)  Solutions of Maxwell’s equations can be expressed in terms of potentials: 21

Potentials, Lorenz Gauges, and Continuity Equations  In homogeneous, isotropic, unbounded regions  Vector and scalar potentials are related by the Lorenz gauge:  Charge densities are related to current densities by continuity equations: 22

Representations of Fields in Terms of Potentials (Frequency-Domain)  Solutions of Maxwell’s equations can be expressed in terms of potentials: 23

Uniqueness of Solutions to Maxwell’s Equations  Fields satisfying in unbounded regions with specified on boundaries and radiation conditions at infinity are unique specified on boundaries and radiation conditions at infinity are unique.  Fields satisfying Maxwell’s eqs. in bounded regions with lossy media or having Re( α / β ) ≠ 0 are unique. h i R ( / β ) ≠ 0 i  Uniqueness fails at certain discrete frequencies in lossless regions having lossless surface impedance boundary conditions [Re( α / β )=0]; these frequencies oss ess su ace peda ce bou da y co d t o s [ e( α / β ) 0]; t ese eque c es correspond to the cavity resonant frequencies of the bounded region. 24

Generalization of the previous transparencies yields to the following yields to the following • To invert the differential operator is to solve the problem → to do this one has to know (find) the Green function → to do this one has to know (find) the Green function G of the given problem; • If G is known, then the solution is simply obtained by , p y y integration; • Unfortunately, for all problems of practical interest G is unknown → numerical approaches to differential/integral problems are required because of the ignorance of the Green function; ignorance of the Green function; • The numerical problem could be well- or bad-posed; this depends on the properties of the operator; p p p p ; 25

• In general, in integral formulations, use of the proper ( (unknown) G is avoided by resorting to general k ) G i id d b ti t l principles ( e.g., equivalence principle), or to integral theorems (e.g., Green theorems); ( g , ); • Integral formulations guarantee the BCs and all the physical properties of the solutions , as a matter of fact, in many cases, the proper integral equations are obtained only at the moment of imposing the BCs (in the finite region); the finite region); • This is why IE are widely used in EM (since radiation conditions / far-field conditions are automatically y satisfied); • Because of the Green kernel, the matrix approximation of the problem is given by a dense matrix . 26

FEM MOM Differential formulation Integral formulation Non linear problems can be dealt Applications to linear problems are ☺ with well known ☺ BCs have to be enforced BCs automatically satisfied ☺ Infinite domains: there is some Infinite domains : no problem problems Dense matrices (problems in ☺ Sparse matrices dealing with very large problems) The integrals to be evaluated are The singularities of the kernel g ☺ ☺ render numerical integration quite d i l i t ti it simple difficult Complex geometry are easily dealt Higher order models have been with with introduced more recently introduced more recently 27

Higher order modelling Higher order modelling Required: Required: • To better represent the geometry of the problem. • To reduce the L 2 ( least squares norm ) of the solution error on regions of interest. l i i f i - It usually improves the convergence of the results and/or reduces the size of the numerical matrices.

Higher order modelling  Regarding the geometry of the problem; need to approximate the curvature of the geometry; d t i t th t f th t need to approximate small details in multiscale problems.  use subsectional bases thereby subdividing the geometry into y g g y simple (curved) elements of small size. 29

Use subsectional bases thereby subdividing the geometry into simple (curved) elements of small size into simple (curved) elements of small size. Pablo Picasso: "Retrato de Ambroise Vollard” (1910). 30

Regarding the geometry of the problem Contribution from specular points (in the optical limit). Points of reflection on the body at which the angle of incidence is equal to the angle of reflection relative to the observation point. l t th l f fl ti l ti t th b ti i t e g the RCS σ of a metallic sphere of radius a is σ = π a 2 e.g., the RCS σ of a metallic sphere of radius a is σ = π a 31

In the high-frequency range, the PEC-sphere “models” shown here poorly model the commonly used “sphere benchmark” (see here poorly model the commonly used sphere-benchmark (see below) NASA Communications Satellite E h P Echo Project (1960-1969) j t (1960 1969) 32

Regarding the geometry of the problem Travelling waves (contribution from points with high curvature) . e.g., estimation of the near-nose-on cross-sections of long, thin bodies. 33

Regarding the geometry of the problem Creeping waves of importance for the analysis in the shadow region (the importance of the creeping waves depends on the dimension in wavelength of the body). dimension in wavelength of the body). 34

Higher order modelling  Regarding the e pansion/testing f nctions sed in the  Regarding the expansion/testing functions used in the numerical application. reduce the L 2 ( least squares norm ) of the solution error on regions of interest; i improve convergence of the results; f th lt reduce the size of the numerical matrices.  use vector functions of high polynomial order. 35

Results from a 1995 paper (16 years old): nose-on incidence on open circular wg. inner length= 30 λ , inner diameter= 5 λ , wall thickness 0.5 λ , profile length= 66.5 λ Magnitude (LHS) and phase in radians (RHS) of the current functions for a truncated circular waveguide at incidence along the z-axis, obtained by solving a system with 816 unknowns. z axis, obtained by solving a system with 816 unknowns. 36

How to g How to get cur How to g How to get cur t curved t curved ed (i.e ed (i.e (i.e., distor (i.e., distor ., distorted) elements ., distorted) elements ted) elements. ted) elements. Source: Zinkiewicz’s book 37

How to g How to get cur t curved / ed / distor distorted elements ted elements. • Any curved cell is obtained by mapping a “parent cell” into the object domain. • The parent cell for triangular patches is a rectilinear triangle. • In object space, the 3 edges are the zero-coordinate lines ξ 1 , ξ 2 , ξ ξ 3 =0. 0 • ξ i is the area coordinate ξ i = A i /A T • Dependency relations: ξ 1 + ξ 2 + ξ 3 = 1 Dependency relations: ξ 1 + ξ 2 + ξ 3 1

How to g How to get cur t curved / ed / distor distorted elements ed elements. • The rectilinear triangle on the right-hand side is r = ξ 1 r 1 + ξ 2 r 2 + ξ 3 r 3 ξ 1 1 ξ 2 2 ξ 3 3 • Where ξ 1 , ξ 2 , ξ 3 are 3 linear “shape functions,” each associated to a different corner node of the cell.

nd or Cur Curved elements; 2 ed elements; 2 nd order distor der distortion ion • Shape function: 2 ξ 3 ( ξ 3 – ½) for node Shape function: 2 ξ 3 ( ξ 3 ½) for node • Shape function: 4 ξ 1 ξ 3 for node • The shape functions of the other nodes are obtained by cyclic permutation of the subscripts permutation of the subscripts. • Interpolatory polynomials easily provide the shape functions.

nd or Cur Curved elements; 2 ed elements; 2 nd order distor der distortion ion Shape functions: Mapping: Mapping:

nd or Cur Curved elements; 2 ed elements; 2 nd order distor der distortion ion Shape functions: Dependency relation, unitary basis vectors, and Jacobian and Jacobian Mapping: Mapping:

Interpolatory polynomials for shape functions . for shape functions Use Lagrange interpolation polynomials written in g g p p y terms of interpolatory polynomials of Silvester.

Interpolatory polynomials of Silvester • Use polynomials of degree i in ξ , where ξ is in the interval [0 1] [0, 1]. • The parameter p indicates the number of uniform subintervals p p into which the interval is divided. • The polynomial is unity at ξ = i/p and has zeros at ξ =0 1/ p 2/ p • The polynomial is unity at ξ = i/p and has zeros at ξ =0, 1/ p , 2/ p , …, ( i -1)/ p    1    i i 1        ( ( p k k ) ) 1 1 i p   R ( p , )  i ! k 0 i   1 i 0  

Silvester polynomials: example for p=1 • There are 2 polynomials: one of 0 th and one of 1 st degree. • Please draw the polynomials on the interval [0, 1]. Pl d h l i l h i l [0 1]         R R ( ( 1 1 , , ) ) 1 1 ; ; R R ( ( 1 1 , , ) ) 0 0 1 1   1 1   i 1      ( p k ) 1 i p   R ( p , )   i ! k 0 i    0 1 1 i i 0  

Silvester polynomials: example for p=1 • There are 2 polynomials: one of 0 th and one of 1 st degree. • Please draw the polynomials on the interval [0, 1]. Pl d h l i l h i l [0 1]      R ( 1 , ) 1 ; R ( 1 , ) 0 1

Silvester polynomials: example for p=2 • Three polynomials: one of 0 th , one of 1 st and one of 2 nd degree. • Please draw these polynomials on the interval [0, 1]. Pl d h l i l h i l [0 1]     2 2 ( 2 1 )             R R ( ( 2 2 , ) ) 1 1 ; ; R R ( ( 2 2 , ) ) ; ; R R ( ( 2 2 , ) ) 0 1 2  1 1 2   1 1   i 1      ( p k ) 1 i p   R ( p , )   i ! k 0 i   1 i 0 

Silvester polynomials: example for p=2 • Please draw these 3 polynomials on the interval [0, 1].     2 2 ( 2 1 )       R ( 2 , ) 1 ; R ( 2 , ) ; R ( 2 , ) 0 1 2  1 1 2

Silvester polynomials: example for p=3 • 4 polynomials; one of 0 th ; one of 1 st ; one of 2 nd , one of 3 rd degree. • Please draw these polynomials on the interval [0, 1]. Pl d th l i l th i t l [0 1]   3       R R ( ( 3 3 , ) ) 1 1 ; ; R R ( ( 3 3 , ) ) ; ; 0 1 1         3 ( 3 1 ) 3 ( 3 1 )( 3 2 )       R R ( ( 3 3 , ) ) ; ; R R ( ( 3 3 , ) ) 2 2 3 3    1 2 1 2 3   1   i 1      ( p k ) 1 i p   R ( p , )   i ! k 0 i    1 1 i i 0 0  

Silvester polynomials: example for p=3 • Please draw these 4 polynomials on the interval [0, 1].  3         R R ( ( 3 3 , ) ) 1 1 ; ; R R ( ( 3 3 , ) ) ; ; 0 1 1         3 ( 3 1 ) 3 ( 3 1 )( 3 2 )       R ( ( 3 , ) ) ; R ( ( 3 , ) ) 2 2 3 3    1 1 2 2 1 1 2 2 3 3

Scalar Lagrangian interpolation on the g g p canonical elements: the segment. • There are two ξ coordinates: ξ • There are two ξ coordinates: ξ 1 , ξ 2 . ξ • These are dependent coordinates ( ξ 1 + ξ 2 =1) α ij ( ξ 1 , ξ 2 )=R i (p, ξ 1 )R j (p, ξ 2 ) with i+j=p is a p-th order Lagrangian polynomial interpolating points within a segment whose normalized coordinates ( ξ 1 , ξ 2 ) are (i/p, j/p). 51

Interpolation on a segment, example for p=1 p g , p p • There are two interpolatory polynomials of 1 st degree There are two interpolatory polynomials of 1 degree. • α ij ( ξ 1 , ξ 2 )=R i (p, ξ 1 )R j (p, ξ 2 ), (with i+j=1). • Please draw the polynomials on the interval ξ 1 =[0 1] • Please draw the polynomials on the interval ξ 1 [0, 1].                 ( ( , ) ) R R ( ( 1 1 , ) ) R R ( ( 1 1 , ) ) 10 1 2 1 1 0 2 1              ( ( , , ) ) R ( ( 1 , , ) ) R ( ( 1 , , ) ) 01 01 1 1 2 2 0 0 1 1 1 1 2 2 2 2 52

Interpolation on a segment, example for p=1 p g , p p                 ( ( , ) ) R R ( ( 1 1 , ) ) R R ( ( 1 1 , ) ) 10 1 2 1 1 0 2 1         ( , ) R ( 1 , ) R ( 1 , ) 01 1 2 0 1 1 2 2 53

Interpolation on a segment, example for p=2 p g , p p • Example for p =2: there are three interpolatrory polynomials of 2 nd degree polynomials of 2 nd degree. • α ij ( ξ 1 , ξ 2 )=R i (p, ξ 1 )R j (p, ξ 2 ), (with i+j=2). • Please draw the polynomials on the interval ξ 1 =[0, 1]. Please draw the polynomials on the interval ξ 1 [0, 1].    2 ( 2 1 )             ( ( , , ) ) R R ( ( 2 2 , , ) ) R R ( ( 2 2 , , ) ) 1 1 20 20 1 1 2 2 2 2 1 1 0 0 2 2 2          ( , ) R ( 2 , ) R ( 2 , ) 2 2 11 1 2 1 1 1 2 1 2    2 ( 2 1 )        ( , ) R ( 2 , ) R ( 2 , ) 2 2 02 1 2 0 1 2 2 2 54

Interpolation on a segment for p=2 p g p    2 ( 2 1 )        ( , ) R ( 2 , ) R ( 2 , ) 1 1 20 1 2 2 1 0 2 2          ( , ) R ( 2 , ) R ( 2 , ) 2 2 11 1 2 1 1 1 2 1 2    2 ( 2 1 )               ( ( , ) ) R R ( ( 2 2 , ) ) R R ( ( 2 2 , ) ) 2 2 02 1 2 0 1 2 2 2 55

Scalar Lagrangian interpolation on a segment on a segment. • For the p-th order Lagrangian polynomial interpolation of a segment there are (p+1) interpolating polynomials of p-th degree. 56

Parameteriza meterization of tion of the points on a triang the points on a triangle le. Recall that the coordinates of a rectilinear triangle may be            , , parameterized as p r r r r r r r r 1 1 1 1 2 2 2 2 3 3 3 3 where r i is a vertex position vector for vertex i . 57

Parametriza metrization of ion of the points on a triang the points on a triangle le. A Lagrange parametrization of order q for a curvilinear triangle can be expressed in terms of interpolating polynomials of the Silvester form as: q                 R R ( ( q q , ) ) R R ( ( q q , ) ) R R ( ( q q , ), ) i i j j k k q q r r r r ijk i 1 j 2 k 3  i , j , k 0 where a triple indexing scheme is used to label the position where a triple indexing scheme is used to label the position vector r ijk interpolating the point with normalized coordinates ( ξ 1 , ξ 2 , ξ 3 )=(i/q, j/q, k/q). ( ξ 1 , ξ 2 , ξ 3 ) ( q, j q, q)

Geometry description f Geometry description f Geometry description for a Geometry description for a r a triang r a triang triangle triangle le. le. Normalized coordinates related by ξ 1 + ξ 2 + ξ 3 =1 Edge vectors derived from the “independent” coordinates ξ 1 and ξ 2 Ed d i d f h “i d d ” di ξ d ξ ℓ i = ∂ r / ∂ ξ i , i=1,2, from which the edge vectors that follow are found: from which the edge vectors that follow are found:

Edg Edge v vecto ectors s of of a triang a triangle le. ℓ 1 = - ℓ 2 , ℓ 2 = ℓ 1 , ℓ 1 ℓ ℓ 3 = ℓ 2 - ℓ 1

Gr Gradient v adient vector ectors of s of a triang a triangle le. The gradient vectors are determined from the edge The gradient vectors are determined from the edge vectors as: “ ξ = ( n x ℓ ) / J “ ξ i = ( n x ℓ i ) / J where n = ℓ 1 x ℓ 2 /J is the unit vector normal to the h ℓ 1 ℓ 2 /J i th it t l t th triangle while J= | ℓ 1 x ℓ 2 | is the Jacobian.

Edg Edge and g and gradien adient v vecto ectors of s of a trian a triangle le. The above definitions for edge, (height) and gradient vectors apply for triangles on curved surfaces with an extended interpretation Triangle tangent to a curvilinear triangle at a point. The curvilinear and (rectilinear) tangent triangles have the same element coordinates, Jacobian, edge vectors, and height vectors at the point of tangency.

Why to use cur Why to use curved ( Why to use cur Why to use curved ( ed ( i.e ed ( i.e i.e. , distor i.e. , distor , distorted) elements , distorted) elements ted) elements? ted) elements? • the same model (that is, the same database) used by structural /mechanical engineers can be used → this is a structural /mechanical engineers can be used → this is a big plus!!! • better approximations of boundaries are obtained with very small error in volume/surface/line models (no rescaling) • usually a smaller number of unknowns is needed whenever usually a smaller number of unknowns is needed whenever higher-order expansion functions are used → shorter computation time 63

Why to use cur Why to use curved ( Why to use cur Why to use curved ( ed ( i.e ed ( i.e i.e. , distor i.e. , distor , distorted) elements , distorted) elements ted) elements? ted) elements? • mixed approaches (FEM + MoM) are facilitated because the expansion functions of the two methods can be chosen the expansion functions of the two methods can be chosen within the “same” set. • there is the possibility to construct singular functions able to model singular current or field behaviors (though applications of this are rather new). But, is there any problem? 64

But, is there any problem? But, is there any problem? • In FEM applications you now have to use quadrature to evaluate the matrix entries, whereas for non-distorted cells there are often closed form results possible. • For MoM applications people are already using quadrature • For MoM applications people are already using quadrature, but the treatment of the Green's function singularity might be different with curved cells. • The main “cost” of curved cells is (1) the additional data structure necessary in the model, (2) the additional computation arising from the quadrature. (2) h ddi i l i i i f h d 65

Vector functions  We first consider the lowest order functions.  Then we move on and consider higher-order vector functions. 66

The family of generalized triangle basis functions (for current representation) Source: Don Wilton’s notes 67

Div Div-Conf Div Div Conf Conf functions on a triangular element Conf functions on a triangular element functions on a triangular element functions on a triangular element Divergence-conforming functions of the Nedelec type maintain only normal continuity across element maintain only normal continuity across element Boundaries (they do not prescribe tangential continuity). 68

Div Div-Conf Div Div Conf Conf functions on a triangular element Conf functions on a triangular element functions on a triangular element functions on a triangular element They eliminates spurious solution of the EFIE operator while discarding the highest order degrees of freedom while discarding the highest order degrees of freedom associated with the nullspace of the divergence operator in the EFIE operator. 69

Divergence-conf Div ence-conforming bases on rming bases on triang triangles les. 1  1    J    ,   • Zeroth-order bases. 1 2 3 3 2 • Three vector basis functions of first order. functions of first order 1 1   ,   J    • They have constant   2 3 1 1 3 normal and linear tangential (CN/LT) tangential (CN/LT) 1 components at element   .   J      edges. 3 1 2 2 1 J 70

Curl-conf l-conforming ba orming bases on tr ses on triang iangles les.     ˆ n : • Zeroth-order bases   • Three vector basis functions Th t b i f ti of first order.                 , , • They have constant 1 1 2 2 3 3 3 3 2 2 tangential and linear normal          , (CT/LN) components at 2 3 1 1 3 element edges          . (prove this by use of eq. 3 1 2 2 1 (48) of 1997 paper). 71

Completenes Compl teness of of z zeroth or th order triangula der triangular bases r bases (w (w (we consider only the cur (we consider only the cur e consider only the curl conf e consider only the curl-conf conforming ones). conforming ones). orming ones). orming ones).              , , 1 1 2 2 3 3 3 3 2 2          , 2 3 1 1 3          . 3 1 2 2 1 • The basis set is incomplete to first order since 6 degrees of The basis set is incomplete to first order since 6 degrees of freedom are required to model linear variations in two independent vector components on a surface. 72

Completenes Compl teness of of z zeroth or th order triangula der triangular bases r bases (w (we consider only the cur (we consider only the cur (w e consider only the curl-conf e consider only the curl conf conforming ones). conforming ones). orming ones). orming ones).              , , 1 1 2 2 3 3 3 3 2 2          , 2 3 1 1 3          . 3 1 2 2 1 • To render the bases first-order complete one must include the p curl-free combinations:    1   , , 1 1 1    , 2 2              . 1 2 2 1 73

Completenes Compl teness of of z zeroth or th order triangula der triangular bases r bases (we consider only the cur (w (w (we consider only the cur e consider only the curl-conf e consider only the curl conf conforming ones). conforming ones). orming ones). orming ones).              , , 1 1 2 2 3 3 3 3 2 2          , 2 3 1 1 3          . 3 1 2 2 1 • Completeness to zeroth-order is proved by noticing that the following linear combinations are able to represent two independent basis vectors on a 2D element (verify this by yourself and express “ ξ 3 ).       , 2 3 1          . 3 2 2 74

Completenes Compl teness of of z zeroth or th order triangula der triangular bases r bases (we consider only the cur (w (w (we consider only the cur e consider only the curl-conf e consider only the curl conf conforming ones). conforming ones). orming ones). orming ones).              , , 1 1 2 2 3 3 3 3 2 2          , 2 3 1 1 3          . 3 1 2 2 1 The Nedelec conditions also require completeness of the The Nedelec conditions also require completeness of the curl to the same order as the bases. Completeness of the curl to zeroth-order follows from (verify this by use of (50, 53) of the 1997 paper) of the 1997 paper). 2             ˆ n n , 1 1 , 2 2 , 3 3  J 75

Higher or Higher order v Higher or Higher order v der vector bases der vector bases ector bases ector bases 76

Higher or Higher or Higher order bases Higher order bases der bases der bases • One can construct higher order bases complete to order p by forming the product of zeroth-order bases with complete f i h d f h d b i h l polynomial factors of order p . • The set of polynomials factors used may take one of several different forms chosen for convenience.       r s t homogeneou s : , r s t p 1 2 3       r s inhomogene ous : , 0 r s p , i j i j       interpolat ory : R (p, ) R (p, ) R (p, ) , r s t p r 1 s 2 t 3        hierarchic al : H ( , , ) , 0 i j k p ijk 1 2 3 77

Interpolatory vector functions I t l t t f ti • For interpolatory vector functions the key idea is to use shifted p y y polynomials of Silvester to move interpolation points away from two of the edges-those along with the tangential (for curl- conf.) and the normal (for div.-conf.) components of the 0 th - ) ( ) p order basis factor vanish. Interpolation nodes for curl- or divergence-conforming bases on triangular elements. Only nodes in basis subset Ω 1 i j k or Λ 1 i j k for p = 3 are shown. 78

Interpolation nodes for curl- or divergence-conforming bases on quadrilateral elements. Only nodes in basis subset Ω 3 ik; j ℓ or Λ 3 ik; j ℓ for p = 2 are shown. p 79

How to move the interpolation points i inside the cell id h ll • We use the shifted polynomial of Silvester 1 ˆ     R ( p , ) R ( p , )  i i 1 p • It is a polynomial of degree (i-1) in ξ , where ξ is in the i l i l f d ( ) i ξ h ξ i i h interval [0, 1] again subdivided in to p uniform subintervals. • The polynomial is unity at ξ = i/p and has zeros at ξ =1/ p , 2/ p , …, ( i -1)/ p. • There is no zero at ξ =0 There is no zero at ξ 0. • For i ¥ 1 one has  p ˆ       R R ( ( p p , ) ) R R ( ( p p , ) ) i i i

How to move the interpolation points How to move the interpolation points inside the cell • Shifted polynomial of Silvester  1    i 1            ( ( p k k ) ) 2 2 i i p 1 1 ˆ   R ( p , )    ( i 1 )! k 1 i   1 i 1  1 1 ˆ     R ( p , ) R ( p , )  i i 1 p   p ˆ    R ( p , ) R ( p , ) • For i ¥ 1 one has i i i

Interpolatory vector functions p y • The polynomials with interpolating nodes as shown in figure are of global order p= 3 and have the form: are of global order p 3, and have the form: ˆ ˆ       R ( p 2 , ) R ( p 2 , ) R ( p 2 , ) 1 j 2 3 i k     i i 0 0 , 1 1 , , p ; j j , k k 1 1 , 2 2 , , p 1 1       with i j k p 2 Interpolation nodes for curl- or divergence-conforming bases on triangular elements Only nodes in basis subset Ω 1 elements. Only nodes in basis subset Ω 1 i j k or Λ 1 or Λ 1 i j k for p = 3 are shown. for p = 3 are shown 82

References References • R.D. Graglia, D.R. Wilton and A.F. Peterson , “Higher order g g interpolatory vector bases for computational electromagnetics,” invited paper, special issue on “Advanced Numerical Techniques in Electromagnetics” IEEE Trans. Antennas Propagat. , vol. 45, no. 3, pp. 329-342, Mar. 1997 . • R.D. Graglia, D.R. Wilton, A.F. Peterson, and I.-L. Gheorma, “Higher order interpolatory vector bases on prism elements,” IEEE Trans. Antennas Propagat vol 46 no 3 pp 442-450 Mar 1998 Propagat. , vol. 46, no. 3, pp. 442-450, Mar. 1998 . • R. D. Graglia, and I.-L. Gheorma , “Higher order interpolatory vector bases on pyramidal elements,” IEEE Trans. Antennas Propagat. , vol. 47, no. 5, May 1999 . 83

Volumetric Elements Volumetric Elements 84

Volumetric Elements 85

Volumetric Elements Volumetric Elements 86

Volumetric Elements 87

Example of results for surface elements: INHOMOGENEOUSLY FILLED WAVEGUIDES   0 3 b  r h 2.5 5 a 2 4 3 4 2 kz/k0 a=2b, h=0.1b 1.5 3 m=1 5  r =10 7 6 8 2 1 6 1 0 1 0.5 0 0 analytical 2 3 0 0 1 2 3 4 5 6 7 8 9 10 + * FEM +, FEM k0*a 88

INHOM. FILLED WG - RELATIVE ERROR 0 10  0  r -1 10 OR P 1 P=1 LATIVE ERRO -2 10 -3 10 10 REL -4 P=2 10 P=3 -5 10 2 3 4 10 10 10 MATRIX DIMENSIONS 89

Normal field component at the air-dielectric interface Normal field component at the air dielectric interface. Fundamental mode Fundamental mode  0 b 15  r h P=3 P=2 a a ya/Eyd 10 a =2b, h =0.2 b P=1 P=1 Ey  r =10, k 0 a =7 10 k 7 5 ~ 1800 UNKNOWNS P 0 P=0 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 P N 0 1641 x 1 1 1873 1873 2 1897 90 3 1681

Image waveguide - - p=0 , (37 incognite), mesh densa - p=3 , (325 incognite), mesh lasca a espansione modale [1] 40 40 4 a` a =1.3 mm, b=1.6 mm, b a` =0.55 mm, b`=0.82 mm  rx =170,  ry =  ry 85 b` 3 35 ENZA GHz z modo 2 30 FREQUE modo 1 25 24 triangoli 291 triangoli 20 19 nodi 166 nodi 0 2 4 6 8 [1] J I Askne E L Kolberg L Pettersson ``Propagation in a waveguide partially filled with [1] J.I.Askne, E.L. Kolberg, L. Pettersson , Propagation in a waveguide partially filled with mm -1 1 K z anisotropic dielectric material``, IEEE Trans. MTT , vol.30, n5, pp.795-799, Maggio 1982 . 91

Singular vector bases Singular vector bases 92

References [1] R.D. Graglia, G. Lombardi, “Singular Higher-Order Complete Vector Bases for Finite Methods,” IEEE Trans. Antennas Propagat., Vol. 52, No. 7, pp. 1672-1685, J l 2004 July 2004 . [2] R.D. Graglia, G. Lombardi, “Hierarchical singular vector bases for the FEM solution of wedge problems ” invited paper for Proceedings of 2004 International solution of wedge problems, invited paper for Proceedings of 2004 International Symposium on Electromagnetic Theory, URSI-Commission B, Pisa, Italia, 23-27 May, 2004 . [3] R.D. Graglia, G. Lombardi, “Singular Higher Order Divergence-Conforming Bases of Additive Kind and Moments Method Applications to 3D Sharp-Wedge Structures,” IEEE Trans. Antennas Propagat., vol. 56, no. 12, pp. 3768-3788, , p g , , , pp , Dec. 2008 . DOI 10.1109/TAP.2008.2007390 93

Circular vaned waveguide – FEM application  =0  =1/2 94 element’s type and conformity

Circular vaned waveguide: eigenmodes  =1/2 •Singular behavior •Numerical precision •Regular mode •Singular mode 95

Modeling capability: very small thickness Double-vaned Circular homogeneous waveguide  =2/3 96

Modeling capability: Multiple singular verteces and curvilinear singular elements curvilinear singular elements  =  /2  =2/3  =     =1/2  =2  /3 2 /3  =3/4 97

The square PEC-plate problem at normal incidence – MoM Application pp The results at left (a c) were The results at left (a, c) were obtained by using the zeroth- order regular base (p = 0) on the dense mesh A. The results at right (b, d) were obtained by using the coarse mesh B and the singular base of order [p = 2, s = 0]. 98

The square PEC-plate problem at normal incidence 99

(10 × 1) PEC-strip Normal incidence Ex Singualar bases p = 2, s = 0 g p , 100