Time Domain Method of Moments

Transcription

1 Time Domain Method of Moments Massahusetts Institute of Tehnology leture notes 1 Introdution The Method of Moments (MoM) introdued in the previous leture is widely used for solving integral equations in the frequeny domain. Yet, some attempts have been made reently at the use of the MoM in the time domain. We shall briefly expose this approah here. 2 Time domain equations The first step is of ourse to write Maxwell s equation and all other relations (onstitutive relations and ontinuity) in time domain: Ē( r, t) = t B( r, t) M( r, t), H( r, t) = t D( r, t) + J( r, t), (1a) B( r, t) = m( r, t), D( r, t) = ρ( r, t), (1b) D( r, t) = ɛē( r, t), B( r, t) = µ H( r, t), (1) J( r, t) + ρ( r, t) =, t M( r, t) + m( r, t) =. t (1d) For the time-domain MoM, it is easier to work with the potentials, and make use of the well-known retarded potentials theory. In view of doing this, we write the definition: H( r, t) = 1 Ā( r, t) µ (2a) Ē( r, t) = φ( r, t) tā( r, t). (2b) Both the vetor potential Ā and the salar potential φ satisfy the wave equation whih, in time-domain domain, writes: 2 2 Ā( r, t) ɛ µ t 2 Ā( r, t) = µ J( r, t), (3a) 2 2 t) φ( r, t) ɛ µ φ( r, t) = ρ( r,. t2 ɛ (3b) These potentials are linked by the time-domain Lorentz gauge: Ā( r, t) + ɛ µ φ( r, t) =. (4) t 1

2 2 Setion 2. Time domain equations We an defined also a time-domain Green s funtion whih satisfies the time-domain salar equation: ( t 2 )g( r, r, t, t ) = δ( r r ) δ(t t ), (5) whih solution is (in free-spae): 1 g( r, r, t, t r r ) = δ(t t r r ) t > t, (6) t < t. From this, the solution to the wave equation for Ā and φ an be written as: Ā( r, t) =µ dv dt J( r, t ) g( r, r, t, t ) = µ dv J( r, t R/) R, (7a) φ( r, t) = dv ρ( r, t R/), (7b) ɛ R where R = r r. These wave equations are known as the time retarded potentials, and essentially say that the potential (either Ā or φ) an be alulated at a given point in spae r and given time t from all previous times. From these equations, we an alulate the spae-time eletromagneti fields: H( r, t) = 1 dv J( r, τ), τ = t R/, (8a) R Ē( r, t) = 1 dv ρ( r, τ) ɛ R µ J( r, τ) t R. (8b) Let us ontinue with the eletri field first: Ē( r, t) = 1 [ 1 dv ɛ R ρ( r, τ) + ρ( r, τ) 1 ] µ R R t J( r, τ). (9) At this point, we need to use the following relations: R = R R, 1 R = R R 3, ρ( r, τ) = τ ρ( r, τ) τ = 1 R τ ρ( r, τ) = 1 R ρ( r, τ). R τ (1) (1a) (1b) We an therefore ontinue with the eletri field as: Ē( r, t) = 1 [ 1 R ɛ R 2 τ ρ( r, τ) + R ] R 3 ρ( r, τ) µ R τ J( r, τ) = 1 [ 1 ɛ τ ρ( r, τ) + 1 ] R R ρ( r, τ) R 2 µ R τ J( r, τ). (11)

3 3 We an perform the same type of alulations for the magneti field using the relation J( r, τ) = 1 R R τ J( r, τ). (12) We get: H( r, t) = 1 dv [ 1 R R 2 τ J( r, τ) R ] R 3 J( r, τ). (13) Upon gathering the expressions for the eletri and magneti field, we eventually get: Ē( r, t) = 1 H( r, t) = 1 {[ 1 τ ρ( r, τ) + 1 ] R R ρ( r, τ) ɛ R 2 µ } R τ J( r, τ) [ 1 τ J( r, τ) + 1 ] R J( r, τ) R R 2. (14a) (14b) Upon using the boundary onditions for the eletri and magneti field, we onstrut the integral equations in a standard way: EFIE: ˆn (Ēi + Ēsat ) = on PEC surfae ˆn Ēi ( r, t) + 1 ˆn ds [...] (15) MFIE: ˆn ( H i + H sat ) = J s. As we have seen before (in a previous lass), this integral equation is expressed in terms of the prinipal value of the integral with a 1/2 additional fator. Thus: 1 2 J( r, t) = ˆn H i ( r, t) + 1 ˆn P ds [...] (16) For the sake of omparison, we an write the MFIE in the frequeny domain and in the time domain: J( r) = 2ˆn H i ( r) + 2ˆn P ds J( r ) g( r, r ) r S, (17a) J( r, t) = 2ˆn H i ( r) + 1 [ 1 2π ˆn P ds τ J( r, τ) + 1 ] R J( r, τ) R R 2. (17b) Note that in the prinipal value, we essentially exlude the part for whih R =. Sine τ = t R/ and R, we always have that τ < t. The time domain equations therefore state that the urrent at loation r and time t is equal to a known term 2ˆn H i ( r, t) plus a term (integral) known from the past history of J. This is the basis for solving the time domain integral equation by iterative methods, the most well-known one being the marhing-on-in-time.

4 4 Setion 3. The marhing-on-in-time tehnique 3 The marhing-on-in-time tehnique 3.1 General equations The integral equation an often be ast in the following form: τ J( r, t) = J i ( r, t) + dv dt K( r, r, t t ) J( r, t ). (18) eq.1 S Note that we have a time integral also as in Eq. ( eq.1 18), J( r, t ) has not yet been set to satisfy any ausality ondition. Hene, we must then impose J i ( r, t) = for t <, r S. In order to apply the MoM, we disretize the urrent both in spae and in time: J( r, t ) = M m =1 n = J p (m, n ) P s ( r r m ) P t (t t n ), (19) where P denotes the simple pulse funtion. In addition, we also apply point-mathing, whih means that we take the following testing funtions: W mn ( r m, t n ) = δ( r r m ) δ(t n t) = δ( r r m ) δ(t t n ), (2) where we take t = min{r mm /}, R mm following example. = r m r m. The method is best illustrated on the 3.2 Example Let us onsider a 1D example governed by the following integral equation: g(x, t) = x K(x, x ) f(x, τ)dx, x [ x, x ], τ = τ(x, x, t) = t x x x Let us hose the following expansion for f:. (21) f(x, τ) a i j P i j (x, τ), (22) where the pulse basis funtions are defined as P i j = 1 for x [x i dx 2, x i + dx 2 ] and t [t j dt 2, t j + dt elsewhere. 2 ] (23) Note that we use the definitions: x i = i d x, t j = j d t, and d x = d t. In order to apply point mathing, we take the following testing funtions: w ij (x, t) = δ(x x i ) δ(t t j ). (24)

5 5 Upon expanding and testing, we get: g(x i, t j ) = g ij = = x x K(x i, x ) (i )dx (i 1 2 )dx dx a i j P i j (x, τ)δ(t t j ) a i j P i j (x, τ)k(id x, x )δ(t t j ). (25) Coming bak to the definition of τ, we write (with the test and the expansion): suh that the oeffiient a i j τ = t j x i x i = jd t i i d x = (j i i )d t, (26) beomes a i,j i i. The integral equation beomes: g ij = a i,j i i (i )dx (i 1 2 )dx dx K(id x, x ). (27) We an define the term and rewrite the previous system as g ij = Z ii a i,j i i Z ii = (i )dx (i 1 2 )dx dx K(id x, x ) (28) = Z ii a ij + Z i,i 1 a i 1,j 1 + Z i,i 2 a i 2,j Z i1 a 1,j i+1 + Z i,i+1 a i+1,j 1 + Z i,i+2 a i+2,j Z 1,N a N,j N+i. (29) In this equation, only the first term involves time step j, all the others terms being at j 1, j 2,... Therefore, we an solve for a ij : a ij = 1 Z ii [ g ij ] Z ii a i,j i i. (3) i i The value of all a ik are known for k < j, so that a ij is ompletely speified in losed form by those and the present value of g ij. This proess is known as a 1D marh-on-in-time approah. Time-domain MoM is nowadays in its early stage and, although it has been suessfully applied to various simple situations, still suffers from numerial instabilities. More work is in progress...