Windowing and Grid Techniques

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EE 533 Elecromagneic Analsis Using Finie Difference Time Domain Lecure #12 Windowing and Grid Techniques Lecure 12 These noes ma conain coprighed maerial obained under fair use rules.

1 EE 533 Elecromagneic Analsis Using Finie Difference Time Domain Lecure #12 Windowing and Grid Techniques Lecure 12 These noes ma conain coprighed maerial obained under fair use rules. Disribuion of hese maerials is sricl prohibied Slide 1 Lecure Ouline Review of Lecure 11 Windowing 2 Grid Technique Dielecric Averaging How o minimie he consequences of limiing he duraion of he simulaion. How o consruc he maerial arras of arbiraril shaped srucures. How o minimie he consequences of fiing devices o a aresian grid. Lecure 12 Slide 2 1

2 Review of Lecure 11 Lecure 12 Slide 3 Mawell s Equaion wih Normalied Elecric Field We normalied he quaniies associaed wih he elecric field 1 E E E 1 D DcD Mawell s equaions became 1 D H c r H E c D E r Lecure 12 Slide 4 2

Revised Flow of Mawell s Equaions ime Updae H from E r H E c Updae D from H 1 D H c & & Updae E from D D E r & Lecure 12 Slide 5 Epanded Mawell s Equaions These are he

3 Revised Flow of Mawell s Equaions ime Updae H from E r H E c Updae D from H 1 D H c & & Updae E from D D E r & Lecure 12 Slide 5 Epanded Mawell s Equaions These are he final form of Mawell s equaions from which we will formulae he 2D and 3D FDTD mehod. E E H c E E H c E E H c H H 1 D c H H 1 D c H H 1 D c D D D E E E Lecure 12 Slide 6 3

i,, k 1, i,, k i, H H H 1 D D,,, 1, i k i k k i,, k i,, k 2 2 2 2 c D D D E E E D D D E i,, k i,, k i,, k E i,, k i,, k i,, k E i,, k i,, k i,, k Lecure 12 Slide 7 Reduce Problems o Two Dimensions

4 Finie Difference Approimaions E E H c E E H c E E H c E E E E c 1, 1 H H i, k i,, k i,, k i,, k i,, k i,, k i,, k 2 2 i,, k i,, k 1 1, H H i,, k i,, k i, k i,, k i,, k 2 2 E E E E c i1,, k i,, k i, 1, k i,, k,,,, E E E i,, k i k i k E H H 2 2 c H H 1 D c H H 1 D c H H 1 D c 1 H H 1,,, 1, i k i k i,, k i,, k i,, k i,, k H H D D c i, k, i, k, 1 i, k, i1, k, H H H H 1 D D c H i,, k i,, k 1, i,, k i, H H H 1 D D,,, 1, i k i k k i,, k i,, k c D D D E E E D D D E i,, k i,, k i,, k E i,, k i,, k i,, k E i,, k i,, k i,, k Lecure 12 Slide 7 Reduce Problems o Two Dimensions Someimes i is possible o describe a phsical device using us wo dimensions. Doing so dramaicall reduces he numerical complei of he problem and is ALWAYS GOOD PRATIE. absorbing B periodic B periodic B absorbing B Lecure 12 Slide 8 4

5 Two Disinc Modes E Mode H E E E E H H H Mode E E E E c E H c H E i, i, i, D E D H 1 D c H E c H i, 1 i,, i E 1, i i,, i E H H H 1 D i, i, 1 i, i, i, H H H i, 1 D D H 2 2 c 2 2 c i, i1, i, i, H H H 1 D i, H H H i, 1 D D 2 2 H 2 2 c c i, i, i, D E D E D E i, i, i, D E Lecure 12 Slide 9 E E E E 1,, i i H H H H i, i, 1 i, E E i, i, i, i, H H 2 2 i, i, i, i, H H 2 2, 1 D 2 c E E E E 1, 1 i i, i, i,, i E c c D i, i, H i E i, i, H H 2 2 i, i, c FDTD Algorihm for E Mode E i, 1 i, i 1, i, E E E E N i N E i, in, i, i, E E N i N & c c H H H H i, i,,,,, E i i i E i, i i, & ime i, i1, i, i, 1 H H H H for i 1 and 1 1, 1, 1, 1 H H H for i 1 and 1 H H H for i1 and 1 1,1 1,1 H H 2 2 for i 1 and 1 i, H i i1,1,1 i 2,1 & D D c E i, i, i, H Simple sof source D D gt,, isrc src isrc src 1 i, i, i, D & & Lecure 12 Slide 1 5

6 Lecure 12 Slide 11 Time Response from FDTD FDTD is a ime domain mehod so i is he ransien response of a device ha is characeried. A Fourier ransform mus be used o calculae frequenc domain, or sead sae, quaniies. refleced field ransmied field source Lecure 12 Slide 12 6

7 Duraion of he Simulaion (i.e. STEPS) Wha happens when he simulaion is sopped a ime s before he simulaion is finished? s s refleced field ransmied field s source Lecure 12 Slide 13 Eample Impulse Response FDTD produces he impulse response h() of a device. In principle, he impulse response eends o infini. h The specrum of he infinie impulse response ma be somehing like his H Lecure 12 Slide 14 7

8 Mahemaical Descripion of Finie Duraion When a simulaion is sopped a ime s, i is us like mulipling he infinie impulse response b a pulse funcion w() wih duraion s. h w This is he resuling infinie impulse response when he simulaion ime is runcaed. h w Lecure 12 Slide 15 A Proper of he Fourier Transform The Fourier ransform of he produc of wo funcions is he convoluion of he Fourier ransform of he wo funcions. In erms of our funcions h() and w(), his is F h w F h F w H W Lecure 12 Slide 16 8

9 Fourier Transform of he Window Funcion The window funcion and is Fourier ransform are: w W Lecure 12 Slide 17 H() is Blurred b he Window Funcion Ever poin H() is esseniall blurred b he window funcion W() due o he convoluion. W H H W Lecure 12 Slide 18 9

10 Severi Trend for Windowing w W h H Long simulaion ime Wide window Narrow window specrum Less blurring High frequenc resoluion H W w W h H Moderae simulaion ime Moderae window Moderae window specrum More blurring Reduced frequenc resoluion H W w h W H Shor simulaion ime Narrow window Wide window specrum Much blurring Poor frequenc resoluion H W Lecure 12 Slide 19 Sampling and Windowing in FDTD Given he ime sep, he Nquis sampling heorem quanifies he highes frequenc ha can be resolved. 1 1 fma Noe: In pracice, ou should se N 2 Nf ma Given he FDTD simulaion runs for STEPS number of ieraions, he frequenc resoluion is 2 fma 1 f STEPS STEPS Therefore, o resolve he frequenc response down o a resoluion of f, he number of ieraions required is 1 1 STEPS f Noes: 1. This is relaed o he uncerain principle. 2. You can calculae kernels for ver finel spaced frequenc poins, bu he acual frequenc resoluion is sill limied b he windowing effec. Your daa will be blurred. Lecure 12 Slide 2 1

11 2 Grid Technique Lecure 12 Slide 21 Wha is he 2 Grid Technique? (1 of 3) Define Device 2 GRID HERE ONLY!! alculae Grid Build Device on Grid + PML alculae Source Iniialie FDTD Done? Updae H from E Handle H Source Updae E from H Handle E Source Updae Transforms Visualie Use of he 2 grid occurs ONLY when building he device. I is no used anwhere else! Lecure 12 Slide 22 11

Device on 1 Grid Device 1 Grid N N d d VERY DIFFIULT STEP!! Lecure 12 Slide 23 Wha is he 2 Grid Technique?

12 Wha is he 2 Grid Technique? (2 of 3) This is he radiional approach for building devices on a Yee grid. I is ver edious and cumbersome o deermine which field componens reside in which maerial. Device on 1 Grid Device 1 Grid N N d d VERY DIFFIULT STEP!! Lecure 12 Slide 23 Wha is he 2 Grid Technique? (3 of 3) The 2 grid echnique simplifies how devices are buil ino he permiivi and permeabili arras. Device N2 = 2*N N2 = 2*N d2 = d/2 d2 = d/2 2 Grid Device on 2 Grid eas 1 Grid N N d d eas eas Device on 1 Grid Lecure 12 Slide 24 12

Recall he Yee Grid 1D Yee Grid E 2D Yee Grids E 3D Yee Grid H E Mode H H E Mode H E H E E H E Mode E H H Mode E E H We define ER a he same poins as E. We define ER a he same poins as E. We define ER a he same poins as E. We define UR a he same poins as H.

13 Recall he Yee Grid 1D Yee Grid E 2D Yee Grids E 3D Yee Grid H E Mode H H E Mode H E H E E H E Mode E H H Mode E E H We define ER a he same poins as E. We define ER a he same poins as E. We define ER a he same poins as E. We define UR a he same poins as H. We define UR a he same poins as H. We define UR a he same poins as H. Lecure 12 Slide Yee Grid for he E Mode i Acual 2D Grid H H E Simplified Grid The field componens are phsicall posiioned a he edges of he cell. The simplified represenaion shows he fields inside he cells o conve more clearl which cell he are in. Lecure 12 Slide 26 i 13

i i i The 2 grid concep is useful because we can creae devices (or PMLs) on he 2 grid wihou having o hink abou where he differen field componens are locaed.

14 The 2 Grid The onvenional 1 Grid Due o he saggered naure of he Yee grid, we are effecivel geing wice he resoluion. I now makes sense o alk abou a grid ha is a wice he resoluion, he 2 grid. i i i The 2 grid concep is useful because we can creae devices (or PMLs) on he 2 grid wihou having o hink abou where he differen field componens are locaed. In a second sep, we can easil pull off he values from he 2 grid where he eis for a paricular field componen. H H E Lecure 12 Slide 27 2 Grid Technique (1 of 9) We define our ordinar 1 grid as usual. The oupu of his sep is he number of cells in he grid N and N and he sie of he cells in he grid d and d. % DEFINE GRID N = 5; N = 6; d = 1; d = 1; 28 14

15 2 Grid Technique (2 of 9) Recall how he various funcions overla ono he grid. Funcions assigned o he same grid cell are in phsicall differen posiions and ma reside in differen maerials as a resul. % DEFINE GRID N = 5; N = 6; d = 1; d = 1; 29 2 Grid Technique (3 of 9) I is like we are geing wice he resoluion due o he saggering of he funcions. In order o sor ou wha values go where, we consruc a 2 grid a wice he resoluion of he original grid. The 2 grid occupies he same phsical amoun of space as he original grid. % DEFINE GRID N = 5; N = 6; d = 1; d = 1; % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; 3 15

9) We sar b building our obec on he 2 grid, ignoring anhing abou or original grid for now.

16 2 Grid Technique (4 of 9) Le s sa we wish o consruc a clinder of radius 2 on our grid. % DEFINE GRID N = 5; N = 6; d = 1; d = 1; % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; 31 2 Grid Technique (5 of 9) We sar b building our obec on he 2 grid, ignoring anhing abou or original grid for now. % DEFINE GRID N = 5; N = 6; d = 1; d = 1; % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; ER2 % REATE IRLE r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.^2 + Y2.^2) <= r^2; 32 16

17 2 Grid Technique (6 of 9) Given he obec on he 2 grid, we erac ER b grabbing values from ER2 ha correspond o he locaions of ER. ER ER2 % DEFINE GRID N = 5; N = 6; d = 1; d = 1; % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.^2 + Y2.^2) <= r^2; % EXTRAT 1X GRID PARAMETERS ER = ER2(2:2:N2,1:2:N2); 33 2 Grid Technique (7 of 9) We hen erac ER b grabbing values from ER2 ha correspond o he locaions of ER. ER ER ER2 % DEFINE GRID N = 5; N = 6; d = 1; d = 1; % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.^2 + Y2.^2) <= r^2; % EXTRAT 1X GRID PARAMETERS ER = ER2(2:2:N2,1:2:N2); ER = ER2(1:2:N2,2:2:N2); 34 17

18 2 Grid Technique (8 of 9) ER Las, we erac ER b grabbing values from ER2 ha correspond o he locaions of ER. ER ER ER2 % DEFINE GRID N = 5; N = 6; d = 1; d = 1; % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.^2 + Y2.^2) <= r^2; % EXTRAT 1X GRID PARAMETERS ER = ER2(2:2:N2,1:2:N2); ER = ER2(1:2:N2,2:2:N2); ER = ER2(1:2:N2,1:2:N2); 35 2 Grid Technique (8 of 9) ER ER % DEFINE GRID N = 5; N = 6; d = 1; d = 1; ER ER, ER, and ER are he oupus of he 2 grid echnique. The are defined on he original 1 grid. % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.^2 + Y2.^2) <= r^2; % EXTRAT 1X GRID PARAMETERS ER = ER2(2:2:N2,1:2:N2); ER = ER2(1:2:N2,2:2:N2); ER = ER2(1:2:N2,1:2:N2); 36 18

ER2 ER ER % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.

19 2 Grid Technique (9 of 9) Afer building ER, ER and ER, he 2 grid is no longer used anwhere. % DEFINE GRID N = 5; N = 6; d = 1; d = 1; All of he 2 grid parameers ma be deleed a his poin because he are no longer needed. ER2 ER ER % 2X GRID N2 = 2*N; N2 = 2*N; d2 = d/2; d2 = d/2; % REATE YLINDER r = 2; a2 = [:N2-1]*d2; a2 = [:N2-1]*d2; a2 = a2 - mean(a2); a2 = a2 - mean(a2); [Y2,X2] = meshgrid(a2,a2); ER2 = (X2.^2 + Y2.^2) <= r^2; % EXTRAT 1X GRID PARAMETERS ER = ER2(2:2:N2,1:2:N2); ER = ER2(1:2:N2,2:2:N2); ER = ER2(1:2:N2,1:2:N2); No more 2 grid!!! 37 MATLAB ode for Parsing Ono 1 Grid ER = ER2(2:2:N2,1:2:N2,1:2:N2); ER = ER2(1:2:N2,2:2:N2,1:2:N2); ER = ER2(1:2:N2,1:2:N2,2:2:N2); UR = UR2(1:2:N2,2:2:N2,2:2:N2); UR = UR2(2:2:N2,1:2:N2,2:2:N2); UR = UR2(2:2:N2,2:2:N2,1:2:N2); E Mode ER = ER2(1:2:N2,1:2:N2); UR = UR2(1:2:N2,2:2:N2); UR = UR2(2:2:N2,1:2:N2); H Mode ER = ER2(2:2:N2,1:2:N2); ER = ER2(1:2:N2,2:2:N2); UR = UR2(2:2:N2,2:2:N2); Lecure 12 Slide 38 19

20 Dielecric Averaging Lecure 12 Slide 39 Wha is Dielecric Averaging? Suppose we have a grid, bu he sie of he device is no an eac ineger number of grid cells. Wha can we do? r 2 1 r1 6 We look a a close up of he cell in quesion and calculae a weighed average. r1 = 6. 3% 7% r 2.5 r2 = 1.,eff r,eff 3% 6. 7% Based on he above, we build our device on he grid as follows: Lecure 12 Slide 4 2

Reason for Dielecric Averaging Dielecric averaging improves he rae of convergence. You can ge awa wih coarser grid resoluion using dielecric averaging.

21 Reason for Dielecric Averaging Dielecric averaging improves he rae of convergence. You can ge awa wih coarser grid resoluion using dielecric averaging. Your simulaions will fun faser and be more memor efficien. Perhaps NRES=1 insead of NRES=2. Lecure 12 Opics Leers, Vol. 31, Issue 2, pp (26) Slide 41 Tpical Saircase Approimaion Represenaion on Grid Phsical Device Device Modeled Lecure 12 Slide 42 21

Represenaion wih Dielecric Averaging Represenaion on Grid wih Dielecric Averaging Phsical

This will arificiall suppress reflecions from our srucures.

a higher resoluion grid Blurred version of he high resoluion device Erac cener value from

22 Represenaion wih Dielecric Averaging Represenaion on Grid wih Dielecric Averaging Phsical Device Device Modeled DO NOT BLUR MORE THAN ONE PIXEL DISTANE. This will arificiall suppress reflecions from our srucures. Lecure 12 Slide 43 Procedure o Perform Averaging Phsical Device Saircase represenaion on a higher resoluion grid Blurred version of he high resoluion device Erac cener value from each low resoluion cell one piel on 1 grid 1 We can use a convoluion o calculae he average dielecric consan in each cell. Lecure 12 Slide 44 22

Animaion of Dielecric Averaging Lecure 12 Slide 45 Produc of Two Funcions Suppose ou have he produc of wo funcions ha ou are approimaing numericall.

23 Animaion of Dielecric Averaging Lecure 12 Slide 45 Produc of Two Funcions Suppose ou have he produc of wo funcions ha ou are approimaing numericall. f a b Slow convergence is encounered whenever one or boh of hese funcions is disconinuous. Suppose a() has a disconinui. Numerical convergence can be significanl improved b smoohing he disconinuous funcion a he disconinui. f a b Lecure 12 Slide 46 23

24 Double Disconinui (1 of 2) Suppose boh a() and b() are disconinuous a he same poin. I is ver difficul o improve he convergence of his problem. There eiss, however, a special case where a() and b() are disconinuous a he same poin, bu heir produc is coninuous. Tha is, f() is coninuous. We can rearrange he equaion so ha onl a single disconinui eiss on each side. 1 f b a Lecure 12 Slide 47 Double Disconinui (2 of 2) We can now improve convergence b smoohing 1/a(). 1 a f b Moving he a() erm o he righ hand side leads o 1 f b 1 a We conclude ha in he double disconinui case where he produc is coninuous, we smooh he reciprocal of a() and hen reciprocae he smoohed funcion. Lecure 12 Slide 48 24

25 Summar of Smoohing When all funcions are coninuous, no smoohing is needed. f a b When one of he funcions has a sep disconinui, convergence is improved b smoohing ha funcion. f a b When boh a() and b() are disconinuous a he same poin, bu heir produc is coninuous, convergence is improved b smoohing he reciprocal of one of he funcions. 1 1 f a b Lecure 12 Slide 49 Smoohing and Mawell s Equaions In Mawell s equaions, we have he produc of wo funcions D E r The dielecric funcion is disconinuous a he inerface beween wo maerials. Boundar condiions require ha E E E 1, 2, E 1 1, 2 2, Tangenial componen is coninuous across he inerface Normal componen is disconinuous across he inerface, bu he produc of E is coninuous. We conclude ha we mus smooh he dielecric funcion differenl for he angenial and normal componens. This implies ha he smoohed dielecric funcion will be anisoropic and described b a ensor. Lecure 12 Slide 5 25

26 High Level Formulaion for Mawell s Equaions Firs, we decompose he elecric field ino angenial and normal componens a an inerface. E E E r r Second, we mulipl ou his equaion. E E E r r r Third, we associae differen dielecric funcions wih he differen field componens. Before smoohing, he are he same. r E E E Finall, we smooh he wo dielecric funcions differenl according o our rules. 1 1 r E E E Lecure 12 Slide 51 Deailed Formulaion (1 of 7) In Mawell s equaions, we have D E The dielecric ensor is Lecure 12 Slide 52 26

This called he normal vecor field. nˆ,, P. Go, T. Schuser, K. Frenner, S. Rafler, W. Osen, Normal vecor mehod for he RWA Lecure 12 wih auomaed vecor field generaion, Op.

27 Deailed Formulaion (2 of 7) The elecric field can be wrien as he sum of parallel and perpendicular polariaions. E E E For an arbiraril shaped device, hese componens can var across he grid. Suppose we could calculae a vecor funcion hroughou he grid ha is normal o all he inerfaces. This called he normal vecor field. nˆ,, P. Go, T. Schuser, K. Frenner, S. Rafler, W. Osen, Normal vecor mehod for he RWA Lecure 12 wih auomaed vecor field generaion, Op. Epress 16(22), (28). Slide 53 Deailed Formulaion (3 of 7) The perpendicular componen of E can be compued from he normal vecor field as follows. E nˆ nˆ E I follows ha he parallel componen of E is E EE E nˆ nˆe Lecure 12 Slide 54 27

28 Deailed Formulaion (4 of 7) In mari noaion, he perpendicular componen can be wrien as E n n E n E n E E nˆnˆ E E n n E n E n E E n n E n E n E 2 n nn nn E 2 nn n nn E 2 nn nn n E Lecure 12 Slide 55 Deailed Formulaion (5 of 7) Le 2 n nn nn ˆ 2 N nn n nn 2 nn nn n The perpendicular and parallel componens are hen E nˆnˆ E Nˆ E E ˆ ENE INˆ E Lecure 12 Slide 56 28

29 Deailed Formulaion (6 of 7) The consiuive relaion can be wrien in erms of he parallel and perpendicular componens of he E field. D E E E E We smooh he dielecric funcions differenl according o our rules for opimum convergence. D E E D E E Lecure 12 Slide Deailed Formulaion (7 of 7) Puing all of his ogeher leads o D E E I Nˆ E Nˆ E I Nˆ Nˆ E We can derive an effecive dielecric ensor from his equaion. ˆ smooh ˆ I N N N ˆ Lecure 12 Slide 58 29

Slide 59 Dielecric Averaging of a Sphere (1 of 2) Given a sphere wih dielecric consan r =5.

30 Summar of Formulaion Given he simulaion problem defined b D E We can improve he convergence rae b smoohing he dielecric funcion according o smooh ˆN 2 n nn nn ˆ 2 N nn n nn 2 nn nn n 1 1 Lecure 12 Slide 59 Dielecric Averaging of a Sphere (1 of 2) Given a sphere wih dielecric consan r =5. in air and in a grid wih N =N =N =25 cells, he dielecric ensor afer averaging is cross secion cross secion Lecure 12 Slide 6 3

31 Dielecric Averaging of a Sphere (2 of 2) A 3D visualiaion is: Lecure 12 Slide 61 Simulaion Eample Recall Eam 1, Problem 1 Response wihou Smoohing 1 db improvemen Response wih Smoohing Lecure 12 Slide 62 31

32 ommens on Dielecric Averaging Even if he original dielecric funcion is isoropic, he averaged dielecric funcion is anisoropic. Anisoropic averaging requires calculaing he normal vecor field. This can be difficul, especiall for arbirar srucures. onvergence sill ends o improve even when onl isoropic averaging is used. smooh The E mode does no require anisoropic averaging of he dielecric funcion. The H mode does no require anisoropic averaging of he permeabili. Lecure 12 Slide 63 32

Advanced FDTD Algorithms

Advanced FDTD Algorithms EE 5303 Elecromagneic Analsis Using Finie Difference Time Domain Lecure #5 Advanced FDTD Algorihms Lecure 5 These noes ma conain coprighed maerial obained under fair use rules. Disribuion of hese maerials