11/14/018 EE 5303 Elecomagneic Analsis Using Finie Diffeence Time Domain Lecue #4 caeing Analsis Lecue 4 These noes ma conain copighed maeial obained unde fai use ules. Disibuion of hese maeials is sicl pohibied lide 1 Lecue Ouline Inoducion FDTD fo caeing imulaions Fomulaion of NFFF Tansfomaion Implemenaion NFFF in Two Dimensions imulaion Eamples Lecue 4 lide 1
11/14/018 Inoducion Lecue 4 lide 3 Animaion of caeing Analsis Finie obec TF/F on all suounding sides Ineesed pimail in he scaeed field Usuall he goal is o quanif scaeing in he fa field. Lecue 4 lide 4
11/14/018 Wha is NFFF Tansfomaion? (1 of ) uppose we wan o know he elecomagneic fields VERY fa awa fom an obec. ow de we do his? Do we us use an eemel lage gid? Obec NO!! Lecue 4 lide 5 Wha is NFFF Tansfomaion? ( of ) Insead, we pefom a small simulaion and use he NFFF echnique o calculae he fields ve fa awa. This is MUC moe efficien! Obec NFFF BIG IMULATION!! Lile simulaion + NFFF Lecue 4 lide 6 3
11/14/018 Repesening he Fa Field Usuall he fa field is ploed in clindical (D) o spheical (3D) coodinaes. Lecue 4 lide 7 Basic Ouline of caeing Analsis ep 1 Build obec on gid ep imulae obec ep 3 Calculae sead sae field ep 4 Pefom NFFF ep 5 Pos pocess NFFF Lecue 4 lide 8 4
11/14/018 FDTD fo caeing imulaions Lecue 4 lide 9 Gid cheme scaeed field oal field PML -lo PML -hi -lo -hi PML Ou souce is bounded on all sides so ha he enie oue poion of he gid is scaeed field. PML Lecue 4 lide 10 5
11/14/018 Recall he Cul Calculaions C C C E E E 1 E E Mode i1, E E i1, 1 C C C E E E Mode 1 E E i1, E E E E i1, 1 E E Noe: he ems in ed idenif ems ha equie modificaion in he TF/F famewok. Lecue 4 lide 11 -lo ide On he -lo side, onl CE and C equie modificaion. C E i1 1,, k C E E i,, k 1 i, E E 1 1 1 i 1,, k i, i 1, i 1, 1 1 i1 i1 1 1 i1, i, 1,sc,, 1 i1, E i, i 1, i1, 1 i1 1,,sc Lecue 4 lide 1 6
11/14/018 -hi ide On he -hi side, onl CE and C equie modificaion. C C E i, E E i 1, i 1, E E i, i 1, i, i, i i 1, Lecue 4 lide 13,sc i 1, i 1, 1 i 1, i 1, i, 1, i 1, 1 E i,,sc -lo ide On he -lo side, onl CE and C equie modificaion. C E 1 1 C 1 E E 1 E E 1 1 1 11 E,sc 1 i1, 1 1 1 1 1 1 i1, 1 1 11 1 1,sc Lecue 4 lide 14 7
11/14/018 -hi ide On he -hi side, onl CE and C equie modificaion. C C E E 1 E 1 E E E,sc 1 1 i1, 1 1, 1 i 1 i1, 1 1,sc Lecue 4 lide 15 Deiving a inusoidal ouce Funcion The geneal epession fo he nomalied elecic field componen of a plane wave is E, ; Pcos k Pcos kk The equivalen epession can be deived fo he magneic field b subsiuing he above epession ino Faada s law. ˆ cos ˆ cos k P k k P k k Lecue 4 lide 16 8
11/14/018 inusoidal ouce Funcions fo he E Mode The E mode has P = P = 0 and we se P = 1. Thus, he souce funcions educe o, ; cos E k k k k k ˆ, ; cos k k k ˆ, ; cos Lecue 4 lide 17 Numeical Equaions fo inusoidal ouce Funcions fo he E Mode In ems of aa indices, he souce funcions ae E E i1,, k,sc T i1 1,,sc T 0.5 i 1,,sc T i,,sc T 0.5 E E 1,sc T cos Tk ik 1 ˆ k cos T 0.5k i 0.5k 1 cos Tk i 1 k kˆ cos T 0.5k i 0.5 k costkik 1 1 1,sc T 0.5 1,sc T,sc T 0.5 ˆ k cos T 0.5 ki k 1 0.5 cos Tkik 1 ˆ k cos T 0.5ki k 0.5 Lecue 4 lide 18 9
11/14/018 Code nippes fo inusoidal ouce % TF/F FOR CURL OF E E0 = RAMP(T); %lo Esc = E0*cos(*pi*f0*T*d - k*nlo*d - k*[nlo:nhi]*d); CE(nlo-1,nlo:nhi) = CE(nlo-1,nlo:nhi) + Esc/d; %hi Esc = E0*cos(*pi*f0*T*d - k*(nhi + 1)*d - k*[nlo:nhi]*d); CE(nhnlo:nhi) = CE(nhnlo:nhi) - Esc/d; %lo Esc = E0*cos(*pi*f0*T*d - k*[nlo:nhi]'*d - k*nlo*d); CE(nlo:nhnlo-1) = CE(nlo:nhnlo-1) - Esc/d; %hi Esc = E0*cos(*pi*f0*T*d - k*[nlo:nhi]'*d - k*(nhi + 1)*d); CE(nlo:nhnhi) = CE(nlo:nhnhi) + Esc/d; % TF/F FOR CURL OF 0 = + (k/k0)*ramp(t); 0 = - (k/k0)*ramp(t); %lo sc = 0*cos(*pi*f0*(T + 0.5)*d - k*(nlo - 0.5)*d - k*[nlo:nhi]*d); C(nlo,nlo:nhi) = C(nlo,nlo:nhi) - sc/d; %hi sc = 0*cos(*pi*f0*(T + 0.5)*d - k*(nhi + 0.5)*d - k*[nlo:nhi]*d); C(nhi+1,nlo:nhi) = C(nhi+1,nlo:nhi) + sc/d; %lo sc = 0*cos(*pi*f0*(T + 0.5)*d - k*[nlo:nhi]'*d - k*(nlo - 0.5)*d); C(nlo:nhnlo) = C(nlo:nhnlo) + sc/d; %hi sc = 0*cos(*pi*f0*(T + 0.5)*d - k*[nlo:nhi]'*d - k*(nhi + 0.5)*d); C(nlo:nhnhi+1) = C(nlo:nhnhi+1) - sc/d; Lecue 4 lide 19 Animaion of CW ouce Lecue 4 lide 0 10
11/14/018 Calculaing k and k (1 of ) The TF/F spans a lage enough amoun of space ha he numeical dispesion due o he Yee gid causes poblems. You mus compensae fo his. Uncompensaed Compensaed Lecue 4 lide 1 Calculaing k and k ( of ) ep 1 Define in dashboad. ep Calculae he efacive inde whee souce is ineced. n sc,sc,sc ep 3 Calculae k and k. k k n cos k k n 0 sc 0 sc ep 3 Calculae fudge faco f. c sin 1 sin 1 sin 0 f k k nsc sin f0 ep 4 Adus k and k. k fk k fk Noe: An alenae mehod is descibed fo he Gaussian souce ha usuall woks bee. Lecue 4 lide 11
11/14/018 1 Lecue 4 lide 3 Analical Equaions fo a Gaussian Pulse Fo a Gaussian pulse, he field componens ae 0 ˆ ˆ k k c, ; ep ˆ, ; ep ˆ, ; ep E k k This ime dela ensues he pulse begins us ouside of he TF/F egion egadless of he angle of incidence. Lecue 4 lide 4 Numeical Equaions fo a Gaussian Pulse Fo a Gaussian pulse, he field componens ae 0 0 0 ˆ ˆ, ; ep ˆ ˆ 0.5 0.5 ˆ, ; ep ˆ ˆ 0.5 0.5 ˆ, ; ep T k i k c E i T T k i k c i T k T k i k c i T k
11/14/018 Code nippes fo Gaussian ouce % TF/F FOR CURL OF E E0 = ep(-((t*d - T0)/au).^); E0 = E0(1::N,1::N); %lo Esc = E0(nlo,nlo:nhi); CE(nlo-1,nlo:nhi) = CE(nlo-1,nlo:nhi) + Esc/d; %hi Esc = E0(nhi+1,nlo:nhi); CE(nhnlo:nhi) = CE(nhnlo:nhi) - Esc/d; %lo Esc = E0(nlo:nhnlo); CE(nlo:nhnlo-1) = CE(nlo:nhnlo-1) - Esc/d; %hi Esc = E0(nlo:nhnhi+1); CE(nlo:nhnhi) = CE(nlo:nhnhi) + Esc/d; % TF/F 0 = ep(-(((t + 0.5)*d - T0)/au).^); 0 = + (k/k0)*0(::n,1::n); 0 = - (k/k0)*0(1::n,::n); %lo sc = 0(nlo-1,nlo:nhi); C(nlo,nlo:nhi) = C(nlo,nlo:nhi) - sc/d; %hi sc = 0(nhnlo:nhi); C(nhi+1,nlo:nhi) = C(nhi+1,nlo:nhi) + sc/d; %lo sc = 0(nlo:nhnlo - 1); C(nlo:nhnlo) = C(nlo:nhnlo) + sc/d; %hi sc = 0(nlo:nhnhi); C(nlo:nhnhi+1) = C(nlo:nhnhi+1) - sc/d; % CALCULATE TIME GRADIENT ACRO X GRID del = *cos(hea); del = *sin(hea); T0 = (cos(hea)*(x + del) + sin(hea)*(y + del))/c0; Lecue 4 lide 5 Animaion of Gaussian Pulse ouce Lecue 4 lide 6 13
11/14/018 Animaion of Gaussian Pulse ouce Wih a caeing Obec Lecue 4 lide 7 Compensaing fo Dispesion fo a Gaussian ouce (1 of ) The TF/F spans a lage enough amoun of space ha he numeical dispesion due o he Yee gid causes poblems. You mus compensae fo his. Uncompensaed Compensaed Lecue 4 lide 8 14
11/14/018 Compensaing fo Dispesion fo a Gaussian ouce ( of ) ep 1 Define in dashboad. ep Calculae he efacive inde whee souce is ineced. n sc,sc,sc ep 3 Calculae k and k. k k0nsccos k k n sin 0 sc ep 4 Calculae fudge faco f. c ep 5 Adus maeial aas ER and UR accoding o f. UR = f*ur; ER = f*er; 1 sin 1 sin 0 f k k nsc sin f0 Noe: This mehod of compensae woks fo he sinusoidal souce as well. Lecue 4 lide 9 Fomulaion of NFFF Tansfomaion Lecue 4 lide 30 15
11/14/018 Elecomagneic Bounda Condiions Magneic Field a an Ineface J, T 1, T s J, T 1, T s nˆ ˆ n1 Js nˆ J 1 s E, ˆn J s E, 1 1 M s Elecic Field a an Ineface E E M, T 1, T s E E M, T 1, T s nˆe ˆ ne1 Ms nˆ E E M 1 s J nˆ s 1 nˆ suface nomal poining fom 1 o M nˆ E E s 1 Lecue 4 lide 31 uface Equivalence Theoem E, ˆn E, M s E, = E, 1 1 J s The fa field E and is compleel descibed b he suface cuens J s and M s flowing aound a closed suface. Lecue 4 lide 3 16
11/14/018 Calculaing he uface Cuens ince we ae onl ineesed in he fa fields E and, we ae fee o le E 1 and 1 be whaeve is convenien. Le E. 1 1 0 ˆn E, M s We can now calculae he suface cuens as J ˆ s n M nˆ E s 0 J s Noe: Calculaing J and M in his wa makes i onl possible o calculae fields ouside of he suface. Lecue 4 lide 33 Veco Poenials We wish o calculae he fields ve fa fom he suface. I is useful o do his using he veco poenials. kr e A J ds 4 R kr e F M ds 4 R R disance fom poin on suface s o he obsevaion poin. Lecue 4 lide 34 17
11/14/018 Fa Field Appoimaion In he fa field, R is eemel lage. cos R fo phase calculaions fo ampliude calculaions Noe: cos ˆ ˆ R Lecue 4 lide 35 Veco Poenials in he Fa Field Using he fa field appoimaion fo R, he equaions fo calculaing he veco poenials become k cos k e cos e A J ds 4 F M cos 4 ds cos k e kcos J e ds 4 k e kcos Me ds 4 Las, we wie hese equaions in spheical coodinaes. k e A,, N, 4 N J e ds kcos, k e F,, L, 4 L M e ds cos, k I is impoan o noe he,, and dependencies in hese equaions. Lecue 4 lide 36 18
11/14/018 Caesian o pheical Coodinaes (1 of ) andad FDTD poduces field quaniies in Caesian coodinaes. This means he cuen ems J s and M s will be in Caesian coodinaes. Usuall fa field analsis is done is spheical coodinaes so hese funcions mus be conveed o spheical coodinaes. J sincos sinsin cos J J coscos cossin sin J J sin cos 0 J N N aˆ N aˆ N aˆ kcos N Jsin cos Jsinsin J cos e ds J kcos N JcoscosJcossinJsine ds J N J sinj cos e ds kcos J Lecue 4 lide 37 Caesian o pheical Coodinaes ( of ) And he calculaion fo L, is M sincos sinsin cos M M coscos cossin sin M M sin cos 0 M L L aˆ L aˆ L aˆ Lecue 4 lide 38 kcos L Msincos M sinsin Mcos e ds M kcos L Mcoscos M cossin Msin e ds M kcos L Msin M cos e ds M 19
11/14/018 N and L Onl Depend on and (1 of 3) N, and L, ae fa field quaniies so he ae independen of he choice of he suface. Le s pick his one fo now. Lecue 4 lide 39 N and L Onl Depend on and ( of 3) The cuen ems J s and M s ae suface cuens and alwas angenial o he suface. J s M s o when we choose o be a sphee, he suface cuens have no adial componens. J M 0 when is a sphee Lecue 4 lide 40 0
11/14/018 N and L Onl Depend on and (3 of 3) We see fom he following equaions ha and ake one he same veco componens as J and M s s. kcos N, J e ds k L, M e ds Theefoe, N and L have no adial componens. N L cos We also ecognie ha his will alwas be he case since he fa field is fied and independen of he choice of. N L 0 alwas Lecue 4 lide 41 Final Epessions fo N and L N, 0, coscos cossin sin kcos N J J J e ds, sin cos kcos N J J e ds L, 0, coscos cossin sin kcos L M M M e ds, sin cos kcos L M M e ds Lecue 4 lide 4 1
11/14/018 Calculaing he Fa Fields (1 of 3) The elecic and magneic fields, E and, ae calculaed fom he veco poenials accoding o 1 A F F 1 E F A A The? ems poduce equaions wih 1/ dependence. In he fa field is ve lage so hese ems vanish. 1 1 A F E F A Lecue 4 lide 43 Calculaing he Fa Fields ( of 3) 1 1 A F E F A The cul ems become k e k e A N F N 4 4 k k e e N N 4 4 k k ke ke Naˆ Naˆ Laˆ Laˆ 4 4 k A aˆ A aˆ k F aˆ F aˆ Lecue 4 lide 44 The elecic and magneic fields ae now k k A aˆ A aˆ F E Faˆ Faˆ A
11/14/018 Calculaing he Fa Fields (3 of 3) k A aˆ A aˆ F We now combine veco componens. And las we pu hese equaions in ems of N and L. k E Faˆ Faˆ A 1 1 A F aˆ A F aˆ E A F aˆ A F aˆ 0 k ke 1 N L 4 k ke 1 N L 4 E 0 E k ke N 4 L E k ke N 4 L Lecue 4 lide 45 Impedance of he fa field medium: Rada Coss ecion The powe in he adiaed field can be calculaed fom he fa field quaniies as 1 * 1 * Pad Re Re E E 1 k N L + N L 4 The ada coss secion (RC) in hee dimensions is RC, lim 4 ad P inc inc P k N L + N L 8P Lecue 4 lide 46 3
11/14/018 Flow of Equaions fo 3D NFFF ep 1 Calculae cuen ems aound a closed suface. J nˆ M nˆ E s ep Loop ove all values of and a) Calculae N and L in he fa field s kcos N JcoscosJcossinJsin e ds kcos N JsinJcos e ds kcos L Mcoscos M cossin Msin e ds kcos L Msin M cos e ds b) Calculae E and in he fa field 0 k ke 1 N L 4 k ke 1 N L 4 E 0 k ke E N L 4 k ke E N L 4 Lecue 4 lide 47 c) Calculae RC o whaeve else ou need k N L N L RC, + 8P inc NFFF in Two Dimensions Lecue 4 lide 48 4
11/14/018 uface Cuens (1 of ) We esic ou analsis o he plane. This means he nomal veco is esiced o he plane. ˆn E, M s nˆ n aˆ n aˆ This also means ha he spheical coodinae paamee is eo. 0 0 Theefoe, we wie N and L onl as funcions of. J s,,, N N L L Lecue 4 lide 49 uface Inegals o Line Inegals (1 of ) Fo he geneal case, we inegae he cuen ems J and M ove a closed suface o obain N() and L() in he fa field. N N ds ds L L ds ds In FDTD, i is mos convenien o seup ou inegaing suface o be a ecangle. oweve, in D he closed suface can be educed o a closed line. Lecue 4 lide 50 5
11/14/018 uface Inegals o Line Inegals ( of ) We mus wie ou inegaions as closed line inegals. ds d Given ha ou inegaing line is a ecangle, we divide his ino fou sepaae odina inegals. d d d 1 1 low side high side 1 1 d d high side low side low high 1 1 low high Lecue 4 lide 51 Oienaion of he uface Cuens (1 of 3) We have defined he suface cuens o be flowing in a common diecion. J, M J, M J, M J, M Lecue 4 lide 5 6
11/14/018 7 Lecue 4 lide 53 Oienaion of he uface Cuens ( of 3) oweve, his is no he inheen diecionali calculaed on he Yee gid., J M, J M, J M, J M Lecue 4 lide 54 Oienaion of he uface Cuens (3 of 3) We have o oae he suface cuen vecos abou he suface nomal., J M, J M, J M, J M ˆn ˆn ˆn ˆn
11/14/018 Roaion of uface Cuen Tems Based on his line of easoning, we mus oae ou suface cuen ems in wo of he line segmen inegals. 1 1 d d d d d 1 1 low side high side high side low side R180 R180 J, M ˆn ˆn J, M Flip sign of J, J, M, and M Flip sign of J, J, M, and M Lecue 4 lide 55 Wie Inegals as andad Inegals The las wo inegals ae calculaed fom b o a. To aive a a moe saighfowad numeical implemenaion, he ode of inegaion is evesed. When his is done, he sign of he inegal changes. 1 1 d d d d d 1 1 low side high side high side low side R180 R180 d dd d 1 R 180 1 1 low side high side high side d 1 low side R 180 Lecue 4 lide 56 8
11/14/018 Final Equaions fo D Analsis Appling all of his o ou oiginal suface inegal equaions fo N() and L(), we ge kcos kcos kcos kcos 1 1 1 1 N J e d J e d J e d J e d low side high side high side low side kcos N JcosJsine d 1 L 1 low side high side kcos cos sin cos sin 1 1 kcos JcosJsin e d kcos J J e d J J e d high side low side kcos kcos kcos kcos Me 1 1 1 1 low side high side high side low side kcos L M M e d cos sin d M e d M e d M e d kcos M cosmsin e d 1 1 low side high side kcos cos sin cos sin 1 1 high side low side Lecue 4 lide 57 kcos M M e d M M e d implificaions fo E and Modes E Mode E E ˆ a aˆ aˆ Mode E Eaˆ Eaˆ aˆ E E 0 E 0 J s na na a a Js na na a M n a n a E a M na na Ea Ea ˆ ˆ ˆ ˆ ˆ ˆ ˆ s J n n aˆ M neaˆ neaˆ s s ˆ ˆ ˆ ˆ ˆ ˆ ˆ s J naˆ naˆ M ne ne aˆ s s J J M 0 M M J 0 Lecue 4 lide 58 9
11/14/018 N(,) and L(,) fo he E Mode Given all of he simplificaions on he pevious slide, ou epessions fo N(,) and L(,) educe o 0 0 kcos kcos kcos kcos + 1 1 1 1 low side high side high side low side N e d e d e d e d N L kcos kcos cos Ee 1 1 low side high side kcos kcos cos 1 1 high side low side L sin E e d d sin E e d E e d Lecue 4 lide 59 N(,) and L(,) fo he Mode Given all of he simplificaions on he pevious slide, ou epessions fo N(,) and L(,) educe o N 0 kcos kcos cos 1 1 low side high side kcos kcos cos 1 1 high side low side N sin e d e d L L sin e d e d kcos kcos kcos kcos E e d E e d E e d E e d 0 1 1 1 1 low side high side high side low side Lecue 4 lide 60 30
11/14/018 E(,) and (,) fo D Analsis Given ha N = L = 0 fo he E mode and N = L = 0 fo he mode, he elecomagneic fa fields fo each mode ae calculaed as E k ke N 4 L E 0 0 E Mode ke 4 k L N E 0 k ke E N L 4 0 Mode ke 4 k L N Lecue 4 lide 61 P ad (,) and RC(,) fo D Analsis E Mode 1 k Pad N L 4 RC Mode k 8P inc inc N L 1 k Pad N L 4 RC k 8P N L Lecue 4 lide 6 31
11/14/018 Inciden Powe P inc The powe in a unifom plane wave is calculaed as 1 E 1 Pinc Assuming a uni ampliude plane wave souce, P inc educes o 0.5 fo E 1 Pinc 0.5 fo 1 Assuming a uni ampliude plane wave souce, P inc educes o P inc 1 0 E Mode wih E 1 P inc 0 Mode wih 1 Lecue 4 lide 63 Flow of Equaions fo E Mode NFFF Loop ove all values of a) Inegae o calculae N and L in he fa field kcos kcos kcos kcos + 1 1 1 1 N e d e d e d e d low side high side high side low side kcos sin 1 low side kcos kcos kcos L E e dcos E e d sin E e d cos E e d 1 1 1 high side high side low side b) Calculae E and in he fa field fom N and L (if desied) ke ke E N L N 4 4 k k L c) Calculae P ad and RC fom N and L (if desied) 1 k k Pad N L RC N L 4 8 Pinc Lecue 4 lide 64 3
11/14/018 Flow of Equaions fo Mode NFFF Loop ove all values of a) Inegae o calculae N and L in he fa field kcos L Ee d 1 kcos kcos kcos kcos 1 1 1 1 N sin e dcos e dsin e dcos e d low side high side high side low side k cos k cos k cos 1 1 1 E e d E e d E e d low side high side high side low side b) Calculae E and in he fa field fom N and L (if desied) k k ke ke L E N L N 4 4 c) Calculae P ad and RC fom N and L (if desied) 1 k k Pad N L RC N L 4 8 Pinc Lecue 4 lide 65 Implemenaion Lecue 4 lide 66 33
11/14/018 ep 1 eup he Gid 3D Gid PML D Gid TF/F Ineface caeing Obec Noe ha he nea field suface is locaed wihin he scaeed field egion. Lecue 4 lide 67 ep imulae o Ge ead ae Field 3D Gid D Gid Calculae sead sae field along The sead sae field fom FDTD is he same daa ha would be geneaed if simulaed b finie diffeence fequenc domain (FDFD). Lecue 4 lide 68 34
11/14/018 ep imulae o Ge ead ae Field Below is MATLAB code fom inside he main FDTD loop ha updaes he Fouie ansfom acoss he enie gid a a single fequenc defined b he kenel K. The Fouie ansfom of he souce a his same fequenc is also calculaed. % Updae Fouie Tansfoms FT_E = FT_E + (K^T*d)*E; FT_ = FT_ + (K^T*d)*; FT_ = FT_ + (K^T*d)*; FT_ = FT_ + (K^T*d)*0(nc,nc); Afe he main FDTD loop has finished, he Fouie ansfoms ae nomalied o souce using he follow code: % NORMALIZE FOURIER TRANFORM FT_E = FT_E/FT_; FT_ = FT_/FT_; FT_ = FT_/FT_; Lecue 4 lide 69 ep 3 Inepolae Field Quaniies Along We need o inepolae he field quaniies in each Yee cell a a common poin. I is convenien o choose he oigin of each cell as he common poin. E E E, k, k, k, k, k, k 1,,,, E i k E i k E i k E i k E i k E i k, 1,,,,, 1,,, 1, 1, 1,,, 1,, i k i k i k i k 4 i k i k i k i k 4 i k i k i k i k 4 1,, 1 1,,,, 1,, 1, 1, 1,,, 1,,, Lecue 4 lide 70 35
11/14/018 ep 3 Inepolae Field Quaniies Along Fo D simulaions, he inepolaion equaions educe o E Mode E i E i,, 1 i i 1,, % INTERPOLATE FIELD TO ORIGIN OF YEE CELL ie = FT_E; i = 0.5*FT_; i(:,:n) = 0.5*(FT_(:,1:N-1)... + FT_(:,:N)); i = 0.5*FT_; i(:n,:) = 0.5*(FT_(1:N-1,:)... + FT_(:N,:)); Mode 1, 1 1,, 1, i i i i 4 Ei1, E E E 1 E E Lecue 4 lide 71 ep 4 Calculae N and L b Inegaing Fields Along We pick a ange of values fo, usuall < <. Fo each value of we inegae E and aound he conou. E Mode Mode kcos kcos kcos kcos + 1 1 1 1 low side high side high side low side kcos kcos kcos kcos sin cos sin cos 1 1 1 1 low side high side high side low side N e d e d e d e d L E e d E e d E e d E e d kcos kcos kcos kcos 1 1 1 1 low side high side high side low side N sin e dcos e dsin e dcos e d kcos k cos k cos k cos L Ee d Ee d Ee d Ee d 1 1 1 1 low side high side high side low side Lecue 4 lide 7 36
11/14/018 Block Diagam fo caeing Analsis (1 of 3) a Iniialie MATLAB Dashboad ouce NFFF Device Gid Calculae Gid N, N, d, d Build Device ER, ER, ER, UR, UR, UR Compue ouce Paamees d, au, ns, ns, T0, k, k, ec. Compensae fo Numeical Dispesion Iniialie FDTD Fields BC s This sep onl calculaes souce paamees, no he souce funcions. Those ae calculaed in he main FDTD loop. Culs Inegaions Lecue 4 lide 73 Block Diagam fo caeing Analsis ( of 3) es Finish Fouie Tansfoms Done? no ime Compue Cul of E CE, CE, CE andle TF/F CE, CE, CE Updae Inegaions Compue Cul of C, C, C andle TF/F C, C, C Updae Inegaions Updae E Updae Fouie Tansfoms Visualie imulaion Updae Updae D This is he odina main loop fo FDTD ecep he TF/F and calculaing Fouie ansfoms whee he NFFF is o be calculaed. Lecue 4 lide 74 37
11/14/018 Block Diagam fo caeing Analsis (3 of 3) Inepolae Fields o Oigin of Yee Cells Denomalie E E E 0 Iniialie NFFF N, N, N, 0 L, L, L, 0 E, E, 0,, 0 P, RC, 0 ad All following calculaions should use he inepolaed and denomalied field values. Inegae Aound Peimee N, N, N, L, L, L Calculae Fields (if desied) E, E,, Calculae RC (if desied) P ad,rc Done? no & es how Resuls Finish Lecue 4 lide 75 igh Level Code fo Loop % CALCULATE N AND L = 10*ma(,); PI = linspace(-ppnpi); fo nphi = 1 : NPI end % Ne Phi Angle phi = PI(nphi); n = [ cos(phi) ; sin(phi) ]; % lo apeoidal inegaion... % hi apeoidal inegaion... % hi apeoidal inegaion... % lo apeoidal inegaion... Lecue 4 lide 76 38
11/14/018 Eample Code fo Inegaion kcos kcos kcos kcos + 1 1 1 1 N e d e d e d e d low side high side high side low side kcos kcos kcos kcos sin cos Ee dsin Ee dcos Ee d 1 1 1 1 low side high side high side low side L E e d Noe: na and nb ae he lef and igh aa indices of. na and nb ae he op and boom aa indices of. % lo apeoidal inegaion N = 0; L = 0; n = na; fo n = na : nb p = [ a(n) ; a(n) ]; e = ep(-1i*k0*do(n,p)); iw = 1 - ((n==na) + (n==nb))/; N = N + iw*(n,n)*e; L = L + iw*e(n,n)*e; end Nh(nphi) = Nh(nphi) - N*d; Lphi(nphi) = Lphi(nphi) - sin(phi)*l*d; Lecue 4 lide 77 imulaion Eamples Lecue 4 lide 78 39
11/14/018 Eample #1: caeing fom a Dielecic Clinde The Poblem Wha is he RC of a dielecic clinde wih adius 0.5 0 and dielecic consan of = 4.0? 1 1 1 4.0 Lecue 4 lide 79 Eample #1: caeing fom a Dielecic Clinde The imulaion (E Mode) The imulaion ( Mode) PML PML Inegaing suface Inegaing suface TF/F ineface TF/F ineface PML PML PML PML caee caee PML PML Lecue 4 lide 80 40
11/14/018 Eample #1: caeing fom a Dielecic Clinde The uface Cuens (E Mode) Lecue 4 lide 81 Eample #1: caeing fom a Dielecic Clinde The uface Cuens ( Mode) Lecue 4 lide 8 41
11/14/018 Eample #1: caeing fom a Dielecic Clinde N and L Funcions (E Mode) Lecue 4 lide 83 Eample #1: caeing fom a Dielecic Clinde N and L Funcions ( Mode) Lecue 4 lide 84 4
11/14/018 Eample #1: caeing fom a Dielecic Clinde E and Fields (E Mode) Lecue 4 lide 85 Eample #1: caeing fom a Dielecic Clinde E and Fields ( Mode) Lecue 4 lide 86 43
11/14/018 Eample #1: caeing fom a Dielecic Clinde RC (E Mode) RC ( Mode) Lecue 4 lide 87 44