Transfer Matrix Method

Transcription

1 9/6/17 Instucto D. Ramond Rumpf (915) Computational lectomagnetics Lectue #4 Tansfe Mati Method Lectue 4These notes ma contain copighted mateial obtained unde fai use ules. Distibution of these mateials is stictl pohibited Slide 1 Outline Fomulation of 44 mati equation fo 1D stuctues Solution in an LI lae Tansfe matices fo multilae stuctues Tansfe matices ae unstable Fomulation of mati equation fo 1D stuctues Lectue 4 Slide 1

2 9/6/17 Fomulation of 44 Mati quation fo 1D Stuctues Lectue 4 Slide 3 1D Stuctues Sometimes it is possible to descibe a phsical device using just one dimension. Doing so damaticall educes the numeical compleit of the poblem and is ALWAYS GOOD PRACTIC. Region I Reflection Region Region II Tansmission Region Lectue 4 Slide 4

3 9/6/17 3D 1D Using omogeniation Man times it is possible to appoimate a 3D device in one dimension. It is ve good pactice to at least pefom the initial simulations in 1D and onl moving to 3D to veif the final design Phsical Device ffective Medium Appoimation in 1D Lectue 4 Slide 5 3D 1D Using Cicuit Wave quivalence n i, i, i Z i i i i Lectue 4 Slide 6 3

4 9/6/17 4 Lectue 4 Slide 7 Stating Point k k k k k k We stat with Mawell s equations in the following fom. ee we have assumed isotopic mateials and we will use the positive sign convention fo waves. j Positive sign convention Lectue 4 Slide 8 Calculation of the Wave Vecto Components The components k and k ae detemined b the incident wave and ae continuous thoughout the 1D device. The k component is diffeent in each lae and calculated fom the dispesion elation in that lae.,inc,inc,inc,inc sin cos sin sin k k k k,,, i i i k k k k i Lae #

5 9/6/17 k and k Continuous Thoughout Device k k n cos sin k,inc ai,inc k n ai cos k k inc k k,ai,ai ai n 1 k ef 1 -k,ai k,1 k k n k k k k n k,1 1 k n k k, k k k n k, n 3 k 3 k k,3 k k n k,3 3 k tn k,ai k,tn knai k k,ai k Lectue 4 Slide 9 Waves in omogeneous Media A wave popagating in a homogeneous lae is a plane wave. It has the following mathematical fom. jk jk jk jk jk jk jk jk e e e e e e e e Note: e +jk sign convention was used fo popagation in + diection. When we take deivatives of these solutions, we see that jk jk jk jk jk jk e e e jke e e jk jk jk jk jk jk jk jk e e e jk e e e jk jk We cannot sa that jk because the stuctue is not homogeneous in the diection. jk Lectue 4 Slide 1 5

6 9/6/17 Reduction of Mawell s qs. to 1D Given that jk jk Mawell s equations become d jk k d d jk k d jk jk k d jk k d d jk k d jk jk k Note: is the onl independent vaiable left so its deivative is odina. d d Lectue 4 Slide 11 Nomalie the Paametes We nomalie the coodinates (,, and ) and wave vecto components (k, k, and k ) accoding to k k k k k k k k k k Using the nomalied paametes, Mawell s equations become d jk d d jk d jk jk d jk d d jk d jk jk Lectue 4 Slide 1 6

7 9/6/17 Solve fo the Longitudinal Components and We solve the thid and sith equations fo the longitudinal field components and. d jk d d jk d j jk jk k k d jk d d jk d j jk jk k k Lectue 4 Slide 13 liminate the Longitudinal Components We eliminate the longitudinal field tems b substituting them back into the emaining equations. d jk d d jk d j k k d jk d d jk d j k k d k k k d d k k k d k k k d d d d k k k Lectue 4 Slide 14 7

8 9/6/17 Reaange Mawell s quations ee we eaange the tems and the ode of the equations. d k k k d d k k k d d d kk k d k k k d k k k d d d d k k k d d kk k d k k k d Lectue 4 Slide 15 Mati Fom of Mawell s quations The emaining fou equations can be witten in mati fom as d d kk k d k k k d d kk k d d k k k d kk k k kk d d kk k k k k Lectue 4 Slide 16 8

9 9/6/17 BTW fo Full Anisotopic Mateials ˆ kk k j k k jk k jk j k k kk kk k j k k jk k k k jk j k k Note: This is fo the j e sign convention. Lectue 4 Slide 17 Solution in an LI Lae Lectue 4 Slide 18 9

10 9/6/17 Mati Diffeential quation Mawell s equations can now be witten as a single mati diffeential equation. d ψ d Ωψ kk k k k k ψ Ω kk k k k k Lectue 4 Slide 19 Solution of the Diffeential quation (1 of 3) The mati diffeential equation is d ψ d Ωψ This is actuall a set of fou coupled diffeential equations. The sstem of fou equations can be solved as a single mati equation as follows. ψ e Ω ψ This is eas to wite, but how do we compute the eponential of a mati? d d kk k d k k k d d kk k d d k k k d e e e e Lectue 4 Slide 1

11 9/6/17 Functions of Matices (1 of ) It is sometimes necessa to evaluate the function of a mati. f A? It is NOT coect to calculate the function applied to eve element in the mati A individuall. A diffeent technique must be used. f A N f A f A f A f A f A f A f A f A f A 1 N M1 M MN This is moe of an aa opeation than a mati opeation so it is incoect to pefom on a mati. Lectue 4 Slide 1 Functions of Matices ( of ) To calculate f(a) coectl, we fist calculate the eigen vectos and eigen values of the mati A. A V eigen-vecto mati of A D eigen-value mati of A [V,D] = eig(a); D1 D D DM Given the eigen vecto mati V and the eigen value mati D, the function of the mati is evaluated as f D1 1 f D f A V f DV f D f D M f(d) is ve eas to evaluate because D is a diagonal mati so the function onl has to be pefomed individuall on the diagonal elements. Lectue 4 Slide 11

12 9/6/17 Solution of the Diffeential quation (1 of ) We had the following mati diffeential equation and geneal solution dψ Ωψ ψ ψ d Ω e We can now evaluate the mati eponential using the eigen values and eigen vectos of the mati. Ω W eigen-vecto mati λ eigen-value mati e We W Ω λ 1 e λ 1 e e 4 e e 3 Lectue 4 Slide 3 Solution of the Diffeential quation ( of ) The solution to the mati diffeential equation is theefoe ψ Ωψ Ω ψ e ψ λ 1 ψ We W ψ d d c We can combine the unknown initial values () with W -1 because that poduct just leads to anothe column vecto of unknown constants. Ou final solution is then d d ψ Ωψ ψ λ W e c 1 c W ψ Lectue 4 Slide 4 1

13 9/6/17 Intepetation of the Solution ψ λ We c ( ) Oveall solution which is the sum of all the modes at plane. c Column vecto containing the amplitude coefficient of each of the modes. This quantifies how much powe is in each mode. W Squae mati who s column vectos descibe the modes that can eist in the mateial. These ae essentiall pictues of the modes which quantif the elative amplitudes of,,, and. e Diagonal mati descibing how the modes popagate. This includes accumulation of phase as well as decaing (loss) o gowing (gain) amplitude. Lectue 4 Slide 5 Getting a Feel fo the Numbes (1 of ) Fo a lae with = 9. and = 1. (i.e. n = 3.) and a wave at nomal incidence, we will have 1 1 Ω 9 9 This mati has the following eigen vectos and eigen values. j.3 j.3 j.3 j.3 W j3. j3. λ j3. j3. Lectue 4 Slide 6 13

14 9/6/17 Getting a Feel fo the Numbes ( of ) We see that the modes occu as eithe an o pai. This is consistent with plane waves. Due to the nomaliation, the ae 9 out of phase. A sign diffeence indicates fowad and backwad waves. Onl the elative amplitude diffeence between and is impotant hee. j.3 j.3 j.3 j.3 W We know the efactive inde (n = 3.), so the eigen values ae consistent with what we would epect. The signs coespond to fowad and backwad waves. The modes in W onl contain infomation about the elative amplitudes of the field components. j3. j3. λ j3. j3. e n e jncosinc jncos inc 3 The numbes in descibe how the modes accumulate phase in the diection. This is essentiall just the comple efactive inde of the mateial. Lectue 4 Slide 7 Visualiing the Modes j.3 j.3 j.3 j.3 W j3. j3. λ j3. j3. Mode 4 Mode 3 Mode Mode 1 -j.3 Mode 1 Mode j.3 j.3 Mode 3 Mode j.3 Lectue 4 Slide 8 14

15 9/6/17 Tansfe Matices fo Multilae Stuctues Lectue 4 Slide 9 Geomet of an Intemediate Lae Lae i-1 Lae i Lae i+1 i ψi 1 kl i 1 ψ i i ψ i kl i ψ i ψ i1 L L i i 1 L i1 c i1 c i c i1 is a local coodinate inside the i th lae that stats at eo at the lae s left side. i Lectue 4 Slide 3 15

16 9/6/17 Field Relations Field inside the i th lae: i i i, i, i ii ii λ ψ ie i W c i, i, i Bounda conditions at the fist inteface: kl ψ ψi 1 i1 i W c Wc λi 1kL i 1 i1e i1 i i Bounda conditions at the second inteface: ψ klψ i i i1 λikl i ie i i1 i1 W c W c We need to include k in the eponential to nomalie L i-1 because the paamete i-1 epects to multipl a nomalied coodinate. Note: We must equate the field on eithe side of the intefaces and not the mode coefficients c. Lectue 4 Slide 31 The Tansfe Mati The tansfe mati T i of the i th lae is defined as: c T c i 1 i i We stat with the bounda condition equation fom the second inteface and eaange tems. T i λikl i 1 λikl i We c W c c W We c i i i1 i1 i1 i1 i i We then ead off the tansfe mati. T 1 e λikl i i Wi 1Wi Lectue 4 Slide 3 16

17 9/6/17 The Tansfe Mati Method The tansfe mati method (TMM) consists of woking though the device one lae at a time and calculating an oveall (global) tansfe mati. T1 T T3 T4 T5 Reflection Region Tansmission Region T global T5 T4 T3 T T1 The ode of multiplication ma seem backwads hee, but it is not. Recall the definition of the tansfe mati to have this make sense. This is standad mati multiplication. Lectue 4 Slide 33 The Global Tansfe Mati The tansfe mati so fa is not et the tue global tansfe mati because it does not connect the eflection egion to the tansmission egion. It onl connects the amplitude coefficients of Lae 1 to the amplitude coefficients in the tansmission egion. This is a esult of how we defined the tansfe mati. c T c tn global 1 The global tansfe mati must connect the amplitude coefficients in the eflection egion to the amplitude coefficients in the tansmission egion. Bounda conditions at the fist inteface equie W c Wc ef ef 1 1 Solving this fo c 1 ields c W W c ef ef The global tansfe mati is deived b substituting this esult into the fist equation. c T W W c 1 tn global 1 ef ef T T W W T T T T T T W W 1 1 global ef global global 1 ef Lectue 4 Slide 34 17

18 9/6/17 Tansfe Matices ae Unstable Lectue 4 Slide 35 The Multi Lae Poblem The diagam below is focused on an abita lae in a stack of multiple laes. We will be eamining the wave solutions in this i th lae. Lectue 4 Slide 36 18

19 9/6/17 Wave Solutions in i th Lae Recall that the wave vecto can be puel eal (pue oscillation), puel imagina (pue eponential deca), o comple (decaing oscillation). Pue oscillation k k Decaing oscillation k k jk Pue deca Lectue 4 Slide 37 k jk Backwad Waves in i th Lae Due to eflections at the intefaces, thee will also be backwad taveling waves in each of the laes. These can also have wave vectos that ae eal, imagina o comple, so the can oscillate, deca/gow, o both. Lectue 4 Slide 38 19

20 9/6/17 All Waves ae Teated as Fowad Waves The pue tansfe mati method teats all waves as if the ae fowad popagating. Decaing fields associated with backwad waves become eponentiall gowing fields and quickl become numeicall unstable. i i i, i, i ii ii λ ψ ie i W c i, i, i Lectue 4 Slide 39 TMM is Inheentl Unstable Ou wave solution was λ ψ W c e This teats all powe as fowad popagating. We know that backwad waves eist. We also know that decaing fields eist when a wave is evanescent o popagating in a loss mateial. When backwad waves ae decaing and teated as fowad popagating waves, the gow eponentiall. This leads to numeicall instabilit. The TMM is inheentl an unstable method because it teats evething as fowad popagating. Lectue 4 Slide 4

21 9/6/17 The Fi We ae teating all powe as fowad popagating because we did not distinguish between fowad and backwad waves. Cleal, the fist pat of the fi is to distinguish between fowad and backwad popagating waves. This can be accomplished b calculating the Ponting vecto associated with the modes and looking at the sign of the component. Be caeful! We ae using a nomalied magnetic field. j j 1 j i.3 i.3 i.3 i.3 W Lectue 4 Slide 41 Reaange igen Modes Now that we know which eigen modes ae fowad and backwad popagating, we can eaange the eigen vecto and eigen value matices to goup them togethe. i.3 i.3 i.3 i.3 W i3. i3. λ i3. i3. eaange modes i.3 i.3 i.3 i.3 W i3. i3. λ i3. i3. You will also need to adjust the vetical positions of the eigen values so that emains a diagonal mati. Lectue 4 Slide 4 1

22 9/6/17 New Intepetation of the Matices i.3 i.3 i.3 i.3 W We have now patitioned ou matices into fowad and backwad popagating elements. i3. i3. λ i3. i3. Note: Fo anisotopic mateials, all the eigen vectos and eigenvalues ae in geneal unique. λ W W W W W e λ λ e e λ i3. i3. i3. λ i3. Lectue 4 Slide 43 Revised Solution to Diffeential quation The mati diffeential equation and its oiginal solution was dψ Ωψ ψ λ W e c d Afte distinguishing between fowad and backwad popagating waves and gouping them in the matices, we can wite ou solution as ψ λ W e W c W W λ e c We now have sepaate mode coefficients c + and c - fo fowad and backwad popagating modes, espectivel. Lectue 4 Slide 44

23 9/6/17 Fomulation of Mati quation fo 1D Stuctues Lectue 4 Slide 45 Recall Deivation Up to 44 Stat with Mawell s equations fom Lectue. Assume LI. Assume device is infinite and unifom in and diections. jk jk Nomalie and wave vectos k, k, and k. k k k k k k k k k k liminate longitudinal components and b substitution. k k k d jk d k d jk d k jk jk k d jk d d jk d jk jk d d kk k d k k k d k k k d jk k d d jk k d jk jk k d jk d d jk d jk jk d d kk d k k k d k Lectue 4 Slide 46 3

24 9/6/17 Deivation of Two Mati quations We can wite ou fou equations as two mati equations. d d kk d k k k d k d 1 k k k d k k k d kk k d d k k k d d 1 kk k d k k k Note: These equations ae valid egadless of the sign convention because thee is alwas a k multipling anothe k and easing the sign. Lectue 4 Slide 47 Standad PQ Fom We can wite ou two mati equations moe compactl as d 1 k k k d k k k d 1 kk k d k k k 1 kk k P k k k 1 kk k Q k k k d d P d d Q Note: We will see this same PQ fom again fo othe methods like MoL, RCWA, and waveguide analsis. TMM, MoL, and RCWA ae all implemented the same afte P and Q ae calculated. Lectue 4 Slide 48 4

25 9/6/17 Mati Wave quation Ou two govening equations ae d d P d q. (1) q. () d Q We can now deive a mati wave equation. Fist, we diffeentiate q. (1) with espect to. d d d d d P d d d P d d Second, we substitute q. () into this esult. d d PQ d d Ω Ω PQ Lectue 4 Slide 49 Numeical Solution (1 of 3) The sstem of equations to be solved is d d Ω Ω PQ This has the geneal solution of e a e Ω Ω a a popotionalit constant of fowad wave a popotionalit constant of backwad wave No mode soting! ee, we solved a second ode diffeential equation whee the modes we calculate ae all popagating in a single diection. We simpl wite them twice fo fowad and backwad waves and thus the ae automaticall distinguished. Befoe we solved a fist ode diffeential equation that lumped fowad and backwad modes togethe. Lectue 4 Slide 5 5

26 9/6/17 Numeical Solution ( of 3) Recall that f 1 A V f DV We can use this elation to compute the mati eponentials. e We W e We W Ω λ 1 Ω λ 1 W igen-vecto mati of Ω λ igen-value mati of Ω So the oveall solution can now be witten as We W a We W a λ 1 λ 1 e λ 1 e 1 e e e N N e e Lectue 4 Slide 51 Numeical Solution (3 of 3) So the oveall solution can now be witten as λ 1 1 e λ W W e a W W a c c The column vectos a + and a ae popotionalit constants that have not et been detemined. The eigen vecto mati W multiplies a + and a to give anothe column vecto of undetemined constants. To simplif the math, we combine these poducts into new column vectos labeled c + and c. We c We c λ λ Lectue 4 Slide 5 6

27 9/6/17 Solution fo the Magnetic Field (1 of ) The magnetic field has a solution of the same fom, but will have its own eigen vecto mati V to descibe its modes. λ λ Ve c Ve c Since the electic and magnetic fields ae coupled and not independent, we should be able to compute V fom W. Fist, we diffeentiate the above solution with espect to. d d λ λ Vλe c Vλe c We ae fee to choose an sign we wish because it can be accounted fo in c -. We put the minus sign in the solution hee so that both tems in the diffeentiated equation will be positive. You will see soon wh this is desied. Lectue 4 Slide 53 Solution fo the Magnetic Field ( of ) We now have d e d Vλ c Vλe c d d λ λ Recall fom pevious slides that Q Combining these esults leads to and Vλe c Vλe c Q We c We c λ λ λ λ λ λ QWe c QWe c Lectue 4 Slide 54 We c We c λ λ Compaing the tems on the left and ight sides of this equation shows that 1 Vλ QW V QWλ 7

28 9/6/17 Combined Solution fo and lectic Field Solution We c We c λ λ c amplitude coefficients of fowad wave c amplitude coefficients of backwad wave W eigen-vecto mati λ diagonal eigen-value mati Magnetic Field Solution λ λ 1 Ve c Ve c V QWλ Combined Solution ψ λ e W W c λ V V e c Does this equation look familia? This is the same equation we had on fo the 44 appoach afte we soted the modes. Lectue 4 Slide 55 Two Paths to Combined Solution 4 4 Mati Sot igen Modes kk kˆ j k k jk k k jk j k k kk jk j k k kk k k k k jk j k k W W W W W e λ e λ λ e Anisotopic Mawell s quations Field Solution λ k W W e c ψ V V λ e c k λ W We c ψ λ V V e c Isotopic o 1 kk k P diagonall k k k anisotopic No soting! 1 kk k Q k k k Lectue 4 PQ Method Slide 56 8