(整理)传输矩阵法

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6.6 传输散射矩阵

2 ZZ Z Z Z Z Z V 0 0 10 2 0 1 2 0 1 2 00 2 S = = j = j 2 1 + 2 2 Z Z V Z + Z Z Z + Z Z + 0 2 1V 0 2 0 0 10 2 0 0 10 2 = 0
2
即
S21 = - j
2Z0 Z01Z02 Z + Z01Z02
T abc d abc d 1 1 1 T 1 2 【例6.6-1】证明 T T abc d abc d 2 2 1 2 2
证明：
V1 aV2 bI 2 （归一化的ABCD矩阵） I1 cV2 dI 2
轾轾 b T T 1 B 1 1 B 1 2B 犏 =犏犏犏 a T T 1 B 2 1 B 2 2B 臌臌
轾 a 2B 犏犏 b 2B 臌
则两级联之间的入射波和出射波的关系为：
轾轾轾轾轾轾 b T T a TT 1 A 1 1 AT 1 2 A 1 1 BT 1 2 B 2 B 1 1 1 2a 2 B 犏犏犏犏犏犏 = = 犏犏犏犏犏犏 a T T b TTb 1 A 臌 2 1 AT 2 2 A 2 1 BT 2 2 B 2 B 臌 2 1 2 2 2 B 臌臌臌臌
又 S 2 1=
VL+ j (1- GL ) = V1
Z 1 +G ( 1 +S ) 0 1 ( L) 1 1 \ S 2 1= Z j( 1 -G 0 2 L)
而
2 Z0 - Z01Z02 S11 = 2 Z0 + Z01Z02
Z 02 - Z 0 GL = Z 02 + Z 0

TransferMatrixMethod：传递矩阵法

Transfer Matrix MethodG.Eric Moorhouse,University of WyomingReference:Transfer Matrix MethodI.M.Gessel and R.P.Stanley,‘Algebraic Enu-meration’,in Handbook of Combinatorics Vol.2, ed.R.L.Graham et al.,Elsevier,1995,pp.1021–1061.References:Dimensions of CodesN.Hamada,‘The rank of the incidence matrixof points and d-ﬂats inﬁnite geometries’,J. Sci.Hiroshima Univ.Ser.A-I32(1968),381–396.M.Bardoe and P.Sin,‘The permutation mod-ules for GL(n+1,q)acting on P n(q)and F n+1q’,to appear in JLMS./~sin/preprints/hamada.dviG.E.Moorhouse,‘Dimensions of Codes from Finite Projective Spaces’(as html and as Maple worksheet)/~moorhous/src/hamada.html /~moorhous/src/hamada.mwsProblem1Let S k be the set of‘words’of length k consist-ing of‘a’s and‘b’s,with no two consecutive ‘b’s.Determine F k=|S k|.F0=1‘’F1=2‘a’‘b’F2=3‘aa’‘ab’‘ba’F3=5‘aaa’‘aab’‘aba’‘baa’‘bab’F4=8‘aaaa’‘aaab’‘aaba’‘abaa’‘abab’‘baaa’‘baab’‘baba’etc.This gives all but theﬁrst term of the Fibonacci sequence1,1,2,3,5,8,13,21,34,55,89,144,...To ﬁnd a formula for F k ,we work instead with the generating function∞k =0F k t k =1+2t +3t 2+5t 3+8t 4+13t 5+···Observe that words w ∈S k correspond to paths of length k ,starting at vertex 1in the digraph12..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................append ‘a’append ‘a’append ‘b’Words not ending in ‘b’Words ending in ‘b’Agenda1.Motivating Problem1(above)2.Counting Walks by the Transfer Matrix Method3.Application to Problem14.Counting Closed Walks5.Counting Weighted Walks in Digraphs withWeighted Edges6.MAPLE Worksheet for Problem17.Application to Coding TheoryThe Transfer Matrix Method Let D be a digraph (directed graph),possibly with loops,having vertices 1,2,3,...,n .Let A =[a ij :1≤i,j ≤n ]be the adjacency matrix of D ;in other words,a ij = 1,if (i,j )is an edge of D ;0,otherwise.A walk of length k in D is a sequence......................................................................................................................................................................................................................................................................................................................................···◦◦◦◦i 0i 1i 2i k→→→→of (not necessarily distinct)vertices such that each ...........................................................................................◦◦i r −1i r →is an edge of D .Counting Walks from i to jLet w ij(k)be the number of walks of length k from vertex i to vertex j in D.Then w ij(k)is the(i,j)-entry of A k.This is readily computed by reading oﬀthe coeﬃcient of t k in the gen-erating function k≥0w ij(k)t k which in turn is the(i,j)-entry of(I−tA)−1=I+tA+t2A2+t3A3+···. Since the(i,j)-entry of(I−tA)−1is of the formpoly.in t of degree≤n−1det(I−tA),w ij(k)satisﬁes a linear recurrencew ij(k+n)=n−1r=0c r w ij(k+r)for all k≥0where det(I−tA)=1−c n−1t−c n−2t2−···−c0t n.The initial conditions w ij(0),w ij(1),..., w ij(n−1)depend on i and j but the recurrence does not.Counting All WalksLet w(k)= n i=1 n j=1w ij(k),the total num-ber of walks of length k.This is the coeﬃcient of t k in the sum of the entries of(I−tA)−1.In particular w(k)satisﬁes the same recurrence as the w ij(k)’s:w(k+n)=n−1r=0c r w(k+r)for all k≥0but with diﬀerent initial conditions.Counting Closed WalksLet w closed(k)= n i=1w ii(k),the total number of closed walks of length k(i.e.starting and ending at the same vertex).This is the coef-ﬁcient of t k in trace((I−tA)−1).In particular w closed(k)satisﬁes the same linear recurrence as the w ij(k)’s and w(k),but again with diﬀerent initial conditions.Here we assumed the initial/ﬁnal vertex to be distinguished,i.e.the walks(i0,i1,i2,...,i k)and (i1,i2,...,i k,i0)are counted as distinct unless all i0=i1=···=i k.ExampleLet F k be the number of‘words’of length k consisting of‘a’s and‘b’s,with no two con-secutive‘b’s.F0=1‘’F1=2‘a’‘b’F2=3‘aa’‘ab’‘ba’F3=5‘aaa’‘aab’‘aba’‘baa’‘bab’F4=8‘aaaa’‘aaab’‘aaba’‘abaa’‘abab’‘baaa’‘baab’‘baba’etc.This gives all but theﬁrst term of the Fibonacci sequence1,1,2,3,5,8,13,21,34,55,89,144,...Observe that F k is the number of paths of length k ,starting at vertex 1in the digraph12.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................append ‘a’append ‘a’append ‘b’Words not ending in ‘b’Words ending in ‘b’A = 111(I −tA )−1=11−t −t 2 1t t 1−tk ≥0F k t k =sum of (1,1)-and (1,2)-entries of (I −tA )−1=1+t 1−t −t 2=1√5 α21−αt −β21−βt=1√5 k ≥0(αk +2−βk +2)t k where α=(1+√5)/2,β=(1−√5)/2.From (1−t −t 2) k ≥0F k t k =1+t we obtainF k = 1,if k =0;2,if k =1;F k −1+F k −2,if k ≥2so by induction,F k is the (k +1)st Fibonacci number.From the series expansion we obtain the explicit formulaF k =αk +2−βk +2√5for k ≥0.Wraparound VersionLet L k(for k≥0)be the number of‘words’of length k consisting of‘a’s and‘b’s with no consecutive‘b’s,and which do not both start and end with‘b’.For technical reasons we will take L0=2.For k≥2,we are simply counting necklaces with a mber and b lack beads having no two consecutive black beads;however,each neck-lace has a distinguished starting point(a knot in its cord)and a distinguished direction(clock-wise or counter-clockwise).L1=1‘a’L2=3‘aa’‘ab’‘ba’L3=4‘aaa’‘aab’‘aba’‘baa’L4=7‘aaaa’‘aaab’‘aaba’‘abaa’‘abab’‘baaa’‘baba’These are the familiar Lucas numbers which satisfy the same recurrence relation as the Fi-bonacci numbers,but a diﬀerent initial condi-tion.Note that L k is the number of closed walks of length k in our digraph.k ≥0L k t k =trace ((I −tA )−1)=2−t 1−t −t 2=11−αt +11−βt=k ≥0(αk +βk )t kFrom (1−t −t 2) k ≥0L k t k =2−t we obtainL k = 2,if k =0;1,if k =1;L k −1+L k −2,if k ≥2From the series expansion we obtain the ex-plicit formulaL k=αk+βk for k≥0.Counting Walks with Weighted Edges As before,D is a digraph (directed graph),pos-sibly with loops,having vertices 1,2,3,...,n .Assign a weight to each edge:...........................................................................................◦◦i j →a ij (Non-edges have weight zero.)Deﬁne the weight of a walk......................................................................................................................................................................................................................................................................................................................................···◦◦◦◦i 0i 1i 2i k →→→→a i 0i 1a i 1i 2a i 2i 3a i k −1i k of length k to be the producta i 0i 1a i 1i 2a i 2i 3···a i k −1i k .Let A =[a ij :1≤i,j ≤n ].Thenw ij(k):=The sum of all weightsof walks in D of length k from vertex i to vertex j=(i,j)-entry of A khas generating function k≥0w ij(k)t k equal to the(i,j)-entry of(I−tA)−1=I+tA+t2A2+t3A3+···as before.ExampleWe have determined the number F k of words of length k consisting of‘a’s and‘b’s,with no two consecutive‘b’s.How many such words contain r‘a’s and(therefore)k−r‘b’s?12.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................a ab A = a b a0 (I −tA )−1=11−at −abt 2 1bt at 1−atThe sum of the (1,1)-and (1,2)-entries is1+bt1−at −abt 2=1+(a +b )t +(a 2+2ab )t 2+(a 3+3a 2b +ab 2)t 3+(a 4+4a 3b +3a 2b 2)t 4+···Thus,for example,among the F 4=8words of length 4,1has 4‘a’s and 0‘b’s;4have 3‘a’s and 1‘b’;3have 2‘a’s and 2‘b’s.Codes from Finite GeometryConsider the projective plane of order 2:•••••••a bc d e f g ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................The binary code of this geometry is the sub-space C ≤F 7(where F ={0,1}mod 2)spanned by the lines:a b c d e f g a b c d e f g C ={0000000,1111111,1101000,0010111,0110100,1001011,0011010,1100101,0001101,1110010,1000110,0111001,0100011,1011100,1010001,0101110}|C|=24;dim C =4The code above is the 1-error correcting binary Hamming code of length 7.The projective plane is constructed from F 3by taking as points and lines the 1-and 2-dimensional subspaces of F 3.•••••••001010100011110111101.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Codes of Finite Projective SpacesLet F be theﬁeld of order p e,p prime.Pro-jective n-space over F has as its points,lines, etc.the subspaces of F n+1of dimension1,2, etc.Problem:Compute the dimension of the code C=C n,p,e,k spanned by the subspaces of codi-mension k.Solution by Hamada’s Formula(the follow-ing theorem)is usually computationally infea-sible.Solution by the Transfer Matrix MethodTheorem(Bardoe and Sin,1999)Deﬁne M(t)=(1+t+t2+···+t p−1)n+1.Let D=D n,p,e,k be the digraph with vertices 1,2,...,k,and the edge from vertex i to vertex j has weight equal to the coeﬃcient of t pj−i in M(t).Thendim C n,p,e,k=1+sum of weights of closed walks of length ein D=1+ coeﬀ.of t ein tr[(I−tA)−1]where A is the k×k matrix whose(i,j)-entry is the weight of edge(i,j)(deﬁned above).Example:Projective Plane of Order2C=binary code spanned by the seven lines (subspaces of codimension k=1)M(t)=(1+t)3=1+3t+3t2+t3A=[3](coeﬃcient of t1in M(t))(I−tA)−1= 1 1−3ttr[(I−tA)−1]=11−3t=1+3t+9t2+27t3+···dim C=1+3=4。

《动力学分析中的传递矩阵法》教学文案

关于剩余矩阵的选取，通俗的讲，剩余矩阵的行数为末端状态矢量为0的行数（方程组右端为0），其列数取初始状态矢量不为0的行数（方程组变量）。这样我们相当于构造了一个齐次方程组，保证此齐次方程组有非零解的条件是，系数矩阵的行列式为0。
二、传递矩阵法计算步骤
2.3 振型分析及稳态响应分析
➢ 各阶固有频率求得，则传递矩阵确定。求得初始状态矢量，将矩阵依次相乘，便可得到各个节点处的状态矢量，利用计算机便可绘制出其振型图。
2L
三、传递矩阵法应用举例
3.2 输液管道的传递矩阵法
三、传递矩阵法应用举例
3.2 输液管道的传递矩阵法
轴向振动微分方程：
直管轴向运动的单元传递矩阵：
44矩阵
三、传递矩阵法应用举例
3.2 输液管道的传递矩阵法
横向振动微分方程：
直管横向运动的单元传递矩阵
44矩阵
三、传递矩阵法应用举例
传递矩阵法原理二、传递矩阵法计算步骤三、传递矩阵法应用举例
一、传递矩阵法原理
传递矩阵法属于一种半解析数值方法。基本思想是把整体结构离散成若干个子单元的对接与传递的力学问题，建立单元两端之间的传递矩阵，利用矩阵相乘对结构进行静力及动力分析。其应用领域涵盖结构的静力分析、动力特性分析（模态分析、稳定性分析）。
同时考虑直管单元的轴向振动和横向振动，则单元的场传递矩阵为：
88矩阵
三、传递矩阵法应用举例
3.2 输液管道的传递矩阵法
弯曲处的点传递矩阵为：
三、传递矩阵法应用举例
3.2 输液管道的传递矩阵法
则场传递矩阵为：
JXB(x)B1(0)
cosx E Ssinx
sinx 1 E Scosx 0

一维光子晶体的传输矩阵编程

一维光子晶体的传输矩阵的编程准备
一维光子晶体的结构参数：
１. 每一层的厚度，（或者是周期长度及层厚度比例）２. 介电常数（或折射率），对于非色散模型，采用常数；对于色散介质，采用合适的色散模型或色散曲线。３. 确定扫描波长范围或扫描频率范围一维光子晶体的场分布传输矩阵１. 结构模型
一维光子晶体的传输矩阵编程
2009.5.12 xilioo wjun.xjtu@
主要内容：
传输矩阵简述 Leabharlann 输矩阵编程准备传输矩阵编程流程图
编程建议
一维光子晶体的传输矩阵的简述
方法简单，计算量小，尤其适用于有限周期的一维光子晶体
利用传输矩阵可以用来研究：１. 结构的透射系数和反射系数２. 态密度３. 场分布４. 色散曲线传输矩阵基础理论的相关文献：王辉，李永平，用特征矩阵法计算光子晶体的带隙结构，物理学报，2001，50(1):2172
传输矩阵的编程建议
可以采用matlab编程，原因有二： 1. matlab计算矩阵很方便，且循环语句很简单２.matlab画图很简单很多同学问如何去研究场分布，这个其实很简单的，首先求得入射参
1 , 而后利用传输矩阵，求得每一层的电场强度系数，再把它共轭相数 r
乘，就得出光场分布了。传输矩阵不仅能求透射系数和反射系数以及场分布，还能求得态密度，以及色散曲线，而后两者能就较容易地进行物理深度上理论分析，提升你的研究高度。
２. 每一层的特征矩阵
３. 总的传输矩阵
一维光子晶体的传输矩阵的编程流程图
对某一频率依照色散模型或色散曲线得出介电常数或折射率得出某一位置的正负两个方向场强系数，得出透射反射系数

基于传输矩阵法的光纤光栅传感器的应变测量能力研究

ＳｕｄｎＤｙａｉｔａｎＭｅｓｅｅｉｉｙ０ｂｅａｇｔｙ０ｎｍｃＳｒｉａｕｒｍｎｔＡｂｌｔｆＦｉｒＢｒｇ
Ｇｒｔｉｓｄ０Ｔｒｎｍｉｓ０ａｒｘＭｅｈｄａｔｎｇＢａｅｎａｓｓｉｎＭｔｉｔＯ
ｃＥｉ，ｕＹｎ—ｈｎＹｈｎ＿ａ，ＡＧＪｘａｇＨＮｘａｎｘｏｇｃｅｇ，ｕｓａｇｊｎＹＮｉｉｉ — ｎ（．ｅｈｎｔｉｌｒｊｃＡａｅ，ａｉａＵｉｅｓｔ０ｅｅｓｅｈ００，ｈｎｓａ４０７，ｈｎ；１Ｍｃｏ０ｒａＰｏｅｔｃｄｍｙＮｔｎｌｎＶｒ－ｆｆｎｅＴｃｎｌｃ０ｙＤＣａ詈ｈ１０３Ｃｉａ２ＴｅＴｉｃｄ珈ｙｏｎｉｅｒＧｅｅａＳａｆＬ０ａｇ４１２，ｈｎ）．ｈｈｒＡａｅ【ｆｇｅｎｒｌｔｆ，ｕｙｎ７０３ＣｉａｄＥｎ
摘要：光纤光栅在不均匀应变场中会由于其反射谱的展宽而带来测量误差。在光纤光栅参数取典型值时，用传输使
矩阵法计算了峰值为３０ｏ￡的正弦应变波下波长为０Ｏ４～．０．２０１ｍ的测量误差。计算结果表明：当正弦波波长大于００７８ｍ，．３由谱峰展宽带来的相对测量误差可低于１％。计算了固定应变波长为００４ｍ时正弦应变波峰值在０～０．２５００￡变化的测量误差。当正弦应变峰值低于２３０￡时，４由谱峰展宽带来的相对测量误差可低于１。并由此算得了混凝％土等几种典型材料在误差小于１时的可测最高频率，％混凝土为７．４１～ｌ８５ｋｚ水泥砂浆为７．０．Ｈ；９４～９．Ｈ；为２６ｋｚ钢

传输线矩阵解

(5-14)
二、传输矩阵的普遍理论
［性质］1. 级联性质如果第Ⅰ个网络的输出端口是第Ⅱ个网络的输入端口，则称这两个网络级联(Cascade)。有
U1 U2 1] I [ A I 1 2
U2 U3 1] I [ A I 2 3
,求输入驻波比。
P
Z 0
C
Z 0 0 . 1 l
L
Z 0 0 . 2 l
Z l
图 5-4
四、应用举例
［解］将系统对Z0归一化
Zl Zl / Z0 2 j 4 Y1 CZ0 2 3 108 20 10 12 50 18850 . Z0 50 0.2653 8 7 L 2 3 10 10 2 1 l1 01 . 36 Y2
(5-18)
二、传输矩阵的普遍理论
4. 互易性质在互易网络中，［A］矩阵的行列式值等于1，即
d e t [A ] 1
5. 阻抗变换性质
A Z 1 1 l A 1 2 Z in A Z 2 1 l A 2 2
(5-19)
(5-20)
三、典型［A］矩阵
四、应用举例
Z = 1 0 0 + j 2 0 0 ， L = 0 . 1 H C = 2 0 P F ， Z = 5 0 ， f 3 0 0 M H z ［例1］如图示， l 0

0
U ( z )e sz dz I ( z )e sz dz
(5-2)
0
现在考虑一段长度为l的传输线段，在这一节，从负载出发的坐标用z 表示，对式(5-1)左边作 Laplace 变换
dU L sV ( s ) U (0 ) dz dI L sJ ( s ) I (0 ) dz

光线传输矩阵推导过程

光线传输矩阵推导过程光线传输矩阵（Ray Transfer Matrix）是描述光线在光学系统中传输的数学工具，也被称为 ABCD 矩阵。

在光学系统中，常常需要知道光线从一个位置传输到另一个位置，因此需要将光线传输过程用数学描述出来。

光线传输矩阵是一种简便的描述光线传输过程的方法，可以用于计算光学系统的成像性能、衍射现象等。

1. 光线传输是沿直线传播的；2. 光线的传播满足亥姆霍兹方程；3. 光学系统是轴对称的，即沿光路方向上的所有点都是轴对称的。

在这些假设的基础上，可以推导出光线传输矩阵的一般形式。

假设一束光线在一个点 P 处（位置矢量为 [x, y, z]）的方向余弦分别为 l, m, n，那么在向前传播一段距离 d 之后，在点 Q（位置矢量为[x′, y′, z′]）处的方向余弦分别为l′, m′, n′。

下面推导光线传输矩阵。

首先，根据第一条假设，可以得到：x′ = x + dly′ = y + dmz′ = z + dn然后，根据亥姆霍兹方程，可以得到：$$\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} + k^2\psi = 0$$其中，$\psi$ 表示复振幅，$k$ 表示波数，$k = 2\pi/\lambda$。

假设在点 P 处的复振幅为 $\psi_0$，则在点 Q 处的复振幅为：$$\psi = \psi_0e^{ikn′d}$$其中，$n′d$ 表示传输距离。

忽略高阶小量，可以进一步简化为：$$\frac{\partial^2\psi_0}{\partial x^2}dl^2 +\frac{\partial^2\psi_0}{\partial y^2}dm^2 + \frac{\partial^2\psi_0}{\partialz^2}dn^2 + 2\frac{\partial^2\psi_0}{\partial x\partial y}dldm +2\frac{\partial^2\psi_0}{\partial x\partial z}dldn +2\frac{\partial^2\psi_0}{\partial y\partial z}dmdn + k^2\psi_0 = 0$$接下来，定义一个 2x2 的矩阵 A 和一个 2x2 的矩阵 B，它们分别表示在点 P 和点Q 处的光线倾斜角（slopes）和射线高度（heights）：然后，将光线传输方程改写为矩阵形式：$$\begin{bmatrix} l′ \\ m′ \end{bmatrix} = M\begin{bmatrix} l \\ m\end{bmatrix}$$其中，$M$ 表示光线传输矩阵，它可以通过将以上方程变形得到：根据矩阵乘法的定义，可以将 $M$ 表示为：其中，$$\begin{aligned} A_{11} &= 1 + \frac{\partial l}{\partial z}d +\frac{\partial^2\psi_0}{\partial x\partial z}d\frac{\partial l}{\partial z}d \\ A_{12} &= d + l\frac{\partial^2\psi_0}{\partial x\partial z}d \\ A_{21} &=\frac{\partial m}{\partial z}d + m\frac{\partial^2\psi_0}{\partial y\partial z}d \\ A_{22} &= 1 + \frac{\partial m}{\partial z}d + \frac{\partial^2\psi_0}{\partial y\partial z}d\frac{\partial m}{\partial z}d \\ B_1 &= x +l\frac{\partial\psi_0}{\partial z}d +\frac{1}{2}\left(\frac{\partial^2\psi_0}{\partial x^2}l^2 +\frac{\partial^2\psi_0}{\partial y\partial x}lm +\frac{\partial^2\psi_0}{\partial x\partial z}ldx\right)d \\ B_2 &= y +m\frac{\partial\psi_0}{\partial z}d +\frac{1}{2}\left(\frac{\partial^2\psi_0}{\partial y\partial x}lm +\frac{\partial^2\psi_0}{\partial y^2}m^2 + \frac{\partial^2\psi_0}{\partialy\partial z}mdy\right)d \\ C_1 &= -\frac{1}{2}\frac{\partial^2\psi_0}{\partial x\partial y}d^2lm \\ C_2 &= -\frac{1}{2}\frac{\partial^2\psi_0}{\partialy\partial x}d^2lm \end{aligned}$$可以看到，光线传输矩阵 $M$ 的每一项都可以用初始光线的方向余弦和位置来表示。

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