03964.com

文档资料库 文档搜索专家

文档资料库 文档搜索专家

X. RESISTANCE MATRIX

If both dielectric and conducting losses exist in a transmission line, the telegraph equation system will be revised as ? dI ( z ) = ([G ] + jω [C ])V ( z ) dz dV ( z ) = ([R ] + jω [L ])I ( z ) dz

?

where the resistance matrix

[ R]

accounts for the conducting loss.

Let the guided current and voltage waves in +z direction in a transmission line be expressed as I (z ) = I1 , I 2 ,?, I M c V ( z ) = V1 , V2 , ? , VM c where M c is the number of conductors,

[

]

T

? = I exp(? γz ) ? = V exp(? γz )

[

]

T

? ? I and V are the complex current amplitude vector and complex voltage

amplitude vector, respectively,

I i and Vi are the current and voltage at the i-th conductor, respectively,

γ = α + jβ is the propagation constant of the transmission line,

α is the attenuation constant, and β is the phase constant.

Substitution of these two expressions into the telegraph equations gives

? ? γI exp(? γz ) = ([G ] + jω [C ])V exp(? γz ) ? ? γV exp(? γz ) = ([R ] + jω [L ])I exp(? γz )

or

? ? γI = ([G ] + jω [C ])V ? ? γV = ([R ] + jω [L ])I

44

That will lead to

([G ] + jω [C ])([R ] + jω [L])I? = γ 2 I?

? ? ([R ] + jω [L])([G ] + jω [C ])V = γ 2V

If the transmission line is lossless, i.e.

[R] = [G ] = 0 ,

then

γ = α + jβ = j β

( jω )2 [C ][L]I?0 = ( jβ )2 I?0

? ? ( jω )2 [L][C ]V0 = ( jβ )2 V0

giving the following eigenvalue equations:

[C ][L]I?0 = v ?2 I?0 p

? [L][C ]V0 = v ?2V?0 p

? ? where I 0 and V0 represent, respectively, the complex current amplitude

vector and the complex voltage amplitude vector for the lossless line, and these two vectors should be real since the transmission line does not have any losses. v p denotes the phase velocity of the guided wave along the line. vp =

ω β

Since the transmission line is composed of M c conductors, these eigenvalue

?( ?( ?( equations will produce M c eigenvectors for current, I 01) , I 02 ) ,?, I 0M c ) , and ? ? ? M c eigenvectors for voltage, V0(1) ,V0(2 ) , ?,V0( M c ) :

T ( ( ( ?( I 0i ) = I 0i ) (1), I 0i ) (2 ), ? , I 0i ) (M c ) T ? V0(i ) = V0(i ) (1), V0(i ) (2 ), ? , V0(i ) (M c )

[

]

[

]

i = 1,2, ? , M c

45

Since [L ] and [C ] are real symmetrical, the matrices [C ][L ] and [L ][C ] should be mutually transpose and complex conjugate. In fact,

[([C ][L]) ]

T ?

= ([C ][L ]) = [C ] [L ]

T T

(

T T

)

= [L ][C ]

where the superscript T symbolizes the transpose and * the conjugate. It follows that the eigenvectors of [C ][L ] and the eigenvectors of [L ][C ] should be mutually orthogonal, namely bi-orthogonal,

?( ? ? I 0i ) ,V0( j ) = V0( j )

( )

?T

?( ?( I 0i ) = I 0i )

( )

?T

? ?( V0( j ) = I 0i )

( )

?T

? V0(i )δ ij

where δ ij the is Kronecker Delta， ，

δ ij = ?

?1 i = j ?0 i ≠ j

It also follows that the eigenvalues of [C ][L ] and the eigenvalues of [L ][C ] should be mutually complex conjugate. Since the phase velocity v p = ω β is real, these two matrices, [C ][L ] and [L ][C ] , possess the same eigenvalues,

v ?2 (1), v ?2 (2),?, v ?2 (M c ) p p p

Therefore, this transmission line system composed of M c conductors and one ground plane possess M c propagating modes.

?( ?( ?( In a lossless line system, all eigenvectors, either I 01) , I 02 ) ,?, I 0M c ) and ? ? ? V0(1) ,V0(2 ) , ?,V0( M c ) , should be real.

For a lossless system,

[ R ] = [G ] = 0

γ = α + jβ = jβ

? ? ? ? I = I 0 and V = V0

46

the earlier mentioned formulas

? ? γI = ([G ] + jω [C ])V ? ? γV = ([R ] + jω [L ])I

are reduced to ? ? I 0 = v p [C ]V0 ? ? V0 = v p [ L ] I 0 If dissipation is involved in a transmission line, then the complex power propagated along the longitudinal direction, i.e. + z axis, is

? P = ∑ Vm I m = I ? V T m =1 Mc

( )

and the average power is

PT = Re(P ) = Re I ? V

T

{(

)

}

The power loss per unit length of the line system is

PL = ? ?PT ?z

? ?P ? = Re?? ? ? ?z ?

? ? = Re ?? I ? ? ?

( )

T

T

?V ? I ? ? ?z ?z

( )

T

{( = Re{ I (

T

? ? V? ? ?

?T

= Re I ?

) ([R] + jω [L])I + [([G ] + jω [C ])V ]

? T ? T

V

}

V

? T

) ([R] + jω [L])I + (V ) ([G] + jω [C ])

c d

?T

}

= I?

( ) [R]I + (V ) [G ]V = P + P

where Pc and Pd denote, respectively, the conducting loss per unit length and the dielectric loss per unit length,

Pc = I ?

( ) [R]I

T T

Pd = V ?

( ) [G ]V

47

It will be proven below that the attenuation constant α in a dissipated transmission line is given by

α = Re(γ ) =

P P PL = c + d = αc + αd 2 PT 2 PT 2 PT

where α c and α d indicate the attenuation constant for conductors and the attenuation constant for dielectrics , respectively. P I ? [R ]I αc = c = 2 PT 2 Re I ? T V

T

( ) {( )

? T ?

}

αd =

[Proof] Since

Pd 2 PT

(V ) [G ]V = 2 Re {( I ) V }

T

I (z ) = I1 , I 2 ,?, I M c V ( z ) = V1 , V2 , ? , VM c then

[

]

T

? = I exp(? γz ) ? = V exp(? γz )

[

]

T

Mc ? T T ? ? P = ∑ Vm I m = I ? V = I ? V exp ( ?2α z ) m =1

( )

( )

PT = Re ( P ) = Re I ?

PL = ? ?PT ?z

? V } exp ( ?2α z ) } ? ? = ? Re {( I ) V } ( ?2α ) exp ( ?2α z ) = 2α P

T

{( )

?

? V = Re I ?

{( )

T

T

T

hence

α=

PL = αc + αd 2 PT

( I ) [ R ] I + (V ) [G ]V = 2 Re {( I ) V } 2 Re {( I ) V }

? T ? T ? T ? T

To calculate the resistance matrix, the attenuation constant for conductors is consider only,

P I ? [R ]I αc = c = 2 PT 2 Re I ? T V

T

( ) {( )

}

The numerator in this quotient is the dissipated conducting power per unit length, Pc , which, according to the perturbation theory, is approximately

48

equal to

Pc = I ?

( ) [R]I ≈ ∑ R ∫ ( J

T

Mc j =1 s l j

z

) j dl

2

M c is the number of conductors,

l j is the contour of cross section of the i-th conductor, R s is the surface resistance of conductors,

Rs =

πf ? σ

( Jz ) j

is the current density of the j-th conductor, which is approximately equal to

( Jz ) j =

(σ F ) j

1

? 0ε 0

(σ F ) j ， j = 1, 2,? , M c

is the free charge density on the j-th conductor.

If each of M c conductors is driven by the elements of the i-th eigenvector,

i i i i ?i V0( ) = ?V0( ) (1) , V0( ) ( 2 ) ,? , V0( ) ( M c ) ? , here V0( ) ( j ) , j = 1, 2,? , M c , represents ? ?

the voltage between the j-th conductor and the ground, then the free charge

( density on the j-th conductor, σ F ) , can be determined in a way given

i j

( )

earlier, and the surface current density flowing on the j-th conductor, can be found in an above-mentioned formula,

(J( )) ,

i z j

( Jz ) j =

1

? 0ε 0

(σ F ) j ,

j = 1, 2,? , M c

The average power transmitted along the transmission line, PT , appeared in the quotient expression of α c is approximately equal to ? PT = Re ? I ? ? ?

( )

T

? ?? V ? ≈ PT 0 = Re ? I 0 ? ? ? ?

( )

T

? ?? V0 ? ≈ I 0 ? ?

( )

T

? V0

49

Therefore the attenuation constant for conductors, α c , becomes

αc =

Pc 2P T

? T T

( I ) [ R] I = 2 Re {( I ) V }

?

≈

∑R ∫

j =1 s

Mc

?? T ? 2 I 0 V0

( )

lj

J 2 dl j

Since there exist M c current and voltage modes in the transmission line,

?( ?( ?( ? ? ? I 01) , I 02 ) ,?, I 0M c ) and V0(1) ,V0(2 ) , ?,V0( M c ) , then there are M c attenuation

constants, α c(1) , α c(2 ) , ? , α c( M c ) ,

Pc( i ) 2 PT

given by

α c(i ) =

(i )

≈

∑ Rs ∫ J z( )

i j =1 lj i? T 0

Mc

( ) dl ?( ) ? ( ) ， i = 1, 2,? , M 2(I ) V

2 j i 0

c

To calculate the resistance matrix alone, a transmission line is assumed to have conducting loss only, viz. [G ] = 0 , the telegraph equation for voltage

? ? ([R ] + jω [L])([G ] + jω [C ])V = γ 2V

is reduced to

? ? ([R ] + jω [L])( jω [C ])V = (α + jβ )2 V

? ? For a low loss transmission line, V ≈ V0 , the above equation becomes ? ? ([R] + jω [L])( jω [C ])V0 = (α + jβ )2 V0

and the equality of imaginary parts of both sides gives rise to

2 ? ? Im([R ] + jω [L])( jω [C ])V0 = Im(α + jβ ) V0

namely

? ? ω [R ][C ]V0 = 2αβV0

50

? ? It is inferred from I 0 = v p [C ]V0 that

[ R ] I?0 = 2αV?0

If the dielectric loss is not considered for the time being, [G ] = 0 , then the attenuation constant for dielectrics becomes zero,

αd

(V ) [G ]V = 0 = 2 Re {( I ) V }

? T ? T

and the attenuation constant α is simplified as

α = αc + αd = αc ≈

∑R ∫

j =1 s

Mc

?? T ? 2 I 0 V0 is transferred as

( )

lj

J 2 dl j

As a result, the equation

[ R ] I?0 = 2αV?0

[R]I?0 = 2α cV?0

As mentioned above, there are M c modes for α c , α c(1) , α c(2 ) , ? , α c( M c ) , which give rise to M c equations,

[R]I?0(i ) = 2α c(i )V?0(i ) ,

i = 1,2,?, M c

?( ?( ?( I 01) , I 02 ) ,?, I 0M c ) are modal currents given by

[C ][L]I?0 = v ?2 I?0 p

? ? ? V0(1) ,V0(2 ) , ?,V0( M c ) are modal voltages given by

? [L][C ]V0 = v ?2V?0 p

α c(1) ,α c(2 ) ,? ,α c( M

c

)

are modal attenuation constants given by

α c(i ) ≈

∑ Rs ∫ J z( )

i j =1 lj i ? T 0

Mc

( ) dl ?( ) ? ( ) ， i = 1, 2,? , M 2( I ) V

2 j i 0

c

51

J z(1) , J z( 2) ,? J z(

Mc )

are modal current densities given by

(J( ))

i z

j

=

1

? 0ε 0

(σ ( ) ) ,

i F j

i, j = 1, 2, ? , M c

( ( ( σ F1) , σ F2) ,? , σ FM

c

)

are modal charge densities determined by putting each

?i element of modal voltage V0( ) , i = 1, 2,? , M c , to the each conductor.

Each of the following M c matrix equations

[R]I?0(i ) = 2α c(i )V?0(i ) ,

i = 1,2,?, M c

contains M c algebraic equations, and therefore there are M c2 equations. Let i = 1 , the first matrix equation becomes

[ R ] I?0(1) = 2α c(1)V?0(1)

where

? R11 ? R [ R ] = ? 21 ?? ? ? RM c 1 ?

R12 R22 ? RM c 2

R1M c ? ? ? R2 M c ? ? ? ? ? ? RM c M c ? ? ?

T

(1 (1 (1 ?(1 I 0 ) = ? I 0 ) (1) , I 0 ) ( 2 ) ,? , I 0 ) ( M c ) ? ? ?

1 1 1 ?1 V0( ) = ?V0( ) (1) , V0( ) ( 2 ) ,? , V0( ) ( M c ) ? ? ? in which the first equation of the first matrix equation is T

( ( ( R11 I 0 ) (1) + R12 I 0 ) ( 2 ) + ? + R1M c I 0 ) ( M c ) = 2α c( )V0( ) (1)

1 1 1 1 1

Take i = 1, 2,? , M c , all the first equations for each of the above M c matrix equations are similarly given by

( ( ( R11 I 0 ) (1) + R12 I 0 ) ( 2 ) + ? + R1M c I 0 ) ( M c ) = 2α c( )V0( ) (1)

1 1 1 1 1

( R11 I 0

2)

(1) + R12 I 0( 2) ( 2 ) + ? + R1M I 0( 2) ( M c ) = 2α c( 2)V0( 2) (1)

c c c c c

…………………………………………………………

( R11 I 0

Mc )

(1) + R12 I 0( M ) ( 2 ) + ? + R1M I 0( M ) ( M c ) = 2α c( M )V0( M ) (1)

c

52

The simultaneous equations can be solved for the first row of the resistance matrix

[ R] ,

R11 , R12 ,? , R1M c

All other rows of resistance matrix

[ R] [ R]

can be determined in a similar way, and therefore the is finally found,

XI. XI. GROUND LOSS

If there are two ground planes in a transmission line system, then the upper ground is considered to be an additional conductor and its loss computation has been described previously. Therefore only one ground plane need to be considered here. The z-directed narrow strip current J j ?l j of the j-th conductor produces a magnetic field on the ground plane,

? ? H j = ? H j? = ? J j ?l j 2πρ j , j = 1, 2,? , N

The magnetic field has a component tangential to the ground plane,

? J ?l ? ? y ? ? ? j j ?? j H jt = xH jx = xH j? cos ? j = x ? ? 2πρ ? ? ρ j ?? j ? ? ? ? ?

y

J z ?l j

( xi , 0 )

ρj

( xj , yj )

?j

Hj

?j

x

Ground Loss

53

The current of all conductors will produce a tangential magnetic field at the point

( xi , 0 )

on the ground,

? H t ( xi , 0 ) = ∑ H jt = x

j =1

N1

1 2π

∑

j =1

N1

(x ? x )

i j

J j ?l j y j

2

+ y2 j

where N1 is the number of subsections for all conductor-to-dielectric interfaces. In accordance with the method of image, the total tangential magnetic field at the point

( xi , 0 )

due to all conductors will be doubled.

? H tg ( xi , 0 ) = 2 H t ( xi ,0 ) = x

∑ π

j =1

1

N1

(x ? x )

i j

J j ?l j y j

2

+ y2 j

It follows from the boundary condition that the current density at

( xi , 0 )

is

? ? J g ( xi , 0 ) = y × H tg ( xi , 0 ) = ? z

∑ π

j =1

1

N1

(x ? x )

i j

J j ?l j y j

2

+ y2 j

In addition to the conducting loss due to all conductors, Pc ≈ ∑ Rs ∫ J 2 dl j

j =1 lj Mc

the following ground loss needs to be added,

2 Rs ∫ J g dl lg

where lg is the width of ground plane, which is usually taken as 5 to 10 times of the transverse dimension of the transmission line system. After the ground loss is computed, the total power dissipated per unit length is given by

2 Pc ≈ ∑ Rs ∫ J 2 dl + Rs ∫ J g dl j

j =1 lj lg Mc

and the resistance matrix is calculated in the same way as described above. Proposed homework #1 Calculate the resistance per unit length of a parallel-wire transmission line

54

and compare the calculated result with that given by

? Rs ? R= πd ? ? ? ? 2 ( D d ) ?1 ? ? 2D d

where d is the diameter of the wire and D is the spacing between the two wires. Proposed homework #2 Calculate the resistance per unit length of a microstrip line and compare the calculated result with that available in some literatures.

REFERENCES

[1] Cao Wei and Xu Linqin, 《Theory of Electromagnetic Fields and Waves》, 》 Publishing House of Beijing University of Posts and Telecommunications, 1999. (in Chinese) [2] R. F. Harrington, 《Field Computation by Moment Methods》, Macmillan, 》 New York, 1968. [3] Cao Wei, R. F. Harrington, J. R. Mautz, and T. K. Sarkar, “Multiconductor Transmission Lines in Multilayered Dielectric Media,” IEEE Trans., MTT-32, pp.439-450, April 1984. [4] R. F. Harrington and Cao Wei, “Losses on Multiconductor Transmission Lines in Multilayered Dielectric Media,” IEEE Trans., MTT-32, pp.705-710, July 1984. [5] Liao Cheng’en, 《Fundamental of Microwave Techniques》 Xidian University , 》 Press, 1995. (in Chinese) [6] W. T. Weeks, “Calculation of Coefficients of Capacitance of Multiconductor Transmission Lines in the Presence of a Dielectric Interface,” IEEE Trans., MTT-18, Jan. 1970.

55

赞助商链接

相关文章:

- 简易自动电阻测试仪报告
- 待测电阻 恒流源 ADC 采样 单片机系统 自动量程切换 基准
*电阻矩阵*控制面板 ...*19*20 P1.0 VCC P1.1 P0.0 P1.2 P0.1 P1.3 P0.2 P1.4 P0.3...

- 传输矩阵法 免费
- 传输
*矩阵*具有和*电阻*相同的模型特性。 A B (a) E0 E1 (b) 图 1 传输*矩阵*...(18) 将式(18)代入式(17)中可得式(*19*),并将其转为*矩阵*形式(20): + ?...

- 矩阵 说明书
- 视频输入的连接(全中文
*矩阵*系统) 对全中文*矩阵*系统,各视频输入未接有 75*电阻*,...自动切换队列 一个自动切换队列是指一组摄像机输入自动循环地显示在一个单独的...

- 矩阵键盘的设计
*矩阵*键盘的设计_信息与通信_工程科技_专业资料。本实验...通过上拉*电阻*在显示器上显示不同的符号按键 “0-F...RP1 RESPACK-8 U1*19*XTAL1 P0.0/AD0 P0.1/...

- RF开关
*矩阵*和通用继电器的不同配置 板载内存用于确定性扫描...*损耗*,定义如下:文件中找不到关系 ID 为 rId*19*...3)低的插入*损耗*:消除了半导体中扩散*电阻*,极大地...

- 我看矩阵在实际生活中的应用
- 只有成功地应用了数学时,才真正达到了完善的地步”...关键词:
*矩阵*、人口流动、*电阻*电路、密码学、文献管理...于是数集 1,3,3,15,13,16,12,9,*19*,8,5,0...

- 矩阵说明书
- 系统 的视频连接应符合视频输入终结 75 欧姆
*电阻*的...80*矩阵*切换控制系统主机可安装在标准的*19*英寸(EIA)...对列是指将一组摄像机画面顺序地切换到一组连续的...

- 微波技术习题
- R2 R1 λ/4 题 4.
*19*图 4.15 试求在特性...(1)散射参量*矩阵*[S];(2)插入衰减、插入相移; (...*电阻*,因而引入*损耗*,试计算 Q 值。 图 5-44 习题...

- 电力系统分析试卷及答案
- 计算线路 L-3 的功率
*损耗*和电压*损耗*; 4.条件同 ...S 1、试计算变压器绕组*电阻*、漏抗的有名值; 2、...节点导纳*矩阵*,图中给出了各支路阻抗和对地导纳的 ...

- 监控系统矩阵调试经验
- 几乎所有球机或者解码器上都自带 120 欧姆终端
*电阻*,...这个问题通常因为视频分配器对视频 信号*损耗*过大,且...本地的前端,中心*矩阵*可以调看控制任一下级*矩阵*的任...

更多相关标签: