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19.电阻矩阵+地损耗


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.

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