10 Coupled Lines

8/17/2019 10 Coupled Lines

1/16

Coupled transmission linesCoupled transmission lines

When two transmission lines are placed close together, the

propagation in each line is influenced on the other. We talk in this

case of coupled-line propagation

Also in this case every possible solution, represented by a propagatingem field, must be the combination of modes. Each mode define a

configuration of E and H fields in the two lines, with specific

propagation parameters (phase constant, attenuation, characteristic

impedance)


2/16

Circuit model for two coupled linesCircuit model for two coupled lines

V1(z) V2(z)

I1(z)I2(z)

Element dz line 1 Element dz line 2

Electric coupling

Magnetic coupling

C

Ca1

La1

Ca2

La2L

11 11 11 12 22 22 12 11 22 22

11 11 11 12 22 22 12 11 22 22

,

,

L L L L L L

C C L C C L

V j L I j L I V j L I j L I

I j C V j C V I j C V j C V


3/16

Propagation equations

1 1 11 1 12 2

2 2 22 2 12 1

1 1 11 1 12 2

2 2 22 2 12 1

( ) ( )

( ) ( )

( ) ( )

( ) ( )

p

I z dz I z j C dz V z j C V z dz

I z dz I z j C dz V z j C V z dz

V z dz V z j L dz I z j L I z dz

V z dz V z j L dz I z j L I z dz

1 211 1 12 2 12 1 22 2

1 211 1 12 2 12 1 22 2

er 0:dz

dI dI j C V j C V j C V j C V

dz dz

dV dV j L I j L I j L V j L V dz dz

C

Ca1 Ca2

La1 La2

V1(z)

I1(z)

V1(z+dz)

V2(z)

I2(z)

V2(z+dz)

I1(z+dz) I2(z+dz)

L


4/16

Type of solution:

1 1 12 2 2 12 1 2

1 1 12 2 2 12 1 2

V LI L I V L I LI

I CV C V I C V CV

There are only 4 possible solutions for :

12 1212 12 1 1 L C

L L C C LC L C

The first ± sign define the direction of propagation (- toward increasing z). The ± signs

inside the square roots define two different propagation modes (called even and odd).For C12 /C=-L12 /L these modes have the same value of (i.e. the same phase velocity).

That happens when the two lines, taken separately, allow the propagation of a TEM

mode .

Solution for two equal lines (L11=L22, C11=C22)

1,2 0( ) j z

V z V e

If such a solution exists, the phase constant must satisfy the following equations (with

L=L11=L22, C=C11=C22):

2

12

12 121 1 C C

e o

C v v LC L L

C


5/16

For the symmetries, it can be easily understood that the even mode is characterized by the

voltages V 1

(z) and V 2

(z) equal and in phase, while for the odd mode these voltages are

equal and out of phase:

1 2 1 2,e e o oV V V V

As in the case of simple lines, the ratio of incident (or reflected) voltage and current

voltages define the characteristic impedance of the each mode:

12 121 1 1 1

12 12

,ce e e co o o L L L L

Z V I Z V I C C C C

Note that cannot never be Z ce=Z

co; this means that it is not possible to cancel the

reflected waves of both modes at the same time.

A generic solution in the two coupled lines is constituted by a linear combination of the

even and odd modes. For instance, considering only the progressive wave, it has :

1 1 0 1 0 2 2 0 2 0, p pd d

j z j z j z j z

e o e oV z V e V e V z V e V e


6/16

We have seen that, in this case, C 12 /C=-L12 /L, so:

Equal TEM lines (C11=C22=C)

2

12

1 1

1

e ov v v

LC C C

12 12

12 12

1 1,

1 1ce co

C C C C L L Z Z

C C C C C C

Zce e Zco can be expressed as function of the capacitance p.u.l. of the two

modes:

12

12

1

,

1,

cp even

even

cd odd

odd

Z C C C v C

Z C C C

v C


7/16

Being the voltages equal and in phase for the even mode , there is along the symmetry axis

a perfect magnetic wall. For the odd mode, the condition of voltages equal and out of

phase implies the presence of perfect electric wall along the same axis.

For computing the impedances Z cpe Z cd it is then possible to refer to the equivalent lines

obtained imposing the presence of such walls:

Magnetic

WallEven mode

Odd mode

1co

odd

Z v C

1ce

even

Z v C

Physical meaning of even and odd modes

The phase velocity is the same for the two TEM lines. It depends on

the medium filling the structure (dielectric constant r ):

Electric

WallSymmetryaxis

2

12

1

1 r

cv

LC C C


8/16

Circuit model of two coupled lines with finiteCircuit model of two coupled lines with finite

lengthlength

Zce, Zco

L

1 2

3 4

Goal: compute the 4-port Z matrix (or Y, or S).

Hypothesis: equal lines (two symmetry axis)

Evaluation method: matrix eigenvalues


9/16

Eigenvalues evaluationEigenvalues evaluation

Symmetry axis 1

Symmetry axis 2

Exciting the network with an eigenvector, an electric or a

magnetic wall is obtained alon the symmetry axis. Withreference to Z matrix, the exciting currents for eacheigenvector result:

1 2

3 4

1

2

3

4

1, 1, 1, 1

1, 1, 1, 1

1, 1, 1, 1

1, 1, 1, 1

I

I

I

I

Axis 1: Magnetic, Axis 2: Magnetic

Axis 1: Magnetic, Axis 2: Electric

Axis 1:Electric, Axis 2: Magnetic

Axis 1: Electric, Axis 2: Electric


10/16

Evaluation of the eigenvalues using theEvaluation of the eigenvalues using the eigenetworkeigenetwork

EigenvalueEigenvalue Z1:

Z1

/2, Zcp OPEN 1 cot 2cp Z jZ


Z2

/2, Zcp SHORT 2 tan 2cp Z jZ


Z3

/2, Zcd OPEN 3 cot 2cd Z jZ


Z4

/2, Zcd SHORT 4 tan 2cd Z jZ

OPEN

OPEN

SHORT

SHORT


11/16

Evaluation of Z MatrixEvaluation of Z Matrix

From the definition of Z, imposing each eigenvector as excitation,

the four independent elements of Z are obtained:

11 11 12 13 14

0

12 11 12 13 14

0

13 11 12 13 14

0

1

4 11 12 13 14

0

V Z Z Z Z Z I

V Z Z Z Z Z

I V

Z Z Z Z Z I

V

Z Z Z Z Z I

11 1 2 3 4

12 1 2 3 4

13 1 2 3 4

14 1 2 3 4

1

4

1

41

4

1

4

Z Z Z Z Z

Z Z Z Z Z

Z Z Z Z Z

Z Z Z Z Z

Using Z i, the eigenvalues of the other matrices (Y, S) can be

obtained. The above formulas can be then used for computing the

elements of also these matrices


12/16

Expression of Z matrix elementsExpression of Z matrix elements

11

12

13

cot tan cot tan4 2 2 2 2

cot tan cot tan4 2 2 2 2

cot tan cot t4 2 2 2

cp cp cd cd

cp cp cd cd

cp cp cd cd

j Z Z Z Z Z

j Z Z Z Z Z

j Z Z Z Z Z

14

an2

cot tan cot tan

4 2 2 2 2

cp cp cd cd

j Z Z Z Z Z


13/16

11

12

13

14

cot

2

1

2 sin

cot2

12 sin

cp cd

cp cd

cp cd

cp cd

Z Z Z j

Z Z Z j

Z Z Z j

Z Z Z j

Compact expressions of Z and Y elementsCompact expressions of Z and Y elements

11

12

13

14

cot2

1

2 sin

cot2

1

2 sin

cp cd

cp cd

cp cd

cp cd

Y Y Y j

Y Y Y j

Y Y Y j

Y Y

Y j


14/16

Terminating the ports in cc orTerminating the ports in cc or ococ

1

3

2-port network

1 11 1 12 13 3 14 11 1 13 3

3 31 1 32 33 3 34 31 1 33 3

0 0

0 0

I Y V Y Y V Y Y V Y V

I Y V Y Y V Y Y V Y V

11 22 11

12 13

cot ,2

cot2

cp cd

cp cd

Y Y Y Y Y j

Y Y Y Y j

1

4

2-port network

1 11 1 12 13 14 4 11 1 14 4

4 41 1 42 43 44 4 41 1 44 4

0 0

0 0

V Z I Z Z Z I Z I Z I

V Z I Z Z Z I Z I Z I

11 22 11

12 14

cot ,2

1

2 sin

cp cd

cp cd

Z Z Z Z Z j

Z Z

Z Z j


15/16

Special casesSpecial cases

LL=180°

The eigenvalues of Z are [0, ∞, 0, ∞], so those of S result:

Si= [-1, 1, -1, 1]. The scattering matrix elements are then:

11 12 13 140, 1, 0, 0S S S S Note that line 2 is completely uncoupled from line 1 independently on Z0!

Perfect matching at the four ports

There is a value of the load Z0 for which the ports are all matched(S11=S22=S33=S44=0). Moreover this happens independently of L:

0 , ,= Zc p c d Z Z

Note that the matching at the ports does not imply the absence ofreflected waves along the two lines. Actually that happens only at the

ports


16/16

Derivation of the matching conditionDerivation of the matching condition

Eigenvalues of S

1

1

ii

i

jX S

jX

Parameter S11:

11 1 2 3 41

04

S S S S S

1 0

2 0

3 0

4 0

cot 2

tan 2

cot 2

tan 2

cp

cp

cd

cd

jX jZ Z

jX jZ Z

jX jZ Z

jX jZ Z

Eigenvalues of Z

There are only two cases where the above condition can be satisfied

independently on L, i.e.:

1 2

3 4

0

0

S S

S S

1 2

3 4

1

1

X X

X X

2 2

0

2 20

cp

cd

Z Z

Z Z

NOT

Admissible

1 4

2 3

0

0

S S

S S

1 4

3 2

1

1

X X

X X

2

0

2

0

cp cd

cp cd

Z Z Z

Z Z Z

Admissible

10 Coupled Lines

Documents

Transcript of 10 Coupled Lines