Post on 06-Jul-2018
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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Evaluation of the eigenvalues using theEvaluation of the eigenvalues using the eigenetworkeigenetwork
EigenvalueEigenvalue Z1:
Z1
/2, Zcp OPEN 1 cot 2cp Z jZ
EigenvalueEigenvalue Z2:
Z2
/2, Zcp SHORT 2 tan 2cp Z jZ
EigenvalueEigenvalue Z3:
Z3
/2, Zcd OPEN 3 cot 2cd Z jZ
EigenvalueEigenvalue Z4:
Z4
/2, Zcd SHORT 4 tan 2cd Z jZ
OPEN
OPEN
SHORT
SHORT
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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
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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
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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
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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
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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
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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