Download abcd hybrid matrix scattering

Microwave Engineering
Cheng-Hsing Hsu
Department of Electrical Engineering
National United University
Outline
1. Transmission Line Theory
2. Transmission Lines and Waveguides
General Solutions for TEM, TE, and TM waves ; Parallel Plate waveguide ; Rectangular Waveguide ; Coaxial
Line ; Stripline ; Microstrip
3. Microwave Network Analysis
Impedance and Equivalent Voltages and Currents ; Impedance and Admittance Matrices ; The Scattering
Matrix ; ABCD Matrix ; Signal Flow Graphs ; Discontinuties and Model Analysis
4. Impedance Matching and Tuning
Matching with Lumped Elements ; Single-Stub Tuning ; Double-Stub Tuning ; The Quarter-Wave Transformer ;
The Theory of Small Reflections
5. Microwave Resonators
Series and Parallel Resonant Circuits ; Transmission Line Resonators ; Dielectric Resonators
6. Power Dividers and Directional Couplers
Basic Properties of Dividers and Couplers ; The T-Junction Power Divider ; The Wilkinson Power Divider ;
Coupled Line Directional Couplers ; 180o hybrid
7. Microwave Filters
Periodic Structure ; Filter Design by the Insertion Loss Method ; Filter Transformations ; Filter Implementation ;
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6. Power Dividers and Directional Couplers
Basic Properties of Dividers and Couplers ;
The T-Junction Power Divider ;
The Wilkinson Power Divider ;
Coupled Line Directional Couplers ;
180o hybrid
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Basic Properties of Dividers and Couplers
For the scattering matrix theory, some basic properties of three and four port networks have
been discuss and drive.  isolation , coupling, and directivity, which are useful quantities for
the characterization of couplers and hybrids.
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Three-port Network (T-junctions)
The simplest type of power divider is a T-junction, which is three-port network with two inputs and
one output. -> the scattering matrix of an arbitrary three network has nine independent elements.
S11


S 
S 21


S 31

S12
S 22
S 32
S13 
S 23 

S 33 

0

S 
S12


S13

Matched , Reciprocal
S12
0
S 23
S13 
S 23 

0

If the component is passive and contains no anisotropic materials, then it must be reciprocal and its [S] matrix
must be symmetric (Sij=Sji).
A junction that is lossless and matched at all ports to avoid the power loss.
However, three-port network cannot be lossless, reciprocal, and matched at all ports.
If the network is also lossless, then energy conservation requires that the scattering matrix be unitary =>
2
2
2
2
S12 S13 1
S12 S 23 1 from last three equation at least two of the three parameters (S12, S13, S23)
must be zero, but this condition will always be inconsistent with one of
2
2
*
S13 S 23 1
S13 S 23 0
first three equation. => if one of these three condition is relaxed, a
physical realizable device is possible
S * S 0
S * S 0
23 12
12 13
Power division and combining. (a) Power division. (b) Power combining
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If the three-port network is nonreciprocal, then SijSji, and the conditions of input matching at
all ports and energy conservation can be satisfied. => a circulator
Any matched lossless three-port network for a circulator must be nonreciprocal. => The [S] matrix a matched
three-port network has the following form:
0

S 
S 21


S31

S12
0
S32
S13  Then if the network is lossless, [S] must be unitary
2
2
2
2
=>
S12 S13 1
S 21 S 23 1
S 23 

2
2
*
0
S31 S32 1
S 31
S32 0

*
S 21
S 23 0
S12* S13 0
These equations can be satisfied in one of two ways, Either : S12 S 23 S 31 0,
Or : S 21 S32 S13 0,
S 21 S32 S13 1
S12 S 23 S31 1
This result shows that Sij Sji for ij,which implies that the device must be nonreciprocal.
=> The [S] matrices for the two solution are shown in figure, together with the symbols for the two possible
types of circulators. (port 1to 2, or port 2 to 3, or port 3 to 1 from figure (a) direction)
The two types of circulators and their [S] matrices. (The phase references for the ports are arbitrary.)
(a) Clockwise circulation. (b) Counterclockwise circulation.
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Alternatively, a lossless and reciprocal three-port network can be physically realized if only two of
its ports are matched.
0 S12 S13 

S 
S
0
S
If ports 1 and 2 are these matched ports 
12
23



S13

S 23
S33 

To be lossless, the following unitarity conditions must be satisfied:
2
2
2
S12 S13 1
S13 S 23 S33 1
2
2
2
S12 S 23 1
2
*
*
S 23
S12 S33
S13 0
the (front end) three equations show │S13│= │S23│
=> S13=S23=0 => │S12│=│S33│=1. The signal
*
S12* S13 S 23
S 33 0 S13* S 23 0
flow graph seen the network actually consists of
two separate components, one a matched two-port line and the other a totally mismatched one-port.
Finally, if the three-port network is allowed to be lossy, it can
be reciprocal and matched at all ports => resistive divider
In addition, a lossy three-port can be made to have isolation
between its output ports
A reciprocal, lossless three-port network matched at ports 1 and 2.
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Four-port Networks (Directional Couplers)
The [S] matrix of a reciprocal four-port network matched at all ports has the following form:
0

S

S 12

S13

S14

S12
0
S13
S 23
S 23
0
S 24
S 34
S14  If the network is lossless, 10 equations result from the unitarity,
or energy conservation, condition.
S 24 

S34 

0 
Let us consider the multiplication of row 1 and row 2, and the multiplication of row 4 and row 3
*
 S13* S 23 S14* S 24 0 and S14* S13 S 24
S 23 0
*
*
multiply first equation by S 24
and second equation by S13
, and subtract to obtain

S14* S13 S 24
2
2
0
(1)
Similarly, the multiplication of row 1 and row 3, and the multiplication of row 4 and row 2,
*
*
S12
S 23 S14* S 34 0 and S14* S12 S 34
S 23 0
multiply first equation by S12 and second equation by S34 , and subtract to obtain

S 23 S12 S 34
2
2
0
(2)
If S14 S 23 0 (1) and (2) to be satisfied 
directional coupler
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Then the self - products of the rows of the unitary 
S matrix yield the following equation
2
2
S12 S13 1
2
2
S12 S 24 1
2
2
S13 S34 1
2
2
S 24 S 34 1
which imply that S13 S 24 , and that S12 S34
Further simplification can be made by choosing the phase references on three of four ports.
choose S12 S34 α, S
ej, and S 24 β
ej
13 β
where αand βare real, and are phase constants to be determined.
*
*
Dot product of rows 2 and 3 gives S12
S13 S 24
S 34 0 , θ2nπ
If we ignore integer multiples of 2, there are two particular choices that commonly occur in practice :
( I ) Symmetrical Coupler /2
( II ) Antisymmetrical coupler 0, 
0
α

S 
jβ

0
α
0
0
jβ
jβ
0
0
α
0
jβ

α

0
0


α

S 

β

0

α
0
0
β
β
0
0
α
0 
β

α

0 
Noted : two couplers differ only in the choice of reference planes.
Also, the amplitude αand βare not indenpendent
 α2 β2 1
thus, apart from te phase references, an ideal directional coupler has only one degree of freedom.
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C 20 log dB

D 20 log
dB
S14
I 20 log S14 dB
I D C dB
coupling factor S13 β2
2
The ideal coupler would have infinite
2
directivity and isolation (S14=0). Both and  through port with the coefficient S12 2 1 β2
could be determined from the coupler factor, C. no power is delivered to port 4 (isolation port )
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Hybrid couplers are special cases of directional couplers, where the coupling factor is 3 dB,
=> = = 1/(2)1/2
 There are two types of hybrids
 Quadrature hybrid (90o hybrid) => 90o phase shift between ports 2 and 3 (= = / 2)
when fed at port 1, and is an example of a symmetrical coupler.
 Magic-T hybrid or rat-race hybrid (180o hybrid) => 180o phase difference between ports 2
and 3 when fed at port 4, and is an example of an antisymmetrical coupler.
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Directivity Measurement
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The T - Junction Power Divider
The T-junction power divider is a simple three-port network that can be used for power division
or power combining, and can be implemented in virtually any type of transmission line medium
=> commonly used in waveguide, microstrip or stripline structure. (loss or lossless junctions)
(a) E plane waveguide T. (b) H plane waveguide T. (c) Microstrip T-junction
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Lossless Divider
The lossless T-junction can all be modeled as a junction of three transmission lines. => there are
fringing fields and higher order modes associated with the discontinuity at such a junction, leading
to stored energy that can be accounted for by a lumped susceptance, B. => divider to be matched to
the input line of the characteristic impedance of Zo.
If the transmission lines are
assumed to be lossless (or of low
loss), then the characteristic
impedances are real.
=> assume B = 0
=> 1/Z1+1/Z2 = 1/Zo
If B is not negligible, some type of
reactive tunning element can usually
be added to the divider to cancel
this susceptance, at least over a
narrow frequency range.
There will be no isolation between the
two output ports, and there will be a
mismatch looking into the output ports.
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Ex: A lossless T-junction power divider has a source impedance of 50 . Find the
output characteristic impedances so that the input power is divided in a 2:1 ratio.
Compute the reflection coefficients seen looking into the output ports.
1 Vo2
Pin 
2 Zo
1 Vo2 1
P1 
 Pin
2 Z1 3
1 Vo2 2
P2 
 Pin
2 Z2 3
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Resistive (lossy) Divider
If a three-port divider contains lossy components it can be made to be matched at all ports, although
the two outputs may not be isolated. => using lumped-element resistors to fit the circuit for the
divider. => An equal-split (-3dB) divider is form, but unequal power division ratios are also possible.
Assuming that all ports are terminated in the characteristic impedance Zo, the impedance Z, seen
looking into the Zo/3 resistor followed by the output line, is => Z = Zo/3 + Zo = 4Zo/3
 the input impedance of the divider is Zin = Zo/3 + 2Zo/3 = Zo
 shows that the input is matched to the feed line => the network is symmetric from all three
ports, the output ports are also matched. => S11 = S22 = S33 = 0.
If the voltage at port 1 is V1, then by voltage division the voltage V at the center
of the junction is V V1
2Z o 3
2
 V1
Z o / 3 2Z o 3 3
V2 V3 V
Zo
3
1
 V  V1
Z o Z o / 3 4
2
=> S21=S31=S23=1/2, which is -6 dB below the input power
level => is reciprocal, scattering matrix is symmetric
Half of the supplied power is dissipated in the resistors.
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The Wilkinson Power Divider
Due to Lossless T-junction divider is not being matched at all ports and doesn’
t have any isolation
between output ports, and resistive divider can be matched at all ports (but even though it is not
lossless) and still doesn’
t have isolation.
 However, a lossy three-port network can be made having all ports matched with isolation
between the output ports.
 The Wilkinson power divider is such a network, with the property of being lossless when the
output ports are matched; that is, only reflected power is dissipated. (can be made with arbitrary
power division, but consider the equal-split 3dB case.)
Analyze this circuit by
reducing it to two simpler
circuits driven by symmetric and
antisymmetric sources at the
output ports. =>even-odd mode
analysis technique.
The Wilkinson power divider. (a) An equal-split Wilkinson power divider in microstrip form.
(b) Equivalent transmission line circuit.
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Even-odd Mode Analysis
Normalize all impedances to the characteristic impedance Zo, and redraw the circuit with voltage
generation at the output ports as shown in below figure. (A form that is symmetric across the
midplane; the two source resistors of normalized value 2 combine in parallel to give a resistor of
normalized value 1. => representing the impedance of a matched source.)
 quarter-wave lines have a normalized value of Z ; shunt resistor has a normalized value of r
=> the equal-split power divider, these values should be Z = (2)1/2 and r = 2.
we define two separate modes of excitation for the circuit of following figure.
Even-mode : Vg2 = Vg3 = 2Vo ; Odd-mode : Vg2 = -Vg3 = 2Vo
Using superposition of these two-mode => effectively have an excitation of Vg2 = 4Vo, Vg3 = 0,
from which we can find the S-parameters of network.
The Wilkinson power divider circuit in normalized and symmetric form.
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Even-mode
Even-mode excitation, Vg2 = Vg3 = 2Vo, and so V2e = V3e and there is no current flow through the
r/2 resistors or the short circuit between the inputs of two transmission lines at port 1.
From the open circuit to form even-mode analysis, the impedance looking into port 2
=> Zine = Z2 / 2 since the transmission line looks like a quarter-wave transformer.
=> If Z = (2)1/2, port 2 will be matched for even mode excitation => V2e = Vo since Zine = 1.
If we let x = 0 at port 1 and x = -/4 at port 2,
V(x)=V+(e-jx+ejx)
V2e = V(-/4) = jV+(1-)=Vo  S22e=0
V1e = V(0) = V+(1+) = jVo (+1)/(-1)
The reflection coefficient is seen at port 1
 = [2-(2)1/2] / [2+(2)1/2] , V1e = -jVo(2)1/2S12e= -j(2)1/2
Symmetry of port 2 and 3=> V3e=Vo ; S33e=0 ; S13e = -j(2)1/2
Bisection of the circuit. (a) Even-mode excitation. (b) Odd-mode excitation.
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Odd-mode
Odd-mode excitation, Vg2 = -Vg3 = 2Vo, and so V2o = V3o and there is a voltage null along the
middle of the circuit. Thus this circuit by grounding it at two points on its mid-plane. Looking into
port 2, an impedance of r/2, since the parallel-connected transmission line is /4 long and shorted
at port 1, and looks like an open circuit at port 2.
Thus, port 2 will be matched for odd mode excitation if we select r =2. => V2o = Vo and V1o=0
 excitation all power is delivered to the r/2 resistors, with none going to port 1.
Finally, we must find the input impedance at port 1 of the Winkinson divider when ports 2 and 3
are terminated in matched loads.
The resulting circuit is shown in following figure, it is seen that
similar to an even mode of excitation, since V2=V3. => no current
flows through the resistor of normalized value 2, so it can be
removed. We now have the parallel connection of two quarterwave transformers terminated in loads of unity (normalized)
=> The input impedance is then Zin = [(2)1/2]2/2 =1
Analysis of the Wilkinson divider to find S11. (a) The terminated
Wilkinson divider. (b) Bisection of the circuit in (a).
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Summary, we can establish the following S parameters for the Wilkinson divider :
S11 0 ( Z in 1 at port 1) ; S 22 S33 0 (ports 2 and 3 matched for even and odd modes)
V1e V1o
S12 S 21  e
j
V2 V2o
S13 S 31 -j
2 (symmetry due to reciprocity)
2 (symmetry of ports 2 and 3)
S 23 S32 0 (due to short or open at biection)
When then dividers is driven at port 1 and the outputs are matched, no power is
dissipated in the resistor. => the divider is lossless when the outputs are matched; only
reflected power from the ports 2 or 3 is dissipated in the resistor.  since S23 = S32 =
0 -> ports 2 and 3 are isolated.
Photograph of a four-way corporate power divider network
using three microstrip Wilkinson power dividers.
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Ex. Design an equal-split Wilkinson power divider for a 50 system impedance at frequency f0, and
plot the return loss (S11), insertion loss (S21=S31), and isolation (S23=S32) versus frequency from 0.5f0
to 1.5f0.
A characteristic impedance of quarter-wave transmission lines in divider => Z=Z0(2)1/2 = 70.7 ,
and the shunt resistor a value of R = 2Z0 = 100 
Frequency response of an equal-split Wilkinson power divider. Port 1 is the input port; ports 2 and 3 are the output ports
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The quadrature (90o) Hybrid
Quadrature hybrid are 3 dB directional couplers with 90o phase difference in the outputs of the
though and coupled arm. (often made in microstrip or stripline) -> branch-line hybrid
 analyze the operation of the quadrature hybrid using an even-odd mode decomposition
technique similar to that used for the Wilkinson power divider.
 The branch-line coupler with all port matched, the power entering port 1 is evenly divided
between ports 2 and 3, with a 90o phase shift between these outputs.
=> No power is coupled to port 4. => the [S] matrix

Note:
1. has a high degree symmetry, as any port can be used as the input port
2. output ports always be on opposite side of the junction from the input port
3. isolated port remains on the same side as the input port.
4. this symmetry is reflected in the scattering matrix as each row can be obtained as a transposition
of the first row.
Geometry of a branch-line coupler.
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Even-odd Mode Analysis
The schematic circuit of the branch-line coupler in normalized form as the following figure. =>
assume that a wave of unit amplitude A1 = 1 is incident at port 1.
Because of the symmetry or antisymmetry of the excitation, the four port network can be decomposed
into a set of two decoupled two-port network. => since the amplitudes of the incident waves for these
two-ports are ±1/2, the amplitudes of the emerging wave at each port of the branch line hybrid can be
expressed as : B1 = e / 2 + o / 2 ; B2 = Te / 2 + To / 2 ; B3 = Te / 2 –To / 2 ; B4 = e / 2 - o / 2
Circuit of the branch-line hybrid coupler in a normalized form.
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Firstly, consider the calculation of e and Te , for the even-mode two-port circuit. done by
multiplying the ABCD matrices of each cascade component in the circuit.
=>
shunt
Y=j
/4TL
shunt
Y=j
From ABCD parameters can be convert to S-parameters (Z0 =1)
Similarly, with the odd-mode.
=>
 B1 = 0 (port 1 is matched)
B2 = -j / (2)1/2 (haft-power, -90o phase shift from 1 to 2)
B3 = -1 / (2)1/2 (haft-power, -180o phase shift from 1 to 3)
B4 = 0 (no power to power 4)
Using multi-section BL hybrid => increase BW.
Decomposition of the branch-line coupler into even- and
odd-mode excitations. (a) Even mode (e). (b) Odd mode (o).
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Coupled Line Directional Couplers
When two unshielded transmission lines are close together, power can be coupled between the
lines due to the intersection of the electromagnetic fields of each other. Consist of three
conductors in close proximity  In addition, coupled transmission lines are usually assumed to
operate in TEM mode. rigorously valid for stripline structures and approximately valid for
microstrip structures. ( a three-wire line can support two distinct propagating modes -> can be
used to implement directional couplers, hybrids, and filters.)
Various coupled transmission line geometries. (a) Coupled stripline (planar, or edge-coupled).
(b) Coupled stripline (stacked, or broadside-coupled). (c) Coupled microstrip.
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Coupled Line Theory
If we assume TEM propagation, then the
electrical characteristics of the coupled
lines can be completely determined from the
effective capacitances between the lines and
the velocity of propagation on the line.
If the strip conductors are identical in
size and location relative to the ground
conductor, then C11=C22.
A three-wire coupled transmission line and its equivalent capacitance network.
 Consider two special types of excitations
for the coupled line:
even-mode, where the currents in the strip
conductors are equal in amplitude and in
the same direction.
odd-mode, where the currents in the strip
conductors are equal in amplitude but in
opposite direction.
Even- and odd-mode excitations for a coupled line, and the resulting equivalent
capacitance networks.(a) Even-mode excitation. (b) Odd-mode excitation.
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Even - mode :
electric field has even symmetry about the center line,
and no current flows between the two strip conductors.
Lead to the equivalent circuit shown, where C12 is effectively open - circuited.
the resulting capacitance of either line to ground for the even mode is Ce C11 C22
assuming that the two strip conductors are identical in size and location
characteristic impedance for even - mode is Z 0 e 
LCe
L
1


Ce
Ce
v p Ce
Odd mode :
electric field lines have an odd symmetry about the center line,
and a voltage null exists between the two strip conductors.
imageine this as a ground plane through the middle of C12
effective capacitance between either strip conductor and ground is Co C11 2C12 C22 2C12
1
the characteristic impedance for the odd - mode is Z 0 o 
v p Co
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For symmetry coupled stripline (support purely TEM mode), the design graph can be used to
determine the necessary strip widths and spacing for a given set of characteristic impedance, Z0e,
Z0o, and dielectric constant.
For microstrip, the results do not scale with dielectric constant, so design graphs must be made for
specific values of dielectric constant.
Another difficulty with microstrip coupled lines is the fact that the phase velocity is usually
different for the two modes of propagation => since the two modes operate with different field
configurations in the vicinity of the air-dielectric interface=> degrading effect on coupler
directivity.
Normalized even- and odd-mode Even- and odd-mode characteristic
characteristic impedance design impedance design data for coupled microstrip
data for edge-coupled striplines. lines on a substrate with 
r = 10.
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Design of Coupled Line Couplers
Apply an even-odd mode analysis to a length of coupled line to arrive a the design equations.
This four-port network is terminated in the impedance Z0 at three of its ports, and driven with a
voltage generator of 2Vo and internal impedance Z0 at port 1. => coupler can be designed with
arbitrary coupling such that the input (port 1)is matched, while port 4 is isolated. Port 2 is the
through port, and port 3 is the coupled port . => a ground conductor is understood to be common to
both strip conductors.
A single-section coupled line coupler. (a) Geometry
and port designations. (b) The schematic circuit.
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The input impedance at port 1 of the coupler can thus be expressed as
=> Zin = V1/I1 = (V1e+V1o)/(I1e+I1o)
=> Zin = Z0
 Port 1 is matched.
Now V1=V0 => the voltage at port 3 is
V3 = V3e + V3o = V1e –V1o = V0[(Zine)/(Zine+Z0) –(Zino)/(Zino+Z0)]
Z ine
Z 0 jZ 0e tan θ
where e

Z in Z 0 2Z 0 j 
Z 0e Z 0o 
tan θ
Z ino
Z 0 jZ 0o tan θ

Z ino Z 0 2Z 0 j 
Z 0e Z 0o 
tan θ
Decomposition of the coupled line coupler circuit into evenand odd-mode excitation. (a) Even mode. (b) Odd mode.
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If the characteristic impedance, Z0, and the voltage coupling coefficient, C, are specific,
then the following design equations for the required even- and odd-mode characteristic
impedances can be easily derived.
=>
Z 0 e Z 0
1 C
1 C
Z 0 o Z 0
1 C
1 C
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Lange coupler
Generally the coupling in a coupled line coupler is too loose to achieve coupling factors of 3 dB or 6
dB. Increasing the coupling between edge-coupled lines is to use several lines parallel to each other,
so that the fringing fields at both edges of a line contribute to the coupling.
=> Lange coupler, four coupled lines are used with interconnections to provide tight coupling, and
can easily achieve 3 dB coupling ratios, with an octave or more bandwidth. => There is a 90o phase
difference between the output lines (ports 2 and 3) => Lange coupler is a type of quadrature hybrid.
The main disadvantage of the Lange coupler is difficult to fabricate the
necessary bonding wires across the lines, and the lines are very narrow
This type of coupled line is also referred to as interdigitated; can be used for
filter design.
The Lange coupler. (a) Layout in microstrip form.
(b) The unfolded Lange coupler.
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The unfolded Lange coupler operates essentially the same as the original Lange coupler, but is easier
to model with the equivalent circuit. => consist of a four-wire coupled line structure => all the lines
have the same width and spacing.
If the reasonable assumption that each line couples only to its nearest neighbor, and ignore more
distant couplings => effectively have a two-wire coupled line circuit.
 If we can drive the even- and odd- mode characteristic impedances, Ze4, Zo4, of the four-wire circuit
in terms of Z0e, and Z0o, the even- and odd-mode characteristic impedances of any adjacent pair of
lines.
The capacitance of the four lines to ground are different depending on whether the line is on the
outside (1 and 4), or on the inside (2 and 3) => Cin = Cex –(CexCm) / (Cex+Cm)
Effective capacitance networks for the unfolded Lange coupler
Equivalent circuits for the unfolded Lange coupler. (a) Four-wire coupled equivalent circuits. (a) Effective capacitance for the four-wire model.
(b) Effective capacitance for the two-wire model.
line model. (b) Approximate two-wire coupled line model.
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For even mode excitation, all four conductors are at the same potential, so Cm has no effect and the
total capacitance of any line to ground is Ce4 = Cex + Cin.
For odd-mode excitation, electric walls effectively exist through the middle of each Cm, so the
capacitance of any line to ground is Co4 = Cex + Cin + 6Cm.
 even- and odd-mode characteristic impedances are Ze4 = 1 / (vpCe4) ; Zo4 = 1 / (vpCo4)
Consider any isolated pair of adjacent conductors in the four-line mode; the effectively capacitances
are Ce = Cex ; Co = Cex + 2Cm
 The even-odd mode capacitances of the four-wire line in terms of two-wire coupled line:
C
3Ce Co 
C 
3Co Ce 
Ce 4  e
; Co 4  o
Ce Co
Ce Co
Z Z 0e
Z Z 0e
Since Z 0 1 v p C  Z e 4  0 o
Z 0e ; Z o 4  0o
Z 0o
3Z 0 o Z 0e
3Z 0 e Z 0 o
where Z 0 o , Z 0 e are the odd - and even - mode characteristic impedances of the two - conductor pair.
Z 0e Z 0o 
Z 0 o Z 0 e 



3Z 0 o Z 0 e 
3Z 0 e Z 0o 
2
Z0  Z e 4 Z o 4


Z e 4 Z o 4
3 Z 02e Z 02o
while the voltage coupling coefficient is C 

Z e 4 Z o 4 3 Z 02e Z 02o 2 Z 0 e Z 0 o

4C 3  9 8C 2
Z 0 e 
Z0
2C 
1 C 
/ 1 C 
;
Z 0o

4C 3  9 8C 2

Z0
2C 
1 C 
/ 1 C 
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The 180o hybrid
The 180o hybrid junction is a four-port network with a 180o phase shift between the two output ports.
A signal applied to port 1 will be evenly split into two in-phase components at port 2 and 3, and port
4 will be isolated. => If the input is applied to port 4, it will be equally split into two components with
a 180o phase difference at ports 2 and 3, and port 1 will be isolated.
When operated as a combiner, with input signals applied at ports 2 and 3, the sum of the inputs will
be formed at port 1, while the difference will be formed at port 4. => ports 1 and 4 are referred to as
the sum and difference ports, respectively.
The scattering matrix for the ideal 3 dB 180o hybrid
=>
unitary and symmetry
Hybrid junctions. (a) A ring hybrid, or rate-race,
Symbol for a 180° hybrid junction.
in microstrip or stripline form. (b) A tapered coupled
line hybrid. (c) A waveguide hybrid junction, or magic-T.
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Even-odd Mode Analysis
Consider a unit amplitude wave incident at port 1 (the sum port). => the ring junction this wave will
divide into two components, which both arrive in phase at ports 2 and 3, and 180o out of phase at port 4.
Using the even-odd mode analysis technique, decompose this case into a superposition of the two
simpler circuits and excitations.
 Amplitudes of the scattered waves :
B1 = e / 2 +o / 2 ; B2 = Te / 2 + To / 2 ; B3 = e / 2 - o / 2 ; B4 = Te / 2 –To / 2
Even- and odd-mode decomposition of the ring hybrid when port 1
is excited with a unit amplitude incident wave. (a) Even mode. (b) Odd mode.
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Using the ABCD matrix for even- and odd-mode two-port circuits to evaluate the required reflection
and transmission coefficients.
 B1 = 0 ; B2 = -j / (2)1/2 ; B3 = -j / (2)1/2 ; B4 = 0
=> the input port is matched, port 4 is isolated, and the input power is evenly divided and in
phase between ports 2 and 3.
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Consider a unit amplitude wave incident at port 4 (the difference port). => the two wave components on
the ring will arrive in phase in phase at ports 2 and 3, with a net phase difference of 180o between these
ports.
Amplitudes of the scattered waves :
B1 = Te / 2 –To / 2; B2 = e / 2 - o / 2; B3 = Te / 2 + To / 2; B4 = e / 2 +o / 2
Even- and odd-mode decomposition of the ring hybrid when port 4 is
excited with a unit amplitude incident wave. (a) Even mode. (b) Odd mode.
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Using the ABCD matrix for even- and odd-mode two-port circuits to evaluate the required reflection
and transmission coefficients.
A B  1 j 2 
A B   1


 ; 



C D
C D
1 


j 2
j 2
e
o
j
j
j
j
e  , Te  , o  , To 
2
2
2
2
j
j
B1 0, B2  , B3  , B4 0
2
2
j 2

1 
 Input port is matched, port 1 is isolated, and the input power is evenly divided into
ports 2 and 3 with a 180o phase difference.
The bandwidth of the ring hybrid is limited by the frequency dependence of the ring
lengths, bur is generally on the order of 20~30%
=> increasing the BW  additional sections or a symmetric ring circuit.
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