directional coupler...a directional coupler is a 4-port network exhibiting: •all ports matched on...
TRANSCRIPT
Directional Coupler
4-port
Network
1 2
3 4
A directional coupler is a 4-port network exhibiting:
• All ports matched on the reference load (i.e. S11=S22=S33=S44=0)
• Two pair of ports uncoupled (i.e. the corresponding Si,j parameters are zero).
Typically the uncoupled ports are (1,3) and (2,4) or (1,4) and (2,3)
Characteristic Parameters
It is here assumed that the uncoupled ports are (1-4) and (2-3). The S
parameters the ideal coupler must exhibit are:
S11=S22=S33=S44=0, S14=S41=S23=S32=0.
In this case the ports (1-2), (1-3), (3-4), (2-4) are coupled.
Let define C (Coupling) as:
C=|S13|2, CdB=-20 log(|S13|)
If the network is assumed lossless and reciprocal, the unitary condition
of the S matrix determines the following relationships:2 2 2 2 2
11 12 13 14 12 12
2 2 2 2 2 2 2 2
21 22 23 24 12 24 24 13
2 2 2 2 2
31 32 33 34 34 34 12
1 1 1
1 1
1 1 1
S S S S S C S C
S S S S S S S S C
S S S S C S S S C
+ + + = + = = −
+ + + = + = = =
+ + + = + = = = −
Then, exciting the network at port 1 (2), the output power is divided between ports
3 and 2 (4 and 1) according the factors C and 1-C. No power comes out of port 4
(3)
Further implications of lossless condition
(symmetric structure)
( ) ( )13 3412 24* *
12 24 13 34 12 24 13 34
12 24 13 34
0
jjS S S S S S e S S e
+ − −+ = =
− = + −
12=24±/2
S12=S34, S13=S24We assume that the network is symmetric:
It can be demonstrated that it is sufficient to impose the matching
condition to the four ports of a lossless and reciprocal network to get a
directional coupler
Then 12=34 and 13=24. From the previous condition:
This means that the outputs are in quadrature each other.
Parameters of a real directional coupler
In a real coupler the matching at the ports is not zeros at all the frequencies. It
is then specified the minimum Return Loss in the operating bandwidth.
The coupling parameter C, it is in general referred to the port with the smallest
coupling.
Moreover, in a real device a not zero power arrives to the uncoupled port. To
characterize this unwanted effect the isolation parameter (I) is introduced:
I=Power to the coupled port/Power to the uncoupled port
For the coupler considered in the previous slides (assuming port 3 that with
the lowest coupling):
2 2
13 14I S S=
Note that for the ideal coupler I is infinite
Use of the directional coupler (1)
Measure of the reflection coefficient
CViVr
GLLine, Zc
13 24,
i L r L L
r L LL
i LL
V V S j C V V V S j C V
V j C V V
V Vj C V
+ + − −
− −
++
= =
= = = G
121, 1 1C S C = −
V+
V-
LV +
LV −
LV V+ +
1 2
3 4
Use of the directional coupler (2)
Power divider (C=3 dB)
C=3dB
Pin 1 2
3 4
Pout= Pin /2
Pout= Pin /2
Ports closed on the
reference load
2 2 1 12 1 2 1
3 3 1 13 1 3 1
1 1,
2 2
1 1
2 2
V V V S V P P
V V V S V P P
− +
− +
= = = =
= = = =
Use of the directional coupler (3)
Power combiner (C=3 dB)
1 1 2 12 3 13 4
22
1
1 2 3
0 0
, 02 2 2 2 2
1 1
2 4
g g g
g
V V VV V V S V S j j j V
VVP P P
R R
− + +
= = + = + = =
= = = +
1 2
3 4Vg
V1 jVg
V+2=jVg/2
V+3=Vg/2
2
2 3
0
12 34
24 13
1
8
1
2
2
gVP P
R
S S
jS S
= =
= =
= =
Only if P2 e P3 are equal
C=3dB
Vout11 2
3 4
VA
VB
Use of the directional coupler (4)
Sum and difference of voltages (C=3 dB)
( )1 1 12 13
1,
2out A B A BV V V S V S V V= = + = +
12=13= 24=0
→ 34=
0°
0° 0°
180°Vout2
( )2 4 24 34
1
2out A B A BV V V S V S V V= = + = −
Ports closed on the
reference load
Use of the directional coupler (5)
Balanced Amplifier
A
A
0°
90°
C=3 dB
90°
0°
12
a
AV +
12
a
AjV +
1a
2a
3a
4b
2b
3bC=3 dB
Gain:
2 1 3 1
2 34 1
2 , 2
2 2
b a b a
b b outb a
in
V A V V j A V
V V PV j j A V A
P
+ + + +
− +
= =
= + = =
Reflection:
Gin
Gin
( )
2 1 2 1 3 1
3 1 1 3 2 1 1
1
1
2 , 2 , 2 ,
12 , 2 2
2
0
a a a in a a a
a in a a a a in a in a
ain
a
V V V V V jV
V j V V jV V V V
V
V
− + + + − +
+ + − + + + +
−
+
= = G =
= G = + = −G +G
G = =
2 1 2a aV V− +=1aV +
2aV +
13
2
aa
jVV
+− =
3aV +
4bV −
Coupled TEM 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.
There are two possible modes of propagation which are called even
and odd (symmetrical structure). Each mode is characterized by its
characteristic impedance whose meaning is illustrated in the next slide
Modes in TEM coupled lines
Even mode (Zce)
Magnetic wall (open)
Electric wall (short)
Odd mode (Zco)
open
Example:
Coupled coax Even mode
Line (Zce)
Odd mode
Line (Zco)
short
Symmetry
axis
Zce, Zco
Symmetry
axis
Circuit model of two coupled lines with finite
length
Zce, Zco
bL
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
Eigenvalues and Eigenvectors of a matrix
The eigenvalues Sl of a square matrix S are the solutions of the equation:
det 0Sl− =S U
The eigenvectors xl associated to S represent the solution of the
homogenous system of equations:
x S xl l l =S
A matrix of order n has n eigenvalues and n eigenvectors (each eigenvectors
has n elements). The eigenvectors are defined up to a constant.
Properties
If a N-port network is excited with a vectors of currents representing an
eigenvector of Z, you see the same impedance at all ports, and its value is
just the eigenvalue. The same holds for all the other matrices (Y, S, …)
If there are symmetry axis in the network, the eigenvalues can be derived by
on suitably defined circuits (eigencircuits). The eigenvectors are obtained by
induction (the excitations must determine either an open or a short along the
symmetry axis).
Once the eigenvalues and eigenvectors are known, simple equations define
the elements of the corresponding matrices.
Example: 2-port network
2-port
Network 11 12
12 22
S S
S S
Let assume that the two eigenvectors are known (we have assigned arbitrarily
the largest element of each xi equal to 1
1
1
1x
+=Eigenvector 1: 2
2
1x
+=Eigenvector 2:
1 1
G1G1
2-port
Network
b1b2
1 2
G2G2
2-port
Network
b1b2
1 21
1
(Eigenvalue 1)1
b b
= = G
1 22
2
(Eigenvalue 2)1
b b
= = G
1 11 12 1
2 11 22 1
1 21
1
1
1
1
b s s
b s s
b b
= +
= +
= = G
2 1 1 2 1 1 2 2 1 211 22 12 21
2 1 1 2 1 2
, , 1 1
s s s s
G − G G − G G −G= = = =
− − −
Relationship between eigenvalues of S, Y, Z
0 0
0 0
Z Z Y YS
Z Z Y Y
l ll
l l
− −= =
+ +0
1 1
1
SZ Z
S Y
ll
l l
+= =
−
2-port network (cont.)
1 11 12 2
2 11 22 2
1 22
2
1
1
1
b s s
b s s
b b
= +
= +
= = G
Eigenvector 1: Eigenvector 2:
Evaluation of the Eigenvectors
In general there is no way to derive the eigenvectors of a generic N-port
network. Only is the network is symmetric we can deduct the eigenvectors by
exploiting the effect produced by the eigenvectors excitation.
Let’s consider for instance the case of a symmetric 2-port (S11=S22). Being the
network symmetric we can obtain the same reflection coefficient at port 1 and
port 2 only when the incident waves at the two ports have the same magnitude.
As a consequence the only possible values for 1 and 2 are 1 and -1:
Symmetric
2-port
Symmetric
2-port
1 1
1
1ex
+=+
Eigenvector 1 (Even):
Symmetric
2-port
Symmetric
2-port
1 -1
1
1ex
+=−
Eigenvector 2 (Odd ):
GeGe Go
Go
11 22 12 21 11 12 11 12 , , , 2 2
e o e oe os s s s s s s s
G +G G −G= = = = G = + G = −
If the first eigenvector is considered , two equal excitations (amplitude and
phase) at the two ports determines, for the network symmetry, an open circuit
along the symmetry axis. The first eigenvalue is the imput reflection coefficient
(or impedance or admittance) of the even eigenetwork (one port):
Eigenetwork 1
(even)
Short circuit
Ge
The second eigenvector consists in two equal but opposite excitations: a short
circuit is then determined along the symmetry axis. The second eigenvalue can
be computed from the odd eigenetwork (one port):
Eigenetwork 2
(odd)
Open circuit
Go
2-port symmetric network (cont.)
Eigenvalues evaluation of 2-coupled lines
Symmetry axis 1
Symmetry axis 2
Exciting the network with an eigenvector, an electric or a
magnetic wall is obtained along the two symmetry axis.
With reference to Z matrix, the exciting currents for each
eigenvector 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
l
l
l
l
= + + + +
= + − + −
= + + − −
= + − − +
Axis 1: Magnetic, Axis 2: Magnetic
Axis 1: Magnetic, Axis 2: Electric
Axis 1:Electric, Axis 2: Magnetic
Axis 1: Electric, Axis 2: Electric
Evaluation of the eigenvalues using the eigenetwork
Eigenvalue Zl1:
Zl1
/2, Zce OPEN ( )1 cot 2ceZ jZl = −
Eigenvalue Zl2:
Zl2
/2, Zce SHORT ( )2 tan 2ceZ jZl =
Eigenvalue Zl3:
Zl3
/2, ZcoOPEN ( )3 cot 2coZ jZl = −
Eigenvalue Zl4:
Zl4
/2, Zco SHORT ( )4 tan 2coZ jZl =
OPEN
OPEN
SHORT
SHORT
Evaluation 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
14 11 12 13 14
0
VZ Z Z Z Z
I
VZ Z Z Z Z
I
VZ Z Z Z Z
I
VZ Z Z Z Z
I
l
l
l
l
= = + + +
= = − + −
= = + − −
= = − − +
( )
( )
( )
( )
11 1 2 3 4
12 1 2 3 4
13 1 2 3 4
14 1 2 3 4
1
4
1
4
1
4
1
4
Z Z Z Z Z
Z Z Z Z Z
Z Z Z Z Z
Z Z Z Z Z
l l l l
l l l l
l l l l
l l l l
= + + +
= − + −
= + − −
= − − +
Using Zli, 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
Expression 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
ce ce co co
ce ce co co
ce ce co co
jZ Z Z Z Z
jZ Z Z Z Z
jZ Z Z Z Z
= − + − +
= − − − −
= − + + −
14
an2
cot tan cot tan4 2 2 2 2
ce ce co co
jZ Z Z Z Z
= − − + +
( )( )
( )
( )
( )( )
( )
( )
11
12
13
14
cot2
1
2 sin
cot2
1
2 sin
ce co
ce co
ce co
ce co
Z ZZ j
Z ZZ j
Z ZZ j
Z ZZ j
+= −
+= −
−= −
−= −
Compact expressions of Z and Y elements
( )( )
( )
( )
( )( )
( )
( )
11
12
13
14
cot2
1
2 sin
cot2
1
2 sin
ce co
ce co
ce co
ce co
Y YY j
Y YY j
Y YY j
Y YY j
+= −
+=
−= −
−=
Special cases
=bL=180°
The eigenvalues of Z are [0, ∞, 0, ∞], so those of S result:
Sli= [-1, 1, -1, 1]. The scattering matrix elements are then:
11 12 13 140, 1, 0, 0 S S S S= = − = =
Note that line 2 is completely uncoupled from line 1!
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) independently on bL:
0 = Z ce coZ Z
Note that the matching at the ports does not imply the absence of
reflected waves along the two lines. It means that the reflected waves
are canceled only at the ports
( ) ( ) ( ) ( ) cot 2 , tan 2 , cot 2 , tan 2ce ce co coZ j Z Z Z Zl = − −
Derivation of the matching condition
Eigenvalues of S
0
0
| i
i
i
jX ZS
jX Z
ll
l
−=
+Parameter S11:
( )11 1 2 3 4
10
4S S S S Sl l l l= + + + =
There are only two cases where the above condition can be satisfied
independently on =bL, i.e.:
( )
( )
1 2
3 4
0
0
S S
S S
l l
l l
+ =
+ =
2
1 2 0
2
3 4 0
X X Z
X X Z
l l
l l
= −
= −
2 2
0
2 2
0
ce
co
Z Z
Z Z
=
=
NOT
Admissible
( )
( )
1 4
2 3
0
0
S S
S S
l l
l l
+ =
+ =
2
1 4 0
2
3 2 0
X X Z
X X Z
l l
l l
= −
= −
0
0
ce co
ce co
Z Z Z
Z Z Z
=
=
Admissible
( ) ( ) ( ) ( ) cot 2 , tan 2 , cot 2 , tan 2ce ce co co iZ j Z Z Z Z jXl l= − − =
Coupled TEM lines as directional coupler
0 = Z ce coZ Z
Zce, Zco
bL
1 2
3 4
Using the admissible condition we obtain the match at all port imposing:
This condition also implies that port 4 is uncoupled :
( )
( )
1 4
2 3
0
0
S S
S S
l l
l l
+ =
+ =
( )14 1 2 3 4
10
4S S S S Sl l l l= − − + =
The maximum value |S13|2 defines the coupling: C=(|S13|
2)max
It is obtained for bL=/2 :
( ) ( )22 2
2
13 1 2 3 4 1 2maxmax max
1 1
4 2
ce co
ce co
Z ZS S S S S S S
Z Zl l l l l l
−= + − − = + =
+
Variation of frequency
Matching and Isolation are frequency independent and equal to zero and
infinity respectively (ideal lossless TEM line).
The coupling varies with the frequency according to the following expression:
( )( )
maxmax2
max
, 1 (1 ) cot
ce co
ce co
C Z ZC C
C Z Z
− = =
+ − +
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
f/f0
C/C
max
Cmax=0.01-0.1
B0.5 dBFor Cmax<0.1 the variation of C
is practically independent on
Cmax.
Note that the band for
C/Cmax< 0.5 dB is about 0.44 f0
f0 is the frequency for which bL=/2
Practical Restrictions
They concern mainly the maximum value of Cmax. In fact, increasing Cmax, the
lines become closer and closer until the practical implementation is no more
possible with sufficient accuracy. Typically the maximum value of Cmax must be
lower than 0.1 (CdB=10)
Example: Design a stripline coupler with C=0.1 with Z0=50 W using the
following figures reporting the values of
as a function of S and W (lines separation and width). Frequency: 1 GHz
max 0, ce coce co
ce co
Z ZC Z Z Z
Z Z
−= =
+
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
S (mm)
Cmax
0
0.025
0.05
0.075
0.1
0.125
0.15
0.175
0.2
Re(Eqn(Cmax))Output Equations
W=9.5
W=12
W=11.18
S=0.195
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
S (mm)
Z0
45
50
55
60
65
Re(Eqn(Z0))
Output Equations W=9.5
10
10.5
11
11.5
12
W=11.18
S=0.195
Solution:
We draw the line C=0.1 on the first graph and the line Z0=50 on the second graph.
A point on each of these lines has to be then found, which is characterized by the
same pair of values (S, W) .
10 mm
11.18 mm 11.18 mm
0.19565 mmer=1
Note: If a coupler dimensioned for the requested C but with a different value
of Z0 is available, impedance transforming networks can be used in place of
redesigning a new coupler.
0 2
75 mm
LL
c
L
b = =
=
C=0.1
Z’0
Z0
Z0
Z0
Z0
Z’0
Z’0
Z’0
Z’0
Couplers with quasi-TEM LinesIf quasi-TEM coupled lines are considered (Microstrip), the phase velocity of the
even and odd modes are not exactly coincident. Strictly speaking, that would
not allow to apply the model here assumed for the characterization of the
directional coupler.
In the practice, until the difference between the two velocities is not too large,
the same phase velocity can be assumed for both modes (equal to the average
of the actual values), assuming the lines as TEM. There are however some
differences comparing the performances with the ideal TEM coupler (a perfect
matching is no more possible)
800 850 900 950 1000 1050 1100 1150 1200
Frequency (MHz)
Microstrip Coupler
-40
-35
-30
-25
-20
-15
-10
-5
0
1000.3 MHz-35.83 dB
999.18 MHz-26.38 dB
1000.1 MHz-10.03 dB
S21
S31
S41
S11
1.575 mm
2 mm 2 mm
0.4037 mmer=2.33
0
, ,
0, ,
0.1 0.4037 mm
74.12
1.97, 1.71
1.84, 2
55.24 mm
cp cd
eff p eff d
eff medio eff medio
m S
Z Z Z
LL
c
L
e e
e b e
= =
= =
= =
= = =
=
Coupler with lumped couplingsTo realize couplers with a large coupling (C<10 dB) structure with lumped
couplings are used. In planar technology two kinds of such couplers are
employed, the branch-line and the rat-race.
Branch-line
Y’c,
Y’c,
Y’’ c
,
Y’’ c
,
1 2
3 4
0 L=2 4
l =
Evaluation of the eigenvalues
Y’c, °
Y’’ c
, °
1 Y’c, °
Y’’ c
, °
1
Y’c, °
Y’’ c
, °
1 Y’c, °
Y’’ c
, °
1
Y’c, °
Y’’ c
, °
1 Y’c, °
Y’’ c
, °
1
Y’c, °
Y’’ c
, °
1 Y’c, °
Y’’ c
, °
1
( )( ) ( )
( ) ( )
1
1 0 0
tan 45 c c c c
s s
Y j Y Y j Y Y
S Y jB Y jB
l
l
= + = +
= − +
( ) ( )
( )
( ) ( )
2
2 0 0
tan 45 tan 45c c
c c
d d
Y jY jY
j Y Y
S Y jB Y jB
l
l
= − +
= − −
= + −
( ) ( ) ( )
( ) ( )
3
3 0 0
tan 45 cot 45c c c c
d d
Y jY jY j Y Y
S Y jB Y jB
l
l
= − = −
= − +
( ) ( )
( )
( ) ( )
4
4 0 0
tan 45 cot 45c c
c c
s s
Y jY jY
j Y Y
S Y jB Y jB
l
l
= − −
= − +
= + −
s c cB Y Y = + d c cB Y Y = −
Coupling conditions
All ports matched, ports 1-3/2-4 uncoupled:
1 2 3 40, 0S S S Sl l l l+ = + =
S11=0 and S13=S24=0
This condition is actually feasible, giving in fact:
2
0
1 s dB B
Y=
2 2 2
0c cY Y Y − =
( )
( )
11 1 2 3 4
13 1 2 3 4
10
4
10
4
S S S S S
S S S S S
l l l l
l l l l
= + + + =
= + − − =
Then, being f12=-/2 → f14=. The parameter bs must be then >1 for having
S14 negative. By imposing now the requested coupling (|S14|2=C), we can
obtain the requested value of bs:
( ) ( )
( ) ( )
12 1 2 3 4 1 4 2
2
14 1 2 3 4 1 4 2
0
21 1
4 2 1
11 1,
4 2 1
s
s
s ss
s
jbS S S S S S S
b
b BS S S S S S S b
b Y
l l l l l l
l l l l l l
−= − + − = − =
+
−= − − + = + = =
+
For S12 e S14 we have:
Moreover, the unitary condition of S implies the following relation between
the phases of S12 and S14:
12 142
f f− =
2
14 2
0
1 1,
1 1
s c cs
s
b Y YCS C b
b YC
− ++− = = = =
+ −
Taking into account the first found condition ( 2 2 2
0 c cY Y Y − = ) we
finally obtain the expressions of Y’c e Y’’c:
0 0
1,
11c c
CY Y Y Y
CC = =
−−
Branch-line coupler with C=0.5 (3 dB)
The couplers with C=3 dB are identified with the word Hybrid.
To realize an hybrid of branch-line type the characteristic impedances of the
lines must be:
( )0 0 0
0.51 0.5 35.35 , 50 50
0.5c cZ Z Z Z Z = − = W = = W = W
Practical restrictions
One can easily verify that, with C tending to 0: Z’c→Z0 and Z’’c→∞. In the
practice, even with con C=0.1 (10 dB) the corresponding value of Z’’c is
very difficult to realize (Z’’c =3Z0). Usually, C must be between 3 and 6 dB.
Frequency dependence
For this device, both matching and isolation vary with frequency (the
nominal value is obtained at the frequency where the length of the lines is
l0/4). Also the coupling depends on f (the max is again at f0).
The bandwidth for a given value of maximum coupling increases with Cmax;
Usually, the frequency variation of matching and isolation is more
pronounced than that of the coupling.
Branch Line: a summary
11 33 22 44 13 24
2 2
14 23
2 2
12 34
0, 0
1
S S S S S S
S S C
S S C
= = = = = =
= =
= = −
Design equations:
S parameters obtained:
14 23 12 34, 1S S C S S j C= = − = = − −
For C=0.5 (3dB), it has (assuming Z0=1/Y0=50W):
Conditions to be imposed:
0 0
1,
11c c
CY Y Y Y
CC = =
−−
( )0 0 0
0.51 0.5 35.35 , 50 50
0.5c cZ Z Z Z Z = − = W = = W = W
=0180°
-90°cY
cY
cY cY
1 2
3 4
800 850 900 950 1000 1050 1100 1150 1200
Frequency (MHz)
Branch-line C=3dB
-50
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
DB(|S(1,1)|)Schematic 1
DB(|S(2,1)|)Schematic 1
DB(|S(3,1)|)Schematic 1
DB(|S(4,1)|)Schematic 1
Adattamento
Isolamento
Accoppiamento
Rat-race coupler
cY cY
1 3
2 4
cY
cY
0 4l
0 4l0 4l
03 4l
There is only one symmetry axis (vertical)
It is anyhow possible to still use the eigenvalues method for finding the
dimensioning equations
P1
P2
P3
P4
cY
cY
cY cY
Conditions to be imposed:
11 33 22 44 23 14
2 2
13 24
2 2
12 21
0, 0
1
S S S S S S
S S C
S S C
= = = = = =
= =
= = −
Design equations:
0 01 , c cY Y C Y Y C = − =
S parameters obtained:
13 24 12 34, , 1S j C S j C S S j C= − = = = − −
1
2 4
3
=0
cY cY cY
cY
-90°
-90°
1
2 4
3
=0
cY cY cY
cY
-90°
90°
For C=0.5 (3dB), we obtain (assuming Z0=1/Y0=50W):
0 0 0 070.707 , 70.7071 0.5 0.5
c c
Z Z Z ZZ Z
C C = = = W = = = W
−
Parameters dependence on f
0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2
Frequency (GHz)
Rat-race C=3dB
-50
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
DB(|S(1,2)|)
RatRace
DB(|S(1,4)|)
RatRace
DB(|S(1,3)|)
RatRace
DB(|S(1,1)|)
RatRace
Accoppiamento
Adattamento
Isolamento
• The bandwidth is larger than that of the branch-line
• With the decreasing of the coupling the bandwidth increases
• The practical feasibility limits the coupling between about 3 and 8 dB
Couplers with lumped elementsIn some cases, the practical implementation of planar couplers with C in the
range 10-15 dB may be difficult using distributed elements. A possible
alternative is represented by a structure similar to the branch line but employing
lumped components (for the most critical elements).
Le consider the following (symmetrical) 4-port configuration, constituted by
lumped susceptances (also the eigen-networks are shown in the figure):
Ba1 2
3 4BaB
b
Bb
Br
Br
Br
Br
Br
Br
2Ba
Br
2Bb
Br
2Ba
2Bb
1Sl
2Sl
3Sl
4Sl
Eigenvalues expressions (bi=Bi/Yo):
( )
( )
( )
( )
( )
( )1 2 3 4
1 2 1 2 1 2 21, , , ,
1 1 2 1 2 1 2 2
r a r b r a br
r r a r b r a b
j b b j b b j b b bjbS S S S
jb j b b j b b j b b bl l l l
− + − + − + +−= = = =
+ + + + + + + +
Now impose S11=0 and S13=0 (uncoupled ports 1-3):
1 2 3 40, 0S S S Sl l l l+ = + =( )
( )
11 1 2 3 4
13 1 2 3 4
10
4
10
4
S S S S S
S S S S S
l l l l
l l l l
= + + + =
= + − − =
( )r a bb b b= − +
Imposing the coupling C=|S14|2:
( ) ( )( )12 1 2 3 4 1 3 2 2
41 1
4 2 1 2
b
a b a
j bS S S S S S S
b b j bl l l l l l= − − + = − =
− + −
If we assume S12 real, ba2-bb
2=1, and: 13b
a
bS C
b= =
Finally:
( )1
, , 1 1
a b r a b
Cb b b b b
C C= = = − +
− −
Example of implementation
We observe that Ba and Bb are always positive, so can be implemented with
capacitances. Br is instead negative and can be realized with a short circuited
stub with characteristic impedance Z0.
Assume CdB=10 dB (C=0.1), f0=945 MHz and Y0=1/50. From the previous
formulas we get: ba=1.054, bb=0.3333, br=-1.387.
The capacitances implementing Ba and Bb are given by: Ca=3.55 pF, Cb=1.12
pF. The susceptance Br is realized with a short circuited stub with Zc=50 W and
bL=atan(-1/br)=35.78°. The final circuit is shown below together with the
expressions of S12 and S14
( ) ( )
( ) ( )
( ) ( )
2 2
12 0 122 2
0
14 0 142 2
1 1, for =
1 2
2, for =
1 2
a b
aa b a
b b
aa b a
C CS S j
CC C j C
j C CS S
CC C j C
+ −= → =
− + −
= → = −− + −
1 2
3 4
Ca
Ca
CbCb
=090°
180°
Frequency response of the lumped coupler
850 875 900 925 950 975 1000 1025 1050
Frequency (MHz)
Lumped Coupler
-50
-40
-30
-20
-10
0
DB(|S(1,1)|)Test
DB(|S(1,2)|)Test
DB(|S(1,3)|)Test
DB(|S(1,4)|)Test
S14
S12
S11
S13
3-port lossless networks
A 3-port reciprocal lossless network cannot be matched at
the 3 ports (i.e. S11=S22=S33=0 not possible).
In fact, imposing the unitary of S:
2 2
12 13
2 2 2 2 2
12 23 12 13 23
2 2
13 23
1
1 0.5
1
S S
S S S S S
S S
+ =
+ = = = =
+ =
*
13 23
*
12 23 12 23 23
*
12 13
0
0 =0 or 0 or 0
0
S S
S S S S S
S S
=
= = =
=
These conditions are
incompatible so it is
impossible to have
S11=S22=S33=0
Is it still possible to realize a power splitter
with a 3-port?
Possible solution: a 3 dB hybrid with the uncoupled
port closed on a matched load
C=3dB
Pin
Z0
2
4 3Pout= Pin /2
Pout= Pin /2
Circuito a 3 porte
Porta
disaccoppiata
The 3-port network is lossy
due to the presence of Z0.
This resistor however does
not dissipate power
because the port 4 is
uncoupled. Pin is then split
between ports 2 and 3
without losses
A 3-port divider: the Wilkinson network
cZ
1
2
3
RW
0 4l
0 4l
cZ
0Z
0Z
0Z
Due to the presence of RW the 3-
port network is lossy, so the
condition S11=S22=S33=0 can be
imposed
2-port network obtained by closing port 1 with Z0:
2 3
0 4l0 4l
cZ
0Z
cZ
RW
S22=S33 and S23 can be computed thorugh the eigenvalues of this network
Ze
0 4l02Z
cZZo
0 4l
2
WR
cZ
2 3
0 4l0 4l
cZ
02Z
cZ
02Z
2
WR
2
WR
2 3
0 4l0 4l
cZcZ
02Z
cZcZ
02Z
2
WR
2
WR
Even Network
2 2 2
0
2 2
0 0
2,
2 2
c ce e
c
Z Z ZZ
Z Z Z
−= G =
+0
0
2,
2 2
W Wo o
W
R R ZZ
R Z
−= G =
−
22 230, 0 0, 0 2 2
e o e oe oS S
G +G G −G= = = = G = G =
For Z0=50W→ Zc=70.7W, RW=100W
Odd Network
0 0
2 , 2c WZ Z R Z
= =
Evaluation of S11
Exciting port 1 with ports 2 and 3 closed with Z0, there is no current through
RW, so it can be discarded.
cZ
0 4l
0 4l
cZ
inZ
0Z
0Z
2 2 2
011 2 2
0 0
21,
2 2
c cin in
c
Z Z ZZ S
Z Z Z
−= G = =
+
Being 02cZ Z= also S11=0.
Evaluation of S21= S31
Power entering port 1 is not reflected and there is there is no dissipation in RW.
So the power is all transferred to ports 2 e 3; for the symmetry, the power
exiting at each port is the half of Pin:
2 2
21 31 0.5S S= =
The phase for both parameters is -90°.
Frequency response
0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2
Frequency (GHz)
Divisore di Wilkinson
-50
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
DB(|S(1,1)|)
Ideale
DB(|S(2,1)|)
Ideale
DB(|S(3,1)|)
Ideale
Adattamento
Trasmissione
• The bandwidth for RL=20 dB is about 40% of f0
• Transmission (|S21|=|S31|) is independent on frequency
• Dissipation in RW is zero provided that the load at ports 2 and 3 is the
same
Microstrip implementation
RW
The very small size of Rw (pseudo
lumped component) must be
accounted for. The two output lines
must be enough close to allow the
connection of Rw.
On the other hand is not advisable to have
the output lines too close each other
because an unwanted coupling may arise.
So diverging lines are often used.
Example of derivation of S parameters from
the eigenvalues of S matrix
jBZc Zc
f f
jBZc Zc
f fj2BZc
f
Zc
f
Even eigenetwork
Odd eigenetwork
jB/2
( ) ( )
( ) ( )
2exp 2 exp 2
2
exp 2 exp 2
ce B
c
o cc
Y j Bj j
Y j B
j j
f f
f f
−G = G − = −
+
G = G − = − −
( ) ( )
( ) ( )
11 22
12 21
21exp 2 1 exp 2
2 2 2 2
2 21exp 2 1 exp 2
2 2 2 2
e o c
c c
e o c c
c c
Y j B jBS S j j
Y j B Y jB
Y j B YS S j j
Y j B Y jB
f f
f f
G +G −= = = − − = − −
+ +
G −G −= = = − + = −
+ +
Ge
Go