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Transmission Lines
© Amanogawa, 2006 – Digital Maestro Series 192
Single stub impedance matching
Impedance matching can be achieved by inserting another transmission line (stub) as shown in the diagram below
Z A = Z 0
dstub
Z R Z 0
Lstub
Z 0S
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Transmission Lines
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There are two design parameters for single stub matching:
The location of the stub with reference to the load dstub
The length of the stub line Lstub
Any load impedance can be matched to the line by using singlestub technique. The drawback of this approach is that if the load ischanged, the location of insertion may have to be moved.
The transmission line realizing the stub is normally terminated by ashort or by an open circuit. In many cases it is also convenient toselect the same characteristic impedance used for the main line,although this is not necessary. The choice of open or shorted stubmay depend in practice on a number of factors. A short circuited
stub is less prone to leakage of electromagnetic radiation and issomewhat easier to realize. On the other hand, an open circuitedstub may be more practical for certain types of transmission lines,for example microstrips where one would have to drill the insulatingsubstrate to short circuit the two conductors of the line.
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Transmission Lines
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Since the circuit is based on insertion of a parallel stub, it is more
convenient to work with admittances, rather than impedances.
Y A = Y 0
dstub
Y R= 1/ Z RY 0= 1/ Z 0
Lstub
Y 0S
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Transmission Lines
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For proper impedance match:
( )stub stub 00
1d AY Y Y Y Z
= + = =
dstub
Y R= 1/ Z RY (dstub)
Lstub
Y 0S
Y stub
+
Input admittance
of the stub line
Line admittance at location
dstub before the stub is applied
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Transmission Lines
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In order to complete the design, we have to find an appropriate
location for the stub. Note that the input admittance of a stub isalways imaginary (inductance if negative, or capacitance if positive)
stub stubY jB
A stub should be placed at a location where the line admittance hasreal part equal to Y 0
( ) ( )stub 0 stubd dY Y jB+
For matching, we need to have
( )stub stubd B B−
Depending on the length of the transmission line, there may be anumber of possible locations where a stub can be inserted for impedance matching. It is very convenient to analyze the possiblesolutions on a Smith chart.
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Transmission Lines
© Amanogawa, 2006 – Digital Maestro Series 197
1
-1
0
0.2
-0.2
21
-0.5
0.5
-3
3
2
-2
z
y R
Loadlocation
First location
suitable for stub insertion
Second locationsuitable for stubinsertion
y(dstub1)
y(dstub2)
Unitaryconductance
Constant
|Γ(d)| circle
θ2
θ1
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Transmission Lines
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The red arrow on the example indicates the load admittance. This
provides on the “admittance chart” the physical reference for theload location on the transmission line. Notice that in this case theload admittance falls outside the unitary conductance circle. If onemoves from load to generator on the line, the corresponding chartlocation moves from the reference point, in clockwise motion,
according to an angle θ (indicated by the light green arc)
42 d dθ = β =
λ
The value of the admittance rides on the red circle whichcorresponds to constant magnitude of the line reflection coefficient,
|Γ(d)|=|Γ R |, imposed by the load.
Every circle of constant |Γ(d)| intersects the circle Re { y } = 1
(unitary normalized conductance), in correspondence of two points.Within the first revolution, the two intersections provide thelocations closest to the load for possible stub insertion.
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Transmission Lines
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The first solution corresponds to an admittance value with positive
imaginary part, in the upper portion of the chart
( ) ( )
( ) ( )
( )
1
1
1
stub 0 stub1
stub stub1 1
1stub
stub
d d
d 1 d
d4
d
Y Y j B
y j b
j B
= +
= +
θ= λ
π
−
Line Admittance - Actual :
Normalized:
Stub Location:
Stub Admittance - Actual :
Norma ( )
( )
( ))
1
1
1
stub
1stub
0 stub
1stub 0 stub
d
1tan ( )
2 d
tan d ( )2
s
s
j b
L
Z B
L Z B
−
−
−
λ
λ
=π
lized:
Stub Length : short
open
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Transmission Lines
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The second solution corresponds to an admittance value with
negative imaginary part, in the lower portion of the chart
( ) ( )
( ) ( )
( )
2
2
2
stub 0 stub2
stub stub2 2
2stub
stub
d d
d 1 d
d4
d
Y Y j B
y j b
j B
= −
= −
θ= λ
π
Line Admittance - Actual:
Normalized:
Stub Location:
Stub Admittance - Actual :
Norma ( )
( )
( ))
2
2
2
stub
1
stub0 stub
1stub 0 stub
d
1
tan ( )2 d
tan d ( )2
s
s
j b
L Z B
L Z B
−
−
λ
− λ
= −π
lized:
Stub Length: short
open
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Transmission Lines
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If the normalized load admittance falls inside the unitary
conductance circle (see next figure), the first possible stub locationcorresponds to a line admittance with negative imaginary part. Thesecond possible location has line admittance with positive imaginary part. In this case, the formulae given above for first andsecond solution exchange place.
If one moves further away from the load, other suitable locations for stub insertion are found by moving toward the generator, atdistances multiple of half a wavelength from the original solutions.
These locations correspond to the same points on the Smith chart.
1
2
stub
stub
d2
d2
n
n
λ= +
λ+
First set of locations
Second set of locations
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Transmission Lines
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1
-1
0
0.2
-0.2
21
-0.5
0.5
-3
3
2
-2
R
z R
Loadlocation
Second location
suitable for stubinsertion
First locationsuitable for stub
y(dstub2)
y(dstub1)
Unitaryconductancecircle
Constant
|Γ(d)| circle
θ2
θ1
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Transmission Lines
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Single stub matching problems can be solved on the Smith chart
graphically, using a compass and a ruler. This is a step-by-stepsummary of the procedure:
(a) Find the normalized load impedance and determine thecorresponding location on the chart.
(b) Draw the circle of constant magnitude of the reflection
coefficient |Γ| for the given load.
(c) Determine the normalized load admittance on the chart. This is
obtained by rotating 180° on the constant |Γ| circle, from theload impedance point. From now on, all values read on the chart
are normalized admittances.
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Transmission Lines
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1
-1
0
0.2
-0.2
21
-0.5
0.5
-3
3
-2
z
y R
(c) Find the normalized loadadmittance knowing that
y R = z(d=λ /4 )From now on the chartrepresents admittances.
(a) Obtain the normalized load
impedance z R=Z R / Z 0 and find
its location on the Smith chart
(b) Draw the
constant |Γ(d)|circle180° = λ /4
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Transmission Lines
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(d) Move from load admittance toward generator by riding on the
constant |Γ| circle, until the intersections with the unitarynormalized conductance circle are found. These intersectionscorrespond to possible locations for stub insertion. CommercialSmith charts provide graduations to determine the angles of rotation as well as the distances from the load in units of
wavelength.
(e) Read the line normalized admittance in correspondence of thestub insertion locations determined in (d). These values will
always be of the form
( )
( )
stub
stub
d 1 top half of chart
d 1 bottom half of chart
y jb
y jb
= +
= −
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Transmission Lines
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1
-1
0 0.2 0.5 5
0.2
-0.2
21
-0.5
-3
3
2
-2
z R
y R
Loadlocation
First location suitable for
stub insertion
dstub1=(θ1/4π)λ
θ1
(d) Move from load towardgenerator and stop at alocation where the realpart of the normalized lineadmittance is 1.
Unitaryconductancecircle
(e) Read here thevalue of thenormalized line
admittancey(dstub1) = 1+jb
First Solution
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1
-1
0
0.2
-0.2
21
-0.5
-3
3
-2
z R
y R
Loadlocation
Second location suitablefor stub insertion
dstub2=(θ2/4π)λ
(e) Read here thevalue of thenormalized line
admittancey(dstub2) = 1 - jb
Unitaryconductance2
(d) Move from loadtoward generator andstop at a locationwhere the real part of the normalized lineadmittance is 1.
Second Solution
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Transmission Lines
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(f) Select the input normalized admittance of the stubs, by taking
the opposite of the corresponding imaginary part of the lineadmittance
( )
( )
stub stub
stub stub
line: d 1 stub:
line: d 1 stub:
y jb y jb
y jb y jb
= + → = −
= − → = +
(g) Use the chart to determine the length of the stub. Theimaginary normalized admittance values are found on the circleof zero conductance on the chart. On a commercial Smith chart
one can use a printed scale to read the stub length in terms of wavelength. We assume here that the stub line has
characteristic impedance Z 0 as the main line. If the stub has
characteristic impedance Z 0S ≠ Z 0 the values on the Smith chart
must be renormalized as
0 0
0 0
' s
s
Y Z jb jb jb
Y Z ± = ± = ±
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1
-1
0
0.2
-0.2
21
-0.5
-3
3
-2
y = ∞ Short circuit
(f) Normalized inputadmittance of stub
ystub = 0 - jb
(g) Arc to determine the length of ashort circuited stub with normalized
in ut admittance - b
0.5
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1
-1
0
0.2
-0.2
21
-0.5
-3
3
2
-2
y = 0Open circuit
0.5
(f) Normalized inputadmittance of stub
ystub = 0 - jb
(g) Arc to determine the length of anopen circuited stub with normalized
in ut admittance - b
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Transmission Lines
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1
-1
0
0.2
-0.2
21
-0.5
-3
3
2
-2
y = ∞ Short circuit
0.5
(f) Normalized inputadmittance of stub
ystub = 0 + jb
(g) Arc to determine the length of ashort circuited stub with normalized
in ut admittance + b
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Transmission Lines
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1
-1
0 0.2 0.5 5
0.2
-0.2
21
-0.5
-3
3
-2
0.5
(f) Normalized inputadmittance of stub
stub = 0 + b
(g) Arc to determine the length of anopen circuited stub with normalized
in ut admittance + b
y = 0Open circuit
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Transmission Lines
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1
-1
0 0.2 0.5 5
0.2
-0.2
21
-0.5
0.5
-3
3
2
-2
y R
First Solution
After the stub is inserted,the admittance at the stublocation is moved to thecenter of the Smith chart,
which corresponds tonormalized admittance 1and reflection coefficient 0(exact matching condition). If you imagine to add
gradually the negativeimaginary admittance of the inserted stub, the totaladmittance would followthe yellow arrow, reachingthe match point when the
complete stub admittanceis added.
matchingcondition
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1
-1
0 0.2 0.5 5
0.2
-0.2
21
-0.5
0.5
-3
3
-2
y R
First Solution If the stub does not havethe proper normalized input
admittance, the matchingcondition is not reached
Effect of a stub withpositive susceptance
Effect of a stub with
negative susceptance of insufficient magnitude
Effect of a stub withnegative susceptance of excessive magnitude
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