Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 1 Lecture Outline Impedance Transformation Techniques Impedance-Admittance Conversion Matching with Lumped Elements Stub Admittances and Shunt Matching Stub Impedances and Series Matching Double Stub Matching
41
Embed
Lecture Outline - eskisehir.edu.tr 509/icerik...Microwave Engineering January 29, 2003 Dr. Wolfgang J.R. Hoefer 1 Microwave Engineering University of Victoria Dr. Wolfgang J.R. Hoefer
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 1
Microwave Engineering
University of VictoriaDr. Wolfgang J.R. HoeferLayout by Dr. Poman P.M. So
Lecture 5
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 1
Lecture Outline
Impedance Transformation Techniques
Impedance-Admittance Conversion
Matching with Lumped Elements
Stub Admittances and Shunt Matching
Stub Impedances and Series Matching
Double Stub Matching
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 2
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 2
Reasons for Impedance Transformation
Maximum power is delivered when the load and generator are matched to the line.Proper input impedance transformation of sensitive receiver components (antenna, LNA, etc.) improves the S/N ratio of the system.Impedance matching in a power distribution network (such as antenna array feed network) will reduce amplitude and phase errors.
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 3
Transforming Network Selection Criteria
Complexity — A simpler impedance transformation network is usually cheaper, more reliable, and less lossy than a more complex design.
Bandwidth — larger BW → increase in complexity.
Implementation — Short-circuited stubs in coax and waveguide. Open-circuited stubs in stripline and microstrip.
Adjustability — tuning screws in waveguides.
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 2
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 2
Reasons for Impedance Transformation
Maximum power is delivered when the load and generator are matched to the line.Proper input impedance transformation of sensitive receiver components (antenna, LNA, etc.) improves the S/N ratio of the system.Impedance matching in a power distribution network (such as antenna array feed network) will reduce amplitude and phase errors.
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 3
Transforming Network Selection Criteria
Complexity — A simpler impedance transformation network is usually cheaper, more reliable, and less lossy than a more complex design.
Bandwidth — larger BW → increase in complexity.
Implementation — Short-circuited stubs in coax and waveguide. Open-circuited stubs in stripline and microstrip.
Adjustability — tuning screws in waveguides.
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 3
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 4
A lossless network matching an arbitrary load impedance to a transmission line.
To avoid unnecessary power loss, matching network is ideally lossless.The impedance looking in to the matching network is Zo.Reflections are eliminated on the transmission line to the left of the matching network.There will be multiple reflections between the matching network and the load.
MatchingNetwork
LoadZL
Zo
Zo
A Lossless Matching Network
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 5
Matching with Lumped Elements
L section matching network.(a) Network for zL inside the 1+jx circle (i.e. RL>Zo).(b) Network for zL outside the 1+jx circle (i.e. RL<Zo).ZL must have non-zero real part.
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 4
A lossless network matching an arbitrary load impedance to a transmission line.
To avoid unnecessary power loss, matching network is ideally lossless.The impedance looking in to the matching network is Zo.Reflections are eliminated on the transmission line to the left of the matching network.There will be multiple reflections between the matching network and the load.
MatchingNetwork
LoadZL
Zo
Zo
A Lossless Matching Network
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 5
Matching with Lumped Elements
L section matching network.(a) Network for zL inside the 1+jx circle (i.e. RL>Zo).(b) Network for zL outside the 1+jx circle (i.e. RL<Zo).ZL must have non-zero real part.
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 11/52
Matching with lumped elements (L networks)
Simplest type matching is L-section with 2 reactive elementsTwo possible configurations:
(a): network for zL within 1+jx circle(b): network for zL outside 1+jx circle
Reactive elements: capacitor or inductorParasitics limit usable frequency rangeSolutions analytical or using Smith Chart
ELEC344, Kevin Chen, HKUST 1
Design an L section matching network to match a series RCload with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, ata frequency of 500 MHz.
Solution:
Step 1: Convert the load impedance to admittance bydrawing the SWR circle through the load, and a straight linefrom the load through the center of the Smith Chart.
Step 2: Move the load impedance to the impedance circle of1+ jx (done in admittance Smith Chart) -- add j 0.3 insusceptance ELEC344, Kevin Chen, HKUST 2
Step 1
Step 2
Step 3: Convert back toimpedance.
Step 4: Move to the centerof the Smith Chart byadding an series inductor
Step 3
Step 4
ELEC344, Kevin Chen, HKUST 3
pFfZ
bC 92.0
2 0
==π
Therefore we have b = 0.3, x = 1.2 (check this result with theanalytic solution). Then for a frequency at f = 500 MHz,
we have
nHf
xZL 8.38
20 ==
π
Is there another solution?
ELEC344, Kevin Chen, HKUST 4
There are two solutions for the matching networks. In thiscase, there is no substantial difference in bandwidthbetween the two solutions.
ELEC344, Kevin Chen, HKUST 1
Lect. 13: Impedance Matching (2)
Smith Chart Solution 2 (not using combined ZY SmithChart)Example 5.1 on Page 254 of Pozar
Design an L section matching network to match a series RCload with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, ata frequency of 500 MHz.
Solution:
Step 1: Convert the load impedance to admittance bydrawing the SWR circle through the load, and a straight linefrom the load through the center of the Smith Chart.
Step 2: Move the load impedance to the impedance circle of1+ jx (done in admittance Smith Chart) -- add j 0.3 insusceptance ELEC344, Kevin Chen, HKUST 2
Step 1
Step 2
Step 3: Convert back toimpedance.
Step 4: Move to the centerof the Smith Chart byadding an series inductor
Step 3
Step 4
ELEC344, Kevin Chen, HKUST 3
pFfZ
bC 92.0
2 0
==π
Therefore we have b = 0.3, x = 1.2 (check this result with theanalytic solution). Then for a frequency at f = 500 MHz,
we have
nHf
xZL 8.38
20 ==
π
Is there another solution?
ELEC344, Kevin Chen, HKUST 4
There are two solutions for the matching networks. In thiscase, there is no substantial difference in bandwidthbetween the two solutions.
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 4
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 6
Smith Chart Solution 1
ZL=200–j100Ω, Zo=100Ω, fo=500MHz.
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
ZL
jX
jBZo
Plot zL =2–j1
Draw SWR and y =1 circles
Convert zL to yL
Add shunt susceptance to yL
Add series reactance
Convert y to z
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 7
Smith Chart Solution 2
ZL=200–j100Ω, Zo=100Ω, fo=500MHz.
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
ZL
jX
jBZo
Draw SWR and y =1 circles
Add shunt susceptance to yL
Add series reactance
Convert y to z
Convert zL to yL
Plot zL =2–j1
ELEC344, Kevin Chen, HKUST 1
Lect. 13: Impedance Matching (2)
Smith Chart Solution 2 (not using combined ZY SmithChart)Example 5.1 on Page 254 of Pozar
Design an L section matching network to match a series RCload with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, ata frequency of 500 MHz.
Solution:
Step 1: Convert the load impedance to admittance bydrawing the SWR circle through the load, and a straight linefrom the load through the center of the Smith Chart.
Step 2: Move the load impedance to the impedance circle of1+ jx (done in admittance Smith Chart) -- add j 0.3 insusceptance ELEC344, Kevin Chen, HKUST 2
Step 1
Step 2
Step 3: Convert back toimpedance.
Step 4: Move to the centerof the Smith Chart byadding an series inductor
Step 3
Step 4
ELEC344, Kevin Chen, HKUST 3
pFfZ
bC 92.0
2 0
==π
Therefore we have b = 0.3, x = 1.2 (check this result with theanalytic solution). Then for a frequency at f = 500 MHz,
we have
nHf
xZL 8.38
20 ==
π
Is there another solution?
ELEC344, Kevin Chen, HKUST 4
There are two solutions for the matching networks. In thiscase, there is no substantial difference in bandwidthbetween the two solutions.
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 4
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 6
Smith Chart Solution 1
ZL=200–j100Ω, Zo=100Ω, fo=500MHz.
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
ZL
jX
jBZo
Plot zL =2–j1
Draw SWR and y =1 circles
Convert zL to yL
Add shunt susceptance to yL
Add series reactance
Convert y to z
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 7
Smith Chart Solution 2
ZL=200–j100Ω, Zo=100Ω, fo=500MHz.
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
ZL
jX
jBZo
Draw SWR and y =1 circles
Add shunt susceptance to yL
Add series reactance
Convert y to z
Convert zL to yL
Plot zL =2–j1
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 5
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 8
Smith Chart Solutions 1&2
nH 8.382
pF 92.02
===
===
fπxZ
ωxZL
fZπb
ωbYC
oo
o
o
→= jXZ
pF 61.22
11
1
=−=−=
==−=
ofxZπωXC
ωCjjBX
jY
→= jBY
nH 1.462
1
1
=−=−=
==−=
fbπZ
ωBL
ωLjjXB
jZ
o
2.1 ,3.0 == xb 2.1 ,7.0 −=−= xb
ZL
L
CZo ZL
C
LZo
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 9
Single Stub MatchingProblems of Matching with Lumped Elements:
Lumped element impedance matching is not always possible or easily realizable.
Solutions:A section of open-circuited or short-circuited transmission line (a “stub”) connected in parallel or in series with the feed line at a distance from the load can be used.The tuning parameters are the distance from the load (d) and the length of the stub (l).
ELEC344, Kevin Chen, HKUST 1
Lect. 13: Impedance Matching (2)
Smith Chart Solution 2 (not using combined ZY SmithChart)Example 5.1 on Page 254 of Pozar
Design an L section matching network to match a series RCload with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, ata frequency of 500 MHz.
Solution:
Step 1: Convert the load impedance to admittance bydrawing the SWR circle through the load, and a straight linefrom the load through the center of the Smith Chart.
Step 2: Move the load impedance to the impedance circle of1+ jx (done in admittance Smith Chart) -- add j 0.3 insusceptance ELEC344, Kevin Chen, HKUST 2
Step 1
Step 2
Step 3: Convert back toimpedance.
Step 4: Move to the centerof the Smith Chart byadding an series inductor
Step 3
Step 4
ELEC344, Kevin Chen, HKUST 3
pFfZ
bC 92.0
2 0
==π
Therefore we have b = 0.3, x = 1.2 (check this result with theanalytic solution). Then for a frequency at f = 500 MHz,
we have
nHf
xZL 8.38
20 ==
π
Is there another solution?
ELEC344, Kevin Chen, HKUST 4
There are two solutions for the matching networks. In thiscase, there is no substantial difference in bandwidthbetween the two solutions.
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 5
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 8
Smith Chart Solutions 1&2
nH 8.382
pF 92.02
===
===
fπxZ
ωxZL
fZπb
ωbYC
oo
o
o
→= jXZ
pF 61.22
11
1
=−=−=
==−=
ofxZπωXC
ωCjjBX
jY
→= jBY
nH 1.462
1
1
=−=−=
==−=
fbπZ
ωBL
ωLjjXB
jZ
o
2.1 ,3.0 == xb 2.1 ,7.0 −=−= xb
ZL
L
CZo ZL
C
LZo
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 9
Single Stub MatchingProblems of Matching with Lumped Elements:
Lumped element impedance matching is not always possible or easily realizable.
Solutions:A section of open-circuited or short-circuited transmission line (a “stub”) connected in parallel or in series with the feed line at a distance from the load can be used.The tuning parameters are the distance from the load (d) and the length of the stub (l).
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 31/52
Single-stub matching
Single open or short circuited length TL (stub) connected either in shunt or series at certain distance from loadNo lumped components are required, convenient for MICShunt stub often preferred (especially in stripline or µstrip)Parameters: distance series line and value susceptance (or reactance) provided by shunt or series stubShunt: d choosen that after line admittance into line (Y) is Y0+jB, then stub susceptance = -jB; for series d such that Z into line has form Z0+jX
shunt stub series stub
Single-stub tuningA single open-circuited or short-circuited length of transmission line (i.e. a stub) can connect in either parallel or series with the main feed line to achieve impedance matching.
The two adjustable parameters are the distance, d, from the load to the stub position, and the value of susceptance or reactance provided by the shunt or series stub.
( = Z0+jX @ the stub )
-jB
( = Y0+jB @ the stub )
-jX
Single-stub tuningProper length of both open or shored transmission line can provide any desired value of reactance or susceptance.
For a given susceptance or reactance, the difference in lengthsof an open- or short-circuited stub is λ/4.
For microstrip or stripline, open-circuited stubs are easier to fabricate.
For lines like coax or waveguide, however, short-circuited stubsare usually preferred. (open-circuited stubs tend to radiate)
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 6
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 10
Matching Operations:Select d, so that y= 1+jb.Select l, so that the stub susceptance is –jb.
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 11
Example 5.2 — Solution 1
ZL=15+j10Ω, Zo=50Ω, fo=2GHz.
Draw SWR and y =1 circles
Convert zL to yL
Transform yL to y1, d=0.044λAdd shunt susceptance to y1
YL
d
YoYo
Yo
Open orshortedstub
ZY 1=
l
Plot zL =0.3+j0.2
b=–1.33
l=0.147λ b=1.33
The stub length is: l=0.147λ
y=0
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 32/52
Single-stub Shunt Matching
For shunt stub in microstrip or stripline open stub is preferred (no VIA hole needed), for waveguide, coax and also to apply DC-bias short stub often more convenient
? For ZL=15+j10 Ω, design two single-stub shunt tuning networks to match load to 50Ω line (3rd Ed. book has different example)First plot normalized load impedanceConvert to admittance by imagingPlot 1+jb circle on Y-chart (1+jx on Z)Turn on SWR circle leads to two intersections (y1 & y2)Distance d from load to stub given by WTG scaleSusceptance stub given by normalized admittance y1 & y2Length stub for given b, determined on Smith chart (start from short!).
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 6
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 10
Matching Operations:Select d, so that z= 1+jx.Select l, so that the stub susceptance is –jx.
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 33/52
Single-stub shunt matching on Smith Chart
2.03.01015
jzjZ
L
L+=
Ω+=
1+jb circle
λλ
387.0171.0284.05.0044.0284.0328.0
2
1=+−=
=−=
dd
33.1133.11
2
1jyjy
+=
−=
λλ
353.0147.0
2
1=
=
ll
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 34/52
Two solutions single-stub shunt matching
Solution leading to shortest length transmission lines has clearly better bandwidthShorter TL reduce frequency variation match and also in practice will reduce lossesAnalytical solution in book (p. 231-232)
Single-stub tuningShunt stubs – Smith chart
For a load impedance ZL=60-j80, design two single-stub shunt tuningnetworks to match this load to a 50Ω line at 2 GHz.
1.2 1.6Lz j= −
0.3 0.4Ly j= +
1
2
1
2
1
2
0.176 0.065 0.1100.325 0.065 0.260
1.00 1.471.00 1.47
0.0950.405
dd
y jy j
ll
λλ
λλ
= − == − =
= += −
==
d1d2
y1=1+j1.47
y2=1-j1.47zL
yL
-jb=1+jb
s.c.
Single-stub tuningAt 2 GHz, ZL=60-j80 can be modeled as a series combination of R=60Ωand C=0.995 pF. The frequency response is then
1
2
1
2
0.110 0.260
0.095 0.405
dd
ll
λλ
λλ
==
==
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 7
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 12
Example 5.2 — Solution 2
ZL=15+j10Ω, Zo=50Ω, fo=2GHz.
Draw SWR and y =1 circles
Convert zL to yL
Transform yL to y1, d=0.387λAdd shunt susceptance to y1
YL
d
YoYo
Yo
Open orshortedstub
ZY 1=
l
Plot zL =0.3+j0.2
l=0.353λ b=–1.33
b=1.33
The stub length is: l=0.353λ
y=0
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 13
Matching Operations:Select d, so that z= 1+jx.Select l, so that the stub susceptance is –jx.
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 35/52
Single-stub Series Tuning
? Match ZL=100+j80 Ω to 50Ω line using single series open-circuit stub
First plot normalized load impedance
Plot 1+jx circle on Z-chart
Turn on SWR circle gives 2 intersections (z1 & z2)
Distance d1 & d2 from load to stub given by WTG scale
Reactance stub given by normalized impedance z1 & z2
Length stub for given b, determined on Smith chart starting from open point on Z-chart.
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 8
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 14
Example 5.3 — Solution 1
ZL=100+j80Ω, Zo=50Ω, fo=2GHz.
Draw SWR and z =1 circles
Transform zL to z1, d=0.120λAdd series reactance to z1
Plot zL =2+j1.6
ZL
d
ZoZo
Zo
Open orshortedstub
YZ 1=l
b=–1.33
l=0.397λ b=1.33
The stub length is: l=0.397λ
z=∞
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 15
Example 5.3 — Solution 2
ZL=100+j80Ω, Zo=50Ω, fo=2GHz.
Draw SWR and z =1 circles
Transform zL to z1, d=0.463λAdd series reactance to z1
Plot zL =2+j1.6
ZL
d
ZoZo
Zo
Open orshortedstub
YZ 1=l
b=–1.33 l=0.103λ
b=1.33
The stub length is: l=0.103λ
z=∞
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 8
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 14
Example 5.3 — Solution 1
ZL=100+j80Ω, Zo=50Ω, fo=2GHz.
Draw SWR and z =1 circles
Transform zL to z1, d=0.120λAdd series reactance to z1
Plot zL =2+j1.6
ZL
d
ZoZo
Zo
Open orshortedstub
YZ 1=l
b=–1.33
l=0.397λ b=1.33
The stub length is: l=0.397λ
z=∞
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 15
Example 5.3 — Solution 2
ZL=100+j80Ω, Zo=50Ω, fo=2GHz.
Draw SWR and z =1 circles
Transform zL to z1, d=0.463λAdd series reactance to z1
Plot zL =2+j1.6
ZL
d
ZoZo
Zo
Open orshortedstub
YZ 1=l
b=–1.33 l=0.103λ
b=1.33
The stub length is: l=0.103λ
z=∞
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 36/52
Single-stub series matching on Smith Chart
1+jx circle6.12
80100jz
jZ
L
L
+=
Ω+=
λλ
463.0172.0208.05.0120.0208.0328.0
2
1
=+−=
=−=
dd
33.1133.11
2
1
jzjz
+=
−=
λλ
103.0397.0
2
1
=
=
ll
Dr. Y. Baeyens E4318-Microwave Circuit Design L.5 – 37/52
Two solutions single-stub series matching
Lengths for both solutions approx. similar, so no big difference in bandwidthSeries matching not so convenient, requires seperate connection to conductor and groundAnalytical solution: see Pozarp.234-235
Single-stub tuningSeries stubs Series stubs –– Smith chart Smith chart
For a load impedance ZL=100+j80, design two single-stub series tuningnetworks to match this load to a 50Ω line at 2 GHz.
2 1.6Lz j= +
1
2
1
2
1
2
(0.328 0.208) 0.120
(0.5 0.208) 0.172 0.463
1.00 1.331.00 1.33
0.3970.103
d
d
z jz j
ll
λ
λ
λλ
= −== − +=
= −= +
==
d1
d2
z1=1-j1.33
z2=1+j1.33
zL
o.c.
=1+jx-jx
Single-stub tuningAt 2 GHz, ZL can be modeled as a series combination of R=100Ω and L=6.37 nH. The frequency response is then
To derive formulas for d and l, let the load impedance be written as .
The impedance Z down a length, d, of line from the load is
where . The admittance at this point is thus
For matching reason, the d (which implies t) is chosen so that .
LLLL XRYZ +== /1
dt βtan=
00 /1 ZYG ==
Single-stub tuningThis results in a quadratic equation for t:
Solving for t gives
Thus, the two principal solutions for d are
02/ ZXt L−= 0for LR Z=
Single-stub tuningFor the required stub lengths, first use t in (5.8b) to find the stub susceptance, and Bs = -B.
Then, for an open-circuit stub,
While for a short-circuited stub,
If the resultant length is negative, λ/2 can be added to give a positive result.
Single-stub tuningSeries stubs Series stubs –– analytical solutionanalytical solution
The input impedance Zin = Rin +j Xin down a length d from the load can be first evaluated.
The matching condition is Rin= Z0 at the stub (let ),
and the two principal solutions for d are then
The required stub lengths arefor an open-circuited stub.
for a short-circuited stub. where X was obtained by substituting t into (5.13b)
0for LG Y=02/ YBt L−=
dt βtan=
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 9
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 16
Double Stub MatchingDisadvantages of Matching with Single Stub:
A variable length of line between the load and the stub is needed.This would be a problem if an adjustable tuner was desired.
Solution:Double stub matching — two tuning stubs in fixed positions.Adjustable stubs are usually connected in parallel to the main feed line.Double stub tuner cannot match all load impedances.
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 17
Matching Operations
Double Shunt Stub Matching
Select l1, so that y1 lies on the rotated 1+jb circle; the amount of rotation is dwavelengths towards the load. In practice, d=λ/8 or 3λ/8.Transform y1 toward the generator through a length d; the new admittance, y2 =1+jb2, lies on the 1+jb circle.Select l2, so that the stub susceptance is –b2.
d
Y’LYoYo
Yo
Open orshortedstub
l2
Yo
Yo
Open orshortedstub
LL Z
Y 1=
l1
jB2 jB1
Y1Yo
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 9
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 16
Double Stub MatchingDisadvantages of Matching with Single Stub:
A variable length of line between the load and the stub is needed.This would be a problem if an adjustable tuner was desired.
Solution:Double stub matching — two tuning stubs in fixed positions.Adjustable stubs are usually connected in parallel to the main feed line.Double stub tuner cannot match all load impedances.
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 17
Matching Operations
Double Shunt Stub Matching
Select l1, so that y1 lies on the rotated 1+jb circle; the amount of rotation is dwavelengths towards the load. In practice, d=λ/8 or 3λ/8.Transform y1 toward the generator through a length d; the new admittance, y2 =1+jb2, lies on the 1+jb circle.Select l2, so that the stub susceptance is –b2.
d
Y’LYoYo
Yo
Open orshortedstub
l2
Yo
Yo
Open orshortedstub
LL Z
Y 1=
l1
jB2 jB1
Y1Yo
Microwave Engineering January 29, 2003
Dr. Wolfgang J.R. Hoefer 10
Dr. W.J.R. Hoefer ELEC 454 Microwave Engineering 18