Microwave Engineering
January 29, 2003
Microwave Engineering
University of Victoria
Dr. Wolfgang J.R. Hoefer
Layout by Dr. Poman P.M. So
Lecture 5
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
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
1
1
Microwave Engineering
January 29, 2003
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
2
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.
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
3
2
Microwave Engineering
January 29, 2003
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
2
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.
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
3
2
Microwave Engineering
January 29, 2003
A Lossless Matching Network
Matching
Network
Zo
Load
ZL
Zo
 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.
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
4
Matching with Lumped Elements
jX
(a)
jB
Zo
jX
(b)
ZL
B=
X L ± RL Z o RL2 + X L2 − Z o RL
RL2 + X L2
X =
Z
1 X LZo
+
− o
B
RL
BRL
jB
Zo
B=±
(Z o − RL )
ZL
RL
Zo
X = ± RL (Z o − RL ) − X L
 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.
D.M. Pozar, Microwave Engineering, 2nd Edition, section 5.1, p.p. 252, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
5
3
Microwave Engineering
January 29, 2003
A Lossless Matching Network
Matching
Network
Zo
Load
ZL
Zo
 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.
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
4
Matching with Lumped Elements
jX
(a)
jB
Zo
jX
(b)
ZL
B=
X L ± RL Z o RL2 + X L2 − Z o RL
RL2 + X L2
X =
Z
1 X LZo
+
− o
B
RL
BRL
jB
Zo
B=±
(Z o − RL )
ZL
RL
Zo
X = ± RL (Z o − RL ) − X L
 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.
D.M. Pozar, Microwave Engineering, 2nd Edition, section 5.1, p.p. 252, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
5
3
Matching with lumped elements (L networks)
Š Simplest type matching is L-section with 2 reactive elements
Š Two possible configurations:
„ (a): network for zL within 1+jx circle
„ (b): network for zL outside 1+jx circle
Š Reactive elements: capacitor or inductor
Š Parasitics limit usable frequency range
Š Solutions analytical or using Smith Chart
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 11/52
Step 3: Convert back to
impedance.
Step
2
Design an L section matching network to match a series RC
load with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, at
a frequency of 500 MHz.
Step
1
Solution:
Step 4: Move to the center
of the Smith Chart by
adding an series inductor
1
Therefore we have b = 0.3, x = 1.2 (check this result with the
analytic solution). Then for a frequency at f = 500 MHz,
we have
C=
b
= 0.92 pF
2πfZ 0
L=
xZ 0
= 38.8nH
2πf
3
4
Step 2: Move the load impedance to the impedance circle of
1+ jx (done in admittance Smith Chart) -- add j 0.3 in
ELEC344, Kevin Chen, HKUST
susceptance
ep
St
ep
St
Step 1: Convert the load impedance to admittance by
drawing the SWR circle through the load, and a straight line
from the load through the center of the Smith Chart.
ELEC344, Kevin Chen, HKUST
2
There are two solutions for the matching networks. In this
case, there is no substantial difference in bandwidth
between the two solutions.
Is there another solution?
ELEC344, Kevin Chen, HKUST
3
ELEC344, Kevin Chen, HKUST
4
Lect. 13: Impedance Matching (2)
Step 3: Convert back to
impedance.
Smith Chart Solution 2 (not using combined ZY Smith
Chart)
Example 5.1 on Page 254 of Pozar
Step
2
Design an L section matching network to match a series RC
load with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, at
a frequency of 500 MHz.
Step
1
Solution:
Step 4: Move to the center
of the Smith Chart by
adding an series inductor
1
Therefore we have b = 0.3, x = 1.2 (check this result with the
analytic solution). Then for a frequency at f = 500 MHz,
we have
C=
b
= 0.92 pF
2πfZ 0
L=
xZ 0
= 38.8nH
2πf
3
4
Step 2: Move the load impedance to the impedance circle of
1+ jx (done in admittance Smith Chart) -- add j 0.3 in
ELEC344, Kevin Chen, HKUST
susceptance
ep
St
ep
St
Step 1: Convert the load impedance to admittance by
drawing the SWR circle through the load, and a straight line
from the load through the center of the Smith Chart.
ELEC344, Kevin Chen, HKUST
2
There are two solutions for the matching networks. In this
case, there is no substantial difference in bandwidth
between the two solutions.
Is there another solution?
ELEC344, Kevin Chen, HKUST
3
ELEC344, Kevin Chen, HKUST
4
Microwave Engineering
January 29, 2003
Smith Chart Solution 1
jX
jB
Zo
ZL
 ZL=200–j100Ω, Zo=100Ω,
fo=500MHz.






Plot zL =2–j1
Draw SWR and y =1 circles
Convert zL to yL
Add shunt susceptance to yL
Convert y to z
Add series reactance
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
6
Smith Chart Solution 2
jX
Zo
jB
ZL
 ZL=200–j100Ω, Zo=100Ω,
fo=500MHz.






Plot zL =2–j1
Draw SWR and y =1 circles
Convert zL to yL
Add shunt susceptance to yL
Convert y to z
Add series reactance
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
7
4
Lect. 13: Impedance Matching (2)
Step 3: Convert back to
impedance.
Smith Chart Solution 2 (not using combined ZY Smith
Chart)
Example 5.1 on Page 254 of Pozar
Step
2
Design an L section matching network to match a series RC
load with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, at
a frequency of 500 MHz.
Step
1
Solution:
Step 4: Move to the center
of the Smith Chart by
adding an series inductor
1
Therefore we have b = 0.3, x = 1.2 (check this result with the
analytic solution). Then for a frequency at f = 500 MHz,
we have
C=
b
= 0.92 pF
2πfZ 0
L=
xZ 0
= 38.8nH
2πf
3
4
Step 2: Move the load impedance to the impedance circle of
1+ jx (done in admittance Smith Chart) -- add j 0.3 in
ELEC344, Kevin Chen, HKUST
susceptance
ep
St
ep
St
Step 1: Convert the load impedance to admittance by
drawing the SWR circle through the load, and a straight line
from the load through the center of the Smith Chart.
ELEC344, Kevin Chen, HKUST
2
There are two solutions for the matching networks. In this
case, there is no substantial difference in bandwidth
between the two solutions.
Is there another solution?
ELEC344, Kevin Chen, HKUST
3
ELEC344, Kevin Chen, HKUST
4
Microwave Engineering
January 29, 2003
Smith Chart Solution 1
jX
jB
Zo
ZL
 ZL=200–j100Ω, Zo=100Ω,
fo=500MHz.






Plot zL =2–j1
Draw SWR and y =1 circles
Convert zL to yL
Add shunt susceptance to yL
Convert y to z
Add series reactance
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
6
Smith Chart Solution 2
jX
Zo
jB
ZL
 ZL=200–j100Ω, Zo=100Ω,
fo=500MHz.






Plot zL =2–j1
Draw SWR and y =1 circles
Convert zL to yL
Add shunt susceptance to yL
Convert y to z
Add series reactance
D.M. Pozar, Microwave Engineering, 2nd Edition, Example 5.1, p.p. 254, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
7
4
Microwave Engineering
January 29, 2003
Smith Chart Solutions 1&2
L
C
C
Zo
ZL
b = 0.3, x = 1.2
C=
bYo
b
=
= 0.92 pF
ω 2π fZ o
L=
xZ o xZ o
=
= 38.8 nH
ω
2π f
Dr. W.J.R. Hoefer
Zo
L
ZL
b = −0.7, x = −1.2
1
Y = jB → Z = − j = jX = jωL
B
−1 − Zo
L=
=
= 46.1 nH
ωB 2π fb
1
Z = jX → Y = − j = jB = jωC
X
−1
−1
C=
=
= 2.61pF
ωX 2π fxZo
ELEC 454 Microwave Engineering
8
Single Stub Matching
 Problems 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. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
9
5
Lect. 13: Impedance Matching (2)
Step 3: Convert back to
impedance.
Smith Chart Solution 2 (not using combined ZY Smith
Chart)
Example 5.1 on Page 254 of Pozar
Step
2
Design an L section matching network to match a series RC
load with an impedance ZL = 200 - j 100 Ω, to a 100 Ω line, at
a frequency of 500 MHz.
Step
1
Solution:
Step 4: Move to the center
of the Smith Chart by
adding an series inductor
1
Therefore we have b = 0.3, x = 1.2 (check this result with the
analytic solution). Then for a frequency at f = 500 MHz,
we have
C=
b
= 0.92 pF
2πfZ 0
L=
xZ 0
= 38.8nH
2πf
3
4
Step 2: Move the load impedance to the impedance circle of
1+ jx (done in admittance Smith Chart) -- add j 0.3 in
ELEC344, Kevin Chen, HKUST
susceptance
ep
St
ep
St
Step 1: Convert the load impedance to admittance by
drawing the SWR circle through the load, and a straight line
from the load through the center of the Smith Chart.
ELEC344, Kevin Chen, HKUST
2
There are two solutions for the matching networks. In this
case, there is no substantial difference in bandwidth
between the two solutions.
Is there another solution?
ELEC344, Kevin Chen, HKUST
3
ELEC344, Kevin Chen, HKUST
4
Microwave Engineering
January 29, 2003
Smith Chart Solutions 1&2
L
C
C
Zo
ZL
b = 0.3, x = 1.2
C=
bYo
b
=
= 0.92 pF
ω 2π fZ o
L=
xZ o xZ o
=
= 38.8 nH
ω
2π f
Dr. W.J.R. Hoefer
Zo
L
ZL
b = −0.7, x = −1.2
1
Y = jB → Z = − j = jX = jωL
B
−1 − Zo
L=
=
= 46.1 nH
ωB 2π fb
1
Z = jX → Y = − j = jB = jωC
X
−1
−1
C=
=
= 2.61pF
ωX 2π fxZo
ELEC 454 Microwave Engineering
8
Single Stub Matching
 Problems 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. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
9
5
Single-stub matching
shunt stub
series stub
Š Single open or short circuited length TL (stub) connected
either in shunt or series at certain distance from load
Š No lumped components are required, convenient for MIC
Š Shunt stub often preferred (especially in stripline or µstrip)
Š Parameters: distance series line and value susceptance (or
reactance) provided by shunt or series stub
Š Shunt: 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
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 31/52
Single-stub tuning
„
A 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.
-jX
( = Y0+jB @ the stub )
( = Z0+jX @ the stub )
-jB
„
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.
Single-stub tuning
„
„
„
„
Proper 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 lengths
of 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 stubs
are usually preferred. (open-circuited stubs tend to radiate)
Microwave Engineering
January 29, 2003
Shunt Stub Matching
d
Yo
Yo
Yo
Y=
Open or
shorted
stub
YL
1
Z
 Matching Operations:
l
„
„
Select d, so that y= 1+jb.
Select l, so that the stub
susceptance is –jb.
D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(a), p.p. 258, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
10
Example 5.2 — Solution 1
d
Yo
l=0.147λ
Yo
Open or
shorted
stub
Y=
b=1.33
YL
Yo
1
Z
l
y=0
 ZL=15+j10Ω, Zo=50Ω,
fo=2GHz.





Plot zL =0.3+j0.2
Draw SWR and y =1 circles
Convert zL to yL
Transform yL to y1, d=0.044λ
Add shunt susceptance to y1
 The stub length is: l=0.147λ
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
b=–1.33
11
6
Single-stub Shunt Matching
Š 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)
Š First plot normalized load impedance
Š Convert to admittance by imaging
Š Plot 1+jb circle on Y-chart (1+jx on Z)
Š Turn on SWR circle leads to two intersections (y1 & y2)
Š Distance d from load to stub given by WTG scale
Š Susceptance stub given by normalized admittance y1 & y2
Š Length stub for given b, determined on Smith chart (start
from short!).
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 32/52
Microwave Engineering
January 29, 2003
Shunt Stub Matching
d
Yo
Yo
Yo
Y=
Open or
shorted
stub
YL
1
Z
 Matching Operations:
l
„
„
Select d, so that y= 1+jb.
Select l, so that the stub
susceptance is –jb.
D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(a), p.p. 258, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
10
Example 5.2 — Solution 1
d
Yo
l=0.147λ
Yo
Open or
shorted
stub
Y=
b=1.33
YL
Yo
1
Z
l
y=0
 ZL=15+j10Ω, Zo=50Ω,
fo=2GHz.





Plot zL =0.3+j0.2
Draw SWR and y =1 circles
Convert zL to yL
Transform yL to y1, d=0.044λ
Add shunt susceptance to y1
 The stub length is: l=0.147λ
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
b=–1.33
11
6
Microwave Engineering
January 29, 2003
Example 5.2 — Solution 2
d
Yo
b=1.33
YL
Yo
Yo
Open or
shorted
stub
Y=
1
Z
l
y=0
 ZL=15+j10Ω, Zo=50Ω,
fo=2GHz.





Plot zL =0.3+j0.2
Draw SWR and y =1 circles
Convert zL to yL
Transform yL to y1, d=0.387λ
Add shunt susceptance to y1
l=0.353λ
 The stub length is: l=0.353λ
Dr. W.J.R. Hoefer
b=–1.33
ELEC 454 Microwave Engineering
12
Series Stub Matching
d
Zo
Zo
Zo
l
Z=
Open or
shorted
stub
ZL
1
Y
 Matching Operations:
„
„
Select d, so that z= 1+jx.
Select l, so that the stub
susceptance is –jx.
D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(b), p.p. 258, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
13
7
Single-stub shunt matching on Smith Chart
Z L = 15 + j10Ω
z L = 0.3 + j 0.2
1+jb circle
d1 = 0.328 − 0.284 = 0.044λ
d 2 = 0.5 − 0.284 + 0.171 = 0.387λ
y1 = 1 − j1.33
y2 = 1 + j1.33
l1 = 0.147λ
l2 = 0.353λ
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 33/52
Two solutions single-stub shunt matching
Š Solution leading to shortest
length transmission lines has
clearly better bandwidth
Š Shorter TL reduce frequency
variation match and also in
practice will reduce losses
Š Analytical solution in book
(p. 231-232)
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 34/52
Single-stub tuning
Shunt stubs – Smith chart
For a load impedance ZL=60-j80, design two single-stub shunt tuning
networks to match this load to a 50Ω line at 2 GHz.
d1
yL
z L = 1.2 − j1.6
d2
yL = 0.3 + j 0.4
y1=1+j1.47
d1 = 0.176 − 0.065 = 0.110λ
d 2 = 0.325 − 0.065 = 0.260λ
s.c.
zL
y2=1-j1.47
-jb
=1+jb
y1 = 1.00 + j1.47
y2 = 1.00 − j1.47
l1 = 0.095λ
l2 = 0.405λ
Single-stub tuning
„
At 2 GHz, ZL=60-j80 can be modeled as a series combination of R=60Ω
and C=0.995 pF. The frequency response is then
d1 = 0.110λ
d 2 = 0.260λ
l1 = 0.095λ
l2 = 0.405λ
Microwave Engineering
January 29, 2003
Example 5.2 — Solution 2
d
Yo
b=1.33
YL
Yo
Yo
Open or
shorted
stub
Y=
1
Z
l
y=0
 ZL=15+j10Ω, Zo=50Ω,
fo=2GHz.





Plot zL =0.3+j0.2
Draw SWR and y =1 circles
Convert zL to yL
Transform yL to y1, d=0.387λ
Add shunt susceptance to y1
l=0.353λ
 The stub length is: l=0.353λ
Dr. W.J.R. Hoefer
b=–1.33
ELEC 454 Microwave Engineering
12
Series Stub Matching
d
Zo
Zo
Zo
l
Z=
Open or
shorted
stub
ZL
1
Y
 Matching Operations:
„
„
Select d, so that z= 1+jx.
Select l, so that the stub
susceptance is –jx.
D.M. Pozar, Microwave Engineering, 2nd Edition, Figure 5.4(b), p.p. 258, John Wiley & Sons, 1998
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
13
7
Single-stub Series Tuning
? Match ZL=100+j80 Ω to 50Ω line using single series opencircuit stub
Š First plot normalized load impedance
Š Plot 1+jx circle on Z-chart
Š Turn on SWR circle gives 2 intersections (z1 & z2)
Š Distance d1 & d2 from load to stub given by WTG scale
Š Reactance stub given by normalized impedance z1 & z2
Š Length stub for given b, determined on Smith chart starting
from open point on Z-chart.
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 35/52
Microwave Engineering
January 29, 2003
Example 5.3 — Solution 1
d
Zo
l=0.397λ
Zo
Zo
l
Z=
b=1.33
ZL
1
Y
Open or
shorted
stub
z=∞
 ZL=100+j80Ω, Zo=50Ω,
fo=2GHz.





Plot zL =2+j1.6
Draw SWR and z =1 circles
Transform zL to z1, d=0.120λ
Add series reactance to z1
The stub length is: l=0.397λ
b=–1.33
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
14
Example 5.3 — Solution 2
d
Zo
b=1.33
Zo
Zo
l
Z=
ZL
1
Y
Open or
shorted
stub
z=∞
 ZL=100+j80Ω, Zo=50Ω,
fo=2GHz.





Plot zL =2+j1.6
Draw SWR and z =1 circles
Transform zL to z1, d=0.463λ
Add series reactance to z1
The stub length is: l=0.103λ
b=–1.33
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
l=0.103λ
15
8
Microwave Engineering
January 29, 2003
Example 5.3 — Solution 1
d
Zo
l=0.397λ
Zo
Zo
l
Z=
b=1.33
ZL
1
Y
Open or
shorted
stub
z=∞
 ZL=100+j80Ω, Zo=50Ω,
fo=2GHz.





Plot zL =2+j1.6
Draw SWR and z =1 circles
Transform zL to z1, d=0.120λ
Add series reactance to z1
The stub length is: l=0.397λ
b=–1.33
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
14
Example 5.3 — Solution 2
d
Zo
b=1.33
Zo
Zo
l
Z=
ZL
1
Y
Open or
shorted
stub
z=∞
 ZL=100+j80Ω, Zo=50Ω,
fo=2GHz.





Plot zL =2+j1.6
Draw SWR and z =1 circles
Transform zL to z1, d=0.463λ
Add series reactance to z1
The stub length is: l=0.103λ
b=–1.33
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
l=0.103λ
15
8
Single-stub series matching on Smith Chart
Z L = 100 + j 80Ω
z L = 2 + j1.6
1+jx circle
d1 = 0.328 − 0.208 = 0.120λ
d 2 = 0.5 − 0.208 + 0.172 = 0.463λ
z1 = 1 − j1.33
z2 = 1 + j1.33
l1 = 0.397λ
l2 = 0.103λ
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 36/52
Two solutions single-stub series matching
Š Lengths for both solutions approx.
similar, so no big difference in
bandwidth
Š Series matching not so convenient,
requires seperate connection to
conductor and ground
Š Analytical solution: see Pozar
p.234-235
Dr. Y. Baeyens
E4318-Microwave Circuit Design
L.5 – 37/52
Single-stub tuning
Series stubs – Smith chart
For a load impedance ZL=100+j80, design two single-stub series tuning
networks to match this load to a 50Ω line at 2 GHz.
z L = 2 + j1.6
d1 = (0.328 − 0.208)
z2=1+j1.33
= 0.120λ
d 2 = (0.5 − 0.208) + 0.172
zL
d2
o.c.
d1
-jx
z1=1-j1.33
=1+jx
= 0.463λ
z1 = 1.00 − j1.33
z2 = 1.00 + j1.33
l1 = 0.397λ
l2 = 0.103λ
Single-stub tuning
„
At 2 GHz, ZL can be modeled as a series combination of R=100Ω and
L=6.37 nH. The frequency response is then
d1 = 0.120λ
d 2 = 0.463λ
l1 = 0.397λ
l2 = 0.103λ
Single-stub tuning
Stunt stubs – analytical solution
„
To derive formulas for d and l, let the load impedance be written as
Z L =1/ YL = RL + X L .
„
The impedance Z down a length, d, of line from the load is
„
where t = tan β d .
The admittance at this point is thus
„
For matching reason, the d (which implies t) is chosen so
that G = Y0 =1/ Z0.
Single-stub tuning
„
This results in a quadratic equation for t:
„
Solving for t gives
t = −X L / 2Z0
„
Thus, the two principal solutions for d are
for RL = Z0
Single-stub tuning
„
For 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 tuning
Series stubs – analytical 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 t = tanβ d
t = −BL / 2Y0
for
),
GL = Y0
and the two principal solutions for d are then
„
The required stub lengths are
for an open-circuited stub.
for a short-circuited stub.
where X was obtained by substituting t into (5.13b)
Microwave Engineering
January 29, 2003
Double Stub Matching
 Disadvantages 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
16
Double Shunt Stub Matching
d
Yo
Y1
Yo
jB2
Yo
Yo
Yo
YL =
Open or
shorted
stub
l2
Open or
shorted
stub
Yo
jB1
Y’L
1
ZL
l1
Matching Operations
 Select l1, so that y1 lies on the rotated 1+jb circle; the amount of rotation is d
wavelengths 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.
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
17
9
Microwave Engineering
January 29, 2003
Double Stub Matching
 Disadvantages 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
16
Double Shunt Stub Matching
d
Yo
Y1
Yo
jB2
Yo
Yo
Yo
YL =
Open or
shorted
stub
l2
Open or
shorted
stub
Yo
jB1
Y’L
1
ZL
l1
Matching Operations
 Select l1, so that y1 lies on the rotated 1+jb circle; the amount of rotation is d
wavelengths 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.
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
ELEC 454 Microwave Engineering
17
9
Microwave Engineering
January 29, 2003
Example 5.4 — Solution 1
d
Yo
Yo
Yo
jB2
Yo
Shorted
stub
Y1
jB1 YL
Yo
Shorted
stub
l2
l1
 ZL=60–j80Ω, Zo=50Ω, d=λ/8,
fo=2GHz.





Plot yL =0.3+j0.4
Draw conductance circles
yL → y1; b1=1.314
y1 → y2; via the SWR circle.
y2 → 1; b2=3.38
Forbidden Region
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
18
Example 5.4 — Solution 2
d
Yo
Yo
Yo
jB2
Yo
Shorted
stub
l2
Y1
jB1 YL
Yo
Shorted
stub
l1
 ZL=60–j80Ω, Zo=50Ω, d=λ/8,
fo=2GHz.





Plot yL =0.3+j0.4
Draw conductance circles
yL → y1; b1=–0.114
y1 → y2; via the SWR circle.
y2 → 1; b2=–1.38
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
Forbidden Region
ELEC 454 Microwave Engineering
19
10
Microwave Engineering
January 29, 2003
Example 5.4 — Solution 1
d
Yo
Yo
Yo
jB2
Yo
Shorted
stub
Y1
jB1 YL
Yo
Shorted
stub
l2
l1
 ZL=60–j80Ω, Zo=50Ω, d=λ/8,
fo=2GHz.





Plot yL =0.3+j0.4
Draw conductance circles
yL → y1; b1=1.314
y1 → y2; via the SWR circle.
y2 → 1; b2=3.38
Forbidden Region
Dr. W.J.R. Hoefer
ELEC 454 Microwave Engineering
18
Example 5.4 — Solution 2
d
Yo
Yo
Yo
jB2
Yo
Shorted
stub
l2
Y1
jB1 YL
Yo
Shorted
stub
l1
 ZL=60–j80Ω, Zo=50Ω, d=λ/8,
fo=2GHz.





Plot yL =0.3+j0.4
Draw conductance circles
yL → y1; b1=–0.114
y1 → y2; via the SWR circle.
y2 → 1; b2=–1.38
Dr. W.J.R. Hoefer
Dr. Wolfgang J.R. Hoefer
Forbidden Region
ELEC 454 Microwave Engineering
19
10