Impedance Matching and Tuning

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Chapter 5

• Impedance Matching and Tuning

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Why need Impedance Matching

• Maximum power is delivered and power loss is
  minimum.
• Impedance matching sensitive receiver components
  improves the signal-to-noise ratio of the system.
• Impedance matching in a power distribution
  network will reduce amplitude and phase errors.

• Basic Idea

The matching network is ideally lossless and is
placed between a load and a transmission line, to
avoid unnecessary loss of power, and is usually
designed so that the impedance seen looking into
the matching network is Z0. ( Multiple
reflections will exist between the matching
network and the load)

• The matching procedure is also referred to as
  tuning.

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Design Considerations of Matching Network

• As long as the load impedance has non-zero real
  part (i.e. Lossy term), a matching network can
  always be found.
• Factors for selecting a matching network
• 1) Complexity a simpler matching network is
  more preferable
• because it is cheaper, more reliable, and
  less lossy.
• 2) Bandwidth any type of matching network can
  ideally give a perfect
• match at a single frequency. However, some
  complicated design
• can provide matching over a range of
  frequencies.
• 3) Implementation one type of matching
  network may be preferable
• compared to other methods.
• 4) Adjustability adjustment may be required
  to match a variable load
• impedance.

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Lumped Elements Matching

• L-Shape (Two-Element) Matching

• Case 1 ZL inside the 1jx circle (RLgtZ0)

• Use impedance identity method

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Example5.1 Design an L-section matching network
to match a series RC load with an impedance ZL
200-j100?, to a 100? line, at a frequency of 500
MHz?
Solution ( Use Smith chart)
1. Because the normalized load impedance ZL 2-j1
inside the 1jx circle, so case 1 network
is select. 2. jB close to ZL, so ZL ? YL. 3. Move
YL to 1jx admittance circle, jB j 0.3, where YL
? 0.4j 0.5. 4. Then YL ? ZL, ZL ? 1j 1.2. So jX
j 1.2. 5. Impedance identity method derives jB
j 0.29 and jX j 1.22.

• ?

6. Solution 2 uses jB -j 0.7, where YL ? 0.4-j
0.5. 7. Then YL ? ZL, ZL ? 1-j 1.2. So jX -j
1.2. 8. Impedance identity method derives jB -j
0.69 and jX -j 1.22.
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-0.7
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• Use resonator method (Case 1 RSlt1/GL)

Goal ZinRs ?S11(Zin-Rs)/(ZinRs)0
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• Case 2 ZL outside the 1jx circle (RLltZ0)

• Use admittance identity method

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• Use resonator method (Case 2 RSgtRL)

Goal Yin1/Rs ?S11 0
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• Matching Bandwidth

Series-to-Parallel Transformation
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• Case 1 ZL inside the 1jx circle (RLgtZ0)

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• Define QL and Qin for RLC resonator

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Similarly, for case 2 ZL inside the 1jx circle
(RLgtZ0).
Summary
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Example5.2 Design an L-section matching network
to match load impedance RL 2000?, to a RS
50?, at a frequency of 100 MHz?
Solution
Because RS lt1/ GL, so case 1 network is select.

• ?

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?S11
?S22
BW34
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• Three Elements Matching (High-Q Matching)

• Use resonator method for complex load impedance.
• Splitting into two L-shape matching networks.

• Case A ?- shape matching

• Case B T- shape matching

Goal Zin(??0)RS , ?(??0)0
Goal Zin(??0)RS , ?(??0)0
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• Case A ?- shape matching

Conditions RVlt1/GL , RVltRS
P.S. RV Virtual resistance
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• Case B T- shape matching

Conditions RVgtRL , RVgtRS
P.S. RV Virtual resistance
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Example5.3 Design a three elements matching
network to match load impedance RL 2000?, to a
RS 50?, at a frequency of 100 MHz, and to have
BWlt5?
Solution
Case A ?- shape matching
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Splitting into two L-shape matching networks
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Four solutions for ?-shape matching networks
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BW4
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Case B T- shape matching
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Splitting into two L-shape matching networks
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Four solutions for T-shape matching networks
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BW4
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• Cascaded L-Shape Matching (Low-Q Matching)

• Use resonator method for complex load impedance.
• Splitting into two L-shape matching networks.
• Low Q value but large bandwidth.

• Case A

• Case B

Conditions RLlt RVltRS Goal Zin(??0)RS ,
?(??0)0
Conditions 1/GLgt RVgtRS Goal Zin(??0)RS ,
?(??0)0
P.S. RV Virtual resistance
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Splitting into two L-shape matching networks
for case A
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Splitting into two L-shape matching networks
for case B
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Example5.4 Design a cascaded L-shape matching
network to match load impedance RL 2000?, to a
RS 50?, at a frequency of 100 MHz, and to have
BW?60?
Solution
Select 1/GLgt RVgtRS
Splitting into two L-shape matching networks
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Four solutions for Cascaded L-shape matching
networks
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BW61
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• Multiple L-Shape Matching

• Lower Q value but larger bandwidth follows the
  number of L-section increased.

• Conclusions

• Lossless matching networks consist of inductances
  and capacitances but not resistances to avoid
  power loss.
• Four kinds of matching techniques including
  L-shape, ?-shape, T-shape, and cascaded L-shape
  networks can be adopted. Generally larger Q value
  will lead to lower bandwidth.
• A large range of frequencies (gt 1GHz) and circuit
  size may not be realizable.

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Transmission-Line Elements Matching

• Single-Stub Matching

• Easy fabrication in microstrip or stripline form,
  where open-circuit stub is preferable. While
  short-circuit stub is preferable for coax or
  waveguide.
• Lumped elements are not required.
• Two adjustable parameters are the distance d and
  the value of susceptance or reactance provided by
  the shunt or series stub.

• Shunt Stub

• Series Stub

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Example5.5 Design two single-stub (short
circuit) shunt tuning networks to match this load
ZL 60?-j 80? to a 50? line, at a frequency of 2
GHz?
Solution
1. The normalized load impedance ZL 1.2-j1.6.
2. SWR circle intersects the 1jb circle at both
points y1 1.0j1.47 y2 1.0-j1.47.
Reading WTG can obtain d1
0.176-0.0650.11? d2 0.325-0.0650.26?. 3.
The stub length for tuning y1 to 1 requires
l1 0.095?, and for tuning y1 to 1 needs l2
0.405?.
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1. ZL 60?-j 80? at 2 GHz can find R
60?,C0.995pF. 2. Solution 1 is better than
solution 2 this is because both d1 and l1 are
shorter for solution, which reduces the frequency
variation of the match.
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• Analytic Solution for Shunt Stub

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Problem 1 Repeat example 5.5 using analytic
solution.
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Example5.6 Design two single-stub (open circuit)
series tuning networks to match this load ZL
100?j 80? to a 50? line, at a frequency of 2 GHz?
Solution
1. The normalized load impedance ZL 2-j1.6. 2.
SWR circle intersects the 1jx circle at both
points z1 1.0-j1.33 z2 1.0j1.33.
Reading WTG can obtain d1
0.328-0.2080.12? d2 0.672-0.2080.463?. 3.
The stub length for tuning z1 to 1 requires
l1 0.397?, and for tuning z1 to 1 needs l2
0.103?.
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1. ZL 100?j 80? at 2 GHz can find R
100?,L6.37nH.
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• Analytic Solution for Series Stub

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Problem 2 Repeat example 5.6 using analytic
solution.
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• Double-Stub Matching

adjustable tuning

• Variable length of length d between load and stub
  to have adjustable tuning between load and the
  first stub.
• Shunt stubs are easier to implement in practice
  than series stubs.
• In practice, stub spacing is chosen as ?/8 or
  3?/8 and far away 0 or ?/2 to reduce frequency
  sensitive.

• Original circuit

• Equivalent circuit

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• Disadvantage is the double-stub tuner cannot
  match all load impedances. The shaded region
  forms a forbidden range of load admittances.
• Two possible solutions
• b1,b2 and b1,b2 with the same distance d.

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Example5.7 Design a double-stub (open circuit)
shunt tuning networks to match this load ZL
60?-j 80? to a 50? line, at a frequency of 2 GHz?
Solution
1. The normalized load impedance YL 0.3j0.4
(ZL 1.2-j1.6). 2. Rotating ?/8 toward the load
(WTL) to construct 1jb circle can find two
values of first stub b1 1.314 b1
-0.114. 3. Rotating ?/8 toward the generator
(WTG) can obtain y2 1-j3.38 y2 1j1.38.
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4. The susceptance of the second stubs should be
b2 3.38 b2 -1.38. 5. The lengyh of
the open-circuited stubs are found as l1
0.146?, l2 0.482?, or l1 0.204?,
l2 0.350?. 6.ZL 60?-j 80? at 2 GHz can
find R 60?, C0.995pF.
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• Analytic Solution for Double Stub

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Problem 3 Repeat example 5.7 using analytic
solution.
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• Quarter-Wave transformer

• It can only match a real load impedance.
• The length l ?/4 at design frequency f0.
• The important characteristics

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Example5.8 Design a quarter-wave matching
transformer to match a 10? load to a 50? line?
Determine the percent bandwidth for SWR?1.5?
Solution

• ?

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Binomial Multi-section Matching

• The passband response of a binomial matching
  transformer is optimum to have as flat as
  possible near the design frequency, and is known
  as maximally flat.
• The important characteristics

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• Binomial Transformer Design

• If ZLltZ0, the results should be reversed with Z1
  starting at the end.

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Example5.9 Design a three-section binomial
transformer to match a 50? load to a 100? line?
And calculate the bandwidth for ?m0.05?
Solution

• ?

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• Using table design for N3 and ZL/Z02(reverse)
  can find coefficient as 1.8337, 1.4142, and
  1.0907.

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Chebyshev Multi-section Matching

• The Chebyshev transformer is optimum bandwidth to
  allow ripple within the passband response, and is
  known as equally ripple.
• Larger bandwidth than that of binomial matching.
• The Chebyshev characteristics

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• Chebyshev Transformer Design

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Example5.10 Design a three-section Chebyshev
transformer to match a 100? load to a 50? line,
with ?m0.05?
Solution

• ?

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• Using table design for N3 and ZL/Z02 can find
  coefficient as 1.1475, 1.4142, and 1.7429. So
  Z157.37?, Z270.71?, and Z387.15?.

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Tapered Lines Matching

• The line can be continuously tapered instead of
  discrete multiple sections to achieve broadband
  matching.
• Changing the type of line can obtain different
  passband characteristics.
• Relation between characteristic impedance
• and reflection coefficient

• Three type of tapered line
• will be considered here
• 1) Exponential
• 2)Triangular
• 3) Klopfenstein

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Exponential Taper

• The length (L)of line should be greater than
  ?/2(?lgt?) to minimize the mismatch at low
  frequency.

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Triangular Taper

• The peaks of the triangular taper are lower than
  the corresponding peaks of the exponential case.
• First zero occurs at ?l2?

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Klopfenstein Taper

• For a maximum reflection coefficient
  specification in the passband, the Klopfenstein
  taper yields the shortest matching section
  (optimum sense).
• The impedance taper has steps at z0 and L, and
  so does not smoothly join the source and load
  impedances.

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Example5.11 Design a triangular, exponential,
and Klopfenstein tapers to match a 50? load to a
100? line?
Solution

• Triangular taper

• ?

• Exponential taper

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• Klopfenstein taper

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Bode-Fano Criterion

• The criterion gives a theoretical limit on the
  minimum reflection magnitude (or optimum result)
  for an arbitrary matching network
• The criterion provide the upper limit of
  performance to tradeoff among reflection
  coefficient, bandwidth, and network complexity.
• For example, if the response ( as the left hand
  side of next page) is needed to be synthesized,
  its function is given by applied the criterion of
  parallel RC

• For a given load, broader bandwidth ??, higher
  ?m.
• ?m ? 0 unless ??o. Thus a perfect match can be
  achieved only at a finite number of frequencies.
• As R and/or C increases, the quality of the match
  (?? and/or ?m) must decrease. Thus higher-Q
  circuits are intrinsically harder to match than
  are lower-Q circuits.

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