1. Impedance Matching and Tuning
Chapter 5
Impedance Matching and Tuning

2. 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”.

3. 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.

4. Lumped Elements Matching
L-Shape (Two-Element) Matching
Case 1: ZL inside the 1+jx circle (RL>Z0)
Use impedance identity method

5. Solution ( Use Smith chart)
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 1+jx circle,
so case 1 network is select.
2. jB close to ZL, so ZL  YL.
3. Move YL to 1+jx admittance circle, jB =j 0.3, where YL  0.4+j 0.5.
4. Then YL  ZL, ZL  1+j 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.

6. -0.7

8. Use resonator method (Case 1: RS<1/GL)
Goal: Zin=Rs =S11=(Zin-Rs)/(Zin+Rs)=0

10. Case 2: ZL outside the 1+jx circle (RL<Z0)
Use admittance identity method

11. Use resonator method (Case 2: RS>RL)
Goal: Yin=1/Rs =S11= =0

13. Matching Bandwidth
Series-to-Parallel Transformation

14. Case 1: ZL inside the 1+jx circle (RL>Z0)

15. Define QL and Qin for RLC resonator

16. Similarly, for case 2: ZL inside the 1+jx circle (RL>Z0).
Summary

17. 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 <1/ GL, so case 1 network is select. 

18. =S11
=S22
BW=34%

19. 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

20. Conditions: RV<1/GL , RV<RS
Case A: - shape matching
Conditions: RV<1/GL , RV<RS
P.S. RV : Virtual resistance

21. Case B: T- shape matching
Conditions: RV>RL , RV>RS
P.S. RV : Virtual resistance

22. 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 BW<5%?
Solution
Case A: - shape matching

23. Splitting into two “L-shape” matching networks

24. Four solutions for -shape matching networks

25. BW=4%

26. Case B: T- shape matching

27. Splitting into two “L-shape” matching networks

28. Four solutions for T-shape matching networks

29. BW=4%

30. 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: RL< RV<RS
Goal: Zin(=0)=RS , (=0)=0
Conditions: 1/GL> RV>RS
Goal: Zin(=0)=RS , (=0)=0
P.S. RV : Virtual resistance

31. Splitting into two “L-shape” matching networks for case A

32. Splitting into two “L-shape” matching networks for case B

33. Splitting into two “L-shape” matching networks
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/GL> RV>RS
Splitting into two “L-shape” matching networks

35. Four solutions for Cascaded L-shape matching networks

36. BW=61%

37. 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 (> 1GHz) and circuit size may not be realizable.

38. 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

39. 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 1+jb circle at both points
y1 = 1.0+j1.47
y2 = 1.0-j1.47.
Reading WTG can obtain:
d1= =0.11
d2= =0.26.
3. The stub length for tuning y1 to 1 requires
l1 = 0.095,
and for tuning y1 to 1 needs
l2 = 0.405.

40. 1. ZL = 60-j 80 at 2 GHz can find R= 60,C=0.995pF.
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.

41. Analytic Solution for Shunt Stub

42. Problem 1: Repeat example 5.5 using analytic solution.

43. 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 1+jx circle at both points
z1 = 1.0-j1.33
z2 = 1.0+j1.33.
Reading WTG can obtain:
d1= =0.12
d2= =0.463.
3. The stub length for tuning z1 to 1 requires
l1 = 0.397,
and for tuning z1 to 1 needs
l2 = 0.103.

44. 1. ZL = 100+j 80 at 2 GHz can find R= 100,L=6.37nH.

45. Analytic Solution for Series Stub

46. Problem 2: Repeat example 5.6 using analytic solution.

47. Shunt stubs are easier to implement in practice than series stubs.
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

48. Disadvantage is the double-stub tuner cannot match all load impedances
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.

49. 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= 0.3+j0.4 (ZL= 1.2-j1.6).
2. Rotating /8 toward the load (WTL) to construct 1+jb circle can find two values of first stub
b1 = 1.314
b’1 =
3. Rotating /8 toward the generator (WTG) can obtain
y2= 1-j3.38
y’2= 1+j1.38.

50. 4. The susceptance of the second stubs should be
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, C=0.995pF.

51. Analytic Solution for Double Stub

52. Problem 3: Repeat example 5.7 using analytic solution.

53. Quarter-Wave transformer
It can only match a real load impedance.
The length l= /4 at design frequency f0.
The important characteristics

54. 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 

55. 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

56. Binomial Transformer Design
If ZL<Z0, the results should be reversed with Z1 starting at the end.

57. Example5.9: Design a three-section binomial transformer to match a 50 load to a 100 line? And calculate the bandwidth for m=0.05?
Solution 

58. Using table design for N=3 and ZL/Z0=2(reverse) can find coefficient as 1.8337, 1.4142, and 1.0907.

59. 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

61. Chebyshev Transformer Design

62. Example5.10: Design a three-section Chebyshev transformer to match a 100 load to a 50 line, with m=0.05?
Solution 

64. Using table design for N=3 and ZL/Z0=2 can find coefficient as 1
Using table design for N=3 and ZL/Z0=2 can find coefficient as , , and So Z1=57.37, Z2=70.71, and Z3=87.15.

65. 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

66. Exponential Taper
The length (L)of line should be greater than /2(l>) to minimize the mismatch at low frequency.

67. Triangular Taper
The peaks of the triangular taper are lower than the corresponding peaks of the exponential case.
First zero occurs at l=2

68. 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 z=0 and L, and so does not smoothly join the source and load impedances.

69. Example5.11: Design a triangular, exponential, and Klopfenstein tapers to match a 50 load to a 100 line?
Solution
Triangular taper 
Exponential taper

70. Klopfenstein taper

71. 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.