1. EKT 441 MICROWAVE COMMUNICATIONS
CHAPTER 1:
TRANSMISSION LINE THEORY (PART III)

2. TRANSMISSION LINE THEORY PART III
Impedance Matching
Quarter-Wave Transformer
Single stub Tuner
Lumped element tuner
Multi-section transformer

3. Resistive L Section (Resistors)
Matched condition: a) b)
Solving (a) and (b) to give R1 and R2

4. Resistive L Section (Resistors)
Attenuation

5. Example: Resistive L Section (R)
Design a broadband resistive L section matching network to match a 75Ω antenna output to a 50 Ω transmission line. Calculate the attenuation by the network.

6. Example: Resistive L Section (R)
Solution:
Ro1 = 75 Ω; Ro2 = 50 Ω; substituting these values into previous equation;
R1 = 43.3 Ω; R2 = 86.6 Ω

7. Reactive L Section (L&C)
Matched condition: a) b)
Xs = +ve for L; -ve for C
Xp = -ve for L; +ve for C

8. Reactive L Section (L&C)
Solving (a) and (b) to give Xs and Xp

9. Example: Reactive L Section (L&C)
Design a L-section matching network with L and C to match a 50Ω source to a 600 Ω load for a maximum power transfer at 400MHz. Give both solutions

10. Example: Reactive L Section (L&C)
Solution:
RS = 50 Ω; Rp = 600 Ω
Substituting these values into previous eqs:
Xs = Q.RS = 166 Ω
Xp = Rp/Q = 181 Ω

11. Example: Reactive L Section (L&C)
Solution 1:
Xs = +166Ω (inductive); Xp = -181 Ω (capacitive)

12. Example: Reactive L Section (L&C)
Solution 2:
Xs = -166Ω (capacitive); Xp = +181 Ω (inductive)

13. L Section for Complex Impedance
If RS has a reactance of jX’, simply introduce a series equal but opposite sign reactance (-jX’)
The extra reactance -jX’ is combined with jXs to give its final value

14. L Section for Complex Impedance
Similar treatment can be applied to Rp by introducing a parallel jX’ and a parallel –jX’
Basically there are 2 solutions;
T Section
Pi Section
Advantages
R1 can be < or > R2
Variable Q
More L and C elements

15. T Section
Method
Introduce a hypothetical resistance level Rh at jX2 such that Rh > R1 and Rh > R2 (Rh determines the level of Q)
Split jX2 into jX2’ and jX2’’ (in parallel)
Treat T-section as two L-sections (L-1 & L-2) Rh

16. T Section Q of original L section:
Q of T-section if Rh = 5050 Ω (Note Q1 > Q2)
L1 section:
L2 section: Rh

17. Example – T-section
Design a T-section LC matching network to match R1 = 50 Ω, and R2 = 600 Ω at 400 MHz, and Q associated with R1 is 10. Rh

18. Example – T-section For L1 network, select Q1 = 10 .
-ve for C (arbitrarily selected)
+ve for L (opposite to X2)

19. Example – T-section For L2 network; 1. 2. 3. 4.
+ve for L (arbitrarily selected)
-ve for C (opposite to X2’’)
-ve means C

20. Example – T-section 5. 6. 7.

21. Note: Q of L-1 section > Q of 2-element L section (as Rh < R2)
Pi Section
Method
Introduce a hypothetical resistance level Rh at jX2 such that Rh < R1 and Rh < R2 (Rh determines the level of Q)
Split jX2 into jX2’ and jX2’’ (in series)
Treat Pi section as two L-sections (L-1 & L-2) Rh
Note: Q of L-1 section > Q of 2-element L section (as Rh < R2)

22. Example – Pi-section
Design a π-section LC matching network to match R1 = 50 Ω, and R2 = 600 Ω at 400 MHz, and Q associated with R1 is 10.

23. Example – Pi-section For L1 network, select Q1 = 10 .
+ve for L (arbitrarily selected)
-ve for C (opposite to X2’)

24. Example – Pi-section For L2 network; 1. 2. 3. 4.
+ve for L (arbitrarily selected)
-ve for C (opposite to X2’’)
+ve means L

25. Example – Pi-section 5. 6. 7.

26. Summary Resistive L-Section Reactive (LC) L-Section: T-Section
Advantages: Broadband
Disadvantages: Very lossy
Reactive (LC) L-Section:
A: Low loss, simplicity in design
D: Narrow band, fixed Q
T-Section
A: R1 can be < or > R2; variable Q (higher than L-Section)
D: More LC elements
Pi-Section

27. LUMPED ELEMENTS MATCHING (L NETWORKS)
Figure 10: L-section matching networks. (a) Network for zL inside the 1+jx circle. (b) Network for zL outside the 1+jx circle.

28. Impedances for serial lumped elements
Serial circuit
Reactance relationship values
+ve X=2pfL L=X/(2pf)
-ve X=1/(2pfC) C=1/(2pfX)

29. Impedances for parallel lumped elements
Parallel circuit
Susceptance relationship values
+ve B=2pfC C=B/(2pf)
-ve B=1/(2pfL) L=1/(2pfB)

30. Matching using Smith chart
(+ve)
(+ve)
(-ve)
(-ve)

31. LUMPED ELEMENTS MATCHING (L NETWORKS)
Figure 11: Effect of adding a Series L

32. LUMPED ELEMENTS MATCHING (L NETWORKS)
Figure 12: Effect of adding a Parallel L

33. LUMPED ELEMENTS MATCHING (L NETWORKS)
Figure 13: Effect of adding a parallel C

34. LUMPED ELEMENTS MATCHING (L NETWORKS)
Figure 14: Effect of adding a Series C

35. EXAMPLE 1.11
Given the circuit below, visualize using Smith Chart the changes from RL till Zin (f = 2 GHz)

36. SOLUTION TO EXAMPLE 1.11 In this question, it is given that;
Since we are adding a resistor with a shunt capacitor, we need to work in admittance form. Changing RL to gA (normalized form of GA);

37. SOLUTION TO EXAMPLE 1.11
The capacitance value is given as CL = 1.91 pF. The value of bCL is calculated so that the imaginary part of the impedance (RL + jBL) can be determined
Normalizing BCL to get bCL, BCL need to be multiplied with Z0

38. SOLUTION TO EXAMPLE 1.11
Thus, the total admittance from (G + jBCL) is given as yB = j1.2. Since the next element is a inductance connected serially, we need to work in impedance again. Converting;
The amount of movement from the series inductance can be calculated as follow:

39. SOLUTION TO EXAMPLE 1.11
To add the capacitance value of shunt C, the impedance must first be converted back to admittance;
Then calculating the effect of the shunt C;

40. SOLUTION TO EXAMPLE 1.11
The last element in the circuit is the serially connected L with a value 3.98nH, we need to once again convert to impedance before adding the value of xL to determine the zin value
Then calculating the effect of the series L;

41. SOLUTION TO EXAMPLE 1.11 (Cont’d)

42. Example
Design an L-section matching network to match a series RC load with an impedance ZL=200-j100 W, to a 100 W line, at a frequency of 500 MHz.
Solution
Normalized ZL we have : ZL= 2-j1
Parallel L
(-j0.7)
Serial C
(-j1.2)
ZL= 2-j1
Parallel C
(+j0.3)
Serial L
(j1.2)
Solution 1
Solution 2

43. continue
Solution 2 seems to be better matched at higher frequency

44. MULTISECTION TRANSFORMER
This transformer consist of N equal-length (commensurate) sections of transmission lines.
Figure 12: Partial reflection coefficients for a multisection matching transformer

45. MULTISECTION TRANSFORMER
Partial reflection coefficients can be defined at each junction, as follow:
[36]
[37]
[38]