1. Chapter 3 Matching and TuningBy
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
©Prof Syed Idris Syed Hassan

2. Why matching or tuning is important?
To maximize power delivery and minimize power loss.
To improve signal to noise ratio as in sensitive receiver components such as LNA, antenna, etc.
To reduce amplitude and phase error as in distributed network such as antenna array.
Basic idea of impedance matching

3. Concept of maximum power transfer
In lump circuit
Power deliver at ZL is
Power maximum whence ZL = Zo

4. continue In transmission line
The important parameter is reflection coefficient
No reflection whence ZL = Zo , hence
The load ZL can be matched as long as ZL not equal to zero (short-circuit) or infinity (open-circuit)

5. Factors in selecting matching network
Complexity: simpler, cheaper, more reliable and low loss circuit is preferred.
Bandwidth: match over a desirable bandwidth.
Implementation: depend on types of transmission line either cable, stripline, microstripline, waveguide, lump circuit etc.
Adjustability:some network may need adjustment to match a variable load.

6. Matching with lumped elements
The simplest matching network is an L-section using two reactive elements
Configuration 1
Whence RL>Zo
ZL=RL+jXL
Configuration 2
Whence RL<Zo

7. continue
If the load impedance (normalized) lies in unity circle, configuration 1 is used.Otherwise configuration 2 is used.
The reactive elements are either inductors or capacitors. So there are 8 possibilities for matching circuit for various load impedances.
Configuration 2
Configuration 1
Matching by lumped elements are possible for frequency below 1 GHz or for higher frequency in integrated circuit(MIC, MEM).

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

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

10. Lumped elements for microwave integrated circuit
Planar resistor
Chip resistor
Loop inductor
Spiral inductor
Interdigital gap capacitor
Metal-insulator-metal capacitor
Chip capacitor

11. Matching by calculation for configuration 1
For matching, the total impedance of L-section plus ZL should equal to Zo,thus
Rearranging and separating into real and imaginary parts gives us * **

12. continue
Solving for X from simultaneous equations (*) and (**) and substitute X in (**) for B, we obtain
+ve capacitor
-ve inductor
Since RL>Zo, then argument of the second root is always positive, the series reactance can be found as
+ve inductor
-ve capacitor
Note that two solution for B are possible either positive or negative

13. Matching by calculation for configuration 2
For matching, the total impedance of L-section plus ZL should equal to 1/Zo,thus
Rearranging and separating into real and imaginary parts gives us * **

14. continue
Solving for X and B from simultaneous equations (*) and (**) , we obtain
+ve inductor
-ve capacitor
+ve capacitor
-ve inductor
Since RL<Zo,the argument of the square roots are always positive, again two solution for X and B are possible either positive or negative

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

16. Serial L
Matching using lumped
components
Parallel C
Serial C
10W
50W
Parallel L

17. 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

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

19. Single stub-matching Parallel configuration Short-stub matching
Open-stub matching

20. Example Solution Smith Chart
Design two single–stub shunt tuning networks to match a load ZL=15+j10W to 50W at 2GHz. The load consists of a resistor and inductor in series. Plot the reflection magnitude from 1 GHz to 3 GHz for each solution.
Solution
Normalized the load zL=0.3+j0.2
Construct SWR circle and convert load to admittance
Then move from load to generator to meet a circle (1+jb) to obtain two points i.e y1=1-j1.33 and y2=1+j1.33.
The distance d from the load to stub is obtained either of these two points i.e. d1= 0.044l and d2=0.387l.
To improve bandwidth, the stub is chosen as close as possible to the load.
The tuning requires a stub of j1.33 for y1 and –j1.33 for y2.
For open circuit-stub i1 =0.147l and i2=0.353l.
Smith Chart

21. Continue Convert j0.2 to inductance value, thus
Use this value to calculate reflection coefficient

22. Formulas for calculation
Let ZL=RL+jXL, then the impedance at distance d from the load is
Admittance at this point is
Thus, by substituting Zd and separating real and imaginary, we have

23. continue
Equating G = Yo = 1/Zo,and , thus we have
Solving

24. continue
The two principle solution are
To find the stub length,

25. Single stub-matching Serial configuration Short-stub matching i
Open-stub matching i

26. Example Solution Smith Chart
Match a load impedance of ZL=100+j80 to a 50 W line using a single series open-stub.Assuming the load consists of resistor and inductor in series at 2GHz. Plot the reflection coefficient from 1 GHz to 3 GHz.
Solution
Normalized the load zL=2+j1.6
Construct SWR circle
Then move from load to generator to obtain two points on unity circle(1+jx) z1=1-j1.33 and z2=1+j1.33.
The distance d from the load to stub is obtained either of these two points i.e. d1= 0.120l and d2=0.463l.
To improve bandwidth, the stub is chosen as close as possible to the load.
The tuning requires a stub of j1.33 for z1 and –j1.33 for z2.
For open circuit-stub i1 =0.397l and i2=0.103l.
Smith Chart

27. Continue Convert j1.6 to inductance value, thus
Use this value to calculate reflection coefficient

28. Formulas for calculation
Let YL=GL+jBL, then the load admittance at distance d from the load is
Where t = tan bd
Impedance at this point is
Thus, by substituting Yd and separating real and imaginary, we have

29. continue
Equating R = Zo = 1/Yo , thus we have
Solving

30. continue
The two principle solution are
To find the stub length,

31. Double-stub matching
Open or short stubs
The advantage of this technique is the position of stubs
( d and x) are fixed. The matching are done by changing
the length of stubs. The disadvantage of this technique
is not all impedances can be matched.

32. A A’ B B’
A= load admittance
A’=admittance of A at
stub 2
B= Adjust of stub S2 to
bring to the S2 circle
B’=Admittance of B at S1
Then by adjusting stub 1
will bring the admittance to the center of the chart

33. Example
Design a double-stub shunt tuner to match a load ZL=60-j80 to a 50 W line. The stubs are to be short circuited stubs, and are spaced l/8 apart. The load consists of a series resistor and capacitor that match at 2 GHz. Plot the reflection coefficient magnitude versus frequency from 1 GHz to 3GHz.
Solution
Plot normalized load zL=1.2 –j 1.6 W and convert to admittance we have yL= 0.3 +j0.4.
Construct a rotated 1+ j b circle which is a distance d=l/8 from a 1+jb circle. Get two possible points on the rotated 1+jb circle, y1 and y1’ by adding susceptance of the first stubs. In this case we take x=0(match section immediate after the load . I.e b1=1.314 and b1’= The length of stub will be i1=0.396l and i1’=0.232l.
Now transform both point onto 1+jb circle along SWR circles.This bring two solution y2 =1-j3.38 and y2’=1+j1.38.
Then the second stub should be b2= 3.38 and b2’= The length of stub will be i2=0.454l and i2’=0.100l.

34. continue yL y1
y1’ b1
b1’
y2’ y2
b2’ b2
Rotated 1+jb circle

35. continue
The shorter the stubs the wider will be the bandwidth

36. Formulas for calculation
Let YL=GL+jBL, then the load admittance at distance x from the load is
x is distance between stub and load
Admittance at first stub
After transforming to stub 2 , we have
Where t = tan bd

37. continue At this point the ral part of Y2 =Yo ,thus Solving
Since GL’ is real, the quantity in square root must be nonnegative, thus
Simplified to

38. continue So susceptances of stubs are and To find the stub length ,
B either B1 or B2

39. continue
The two principle solution are
To find the stub length,

40. Graphical method
Suitable for load impedance laying inside the unity circle.
Plot ZL normalised to 50 ohm
Find bisector and point intersect the chart axis (I.e A)
Draw circle at center A touching the center of the chart
Obtain R1 and calculated new characteristic impedance of the line
Renormalised the ZL to ZT, thus we have ZL2 . Then plot on the chart.
Obtain x for the length of the matching section by moving towards generator.

41. Graphical method

42. Transmission line transformerZ1
Quarter-wave
transformer
Matching at one frequency .For other frequency the impedance at the input of matching section will be
Where t = tan bd
At matched frequency tan bd = tan (p/2)

43. continue Reflection coefficient Since Zo2=Z1Z2, this reduces to qmqm
p-qm p

44. continue Fractional bandwidth is given by
Reflection coefficient with different ratio of Z1:Z2

45. Example
Design a single section quarter-wave matching transformer to match a 10 W load to a 50 W line at fo =3GHz. Determine the percentage bandwidth for which the SWR < 1.5
For a pure resistive load we can just calculate directly
and the length of transformer is l/4
Then determine reflection coefficient for SWR=1.5
Hence fractional bandwidth is
or 29%

46. Multisection transformer
Binomial multisection
Chebyshev multisection
Two types of transformer
Reflection coefficient for multisection can be written as
Or its magnitude
At center frequency
(i.e =l/4)

47. Binomial transformer
For binomial expansion the reflection coefficient can be written as
where
Note CnN=CN-nN, C0N=1 and C1N=N=CN-1N
Let
then
Solving for
to obtain value of A or
Hence

48. continue
For bandwidth
Therefore
The fractional bandwidth, thus

49. Example
Design a three section binomial transformer to match 50W load to a 100W line
Solution
N=3 , ZL=50W , Z0=100W then
But
Z1=91.7 W
Z2=70.7 W
Z3=54.5 W

50. Chebyshev transformer
Chebyshev polinomial in general
Useful forms of Chebyshev polinomial are

51. continue
For N section,Chebyshev expansion the reflection coefficient can be written as
Solving for
to obtain value of A or
For maximum reflection coefficient magnitude
Therefore or

52. Example
Design a three section Chebyshev transformer to match a 100W load to a 50W line.
Taking Gm=0.05.
Solution * **
Equating * and ** and A=Gm=0.05
cos3q

53. continue
cosq
For symmetrical
and
But
Then
Start n=0
n=1
n=2

54. Simple form of multisection transformers
2-quarter-wave
transformer
3-quarter-wave
transformer

55. Tapered Transmission line transformer
Impedance distribution
Reflection
Exponential Taper
transformer
Triangular Taper
transformer
Klofenstein Taper
transformer
where

56. Transmission line transformer
Nonsynchronous transformer
E.g matching Z1=75 W
to Z2=50 W
qT = 29.3o

58. Complex to Complex Impedance Matching