1. CHAPTER 4
TRANSMISSION
LINES

2. TRANSMISSION LINES 4.1 INTRODUCTION 4.2 SMITH CHART
4.3 IMPEDANCE MATCHING

3. 4.1 INTRODUCTION One of electromagnetic theory application:
Power lines, telephone lines and cable TV.
Develop equation for wave propagation on a transmission lines.
Introduce Smith chart for the study of transmission lines and use it to develop impedance matching networks.

4. INTRODUCTION (Cont’d)
A sinusoidal voltage is dropped across the resistor.
If the supply and resistor is connected by an ideal (negligible length) conductor, it will be in same phase.

5. INTRODUCTION (Cont’d)
When a quarter wavelength is added between the supply and the resistor, the voltage at the resistor is 900 out of phase with the supply voltage.

6. INTRODUCTION (Cont’d)
Transmission lines examples:
Twin lead
Coaxial Cable

7. INTRODUCTION (Cont’d)
Microstrip

8. INTRODUCTION (Cont’d)
The distributed parameters for a differential segment of transmission line.
R’ (resistance/meter) L’ (inductance/meter)
G’ (conductance/meter) C’ (capacitance/meter)
* Primes indicate per unit length or distributed values.

9. INTRODUCTION (Cont’d)
For example, the distributed parameters for a coaxial cable can be determined by using these formulas:
Where,
Conductance of dielectric
Conductor conductivity

10. INTRODUCTION (Cont’d)
If the transmission line is modeled using instantaneous voltage and current,

11. INTRODUCTION (Cont’d)
Telegraphist’s Equation is used to determine the basic characteristics for transmission line, which are:
Propagation constant
Characteristic impedance
Characteristic impedance for
lossless line

12. INTRODUCTION (Cont’d)
Section of transmission line for attenuation calculation:

13. INTRODUCTION (Cont’d)
The power ratio can be expressed as a gain G(dB) on logarithmic scale, called decibel scale.
Or in G(dBm) scale, where to represent absolute power levels with reference to 1mW

14. INTRODUCTION (Cont’d)
Most of practical problems involving transmission lines relate to what happens when the line is terminated.

15. INTRODUCTION (Cont’d)
The load is simply the ratio of the voltage and the current at the load.
It can be rearrange as:

16. INTRODUCTION (Cont’d)
If the load is unequal to the characteristic impedance of the line, the wave must be reflected back to the load. The degree of impedance mismatch is represented by the reflection coefficient at the load,
Range from 0 to 1

17. INTRODUCTION (Cont’d)
Generally, the reflection coefficient at any point along transmission line is ratio of the reflected wave to incident wave. Superposition of this two waves creates a standing wave pattern or voltage standing wave ratio (VSWR) or ratio of max. to min. voltage amplitude.
Range from 1 to ∞

18. INTRODUCTION (Cont’d)

19. INTRODUCTION (Cont’d)
At any point along the transmission line, we can find the ratio of the total voltage to the total current.

20. INTRODUCTION (Cont’d)
It is known as input impedance,
For a special lossless case, it becomes:

21. EXAMPLE 1 A source with 50  source impedance drives
a 50  transmission line that is 1/8 of
wavelength long, terminated in a load
ZL = 50 – j25 . Calculate:
(i) The reflection coefficient, ГL
(ii) VSWR
(iii) The input impedance seen by the source.

22. SOLUTION TO EXAMPLE 1
It can be shown as:

23. SOLUTION TO EXAMPLE 1 (Cont’d)
(i) The reflection coefficient,
(ii) VSWR

24. SOLUTION TO EXAMPLE 1 (Cont’d)
(iii) The input impedance seen by the source, Zin
Need to calculate
Therefore,

25. 4.2 SMITH CHART

26. SMITH CHART (Cont’d)
Graphical tool for use with transmission line circuits and microwave circuit elements.
Only lossless transmission line will be considered.
Two graphs in one ;
Plots normalized impedance at any point.
Plots reflection coefficient at any point.

27. SMITH CHART (Cont’d)
The transmission line calculator, commonly referred as the Smith Chart

28. HOW TO USE SMITH CHART?
The Smith Chart is a plot of normalized impedance. For example, if a Z0 = 50 Ω transmission line is terminated in a load ZL = j100 Ω as below:

29. SMITH CHART (Cont’d)
To locate this point on Smith Chart, normalize the load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2 Ω

30. SMITH CHART (Cont’d)
The normalized load impedance is located at the intersection of the r =1 circle and the x =+2 circle.

31. SMITH CHART (Cont’d)
The reflection coefficient has a magnitude and an angle :
Where the magnitude can be measured using a scale for magnitude of reflection coefficient provided below the Smith Chart, and the angle is indicated on the angle of reflection coefficient scale shown outside the circle on chart.

32. SMITH CHART (Cont’d) Scale for magnitude of reflection coefficient
Scale for angle of reflection coefficient

33. SMITH CHART (Cont’d)
For this example,

34. SMITH CHART (Cont’d)
After locating the normalized impedance point, draw the constant circle. For example, the line is 0.3λ length:

35. SMITH CHART (Cont’d)
Move along the constant circle is akin to moving along the transmission line.
Moving away from the load (towards generator) corresponds to moving in the clockwise direction on the Smith Chart.
Moving towards the load corresponds to moving in the anti-clockwise direction on the Smith Chart.

36. SMITH CHART (Cont’d) To find ZIN, move towards the generator by:
Drawing a line from the center of chart to outside Wavelengths Toward Generator (WTG) scale, to get starting point a at 0.188λ
Adding 0.3λ moves along the constant circle to 0.488λ on the WTG scale.
Read the corresponding normalized input impedance point c, ZNIN = j0.08Ω

37. SMITH CHART (Cont’d) Denormalizing, to find an input impedance,
VSWR is at point b,

38. SMITH CHART (Cont’d)
For Z0 = 50Ω , a  ZL = 0 (short cct) b  ZL = ∞ (open cct) c  ZL = j100 Ω d  ZL = j100 Ω e  ZL = 50 Ω

39. Take out your Smith Chart, pencil and compass!
LETS TRY!!

40. EXAMPLE 2
Repeat Example 1 using the Smith Chart.

41. SOLUTION TO EXAMPLE 2
Locate the normalized load, and label it as point a, where it corresponds to
Draw constant circle.
It can be seen that
and

42. SOLUTION TO EXAMPLE 2 (Cont’d)
Move from point a (at 0.356λ) on the WTG scale, clockwise toward generator a distance λ/8 or 0.125λ to point b, which is at 0.481λ.
We could find that at this point, it corresponds to
Denormalizing it,

44. EXAMPLE 3
The input impedance for a 100 Ω lossless transmission line of length λ is measured as 12 + j42Ω. Determine the load impedance.

45. SOLUTION TO EXAMPLE 3 Normalize the input impedance:
Locate the normalized input impedance and
label it as point a

46. SOLUTION TO EXAMPLE 3 (Cont’d)
Take note the value of wavelength for point a at
WTL scale.
At point a, WTL = 0.436λ
Move a distance 1.162λ towards the load to point b
WTL = 0.436λ λ
= 1.598λ
But, to plot point b, 1.598λ – 1.500λ = 0.098λ
Note: One complete rotation of WTL/WTG = 0.5λ

47. SOLUTION TO EXAMPLE 3 (Cont’d)
(v) Read the point b:
Denormalized it:

49. EXAMPLE 4
On a 50  lossless transmission line, the VSWR is measured as 3.4. A voltage maximum is located 0.079λ away from the load (towards generator). Determine the load.

50. SOLUTION TO EXAMPLE 4 Use the given VSWR to draw a constant circle.
Then move from maximum voltage at WTG = 0.250λ (towards the load) to point a at WTG = 0.250λ λ = 0.171λ.
At this point we have ZNL = 1 + j1.3 Ω,
or ZL = 50 + j65 Ω.

52. EXAMPLE 5 (TRY THIS!)
Use Smith Chart to determine the input impedance Zin of the two line configuration shown as below:

53. ANSWER FOR EXAMPLE 5
ZIN = 65.7 – j 124.7Ω

54. 4.3 IMPEDANCE MATCHING
The transmission line is said to be matched when Z0 = ZL which no reflection occurs.
The purpose of matching network is to transform the load impedance ZL such that the input impedance Zin looking into the network is equal to Z0 of the transmission line.

55. IMPEDANCE MATCHING (Cont’d)
Adding an impedance matching networks ensures that all power make it or delivered to the load.

56. IMPEDANCE MATCHING (Cont’d)
Techniques of impedance matching :
Quarter-wave transformer
Single / double stub tuner
Lumped element tuner
Multi-section transformer

57. QUARTER WAVE TRANSFORMER
The quarter wave transformer matching network only can be constructed if the load impedance is all real (no reactive component)

58. QUARTER WAVE TRANSF. (Cont’d)
To find the impedance looking into the quarter wave long section of lossless ZS impedance line terminated in a resistive load RL:
But, for quarter wavelength,

59. QUARTER WAVE TRANSF. (Cont’d)
So,
Rearrange to get impedance matched line,

60. IMPEDANCE MATCHING (Cont’d)
It much more convenient to add shunt elements rather than series elements  Easier to work in terms of admittances.
Admittance:

61. IMPEDANCE MATCHING (Cont’d)
Adding shunt elements using admittances:
With Smith chart, it is easy to find normalized admittance – move to a point on the opposite side of the constant circle.

63. SHUNT STUB MATCHING NETWORK
The matching network has to transform the real part of load impedance, RL to Z0 and reactive part, XL to zero  Use two adjustable parameters – e.g. shunt-stub.

64. SHUNT STUB MATCHING NET. (Cont’d)
Thus, the main idea of shunt stub matching network is to:
(i) Find length d and l in order to get yd and yl .
(ii) Ensure total admittance ytot = yd + yl = 1 for complete matching network.

65. SHUNT STUB USING SMITH CHART
Locate the normalized load impedance ZNL.
Draw constant SWR circle and locate YNL.
Move clockwise (WTG) along circle to intersect with 1 ± jB  value of yd.
The length moved from YNL towards yd is the through line length, d.
Locate yl at the point jB .
Depends on the shorted/open stub, move along the periphery of the chart towards yl (WTG).
The distance traveled is the length of stub, l .

66. SHORTED SHUNT STUB MATCHING
Generic layout of the shorted shunt stub matching network:

67. EXAMPLE 6 (TRY THIS!)
Construct the shorted shunt stub matching network for a 50Ω line terminated in a load ZL = 20 – j55Ω

68. SOLUTION TO EXAMPLE 6 Locate the normalized load impedance,
ZNL = ZL/Z0 = 0.4 – j1.1Ω
Draw constant circle.
Locate YNL. (0.112λ at WTG)
Moving to the first intersection with the
1 ± jB circle, which is at 1 + j2.0  yd
5. Get the value of through line length, d
 from 0.112λ to 0.187λ on the WTG scale,
so d = 0.075λ

69. SOLUTION TO EXAMPLE 6 (Cont’d)
6. Locate the location of short on the Smith Chart
(note: when short circuit, ZL = 0, hence YL = ∞)
on the right side of the chart with WTG=0.25λ
Move clockwise (WTG) until point jB, which
is at 0 - j2.0, located at WTG= 0.324λ  yl
8. Determine the stub length, l
 0.324λ – 0.25λ = λ

70. SOLUTION TO EXAMPLE 6 (Cont’d)
Thus, the values are:
d = λ
l = λ
yd = 1 + j2.0 Ω
yl = -j2.0 Ω
Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1

72. OPEN END SHUNT STUB MATCHING
Generic layout of the open ended shunt stub matching network:

73. EXAMPLE 7 (TRY THIS!)
Construct an open ended shunt stub matching network for a 50Ω line terminated in a load ZL = j100 Ω

74. SOLUTION TO EXAMPLE 7 Locate the normalized load impedance,
ZNL = ZL/Z0 = j2.0Ω
Draw constant circle.
Locate YNL. (0.474λ at WTG)
Moving to the first intersection with the
1 ± jB circle, which is at 1 + j1.6  yd
5. Get the value of through line length, d
 from 0.474λ to 0.178λ on the WTG scale,
so d = 0.204λ

75. SOLUTION TO EXAMPLE 7 (Cont’d)
6. Locate the location of open end on the Smith
Chart (note: when short circuit, ZL = ∞, hence
YL = 0) on the left side with WTG = 0.00λ
Move clockwise (WTG) until point jB, which
is at 0 – j1.6, located at WTG= 0.339λ  yl
8. Determine the stub length, l
 0.339λ – 0.00λ = λ

76. SOLUTION TO EXAMPLE 7 (Cont’d)
Thus, the values are:
d = λ
l = λ
yd = 1 + j1.6 Ω
yl = -j1.6 Ω
Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1

78. IMPORTANT!!
In both previous example, we chose the first intersection with the1 ± jB circle in designing our matching network. We could also have continued on to the second intersection.
Thus, try both intersection to determine which solution produces max/min length of through line, d or length of stub, l.

79. EXERCISE (TRY THIS!)
Determine the through line length and stub length for both example above by using second intersection.
For shorted shunt stub (example 6):
d = 0.2 λ and l = λ
For open ended shunt stub (example 7):
d = λ and l = λ

80. CHAPTER 4
END