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

2. TRANSMISSION LINE THEORY PART II
The Smith Chart – Intro
Using the Smith Chart
Reflection Coefficient Mag & Angle
VSWR
Impedance Matching
Quarter-Wave Transformer
Single/double stub Tuner
Lumped element tuner
Multi-section transformer

3. THE SMITH CHART - INTRO The Smith Chart parameters:
The reflection coefficient:
Γ = |Γ|ejθ
(|Γ| ≤ 1), (-180°≤ θ ≤180°)
The normalized impedance:
z = Z / Z0
The normalized admittance:
y = 1 / z
SWR and RL scale
Figure 8: The Smith Chart

4. THE SMITH CHART - INTRO
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.

5. THE SMITH CHART - INTRO
The transmission line calculator, commonly referred as the Smith Chart

6. USING THE 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:

7. USING THE SMITH CHART
To locate this point on Smith Chart, normalize the load impedance, ZNL = ZL/ZN to obtain ZNL = 1 + j2 Ω

8. USING THE SMITH CHART
The normalized load impedance is located at the intersection of the r =1 circle and the x =+2 circle.

9. USING THE SMITH CHART
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.

10. USING THE SMITH CHART Scale for magnitude of reflection coefficient
Scale for angle of reflection coefficient

11. USING THE SMITH CHART
For this example,

12. USING THE SMITH CHART
After locating the normalized impedance point, draw the constant circle. For example, the line is 0.3λ length:

13. USING THE SMITH CHART
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.

14. USING THE SMITH CHART 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

15. USING THE SMITH CHART Denormalizing, to find an input impedance,
VSWR is at point b,

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

17. EXAMPLE 1.5
ZL= 50 - j25 and Z0=50 Ohm. Find Zin, VSWR and ΓL using the Smith Chart.

18. SOLUTION TO EXAMPLE 1.5
Locate the normalized load, and label it as point a, where it corresponds to
Draw constant circle.
It can be seen that
and

19. SOLUTION TO EXAMPLE 1.5 (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,

21. EXAMPLE 1.6
The input impedance for a 100 Ω lossless transmission line of length λ is measured as 12 + j42Ω. Determine the load impedance.

22. SOLUTION TO EXAMPLE 1.6 Normalize the input impedance:
Locate the normalized input impedance and
label it as point a

23. SOLUTION TO EXAMPLE 1.6 (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

24. SOLUTION TO EXAMPLE 1.6 (Cont’d)
(v) Read the point b:
Denormalized it:

26. EXAMPLE 1.7
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.

27. SOLUTION TO EXAMPLE 1.7 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.

29. THE IMPEDANCE MATCHING
The important of impedance matching or tuning:
Maximum power is delivered when the load is matched to the line.
The power loss in the feed line is minimized. (increase power handling capability by optimizing VSWR)
Improved the signal-to-noise ratio (SNR). (e.g with controlled mismatch, an amplifier can operate with minimum noise generation)
Reduced the amplitude and phase errors.

30. THE IMPEDANCE MATCHING
Figure 9: A lossless network matching an arbitrary load impedance to a transmission line.
Factors in the matching network selection:
Complexity
Bandwidth
Implementation
Adjustability

31. 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.

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

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

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

35. 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.

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

38. 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,

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

40. EXAMPLE 1.8.5
Calculate the position and characteristic impedance of a quarter wave transformer that will match a load impedance, RL = 15Ω; to a 50 Ω line.

41. SOLUTION TO EXAMPLE 1.8.5 To get the transformer’s impedance, use
To find the position of quarter-wave transformer from the load:
d = 0.25 λ
ZT = Ω
15 Ω
ZT = Ω
Qwave transformer

42. EXAMPLE 1.8.6
A transistor has an input impedance of ZL = 25 Ω, at an operating frequency of 500 MHz. Find:
The length, l
Width, w
characteristic impedance of the quarter-wave parallel plate line transformer for which matching is achieved.
Assume thickness of the dielectric is d = 1 mm, and the relative dielectric constant is εr = 4. Assume that surface resistance, R and shunt conductance, G, can be neglected.

43. EXAMPLE 1.8.6 We can directly apply that
ℓ=λ/4
Z0=50Ω w
Zline
To transistor Zin = 25Ω
Zin ZL
The characteristic of the line is

44. EXAMPLE 1.8.6 (cont) Parameter 2-wire line Coaxial Line
Parallel Plate Line
Unit R
Ω/m L
H/m G
S/m C
F/m

45. EXAMPLE 1.8.6 (Cont) Thus, the width of the line is
From previous table, we find the values for capacitance and inductance as;

46. EXAMPLE 1.8.6 (Cont) The line length l follows from the condition
The input impedance of the combined transmission line and the load is:
Where d = l = λ/4, and the reflection coefficient is given by

47. EXAMPLE 1.8.6 (Cont) Quarter Wave Impedance Matching
Designed to achieve matching at a single frequency/narrow bandwidth
Easy to build and use

48. STUB MATCHING Stub Matching Single stub or Double Stub
Parallel Stub or Series Stub
Open stub or Shorted Stub

49. SINGLE STUB TUNING
Figure 11: Single-stub tuning circuits. (a) Shunt stub. (b) Series stub.

50. 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.

51. 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.

52. 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 .

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

54. EXAMPLE 1.9
Construct the shorted shunt stub matching network for a 50Ω line terminated in a load ZL = 20 – j55Ω

55. SOLUTION TO EXAMPLE 1.9 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

56. SOLUTION TO EXAMPLE 1.9 (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

57. SOLUTION TO EXAMPLE 1.9 (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

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

60. EXAMPLE 1.10
Construct an open ended shunt stub matching network for a 50Ω line terminated in a load ZL = j100 Ω

61. SOLUTION TO EXAMPLE 1.10 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

62. SOLUTION TO EXAMPLE 1.10 (Cont’d)
6. Locate the location of open end on the Smith
Chart (note: when open 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

63. SOLUTION TO EXAMPLE 1.10 (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

65. 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.

66. EXERCISE (TRY THIS!)
Determine the through line length and stub length for both example above by using second intersection.
For shorted shunt stub (example 1.9):
d = 0.2 λ and l = λ
For open ended shunt stub (example 1.10):
d = λ and l = λ