By Anwer jubba

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

2. 2
•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
TRANSMISSION LINE THEORY
PART II

3. 3
THE SMITH CHART - INTRO
Figure 8: The Smith Chart
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

4. 4
• 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.
THE SMITH CHART - INTRO

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

6. 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 = 50 + j100 Ω as below:

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

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

9. 9
The reflection coefficient has a magnitude
and an angle :
Γ
Γ=Γ θj
L e
LΓ
Γ
θ
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.1=ΓL
USING THE SMITH CHART

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

11. 11
For this example,
0
45
7.0 j
j
L
e
e
=
Γ=Γ Γθ
USING THE SMITH CHART

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

13. 13
• 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.
Γ
Γ θj
L e
USING THE SMITH CHART

14. 14
• 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 = 0.175 - j0.08Ω
Γ
Γ θj
L e
USING THE SMITH CHART

15. 15
Denormalizing, to find
an input impedance,
Ω−=
=
475.8
0
jZ
ZZZ
IN
NININ
USING THE SMITH CHART
VSWR is
at point b,
9.5=VSWR

16. 16
For Z0 = 50Ω ,
a  ZL = 0 (short cct)
b  ZL = ∞ (open cct)
c  ZL = 100 + j100 Ω
d  ZL = 100 - j100 Ω
e  ZL = 50 Ω
USING THE SMITH CHART

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

18. 18
SOLUTION TO EXAMPLE 1.5
(i) Locate the normalized load, and label it as
point a, where it corresponds to
(ii) Draw constant circle.
(iii)It can be seen that
5.01 jZNL −=
Γ
Γ θj
L e
0
76
245.0 −
=Γ j
L e and 66.1=VSWR

19. 19
(iv) 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,
07.062.0 jZNIN −=
Ω−= 5.331 jZIN
SOLUTION TO EXAMPLE 1.5 (Cont’d)

20. 20

21. 21
EXAMPLE 1.6
The input impedance for a 100 Ω lossless
transmission line of length 1.162 λ is
measured as 12 + j42Ω. Determine the
load impedance.

22. 22
SOLUTION TO EXAMPLE 1.6
(i) Normalize the input impedance:
(ii) Locate the normalized input impedance and
label it as point a
42.012.0
100
4212
0
j
j
Z
Z
z in
in
+=
+
==

23. 23
(iii) Take note the value of wavelength for point a at
WTL scale.
At point a, WTL = 0.436λ
(iv) Move a distance 1.162λ towards the load to point b
WTL = 0.436λ + 1.162λ
= 1.598λ
But, to plot point b, 1.598λ – 1.500λ = 0.098λ
Note: One complete rotation of WTL/WTG = 0.5λ
SOLUTION TO EXAMPLE 1.6 (Cont’d)

24. 24
(v) Read the point b:
7.015.0 jZNL −=
Denormalized it:
Ω−=
=
7015
0
j
ZZZ NLL
SOLUTION TO EXAMPLE 1.6 (Cont’d)

25. 25

26. 26
EXAMPLE 1.7
On a 50 Ω 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. 27
(i) Use the given VSWR to draw a constant
circle.
(ii) Then move from maximum voltage at
WTG = 0.250λ (towards the load) to point a
at WTG = 0.250λ - 0.079λ = 0.171λ.
(iii)At this point we have ZNL = 1 + j1.3 Ω,
or ZL = 50 + j65 Ω.
SOLUTION TO EXAMPLE 1.7
Γ
Γ θj
L e

28. 28

29. 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. 30
THE IMPEDANCE MATCHING
Factors in the matching network selection:
 Complexity
 Bandwidth
 Implementation
 Adjustability
Figure 9: A lossless network matching an arbitrary load impedance to a
transmission line.

31. 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. 32
Adding an impedance matching networks
ensures that all power make it or delivered to
the load.
IMPEDANCE MATCHING (Cont’d)

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

34. 34
• It much more convenient to add shunt
elements rather than series elements  Easier
to work in terms of admittances.
• Admittance:
Z
Y
1
=
IMPEDANCE MATCHING (Cont’d)

35. 35
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.
Γ
Γ θj
L e
IMPEDANCE MATCHING (Cont’d)

36. 36

37. 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. 38
To find the impedance looking into the quarter
wave long section of lossless ZS impedance line
terminated in a resistive load RL:
ljRZ
ljZR
ZZ
LS
SL
Sin
β
β
tan
tan
+
+
=
,
24
2 πλ
λ
π
β ==lBut, for quarter wavelength,
∞=∴ lβtan
QUARTER WAVE TRANSF. (Cont’d)

39. 39
So,
0
2
Z
R
Z
Z
L
S
in ==
Rearrange to get impedance matched line,
LS RZZ 0=
QUARTER WAVE TRANSF. (Cont’d)

40. 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. 41
To get the transformer’s impedance, use
SOLUTION TO EXAMPLE 1.8.5
39.27)15)(50(0 === LS RZZ
To find the position of quarter-wave transformer
from the load:
d = 0.25 λ
ZT = 27.39 Ω
ZT = 27.39 Ω 15 Ω
Qwave transformer

42. 42
EXAMPLE 1.8.6
A transistor has an input impedance of ZL = 25 Ω,
at an operating frequency of 500 MHz. Find:
a)The length, l
b)Width, w
c)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. 43
EXAMPLE 1.8.6
We can directly apply that
355.35)25)(50(0 === LS RZZ
To transistorw Zline
ZL
Zin
Z0=50Ω
ℓ=λ/4
The characteristic of the line is
ε
µ
w
d
C
L
Z
p
Line ==

44. 44
EXAMPLE 1.8.6 (cont)
Para
mete
r
2-wire line Coaxial Line
Parallel Plate
Line
Unit
R Ω/m
L H/m
G S/m
C F/m
δσπ conda
1






a
D
a
2
cosh
π
µ
( )a
Da
diel
2
cosh
πσ
( )a
Da
2
cosh
πε






+
bacond
11
2
1
δπσ






a
b
ln
2π
µ
( )a
b
diel
ln
2πσ
( )a
bln
2πε
δσcondw
2
w
d
µ
d
w
dielσ
d
w
ε

45. 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;
mm
Z
d
w
rline
p
329.5
0
0
==
εε
µ
mnH
w
d
L
p
/8.235==
µ
mpF
d
w
C
p
/6.188==
ε

46. 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:
mm
LCf
l 967.74
4
1
4
===
λ
( )
( ) )(1
)(1
tan
tan
d
d
Z
djZZ
djZZ
ZZ line
Lline
lineL
linein
Γ−
Γ+
=
+
+
=
β
β
Where d = l = λ/4, and the reflection coefficient is given
by








−
+
−
=Γ=Γ −
d
v
f
j
ZZ
ZZ
ed
plineL
lineLdj πβ 2
2exp)( 2
0

47. 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. 48
STUB MATCHING
Stub Matching
• Single stub or Double Stub
• Parallel Stub or Series Stub
• Open stub or Shorted Stub

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

50. 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. 51
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.
SHUNT STUB MATCHING NET.
(Cont’d)

52. 52
• 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 .

SHUNT STUB USING SMITH CHART
Γ
Γ θj
L e

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

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

55. 55
1. Locate the normalized load impedance,
ZNL = ZL/Z0 = 0.4 – j1.1Ω
2. Draw constant circle.
3. Locate YNL. (0.112λ at WTG)
4. 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
SOLUTION TO EXAMPLE 1.9
Γ
Γ θj
L e

56. 56
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
7. 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 = 0.074 λ

SOLUTION TO EXAMPLE 1.9 (Cont’d)

57. 57
Thus, the values are:
d = 0.075 λ
l = 0.074 λ
yd = 1 + j2.0 Ω
yl = -j2.0 Ω
Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1
SOLUTION TO EXAMPLE 1.9 (Cont’d)

58. 58

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

60. 60
EXAMPLE 1.10
Construct an open ended shunt stub
matching network for a 50Ω line
terminated in a load ZL = 150 + j100 Ω

61. 61
SOLUTION TO EXAMPLE 1.10
1. Locate the normalized load impedance,
ZNL = ZL/Z0 = 3.0 + j2.0Ω
2. Draw constant circle.
3. Locate YNL. (0.474λ at WTG)
4. 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
Γ
Γ θj
L e

62. 62
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
7. 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 = 0.339 λ

SOLUTION TO EXAMPLE 1.10
(Cont’d)

63. 63
Thus, the values are:
d = 0.204 λ
l = 0.339 λ
yd = 1 + j1.6 Ω
yl = -j1.6 Ω
Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1
SOLUTION TO EXAMPLE 1.10
(Cont’d)

64. 64

65. 65
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.
IMPORTANT!!

66. 66
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 = 0.426 λ
For open ended shunt stub (example 1.10):
d = 0.348 λ and l = 0.161 λ
EXERCISE (TRY THIS!)