Course: BEng (Hons) in Electrical and Electronic Engineering

1. EEEC6430310 ELECTROMAGNETIC FIELDS AND WAVES
Smith Chart
FACULTY OF ENGINEERING AND COMPUTER TECHNOLOGY
BENG (HONS) IN ELECTRICALAND ELECTRONIC ENGINEERING
Ravandran Muttiah BEng (Hons) MSc MIET

2. Introduction to Smith Chart
The smith chart is one of the most useful graphical tools for high frequency
circuit applications. The chart provides a clever way to visualize complex
functions and it continuous to endure popularity decades after its original
conception. From a mathematical point of view, the Smith chart is simply a
representation of all possible complex impedances with respect to
coordinates defined by the reflection coefficient.
The domain of definition of the reflection coefficient is a circle of radius 1 in
the complex plane as shown in figure 1. This is also the domain of the Smith
chart.
1
Figure 1
Re Γ
Im Γ
1
1

3. 2
The goal of the Smith chart is to identify all possible impedances on the
domain of existence of the reflection coefficient. To do so, we start from the
general definition of line impedance (which is equally applicable to the load
impedance).
𝑍 𝑑 =
𝑉 𝑑
𝐼 𝑑
= 𝑍o
1 + Γ 𝑑
1 − Γ 𝑑
This provides the complex function 𝑍 𝑑 = 𝑓 Re Γ , Im Γ that we want
to graph. It is obvious that the result would be applicable only to lines with
exactly characteristic impedance 𝑍o.
In order to obtain universal curves, we introduce the concept of normalised
impedance,
𝑧 𝑑 =
𝑍 𝑑
𝑍o
=
1 + Γ 𝑑
1 − Γ 𝑑

4. 3
The normalised impedance is represented on the Smith chart by using
families of curves that identify the normalised resistance, 𝑟 (real part) and
the normalised reactance, 𝑥 (imaginary part),
𝑧 𝑑 = Re 𝑧 + jIm 𝑧 = 𝑟 + j𝑥
Let’s represent the reflection coefficient in terms of its coordinates,
Γ 𝑑 = Re Γ + jIm Γ
Now we can write,
𝑟 + j𝑥 =
1 + Re Γ + jIm Γ
1 − Re Γ − jIm Γ
=
1 − Re2
Γ − Im2
Γ + j2Im Γ
1 − Re Γ
2
+ jIm2 Γ

5. 4
The real part gives,
𝑟 =
1 − Re2
Γ − Im2
Γ
1 − Re Γ
2
+ Im2 Γ
𝑟 Re Γ − 1 2
+ Re2
Γ − 1 + 𝑟Im2
Γ + Im2
Γ +
1
1 + 𝑟
−
1
1 + 𝑟
= 0
𝑟 Re Γ − 1 2
+ Re2
Γ − 1 +
1
1 + 𝑟
+ 1 + 𝑟 Im2
Γ =
1
1 + 𝑟
1 + 𝑟 Re2
Γ − 2Re Γ
𝑟
1 + 𝑟
+
𝑟2
1 + 𝑟 2
+ 1 + 𝑟 Im2
Γ =
1
1 + 𝑟
⇒ Re Γ −
𝑟
1+𝑟
2
+ Im2
Γ =
1
1+𝑟
2
Equation of a circle
Add a quantity equal to zero

6. 5
The imaginary part gives,
𝑥 =
2Im Γ
1 − Re Γ
2
+ Im2 Γ
𝑥2
1 − Re Γ
2
+ Im2
Γ − 2𝑥Im Γ + 1 − 1 = 0
1 − Re Γ
2
+ Im2
Γ −
2
𝑥
Im Γ +
1
𝑥2
=
1
𝑥2
1 − Re Γ
2
+ Im2
Γ −
2
𝑥
Im Γ +
1
𝑥2
=
1
𝑥2
⇒ Re Γ − 1 2
+ Im Γ −
1
𝑥
2
=
1
𝑥2 Equation of a circle
Multiply by 𝑥 and add a
quantity equal to zero

7. 6
The result for the real part indicates that on the complex plane with
coordinate Re Γ , Im Γ all the possible impedances with a given
normalised resistance, 𝑟 are found on a circle with,
Center =
𝑟
1 + 𝑟
, 0
Radius =
1
1 + 𝑟
As the normalised resistance, 𝑟 varies from 0 to ∞, we obtain a family of
circles completely contained inside the domain of the reflection coefficient
Γ ≤ 1 as shown in figure 2.

8. 7
Re Γ
Im Γ
𝑟 = 1
𝑟 = 5
𝑟 = 0
𝑟 = 0.5
𝑟 → ∞
Figure 2

9. 8
The result for the imaginary part indicates that on the complex plane with
coordinate Re Γ , Im Γ all the possible impedances with a given
normalised reactance, 𝑥 are found on a circle with,
Center = 1,
1
𝑥
Radius =
1
𝑥
As the normalised reactance,𝑥 varies from −∞ to ∞, we obtain a family of
arcs contained inside the domain of the reflection coefficient Γ ≤ 1 as
shown in figure 3.

10. 9
Re Γ
Im Γ
𝑥 = 1
𝑥 → ±∞
𝑥 = 0.5
𝑥 = −0.5
𝑥 = −1
Figure 3
𝑥 = 0

11. 10
Basic Smith Chart techniques for loss-less transmission line.
Given 𝑍 𝑑 ⇒ Find Γ 𝑑
Given Γ 𝑑 ⇒ Find 𝑍 𝑑
Given ΓR and 𝑍R ⇒ Find Γ 𝑑 and 𝑍 𝑑
Given Γ 𝑑 and 𝑍 𝑑 ⇒ Find Γ 𝑑 and 𝑍 𝑑
Find 𝑑max and 𝑑min (maximum and minimum locations for the voltage
standing wave pattern
Find the Voltage Standing Wave Ratio (VSWR)
Given 𝑍 𝑑 ⇒ Find 𝑌 𝑑
Given 𝑌 𝑑 ⇒ Find 𝑍 𝑑

12. 11
Given 𝑍 𝑑 ⇒ Find Γ 𝑑
(1) Normalise the impedance,
𝑍 𝑑 =
𝑍 𝑑
𝑍o
=
𝑅
𝑍o
+ j
𝑋
𝑍o
= 𝑟 + j𝑥
(2) Find the circle of constant normalised resistance 𝑟
(3) Find the arc of constant normalised reactance 𝑥
(4) The intersection of the two curves indicates the reflection coefficient in
the complex plane. The chart provides directly the magnitude and the
phase angle of Γ 𝑑 .
Example 1:
Find Γ 𝑑 , given
𝑍 𝑑 = 25 + j100 Ω with 𝑍o = 50 Ω

13. 12
50.906°
(1) Normalisation
𝑧 𝑑 =
25+j100
50
= 0.5 + j2.0
(2) Find normalized
resistance circle
𝑟 = 0.5
(3) Find normalized
reactance arc
𝑥 = 2.0
(4) This vector represents
the reflection
coefficient
Γ 𝑑 = 0.52 + j0.64
Γ 𝑑 = 0.8246
Γ 𝑑 = 0.8885 rad
= 50.906°
0.8246
Figure 4

14. 13
Given Γ 𝑑 ⇒ Find 𝑍 𝑑
(1) Determine the complex point representing the given reflection
coefficient Γ 𝑑 on the chart.
(2) Read the values of the normalised resistance 𝑟 and of the normalized
reactance 𝑥 that correspond to the reflection coefficient point.
(3) The normalized impedance is,
𝑧 𝑑 = 𝑟 + j𝑥
and the actual impedance is,
𝑍 𝑑 = 𝑍o𝑧 𝑑 = 𝑍o 𝑟 + j𝑥 = 𝑍o𝑟 + j𝑍o𝑥

15. 14
Given ΓR and 𝑍R ⇔ Find Γ 𝑑 and 𝑍 𝑑
Note: The magnitude of the reflection coefficient is constant along a loss-
less transmission line terminated by a specified load, since,
Γ 𝑑 = ΓRexp −j2𝛽𝑑 = ΓR
Therefore, on the complex plane, a circle with center at the origin and radius
ΓR represents all possible reflection coefficients found along the
transmission line. When the circle of constant magnitude of the reflection
coefficient is drawn on the Smith chart, one can determine the values of the
line impedance at any location.
The graphical step-by step procedure is,
(1) Identify the load reflection coefficient ΓR and the normalized load
impedance 𝑍R on the Smith chart.

16. 15
(2) Draw the circle of constant reflection coefficient amplitude Γ 𝑑 =
ΓR .
(3) Starting from the point representing the load, travel on the circle in the
clockwise direction, by an angle,
𝜃 = 2𝛽𝑑 = 2
2π
𝜆
𝑑
(4) The new location on the chart corresponds to location 𝑑 on the
transmission line. Here, the values of Γ 𝑑 and 𝑍 𝑑 can be read from
the chart as before.
Example 2:
Given, 𝑍R = 25 + j100 Ω with 𝑍o = 50 Ω
Find 𝑍 𝑑 and Γ 𝑑 for 𝑑 = 0.18 𝜆

17. 16
Circle with constant Γ
Γ 𝑑 = 0.8246 ∠ − 78.7°
= 0.161 − j0.809
Γ 𝑑
ΓR
∠ΓR
𝑧R
𝑧 𝑑
𝑧 𝑑 = 0.236 − j1.192
𝑍 𝑑 = 𝑧 𝑑 × 𝑍o
= 11.79 − j59.6 Ω
𝜃 = 2𝛽𝑑
= 2
2π
𝜆
0.18 𝜆
= 2.262 rad
= 129.6°
Figure 5

18. 17
Given, ΓR and 𝑍R ⇒ Find 𝑑max and 𝑑min
(1) Identify on the Smith chart the load reflection coefficient ΓR or the
normalized load impedance 𝑍R.
(2) Draw the circle of constant reflection coefficient amplitude Γ 𝑑 =
ΓR . The circle intersects the real axis of the reflection coefficient at
two points which identify 𝑑max (when Γ 𝑑 = Real positive) and
𝑑min (when Γ 𝑑 = Real negative).
(3) A commercial Smith chart provides an outer graduation where the
distances normalised to the wavelength can be read directly. The
angles, between the vector ΓR and the real axis, also provide a way to
compute 𝑑max and 𝑑min.
Example 3:
Find 𝑑max and 𝑑min for
𝑍R = 25 + j100 Ω; 𝑍R = 25 − j100 Ω 𝑍o = 50 Ω

19. 18
Im 𝑍R > 0
ΓR
∠ΓR
𝑧R
𝑧 𝑑
Figure 6
2𝛽𝑑max = 50.9°
𝑑max = 0.0707𝜆
2𝛽𝑑min = 230.9°
𝑑min = 0.3207𝜆
𝑍R = 25 + j100 Ω 𝑍o = 50 Ω

20. 19
Im 𝑍R < 0
ΓR
∠ΓR
𝑧R
Figure 7
2𝛽𝑑max = 309.1°
𝑑max = 0.4293𝜆
2𝛽𝑑min = 129.1°
𝑑min = 0.1793𝜆
𝑍R = 25 − j100 Ω 𝑍o = 50 Ω

21. 20
Given, ΓR and 𝑍R ⇒ Find the VSWR
The VSWR is defined as,
𝑉𝑆𝑊𝑅 =
𝑉
max
𝑉min
=
1 + ΓR
1 − ΓR
The normalized impedance at a maximum location of the standing wave
pattern is given by,
𝑧 𝑑max =
1 + Γ 𝑑max
1 − Γ 𝑑max
=
1 + ΓR
1 − ΓR
= 𝑉𝑆𝑊𝑅
This quantity is always real and ≥ 1. The VSWR is simply obtained on the
Smith chart, by reading the value of the (real) normalized impedance, at the
location 𝑑max where Γ is real and positive.

22. 21
The graphical step-by-step procedure is:
(1) Identify the load reflection coefficient, ΓR and the normalised load
impedance, 𝑍R on the Smith chart.
(2) Draw the circle of constant reflection coefficient amplitude, Γ 𝑑 =
ΓR .
(3) Find the intersection of this circle with the real positive axis for the
reflection coefficient (corresponding to the transmission line location
𝑑max).
(4) A circle of constant normalized resistance will also intersect this point.
Read or interpolate the value of the normalized resistance to determine
the VSWR.
Example 4:
Find the VSWR for;
𝑍R1 = 25 + j100 Ω; 𝑍R2 = 25 − j100 Ω 𝑍o = 50 Ω

23. 22
Figure 8
Circle with constant Γ
𝑍R1
𝑍R2
ΓR2
ΓR1 Circle of constant
Conductance
𝑟 = 10.4
𝑧 𝑑max = 10.4
For both loads
𝑉𝑆𝑊𝑅 = 10.4

24. 23
Given 𝑍 𝑑 ⇔ Find 𝑌 𝑑
Note: The normalized impedance and admittance are defined as,
𝑧 𝑑 =
1 + Γ 𝑑
1 − Γ 𝑑
𝑦 𝑑 =
1 − Γ 𝑑
1 + Γ 𝑑
Since,
Γ 𝑑 +
𝜆
4
= −Γ 𝑑
𝑧 𝑑 +
𝜆
4
=
1 + Γ 𝑑 +
𝜆
4
1 − Γ 𝑑 +
𝜆
4
=
1 − Γ 𝑑
1 + Γ 𝑑
= 𝑦 𝑑

25. 24
We know that the equality,
𝑧 𝑑 +
𝜆
4
= 𝑦 𝑑
is only valid for normalized impedance and admittance. The actual values are
given by,
𝑍 𝑑 +
𝜆
4
= 𝑍o ∙ 𝑧 𝑑 +
𝜆
4
𝑌 𝑑 = 𝑌o ∙ 𝑦 𝑑 =
𝑦 𝑑
𝑍o
where,
𝑌o =
1
𝑍o
is the characteristic admittance of the transmission line.

26. 25
The graphical step-by-step procedure is:
(1) Identify the load reflection coefficient, ΓR and the normalised load
impedance, 𝑍R on the Smith chart.
(2) Draw the circle of constant reflection coefficient amplitude, Γ 𝑑 =
ΓR .
(3) The normalized admittance is located at a point on the circle of constant
Γ which is diametrically opposite to the normalised impedance.
Example 5:
Given,
𝑍R = 25 + j100 Ω with 𝑍o = 50 Ω
Find 𝑌R.

27. 26
Circle with constant Γ
𝑧 𝑑 = 0.5 + j2.0
𝑍 𝑑 = 25 + j100 Ω
𝜃 = 180°
= 2𝛽 ∙
𝜆
4
𝑦 𝑑 = 0.11765 − j0.4706
𝑌 𝑑 = 0.002353 − j0.009421 S
𝑧 𝑑 +
𝜆
4
= 0.11765 − j0.4706
𝑍 𝑑 +
𝜆
4
= 5.8824 − j23.5294 Ω
Figure 9

28. 27
The Smith chart can be used for line admittance, by shifting the space reference
to the admittance location. After that, one can move on the chart just reading
the numerical values as representing admittances.
Let’s review the impedance-admittance terminology:
Impedance = Resistance +j Reactance
𝑍 = 𝑅 + j𝑋
Admittance = Conductance +j Susceptance
𝑌 = 𝐺 + j𝐵
On the impedance chart, the correct reflection coefficient is always represented
by the vector corresponding to the normalized impedance. Charts specifically
prepared for admittances are modified to give the correct reflection coefficient
in correspondence of admittance.
Smith Chart For Admittance

29. 28
Figure 10
𝑧 𝑑 = 0.5 + j2.0
𝑦 𝑑 = 0.11765 − j0.4706
Γ
Negative
Inductive
Susceptance
Positive
Capacitive
Susceptance

30. 29
Since related impedance and admittance are on opposite sides of the same
Smith chart, the imaginary parts always have different sign.
Therefore, a positive (inductive) reactance corresponds to a negative
(inductive) susceptance, while a negative (capacitive) reactance corresponds to
a positive (capacitive) susceptance.
Numerically, we have,
𝑧 = 𝑟 + j𝑥
𝑦 = 𝑔 + j𝑏 =
1
𝑟 + j𝑥
𝑦 =
𝑟 − j𝑥
𝑟 + j𝑥 𝑟 − j𝑥
=
𝑟 − j𝑥
𝑟2 + 𝑥2
𝑔 =
𝑟
𝑟2 + 𝑥2
𝑏 = −
𝑥
𝑟2 + 𝑥2

31. 30
Impedance Matching
A number of techniques can be used to eliminate reflections when line
characteristic impedance and load impedance are mismatched. Impedance
matching techniques can be designed to be effective for a specific frequency of
operation (narrow band techniques) or for a given frequency spectrum
(broadband techniques).
One method of impedance matching involves the insertion of an impedance
transformer between line and load.
Impedance
Transformer
𝑍R
𝑍o
In the following, we neglect effects of loss in the lines.
Figure 11

32. 31
A simple narrow band impedance transformer consists of a transmission line
section of length
𝜆
4
.
The impedance transformer is positioned so that it is connected to a real
impedance 𝑍A. This is always positive if a location of maximum or minimum
voltage standing wave pattern is selected.
𝑍R
Figure 12
𝑍B 𝑍A
𝑍o1 𝑍o2 𝑍o1
𝜆
4
𝑑max or 𝑑min

33. 32
Consider a general load impedance with its corresponding load reflection
coefficient,
𝑍R = 𝑅R + j𝑋R
ΓR =
𝑍R − 𝑍o1
𝑍R + 𝑍o1
= ΓR ej∅
If the transformer is inserted at a location of voltage maximum 𝑑max
𝑍A = 𝑍o1
1 + Γ d
1 − Γ d
= 𝑍o1
1 + ΓR
1 − ΓR
If it is inserted instead at a location of voltage minimum 𝑑min
𝑍A = 𝑍o1
1 + Γ d
1 − Γ d
= 𝑍o1
1 − ΓR
1 + ΓR

34. 33
Figure 13
𝑍in 𝑍o
𝐿 =
𝜆
4
𝑍A
Consider now the input impedance of a line of length
𝜆
4
.
Since:
𝑍A = 𝑍o1
1 + Γ d
1 − Γ d
= 𝑍o1
1 − ΓR
1 + ΓR
We have,
𝑍in =
lim
tan 𝛽𝐿 → ∞
𝑍o
𝑍A + j𝑍otan 𝛽𝐿
j𝑍Atan 𝛽𝐿 + 𝑍o
→
𝑍o
2
𝑍A

35. 34
Note that if the load is real, the voltage standing wave pattern at the load is
maximum when 𝑍R > 𝑍o1 or minimum when 𝑍R < 𝑍o1. The transformer can
be connected directly at the load location or at a distance from the load
corresponding to a multiple of
𝜆
4
.
Figure 14
𝑍B 𝑍A = Real
𝑍o1 𝑍o2 𝑍o1
𝜆
4
𝑍R = Real
𝑑1 = 𝑛
𝜆
4
;
𝑛 = 0,1,2

36. 35
If the load impedance is real and the transformer is inserted at a distance from
the load equal to an even multiple of
𝜆
4
then,
𝑍A = 𝑍R
𝑑1 = 2𝑛
𝜆
4
= 𝑛
𝜆
2
but if the distance from the load is an odd multiple of
𝜆
4
,
𝑍A =
𝑍o1
2
𝑍R
𝑑1 = 2𝑛 + 1
𝜆
4
= 𝑛
𝜆
2
+
𝜆
4

37. 36
The input impedance of the impedance transformer after inclusion in the circuit
is given by,
𝑍B =
𝑍o2
2
𝑍A
For impedance matching we need,
𝑍o1 =
𝑍o2
2
𝑍A
𝑍o2 = 𝑍o1𝑍A
The characteristic impedance of the transformer is simply the geometric
average between the characteristic impedance of the original line and the load
seen by the transformer.
Let’s now review some simple examples.

38. 37
Real Load Impedance,
Figure 15
𝑍B 𝑍A
𝑍o1 = 50 Ω 𝑍o2 = ?
𝜆
4
𝑅R = 100 Ω
𝑍B =
𝑍o2
2
𝑅R
= 𝑍o1
𝑍o2 = 𝑍o1𝑅R = 50 × 100 ≈ 70.71 Ω

39. 38
Note that an identical result is obtained by switching 𝑍01 and 𝑅R,
Figure 16
𝑍B 𝑍A
𝑍o1 = 100 Ω 𝑍o2 = ?
𝜆
4
𝑅R = 50 Ω
𝑍B =
𝑍o2
2
𝑅R
= 𝑍o1
𝑍o2 = 𝑍o1𝑅R = 100 × 50 ≈ 70.71 Ω

40. 39
Another real load case,
Figure 17
𝑍B 𝑍A
𝑍o1 = 75 Ω 𝑍o2 = ?
𝜆
4
𝑅R = 300 Ω
𝑍B =
𝑍o2
2
𝑅R
= 𝑍o1
𝑍o2 = 𝑍o1𝑅R = 75 × 300 ≈ 150 Ω

41. 40
Same impedance as before, but now the transformer is inserted at a distance
𝜆
4
from the load (voltage minimum in this case).
Figure 18
𝑍B 𝑍A
𝑍o2 𝑍o1
𝜆
4
𝑅R = 300 Ω
𝜆
4
𝑍o1 = 75 Ω
𝑍A =
𝑍o1
2
𝑅R
=
752
300
= 18.75 Ω
𝑍B =
𝑍o2
2
𝑍A
= 𝑍o1 ⇒ 𝑍o2 = 𝑍o1𝑍A = 75 × 18.75 ≈ 37.5 Ω

42. 41
Complex Load Impedance - Transformer at voltage maximum
Figure 19
𝑍B 𝑍A
𝑍o2 𝑍o1
𝜆
4
𝑍R = 100 + j100 Ω
𝑑max
𝑍o1 = 50 Ω
ΓR =
100 + j100 − 50
100 + j100 + 50
≈ 0.62
𝑍A = 𝑍o
1 + ΓR
1 − ΓR
≈ 213.28 Ω
𝑍o2 = 𝑍o1𝑍A = 50 × 213.28 = 103.27 Ω

43. 42
Complex Load Impedance - Transformer at voltage minimum
Figure 20
𝑍B 𝑍A
𝑍o2 𝑍o1
𝜆
4
𝑍R = 100 + j100 Ω
𝑑min
𝑍o1 = 50 Ω
ΓR =
100 + j100 − 50
100 + j100 + 50
≈ 0.62
𝑍A = 𝑍o
1 − ΓR
1 + ΓR
≈ 11.72 Ω
𝑍o2 = 𝑍o1𝑍A = 50 × 11.72 = 24.21 Ω

44. 43
If it is not important to realize the impedance transformer with a quarter
wavelength line, we can try to select a transmission line with appropriate length
and characteristic impedance, such that the input impedance is the required real
value.
Figure 21
𝑍A
𝑍o2
𝐿
𝑍R = 𝑅R + j𝑋R
𝑍o1
𝑍o1 = 𝑍A = 𝑍o2
𝑅R + j𝑋R + j𝑍02tan 𝛽𝐿
𝑍o2 + j 𝑅R + 𝑗𝑋R tan 𝛽𝐿

45. 44
After separation of real and imaginary parts we obtain the equations,
𝑍o2 𝑍o1 − 𝑅R = 𝑍o1𝑋Rtan 𝛽𝐿
tan 𝛽𝐿 =
𝑍o2𝑋R
𝑍o1𝑅R − 𝑍o2
2
with final solution,
𝑍o2 =
𝑍o1𝑅R − 𝑅R
2
− 𝑋R
2
1 −
𝑅R
𝑍o1
tan 𝛽𝐿 =
1 −
𝑅R
𝑍o1
𝑍o1𝑅R − 𝑅R
2
− 𝑋R
2
𝑋R
The transformer can be realized as long as the result for 𝑍o2 is real.
Note that this is also a narrow band approach.

46. 45
Example 6:
A signal generator has an internal impedance of 50 Ω. It needs to feed
equal power through a lossless 50 Ω transmission line to two separate
resistive loads of 64 Ω and 25 Ω at a frequency of 10 MHz. Design a
Quarter-wave transformer to match the loads to the 50 Ω line. Assume the
phase velocities of the wave travelling on them is 0.5𝑐.
Solution:
As the two quarter-wave transformers are connected in parallel to the 50
Ω line, if equal powers are required to the two loads, the input
impedances of the two branches looking at the junction from 50 Ω line
must be equal to 100 Ω so that when they add together in parallel, the
total impedances is 50 Ω.

47. 46
𝑅o1
𝑅o2
𝑅in1 = 100 Ω
𝑅in2 = 100 Ω
𝑅L1 = 64 Ω
𝑅L2 = 2𝟓 Ω
𝜆
4
𝜆
4
𝑅o = 50 Ω
100 Ω
100 Ω
50 Ω
Figure 22: Quarter-wave transformer.

48. 47
Therefore, 𝑍in1 = 𝑅in1 = 100 Ω
𝑍in2 = 𝑅in2 = 100 Ω
The characteristic impedances 𝑅o1
′
= 𝑅in1 × 𝑅L1 = 100 × 64 = 80 Ω
𝑅o2
′
= 𝑅in2 × 𝑅L2 = 100 × 25 = 50 Ω
𝑉
p = 0.5𝑐 =
𝑐
𝜀r
=
3 × 108
𝜀r
∴ 𝜀r= 4
where c is the speed of light.
Wavelength along the transformers, 𝜆 =
𝜆o
𝜀r
=
30 m
4
= 15 m.
Physical length of the transformers, 𝑙 =
𝜆
4
=
15
4
= 3.75 m.

49. 48
Example 7:
Design a Quarter-wave transformer is to match a 10 Ω resistor to a 50 Ω
line at 2 GHz. The transmission line used is coaxial cable whose dielectric
has a dielectric constant 𝜀r = 2.25. Sketch the final design, specifying the
dimensions and impedance of the transformer.
Solution:
Transformer impedance:
Given 𝑅L = 10, want 𝑅o
′
= 50. So, 𝑅𝜆
4
= 𝑅L𝑅o
′
= 10 × 50 = 22.36.
Transformer’s physical length:
𝜆 =
𝑉
p
𝑓
=
𝑐
𝜀𝑟
𝑓
=
3 × 108
2.25
2 × 109
= 0.1 m
𝑙 =
𝜆
4
=
0.1
4
= 0. 025m

50. 49
Ratio between Outer and Inner Conductors:
For a lossless coaxial cable, 𝑍o =
1
2π
𝜇
𝜀
ln
𝐷
𝑑
=
𝜂o
2π 𝜀r
ln
𝐷
𝑑
where 𝜂o =
𝜇o
𝜀o
≅ 377 Ω is the characteristic impedance of free space,
while D and d are diameters of the inner and outer conductors respectively.
In our case, we have 𝑍o = 22.36, so,
22.36 =
377
2π 2.25
ln
𝐷
𝑑
ln
𝐷
𝑑
= 0.5590
𝐷
𝑑
= 1.7489

51. 50
Figure 23 shows the Quarter-wave transformer design and cross-sectional
view of the coaxial cable.
D
𝑅𝜆
4
= 22.36 𝑍L
𝜀r = 2.25
𝑅o
′ = 50
𝑙 =
𝜆
4
𝐷 = 1.7489𝑑
Figure 23: Quarter-wave transformer
d

52. 51
Single Stub Impedance Matching
Impedance matching can be achieved by inserting another transmission line
(stub) as shown in the diagram below.
Figure 24
𝑍A = 𝑍o
𝑍R
𝑑stub
𝑍o
𝐿stub
𝑍oS

53. 52
There are two design parameters for single stub matching:
(1) The location of the stub with reference to the load 𝑑stub.
(2) The length of the stub line 𝐿stub.
Any load impedance can be matched to the line by using single stub technique.
The drawback of this approach is that if the load is changed, the location of
insertion may have to be moved.
The transmission line realizing the stub is normally terminated by a short or by
an open circuit. In many cases it is also convenient to select the same
characteristic impedance used for the main line, although this is not necessary.
The choice of open or shorted stub may depend in practice on a number of
factors. A short circuited stub is less prone to leakage of electromagnetic
radiation and is somewhat easier to realize. On the other hand, an open
circuited stub may be more practical for certain types of transmission lines, for
example microstrips where one would have to drill the insulating substrate to
short circuit the two conductors of the line.

54. 53
Since the circuit is based on insertion of a parallel stub, it is more convenient to
work with admittances, rather than impedances.
Figure 25
𝑌A = 𝑌o
𝑌R =
1
𝑍R
𝑑stub
𝑌o =
1
𝑍o
𝐿stub
𝑌oS

55. 54
For proper impedance match:
𝑌A = 𝑌stub + 𝑌 𝑑stub = 𝑌o =
1
𝑍o
Input admittance
of the stub line
Line admittance at location
𝑑stub before the stub is applied
𝐿stub
𝑌oS
𝑌R =
1
𝑍R
𝑑stub
𝑌 𝑑stub
𝑌stub
Figure 26

56. 55
In order to complete the design, we have to find an appropriate location for the
stub. Note that the input admittance of a stub is always imaginary (inductance
if negative, or capacitance if positive).
𝑌stub = j𝐵stub
A stub should be placed at a location where the line admittance has real part
equal to 𝑌o,
𝑌 𝑑stub = 𝑌o + j𝐵 𝑑stub
For matching, we need to have,
𝐵stub = −𝐵 𝑑stub
Depending on the length of the transmission line, there may be a number of
possible locations where a stub can be inserted for impedance matching. It is
very convenient to analyze the possible solution on a Smith chart.

57. 56
𝑦R
𝜃2
𝑧R
Figure 27
Load
location
𝜃1
First location
suitable for stub
insertion
Unitary
conductance
circle
Second Location
suitable for stub
insertion
Constant
Γ 𝑑
circle
𝑦 𝑑stub2
𝑦 𝑑stub1

58. 57
The red arrow on the example indicates the load admittance. This provides on
the “admittance chart” the physical reference for the load location on the
transmission line. Notice that in this case the load admittance falls outside the
unitary conductance circle. If one moves from load to generator on the line, the
corresponding chart location moves from the reference point, in clockwise
motion, according to an angle 𝜃 (indicated by the light green arc).
𝜃 = 2𝛽𝑑 =
4π
𝜆
d
The value of the admittance rides on the red circle which corresponds to
constant magnitude of the line reflection coefficient. Γ 𝑑 = Γ𝑅 , imposed
by the load.
Every circle of constant Γ 𝑑 intersects the circle Re 𝑦 = 1 (unitary
normalized conductance), in correspondence of two points. Within the first
revolution, the two intersections provide the locations closest to the load for
possible stub insertion.

59. 58
The first solution corresponds to an admittance value with positive imaginary
part, in the upper portion of the chart.
Line Admittance - Actual: 𝑌 𝑑stub1
= 𝑌o + j𝐵 𝑑stub1
Normalized: y 𝑑stub1
= 1 + j𝑏 𝑑stub1
Stub Location: 𝑑stub1
=
𝜃1
4π
𝜆
Stub Admittance - Actual: −j𝐵 𝑑stub1
Normalized: −j𝑏 𝑑stub1
Stub Length: 𝐿stub =
𝜆
2π
tan−1 1
𝑍oS𝐵 𝑑stub1
⇒ Short
𝐿stub =
𝜆
2π
tan−1
𝑍oS𝐵 𝑑stub1
⇒ Open

60. 59
The second solution corresponds to an admittance value with negative
imaginary part, in the lower portion of the chart.
Line Admittance - Actual: 𝑌 𝑑stub2
= 𝑌o + j𝐵 𝑑stub2
Normalized: y 𝑑stub2
= 1 + j𝑏 𝑑stub2
Stub Location: 𝑑stub2
=
𝜃2
4π
𝜆
Stub Admittance - Actual: j𝐵 𝑑stub2
Normalized: j𝑏 𝑑stub2
Stub Length: 𝐿stub =
𝜆
2π
tan−1
−
1
𝑍oS𝐵 𝑑stub2
⇒ Short
𝐿stub =
𝜆
2π
tan−1
−𝑍oS𝐵 𝑑stub2
⇒ Open

61. 60
If the normalized load admittance falls inside the unitary conductance circle
(see next figure), the first possible stub location corresponds to a line
admittance with negative imaginary part. The second possible location has line
admittance with positive imaginary part. In this case, the formulae given above
for first and second solution exchange place.
If one moves further away from the load, other suitable locations for stub
insertion are found by moving toward the generator, at distances multiple of
half a wavelength from the original solutions. These locations correspond to
the same points on the Smith chart.
First set of locations = 𝑑stub1
+ 𝑛
𝜆
2
Second set of locations = 𝑑stub2
+ 𝑛
𝜆
2

62. 61
𝑧R
𝜃2
𝑦R
Figure 28
𝜃1
First location
suitable for stub
insertion
Unitary
conductance
circle
Second location
suitable for stub
insertion
Constant
Γ 𝑑
circle
𝑦 𝑑stub1
𝑦 𝑑stub2

63. 62
Single stub matching problems can be solved on the Smith chart graphically,
using a compass and a ruler. This is a step-by-step summary of the procedure:
(a) Find the normalized load impedance and determine the corresponding
location on the chart.
(b) Draw the circle of constant magnitude of the reflection coefficient Γ for
the given load.
(c) Determine the normalized load admittance on the chart. This is obtained
by rotating 180° on the constant Γ circle, from the load impedance point.
From now on, all values read on the chart are normalized admittances.

64. 63
𝑦R
𝑧R
(a) Obtain the normalized
load impedance 𝑧R =
𝑍R
𝑍o
and
find its location on the Smith
chart
(b) Draw the
constant Γ 𝑑
circle
(c) Find the normalized
load admittance knowing
that 𝑦R = 𝑧 𝑑 =
𝜆
4
. From
now on the chart
represents admittances.
180° =
𝜆
4
Figure 29

65. 64
(d) Move from load admittance toward generator by riding on the constant
circle, until the intersections with the unitary normalized conductance
circle are found. These intersections corresponds to possible locations for
stub insertion. Commercial Smith charts provide graduations to determine
the angles of rotation as well as the distances from the load in units of
wavelength.
(e) Read the line normalized admittance in correspondence of the stub
insertion locations determined in (d). These values will always be of the
form.
𝑦 𝑑stub = 1 + j𝑏 top half of chart
𝑦 𝑑stub = 1 − j𝑏 bottom half of chart

66. 65
𝑦R
𝜃1
𝑧R
Figure 30
Load
location
(d) Move from load
toward generator and
stop at a location
where the real part of
the normalized line
admittance is 1
Unitary
conductance
circle
First location
suitable for stub
insertion,
𝑑stub1
=
𝜃1
4π
𝜆
(e) Read here the
value of the
normalized line
admittance,
𝑦 𝑑stub1 = 1 + j𝑏

67. 66
𝑦R
𝜃2
𝑧R
Figure 31
Load
location
(d) Move from load
toward generator and
stop at a location
where the real part of
the normalized line
admittance is 1 Unitary
conductance
circle
Second location
suitable for stub
insertion,
𝑑stub2 =
𝜃2
4π
𝜆
(e) Read here the
value of the
normalized line
admittance,
𝑦 𝑑stub2 = 1 − j𝑏

68. 67
(f) Select the input normalized admittance of the stubs, by taking the opposite
of the corresponding imaginary part of the line admittance,
Line: 𝑦 𝑑stub = 1 + j𝑏 → stub: 𝑦 𝑑stub = −j𝑏
Line: 𝑦 𝑑stub = 1 − j𝑏 → stub: 𝑦 𝑑stub = +jb
(g) Use the chart to determine the length of the stub. The imaginary
normalized admittance values are found on the circle of zero conductance
on the chart. On a commercial Smith chart one can use a printed scale to
read the stub length in terms of wavelength. We assume here that the stub
line has characteristic impedance 𝑍o as the main line. If the stub has
characteristic impedance 𝑍oS ≠ 𝑍o the values on the Smith chart must be
re-normalized as,
±j𝑏′
= ±j𝑏
𝑌o
𝑌oS
= ±j𝑏
𝑍oS
𝑍o

69. 68
Figure 32
𝑦 = ∞
Short circuit
(f) Normalized input
admittance of stub,
𝑦stub = 0 − j𝑏
(g) Arc to
determine the
length of a short
circuited stub with
normalized input
admittance, −j𝑏

70. 69
Figure 33
𝑦 = 0
Open circuit
(f) Normalized input
admittance of stub,
𝑦stub = 0 − j𝑏
(g) Arc to
determine the
length of a short
circuited stub with
normalized input
admittance, −j𝑏

71. 70
Figure 34
(f) Normalized input
admittance of stub,
𝑦stub = 0 + j𝑏
(g) Arc to
determine the
length of a short
circuited stub with
normalized input
admittance +j𝑏
𝑦 = ∞
Short circuit

72. 71
Figure 35
(f) Normalized input
admittance of stub,
𝑦stub = 0 + j𝑏
(g) Arc to
determine the
length of a short
circuited stub with
normalized input
admittance +j𝑏
𝑦 = 0
Open circuit

73. 72
𝑦R
Figure 36
After the stub is
inserted, the
admittance at the
stub location is
moved to the
center of the Smith
chart, which
corresponds to
normalized
admittance 1 and
reflection
coefficient 0 (exact
matching
condition).
Matching
condition
If you imagine to
add gradually the
negative
imaginary
admittance of the
inserted stub, the
total admittance
would follow the
yellow arrow,
reaching the
match point when
the complete stub
admittance is
added.
First Solution

74. 73
𝑦R
Figure 37
If the stub does
not have the
proper normalized
input admittance,
the matching
condition is not
reached
First Solution
Effect of a stub
with positive
susceptance
Effect of a stub
with negative
susceptance of
insufficient
magnitude
Effect of a stub with
negative susceptance
of excessive magnitude

75. 74
Double Stub Impedance Matching
Impedance matching can be achieved by inserting two stubs at specified
locations along transmission line as shown in the diagram below.
Figure 38
𝑑stub1
𝐿stub1
𝑌oS1
𝑌A = 𝑌o1
𝑌o =
1
𝑍o1
𝐿stub2
𝑌oS2
𝑌R =
1
𝑍R
𝑑stub2

76. 75
There are two design parameters for double stub matching:
(1) The length of the first stub line 𝐿stub1.
(2) The length of the second stub line 𝐿stub2.
In the double stub configuration, the stubs are inserted at predetermine
locations. In this way, if the load impedance is changed, one simply has to
replace the stubs with another set of different length.
The drawback of double stub is that a certain range of load admittances
cannot be matched once the stub locations are fixed.
Three stubs are necessary to guarantee that match is always possible.

77. 76
The length of the first stub is selected so that the admittance at the location of
the second stub (before the second stub is inserted) has real part equal to the
characteristic admittance of the line.
Figure 39
𝑑stub1
𝐿stub1
𝑌oS1
𝑌′A = 𝑌o1 + j𝐵
𝑌o =
1
𝑍o1
𝑌R =
1
𝑍R
𝑑stub2

78. 77
Figure 40
𝑑stub1
𝐿stub1
𝑌oS1
𝑌A = 𝑌o1 + j𝐵 − j𝐵 = 𝑌o1
𝑌o =
1
𝑍o1
𝑌R =
1
𝑍R
𝑑stub2
𝐿stub2
𝑌oS2
𝑌stub = −j𝐵
The length of the second stub is selected to eliminate the imaginary part of the
admittance at the location of insertion.

79. 78
Figure 41
At the location
where the second
stub is inserted,
the possible
normalized
admittances that
can give matching
are found on the
circle of unitary
conductance on the
Smith chart
The normalized
admittance that we
want at location
𝑑stub2 is on this
circle

80. 79
Figure 42
𝐿stub1
𝑌oS1
𝑌A = 𝑌o1
𝑌o1 =
1
𝑍o1
𝑌R
𝑑stub2
𝐿stub2
𝑌oS2
𝑌R
𝑑stub
Single stub
Double stub
The two approaches solve the same problem

81. 80
If one moves from the location of the second stub back to the load, the
circle of the allowed normalized admittances is mapped into another
circle, obtained by pivoting the original circle about the center of the
chart.
At the location of the first stub, the allowed normalized admittances
are found on an auxiliary circle which is obtained by rotating the
unitary conductance circle counterclockwise, by an angle.
𝜃aux =
4π
𝜆
𝑑stub2 − 𝑑stub1 =
4π
𝜆
𝑑21

82. 81
Figure 43
This is the
auxiliary circle for
distance between
the stubs 𝑑21 =
𝜆
8
+ 𝑛
𝜆
2
𝜃aux

83. 82
Figure 44
This angle of rotation
corresponds to a
distance
𝑑12 = 𝑑stub2 −𝑑stub1
The normalized
admittance that we
want at location
𝑑stub1 is on this
auxiliary circle
Pivot here
𝜃aux

84. 83
Figure 45
This is the
auxiliary circle for
distance between
the stubs 𝑑21 =
𝜆
4
+ 𝑛
𝜆
2
𝜃aux

85. 84
Figure 46
This is the
auxiliary circle for
distance between
the stubs 𝑑21 =
3
𝜆
8
+ 𝑛
𝜆
2
𝜃aux

86. 85
Figure 47
This is the auxiliary
circle for distance
between the stubs
𝑑21 = 𝑛
𝜆
2
This is not a good
choice for double
stub design.
𝜃aux

87. 86
Given the load impedance, we need to follow these steps to complete the
double stub design:
(a) Find the normalized load impedance and determine the corresponding
location on the chart.
(b) Draw the circle of constant magnitude of the reflection coefficient Γ
for the given load.
(c) Determine the normalized load admittance on the chart. This is
obtained by rotating −180° on the constant Γ circle, from the load
impedance point. From now on, all values read on the chart are
normalized admittances.
(d) Find the normalized admittance at location 𝑑stub1 by moving
clockwise on the constant Γ circle.

88. 87
(e) Draw the auxiliary circle.
(f) Add the first stub admittance so that the normalized admittance point
on the Smith chart reaches the auxiliary circle (two possible
solutions). The admittance point will move on the corresponding
conductance circle, since the stub does not alter the real part of the
admittance.
(g) Map the normalized admittance obtained on the auxiliary circle to the
location of the second stub 𝑑stub2. The point must be on the unitary
conductance circle.
(h) Add the second stub admittance so that the total parallel admittance
equals the characteristic admittance of the line to achieve exact
matching condition.

89. 88
𝑦R
𝑧R
(a) Obtain the normalized
load impedance 𝑧R =
𝑍R
𝑍o
and
find its location on the Smith
chart
180° =
𝜆
4
Figure 48
(b) Draw the
constant Γ 𝑑
circle
(c) Find the normalized
load admittance knowing
that 𝑦R = 𝑧 𝑑 =
𝜆
4
. From
now on the chart
represents admittances.
(d) Move to the
first stub location

90. 89
𝑦R
𝑧R
(f) First solution: Add
admittance of first stub to
reach auxiliary circle
Figure 49
(f) Second solution:
Add admittance of
first stub to reach
auxiliary circle
(e) Draw the auxiliary circle

91. 90
(g) First solution:
Map the normalized
admittance from the
auxiliary circle to
the location of the
second stub 𝑑stub2
Figure 50
First solution:
Admittance at
location 𝑑stub2 before
insertion of second
stub
(h) Add the
second stub
admittance

92. 91
(g) Second solution:
Map the normalized
admittance from the
auxiliary circle to
the location of the
second stub 𝑑stub2
Second solution:
Admittance at
location 𝑑stub2 before
insertion of second
stub
(h) Add the
second stub
admittance
Figure 51

93. 92
Example 8:
A 50 Ω lossless transmission line is connected to a load impedance 𝑍L =
35 − j47.5 Ω. Design a short-circuit stub required to match the load at a
frequency of 200 MHz. Assume that the transmission line is a coaxial line
filled with a dielectric material for which 𝜀r = 9.
Solution:
Given 𝑍o = 𝑅o = 50 Ω and 𝑍L = 35 − j47.5 Ω
Normalized 𝑧L =
𝑍L
𝑍o
= 0.7 − j0.95
(i) Enter 𝑧L at point 𝑃1.
(ii) Draw a Γ -circle centered at O with radius O𝑃1.
(iii) Draw straight line from 𝑃1 through O to point 𝑃2
′
on the perimeter,
intersecting the Γ -circle at 𝑃2, which represents 𝑦L. Note 0.109 at 𝑃2
′
on the wavelengths toward generator scale.

94. 93
(iv) Note the two points of intersection of the Γ -circle with the 𝑔 = 1 circle:
At 𝑃3: 𝑦B1 = 1 + j1.2 = 1 + j𝑏B1
At 𝑃4: 𝑦B2 = 1 − j1.2 = 1 + j𝑏B2
(v) Solutions for the position of the stub:
For 𝑃3 (from 𝑃2
′
to 𝑃3
′
) 𝑑1 = 0.168 − 0.109 𝜆 = 0.059𝜆
For 𝑃3 (from 𝑃2
′
to 𝑃4
′
) 𝑑2 = 0.332 − 0.109 𝜆 = 0.223𝜆
(vi) Solutions for the length of the short-circuited stub to provide 𝑦𝑠 = −j𝑏𝐵:
For 𝑃3 𝑦s = −j𝑏B1 = −j1.2
(from 𝑃sc to 𝑃3
′′
) 𝑙1 = 0.361 − 0.250 𝜆 = 0.111𝜆
For 𝑃4 𝑦s = −j𝑏B2 = j1.2
(from 𝑃sc to 𝑃4
′′
) 𝑙2 = 0.139 − 0.250 𝜆 = 0.389𝜆

95. 94
(vii) To compute the physical lengths of the transmission line sections,
we need to calculate the wavelength on the transmission line.
Therefore,
𝜆 =
𝑐
𝜀r
𝑓
=
3 × 108
4
200 × 106
≅ 0.5 m
Thus: 𝑑1 = 0.059𝜆 = 29.5 mm 𝑙1 = 0.111𝜆 = 55.5 mm
𝑑2 = 0.223𝜆 = 111.5 mm 𝑙1 = 0.389𝜆 = 194.5 mm

96. 95
𝑙1
Figure 52
𝑙2
𝑑1
𝑑2
𝑃1
𝑃4
𝑃2
𝑃3
𝑃2
′ 𝑃4
′′
𝑃3
′
𝑃3
′′
𝑃4
′
−j1.2
j1.2
1 + j1.2
1 − j1.2
𝑧L
𝑦L
𝑔 = 1
O

97. 96
(1) Amanogawa, Transmission Lines: Smith Chart, Digital Maestro Series,
2000.
References