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Microwave Circuit Design I
Lecture 9
Topics:
1. Admittance Smith Chart
2. Impedance Matching
3. Single-Stub Tuning
Reading: Pozar pp. 228–235
The Admittance Smith Chart
Since
ZL = Z◦
1+Γ
,
1−Γ
(1)
YL = Y◦
1−Γ
1+Γ
(2)
the following is also true
from which we can define the normalized admittance
Y =
YL 1 − Γ
=
Y◦
1+Γ
(3)
We’ll follow a treatment similar to what we saw for impedances to
create the admittance Smith chart.
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The normalized admittance is a complex number in general. So,
we can write
1 − Γ 1 − u − jv
Y = G + jB =
=
(4)
1+Γ
1 + u + jv
In a manner similar to before, lines of constant conductance in the
admittance plane map to circles defined by
2
1
G
+ v2 =
(5)
u+
G+1
(G + 1)2
in the Γ–plane. These are circles of radius
r=
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1
G+1
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and center at
G
−
,0
G+1
as shown below
Lines of constant susceptance in the admittance plane map to
circles defined by
2
1
1
(1 + u)2 + v +
= 2
(6)
B
B
in the Γ–plane. The circles have radius
1
r=
B
and center
1
−1, −
B
Note the sign of the center. This negative sign implies that negative
susceptance maps to the top half (positive real) of the Γ–
plane and positive susceptance maps to the bottom half
(negative real) of the Γ–plane as shown below
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The admittance Smith chart is shown on the next page.
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Notes:
• All components of the chart operate as before.
• We can use a combination impedance/admittance chart to convert between Γ, Z, and Y simultaneously.
Example) Given Z = 1.6 + j1.2, determine Y and Γ from the
admittance chart.
Solution:
Y = 0.4 − j0.3
(7)
∠Γ = 39◦
|Γ| = 34/73 = 0.466
(8)
(9)
⇒ Γ = 0.466∠39◦
(10)
Impedance Matching
Consider an arbitrary, complex load impedance ZL. We wish to
match that load impedance to a transmission line using some 2–port
matching network:
Define Z = ZL/Z◦ = Normalized Load Impedance which
can be plotted on the Smith chart. Also, let Zin be the normalized
input impedance to the matching network. With these definitions,
• ZL is matched when Zin = 1.0 + j0.0
• YL is matched when Yin = 1.0 + j0.0
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If ZL is connected directly to the transmission line, the input
impedance traverses a circle centered at (0, 0) on the Smith chart
(YL does the same).
Let’s consider the example of Z = 1.6 + j1.2, or Y = 0.4 − j0.3.
• Option 1 (Series T–Line): Attach the load directly to the
feeding transmission line and see if we can choose a specific length
of line to affect an impedance match. For this analysis, we will
plot Y on the admittance Smith chart as shown on the next page.
To obtain perfect matching, we want the input impedance to be
located at the origin, but a single connecting transmission line
only allows us to traverse the circle centered at the origin. There
is no way the single connecting transmission line can improve
the level of match of this load.
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• Option 2 (Shunt Reactance): Use a shunt reactance at the
load. Adding a shunt (parallel) reactance moves us along the
circle of constant G as shown on a subsequent page. In general
we cannot obtain a perfect match, but clearly this scenario differs
from the previous one.
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• Option 3 (Single–Stub Tuner): A series transmission line
followed by a shunt reactance. If we place a length of series transmission line until the effective input impedance intersects with
the line of constant G = 1, then we can add a shunt admittance
of G = −1 to locate the input admittance at the origin of the
Smith chart. This scenario looks like the following:
and it is viewed on the Smith chart on the following page.
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• Option 4 (Double–Stub Tuner): Two shunt reactances
with a fixed series transmission line between them. If we place a
variable shunt reactance in parallel with the load, a fixed series
transmission line followed by a variable shunt reactance at the
input to the matching network, the input admittance can be
tuned to the origin of the Smith Chart in most cases. This
scenario looks like the following:
This option is typically used when variable load impedances are
encountered. When it is unsure what the load impedance will
be it is easier to create variable length stubs than it is to create
variable length series elements. For example, if the double stub
tuner is fabricated using air–core coaxial cable, the stubs are
simply short circuit elements with a plunger to independently
change the lengths of the two shunt elements.
Single–Stub Tuning
Any arbitrary (complex) load impedance can be matched to a specific
transmission line characteristic impedance at a single frequency by
performing the following steps:
1. Compute Y = Z◦/ZL = normalized load admittance
2. Plot Y on the admittance Smith chart
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3. Draw a circle, centered at the origin, that crosses Y
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4. Draw lines from the origin through the following points:
(a) Y
(b) The negative intersection of the constant SWR circle of step
3 and the circle of constant G = 1.0
(c) The positive intersection of the constant SWR circle of step
3 and the circle of constant G = 1.0
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5. d1 is the shortest clockwise distance (in λ) from Y to the line of
step 4b or 4c
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6. d2 is the clockwise distance (in λ) from Y to the remaining line
of step 4b or 4c
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7. The required shunt admittance a distance d1 from the load is
the negative of the susceptance associated with the intersection
of the constant SWR circle and the constant G = 1.0 circle
assocated with step 5
8. The required shunt admittance a distance d2 from the load is
the negative of the susceptance associated with the intersection
of the constant SWR circle and the constant G = 1.0 circle of
step 6
9. To determine the appropriate length of stub to realize a particular shunt admittance, recall,
jZ◦ tan(β`) , short
Zin =
(11)
−jZ◦ cot(β`) , open
−jY◦ cot(β`) , short
Yin =
(12)
jY◦ tan(β`) , open
Yin
−j cot(β`) , short
⇒
=
(13)
j tan(β`) , open
Y◦
(a) for open–circuit stubs:
1
Y
in
`=
λ
tan−1
2π
jY◦
(b) for short–circuit stubs:
1
Y◦
`=
tan−1
λ
2π
jYin
(14)
(15)
Note: ` cannot be negative. If the equations above give a negative
value for `, simply add π to β`. This adds 0.5 to `/λ.
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Example: Design an open–circuit single–stub tuner for a normalized load impedance of ZL = 0.5 − j1.0
Solution:
YL = 1/ZL = 0.4 + j0.8
(16)
From the Smith chart that follows,
d1 = 0.429λ − 0.365λ = 0.064λ
(17)
d2 = 0.5λ − 0.365λ + 0.072λ = 0.207λ
(18)
The input admittances to these lengths of lines are
Yin,d1 = 1 + j1.6 ⇒ Yd1 = −j1.6
Yin,d2 = 1 − j1.6 ⇒ Yd2 = +j1.6
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(19)
(20)
(21)
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Since we have chosen to use an open–circuit stub, the length of
the stubs are
1
−j1.6
`d1 =
tan−1
λ = −0.161λ
(22)
2π
j1.0
Thus,
`d1 = 0.3389λ
and for the other length
j1.6
1
tan−1
λ = 0.161λ
`d2 =
2π
j1.0
`d2 = 0.161λ
and the circuits look like the following:
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(23)
(24)
(25)
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