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Microwave Circuit Design I
Lecture 10
Topics:
1. Double Stub Tuning
Reading: Pozar pp. 235–240‘
Double–Stub Tuning
1. Compute Y = Z◦/ZL = normalized load admittance
2. Draw a line from the origin at an angle equivalent to the distance
d from the load using the “Wavelengths Toward Load” scale
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3. Draw a circle of radius r = 0.5 centered at the midpoint of
this line. To find the radius of the circle, measure the distance
between the origin and the intersection of the G=3 circle and
the real axis.
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4. Plot Y on the admittance Smith chart
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5. Find the two intersections of the line of constant conductance on
which YL resides and this new circle. Label one intersection YL0
and the other YL00.
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Note that the required load admittance at this point is determined by
0
Ym1
= YL0 − YL
00
Ym1
= YL00 − YL
(1)
(2)
6. The input admittance to the series transmission line loaded with
YL0 is the intersection of the line of constant SWR that passes
through YL0 and the constant G = 1 circle.
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7. The input admittance to the series transmission line loaded with
YL00 is the intersection of the line of constant SWR that passes
through YL00 and the constant G = 1 circle.
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8. The required shunt admittances at this point are the negative of
the susceptances associated with these intersections.
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9. To determine the appropriate length of stub to realize the particular shunt admittances required for this procedure, follow step
9. of the single–stub tuner procedure.
Example: Design an open–circuit double–stub tuner for a normalized load admittance of YL = 0.2−j0.5 with a series transmission
line of d = 0.2λ.
Solution:
From the Smith Chart that follows,
YL0 = 0.2 − j0.77
YL00 = 0.2 + j0.1
(3)
(4)
0
Ym1
= (0.2 − j0.77) − (0.2 − j0.5) = −j0.22
(5)
00
Ym1
= (0.2 + j0.1) − (0.2 − j0.5) = +j0.6
(6)
Thus,
The input admittances to the series transmission line with these
equivalent load admittances are
Yin0 = 1.0 + j2.5
Yin00 = 1.0 − j1.8
(7)
(8)
0
Ym2
= −j2.5
(9)
00
Ym2
= +j1.8
(10)
So,
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Since we have chosen to use an open–circuit stub, the length of the
stubs are
1
−j0.22
`0m1 =
tan−1
λ = −0.0344λ
(11)
2π
j1.0
Thus,
`0m1 = 0.4656λ
1
j0.6
=
tan−1
λ = 0.086λ
2π
j1.0
(12)
`0m1 = 0.086λ
−j2.5
1
tan−1
=
λ = −0.189λ
2π
j1.0
(14)
`0m1 = 0.311λ
1
j1.8
=
tan−1
λ=λ
2π
j1.0
(16)
`00m1
(13)
Thus,
`0m2
(15)
Thus,
`00m2
(17)
Thus,
`0m1 = 0.169λ
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Note that not every load impedance can be matched with this procedure. Since the length between stubs is a fixed value, a “forbidden
region” exists based on that length wherein a load impedance cannot
be matched. Consider, for example, a normalized load admittance
of Y = 3.0 − j1.0 with the previous example of d = 0.2λ. This
situation is illustrated as follows:
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For this load admittance, there is clearly no shunt admittance
that may be placed in parallel to locate Y 0 or Y 00 on the circle drawn
here. In fact, every admittance with a real part that does not cross
the circle shown cannot be matched with this method. So, for this
choice of d, the “forbidden region” is shown here.
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