There is one last technique we can look at which is somewhat
more flexible than the single stub matching which we just looked
at. This is called double stub matching! Suppose we have the
following situation, as depicted in the
figure. There is a load of
Z
L
Z
0
=0.2+1.3i
Z
L
Z
0
0.21.3
located at the end of the line, and then some
arbitrary distance away (
0.11λ
0.11
λ
) an adjustable stub. Another (arbitrary)
0.11λ
0.11
λ
from the first stub, there is a
second one. Let's plot
Y
L
Y
0
Y
L
Y
0
on the Smith Chart, and
then, since the stubs are in shunt across the line, switch to
admittance, and find
Y
L
Y
0
Y
L
Y
0
. It is easy to see that
Y
L
Y
0
=1.5+2.3i
Y
L
Y
0
1.52.3
.
The first thing we might as well do is move down to the first
stub, and see what admittance we have there
Figure 3. We go from the load, to the first stub by
rotating on a
circle of constant
radius(constant
|rs|
r
s
)since all we are doing is going from one place on the
line to another. If we call the location on the line of the
first stub "A", then we can see that
Y
A
Y
0
=0.25+0.6i
Y
A
Y
0
0.250.6
.
Now, what can the first stub accomplish? A shorted stub can
create any
imaginary admittance we want,
but can not change the real part of the admittance. Thus, by
adjusting the first stub, we can move around on a circle of
constant real part
0.25
Y
0
0.25
Y
0
, and have any imaginary part we want. This is shown
schematically
here.
Now, where do we want to go? Well, we would like to end up
someplace so that, after we have moved from A to B on the line (gone
from the first stub to the second), we are on the matching
circle. If this were so, then, since we are on the matching circle, we
could use the second stub to match the whole line and we would be
done.
This is tricky now, so you have to pay attention
and think. If I want to find a place which, when moved from A to
B, ends up on the matching circle, then what I should do is take
the matching circle and move it from B to A. That is, if I
rotate the matching circle around
0.175λ
0.175
λ
towards the load, then any place
on that rotated matching circle is guaranteed to end up on the
real matching circle, when we go
0.175λ
0.175
λ
back towards the generator.
OK, so here's what we do. First, we rotate the
matching circle 0.175 around towards the load (go
counterclockwise) Figure 5. Now what we have to do
is somehow get from
Y
A
Y
0
Y
A
Y
0
without stub to someplace on the rotated matching
circle. The only way we can do this is to change the imaginary
part of
Y
A
Y
A
with the stub. Suppose we move as shown in Figure 6. In going from
Y
A
Y
0
Y
A
Y
0
without stub to
Y
A
Y
0
Y
A
Y
0
with stub we have changed the imaginary part from
−(i0.6)
0.6
to
i0.05
0.05
, thus we have added
i0.65
0.65
to the imaginary part of
Y
A
Y
0
Y
A
Y
0
. Thus using our standard method for finding the length
of the first stub, we start at
∞
,
−∞
(the short at the end of the stub) and go around the
outside of the Smith Chart until we find
i0.05
0.05
. To get from one place to the next we went
(0.25+0.09)λ=0.34λ
0.25
0.09
λ
0.34
λ
and so the length of the first stub,
L
1
L
1
should be
0.09λ
0.09
λ
. Now we are at
Y
A
Y
0
Y
A
Y
0
with stub. The next thing we have to do is to rotate another
0.175λ
0.175
λ
towards the generator so that we
can get to stub B. As we do this rotation, we again stay on a
circle of constant radius, because now we
are moving down the transmission line not
adding reactance by using a stub! This rotation is
guaranteed to end us up on the matching
circle because every point on the rotated
circle (the one we start from) is exactly
0.175λ
0.175
λ
towards the load from the
matching circle. As shown here, we
are now at the point
Y
B
Y
0
Y
B
Y
0
without stub
1.0+1.6i
1.01.6
. Thus we need to adjust the length
L
2
L
2
of the second stub to give us
−(i1.6)
1.6
of reactance, so we can move (along a circle of
constant real part = 1.0) into the center
of the Smith Chart. We have to find
the length
L
2
L
2
for the second stub, but that is now easy! (Figure 10)
Thus, by doing
double stub matching, we are able,
by adding the additional degree of freedom of two adjustable
stubs, not to have to specify exactly where the stubs have to be
placed, so they can be in the line before the matching is
attempted.
Here's the whole sequence
of changes that we made. See if you can begin at "Start" and go
through the numbers
0→5
0
5
and get from
Z
L
Z
0
Z
L
Z
0
to the matching point at the center of the Smith
Chart. Remember, when we move from one place to another on the
line, we must stay on a circle of constant radius. When we
change reactance by adjusting a stub, we must move along circles
of constant real part. If you do that, it's easy!
There's just one little problem. What if
Y
A
Y
0
Y
A
Y
0
without stub had ended up as shown in
here. We are on the
ℜ
Y
A
Y
0
|
without stub
=2.0
Y
A
Y
0
|
without stub
2.0
circle. No matter how hard I try, and no matter where I set
L
1
L
1
all I can do is spin around on the little circle as
shown, and I will never end up on
the rotated matching circle, and I won't be able to make a
match! Well, if I add a
third stub...I'll
let you work it out!
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