Just a few odds and ends. Consider the following which is called a "cascaded line"
problem. These are problems where we have two different
transmission lines, with different characteristic
impedances. Since we will give all of the distances in
wavelengths, λ, we will assume that the λ we are
talking about is the appropriate one for the line involved. If
the phase velocities on the two lines is the same, then the
physical lengths would correspond as well. The approach is
relatively straight-forward. First let's plot
Z
L
Z
0
Z
L
Z
0
on the Smith Chart. Then we
have to rotate
0.2λ
0.2
λ
so that we can find
Z
A
Z
0
1
Z
A
Z
0
1
, the normalized impedance at point A, the junction
between the two lines Figure 3.
Thus, we find
Z
A
Z
0
1
=0.32+0.6j
Z
A
Z
0
1
0.320.6
. Now we have to
renormalize the impedance
so we can move to the line with the new impedance
Z
0
2
Z
0
2
. Since
Z
0
1
=300Ω
Z
0
1
300
Ω
,
Z
A
=96+-180j
Z
A
96-180
. This is the load for the second length of line, so let's find
Z
A
Z
0
2
Z
A
Z
0
2
, which is easily found to be
1.9+-3.6j
1.9-3.6, so this can be plotted on the
Smith Chart. Now we have to rotate around
another
0.15λ
0.15
λ
so that we can find
Z
in
Z
0
2
Z
in
Z
0
2
. This appear to have a value of about
0.15+-0.45j
0.15-0.45, so
Z
in
=7.5+-22.5jΩ
Z
in
7.5-22.5
Ω
Figure 5.
There is one application of the cascaded line problem that is
used quite a bit in practice. Consider the following: We assume
that we have a matched line with impedance
Z
0
2
Z
0
2
and we connect it to another line whose impedance is
Z
0
1
Z
0
1
Figure 6. If we connect the two of them
together directly, we will have a reflection coefficient at the
junction given by
Γ=
Z
0
2
−
Z
0
1
Z
0
2
+
Z
0
1
Γ
Z
0
2
Z
0
1
Z
0
2
Z
0
1
(1)
Now let's imagine that we have inserted a section of line with length
l=λ4
l
λ
4
and impedance
Z
m
Z
m
Figure 7. At point A, the junction
between the first line and the matchng section, we can find the
normalized impedance as
Z
A
Z
M
=
Z
0
2
Z
m
Z
A
Z
M
Z
0
2
Z
m
(2)
We take this impedence and rotate around on the Smith Chart
λ4
λ
4
to find
Z
B
Z
M
Z
B
Z
M
Z
B
Z
M
=
Z
m
Z
0
2
Z
m
Z
B
Z
M
Z
m
Z
0
2
Z
m
(3)where we have taken advantage of the fact that when we
go half way around the Smith Chart, the impedance we get is just
the inverse of what we had originally (half way around turns
rs
r
s
into
−rs
r
s
).
Thus
Z
B
=
Z
m
2
Z
0
2
Z
B
Z
m
2
Z
0
2
(4)
If we want to have a match for line with impedence
Z
0
1
Z
0
1
, then
Z
B
Z
B
should equal
Z
0
1
Z
0
1
and hence:
Z
B
=
Z
0
1
=
Z
m
2
Z
0
2
Z
B
Z
0
1
Z
m
2
Z
0
2
(5)
or
Z
m
=
Z
0
1
Z
0
2
Z
m
Z
0
1
Z
0
2
(6)
This piece of line is called a
quarter wave matching
section and is a convenient way to connect two lines of
different impedance.
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