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FEED LINE BASICS 333

Figure 6.7.3 (N+1)-step coax-to-coax transition: a) schematic, b) Reflection coefficient vs.

As the incident wave coming from the line of impedance reaches the step #0 in the cross
0
section AA it splits into two waves: one of magnitude moves back due to reflection from
0
step #1 while the second one with magnitude continues to propagate as incident wave in the
1
line of impedance . The reflection coefficient can be estimated from the expression (3.87)
1
0
⁄
of Chapter 3 and equals to = ( − ) ( 1 + ). We omitted the subscript e in as
1
0
0
0
implying. Assuming that the impedance serge is small enough such as | 1 − | ≪
0
| + | we can come to conclusion that the power of the reflected wave is proportional
1
0
to | | ≪ 1. It is worthwhile to point out that this condition is fulfilled in wide range of
2
0
2
matching circuits. In our case it means that the wave is proportional to = �1 − | | ≅
1
1
0
1 − | | /2 ≅ 1, i.e. it has magnitude practically the same as in line 0. Then the wave
2
0
1
reaches the cross section BB with the phase delay − 1 and splits into two waves: one of
magnitude () = − 1 , where = ( − ) ( + )⁄ , moves back due to
1 1 1 2 1 2 1
reflection from step #2 and the second one continues to propagate as incident wave in the
2
line of impedance . Assuming again that the impedance surge is small enough, we can come
2
to conclusion that = �1 − | ()| ≅ 1, i.e. practically the same as in line 0 and 1.
2
1
2
Meanwhile, the reflected wave () reaches the cross-section AA with extra phase
1
delay − 1 , i.e. in this cross section () = () − 1 = −21 . Assuming one more
1
1
1
time the same approximation of small impedance surge we can establish that two reflected
waves appear in cross section AA as a single reflected wave of magnitude = + −2 1 .
0

1
Continuing this procedure further, we can define the whole reflection coefficient in phasor form
as

−2( 1 + 2 ) −2 ∑ −2 ∑ =0 (6.45)
−2 1

= + + + ⋯ + =1 = ∑ =0

2
0
1
Here = 0, is the propagation constant supposedly, the same in each line section, is the

0
length of n line section between the adjacent steps, and ≪ 1 is the reflection coefficient
th

th
from m step
+1 −
= , = 1,2,3, … , (6.46)

+1 +
Evidently, the expression (6.45) is only the first order approximation, but very useful letting
solve with some restrictions the synthesis problem, i.e. find the number and length of line
sections and the impedance of each of them while the combined reflection coefficient is lower
than a given value over a given frequency band. This inverse task in such general formulation
can be solved only through extensive numerical simulation that is complicated by the lack of
unique solution. There are many different combinations of transition elements that satisfy the