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202 Impedance Matching
Since sinh is an odd function and parameter b must be positive, (6.78) requires
that 0" > 0,. The prototype g, recursion (6.72) was presented as starting at the
g,,= I-ohm end of the network; for this complex source/load case, the I-ohm
end must be the lower decrement end, whether it is the load or the source.
The design procedure is to solve (6.77) and (6.78) and use these values in
(6.70)-(6.72) to obtain the element values; start with g, equal to the reciprocal
of the lesser decrement. The ending real element is again dependent and given
by (6.73) or by rewriting (6.75):
I
gn+l=-g°'
(6.79)
" "
using the greater decrement for 0". The central (n-2) elements in Figure 6.18
constitute the matching network. The prototype elements are numbered as
shown in Figure 6.18 if the load decrement is less than the source decrement;
i.e., the I-ohm end belongs to the lower decrement. If the source decrement is
less than the load decrement, then the source is normalized to I ohm and the
g, values from (6.72) are generated from the source end to the load end.
Example 6.12. Suppose that both source and load terminations included
shunt capacitors with decrements of 1.35 and 1.25, respectively. Find the
lowest-order matching network and its range of passband SWR. Figure 6.18
shows that only odd-degree networks can have shunt capacitors at both ends.
Choosing n = 3, (6.77) and (6.78) yield sinh a = 1.300 and sinh b = 0.050. Since
load reactance 8, is a reciprocal decrement, g, =0.8. Using (6.72),82= 1.0514
and g,=0.7585. By (6.79), the source resistance is &=0.9766 ohms. The SWR
ranges from 1.0240 to 1.1726 according to (6.57), (6.58), and (4.59).
The networks discussed in this section incorporate single-reactance sources
and loads exactly. However, they may not have the least possible SWR m
.,
when both given decrements are less than the source decrement obtained by
the optimal network in the preceding section. When this is the case, the
"optimal" g" reactance or susceptance may be increased (as part of the
matching network) and thereby decrease the decrement to the higher of the
given values (see (6.75)). Therefore, given two values for source and load
decrement, the lesser of the two should be used first in Program B6-3. The
resulting source decrement should then be computed by (6.75); if it is greater
than the given decremenis, the "optimal" network should be used, with g"
increased as described.
Example 6.13. The lesser of the two decrements in Example 6.12 was 1.25,
which is equivalent to load decrement g, = 0.8. Using Program B6-3 (with
n=3, QL=0.8, and BW= 100%, according to (6.49», obtain g,=0.8, g2=
0.9484, g, = 0.6424, and & =0.9211. By (6.75), 0, ~ 1.6900, which is greater
than the 1.35 decrement given. In fact, (6.75) shows that a 1.35 decrement
corresponds to a g, value of 0.8042 for & ~ 0.9211. Therefore, the best solution