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Green's recursive element formula is
Network Elements for Three Source Conditions 201
4 sin[ (2r- I)O]sin[ (2r+ I)O]lg,
g,+, = sinh 2 a + sinh 2 b+ sin 2 (2rO) - 2 sinh a· sinh b· cos(2rO) ,
(6.72)
for r= 1,2•...,n-1. The source series resistance or shunt conductance shown
in Figure 6.18 is
2 sinO
gn+'= g. sinh a + sinh b (6.73)
Note that tbe source resistance or conductance is dependent. This must be
accepted in lowpass networks, but Section 6.5 will show how to provide for
fairly arbitrary source resistance levels in corresponding bandpass networks.
Program B6-3 in Appendix B contains Newton's method (Program B6-2)
without the print statements; it also performs the prototype element calculations
in (6.70)-(6.73).
Example 6.11. Find the prototype element values for an n = 3 network that
optimally matches a load impedance with Q L = 3 over a 50% bandwidth. The
SWR ripple is shown in Table 6.3. Also, as a result of the Newton-Raphson
iterative solution, a=0.8730 and b=0.3163. Program B6-3 continues to compute
g,=1.5, g2=0.8817, g,=1.0561, and &,=0.7229. According to Figure
6.18, when n is odd, the gn + I value is the necessary source resistance. Also,
note that g, is simply the inverse load decrement according to (6.49).
6.4.2. Complex Source and Complex Load. For a complex source and
complex load, the given load decrement is 8, = 1/g,. From (6.59),
8 = sinha-sinh b
J 2sinO '
(6.74)
where 0 is given by (6.71). The source is now assumed to have a single
reactance as well as a resistance. This is another way to assign the single
degree of freedom identified in Section 6.3.2. Figure 6.18 shows that the
source decrement is
8
n
= __I_.
~gn+1
However, using (6.73), the source decrement can also be expressed as
8 = sinh a+ sinh b
n 2sinO .
(6.75)
(6.76)
Given the source and load decrements, simultaneous solution of (6.74) and
(6.76) for a and b is possible:
sinha=(8 n + 8dsinO,
(6.77)
sinh b=(8 n -8,)sinO.
(6.78)