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Narrow-Bond L, T, and Pi Networks ]79
2 '" 9
20-j10 n -3"-1- "
L network
"
ZL '" 6+j12 no
t'igure 6.7.
Problem for Example 6.4 considering both kinds of L-section networks.
Example 6.4. Consider the matching problem illustrated in Figure 6.7, where
Zg=20-jI0 ohms and ZL =6+jI2 ohms, and both types of L sections are to
be used. The parallel equivalent of the source is 2511-j50, where the symbol II
will be used to mean in parallel with. The type-B L section will thus require
solutions from a 6-ohin resistance to a 25-ohm resistance, as obtained in
Example 6.1 (X, = ± 14.05 and X,~ =+= 10.68 ohms reactance; see Figure 6.2).
To minimize confusion, the reactances inside the type-B matching network
will be designated X, and X b
, as shown in Figure 6.8. The load was given in
series form, and its reactance can become a part of the hypothetical matching
element X" as shown in Figure 6.8. Then X b = -1.32 ohms by subtraction.
Use Program A6-1 to find X,: enter -14.05, then -50, and press key D. This
evaluates (6.21) and yields X, = - 19.54 ohms. As a check, convert the load
mesh, 6+jlO.68 ohms, to parallel form (25.01)JjI4.05). Then combine parallel
reactances jI4.0511-jI9.54, using key C, to obtain the equivalent +j50.01
ohms. Figure 6.8 shows that this reactance will be canceled by the source
reactance, leaving a match to the 25-ohm parallel resistance in the source.
Note that the matching network actually used is composed of two capacitors;
jX 1
~
r---,.----,---r-- jX b -----+j12
'----,---
t 25 -j50 jX~
6r!
X 2 + 10.68
Xl -14.05
X h - 1.32
X a
- 19.54
X inp 50
R inp 25
-10.68
+ 14.05
- 22.68
+ 10.97
50
25
Figure 6.8. Solutions for a type-B L-section in Example 6.4.