
534 Network Analysis and Synthesis
Now, substitute the values of A and B in equation (12.14),
Zs
ss
()=+
+
+
↓↓
3
2
1
2
2
1
↓
==
==
∞=
PP P
C
P
R
P
02 1
0
0
2
2
2
3
2
1
2
1
,,
,,
s
F
F
RP
C
P
∞∞
==
×
=
==
1
2
3
1
22
1
4
1
2
2
2
Ω
Ω
The Foster form-I circuit is shown in Figure 12.19.
For Foster form-II, we write
Zs
()
()
()
==
Let us use the partial fractions as follows:
ss
s
s
()
()()
=
=
+
13
(12.15)
Let us find values of A and B
ss
ss
()()
()
++
=
++
13
or s(s + 2) = A (s + 3) + B(s + 1) (12.16)
To find A, substitute s = −1 in equation (12.16),
−1(−1 + 2) = A (−1 + 3) + 0
−1(1) = 2A
or
A =−
Since none of the coefficient of Z(s) should ...