Question 19.3: Obtain the y parameters for the II network shown in Fig. 19.......
Obtain the y parameters for the \prod{} network shown in Fig. 19.14.
■ METHOD 1 To find y_{11} and y_{21} , short-circuit the output port and connect a current source I_{1} to the input port as in Fig. 19.15(a). Since the 8-Ω resistor is short circuited, the 2-Ω resistor is in parallel with the 4-Ω resistor. Hence,
V_{1} = I_{1}(4 \parallel 2) = \frac{4}{3} I_{1} , y_{11} = \frac{I_{1}}{V_{1}} = \frac{I_{1}}{\frac{4}{3}I_{1}} = 0.75 S
By current division,
-I_{2} = \frac{4}{4 + 2} I_{1} = \frac{2}{3} I_{1} , y_{21} = \frac{I_{2}}{V_{1}} = \frac{- \frac{2}{3} I_{1}}{\frac{4}{3} I_{1}} = – 0.5 S
To get y_{12} and y_{22} , short-circuit the input port and connect a current source I_{2} to the output port as in Fig. 19.15(b). The 4-Ω resistor is short-circuited so that the 2-Ω and 8-Ω resistors are in parallel.
V_{2} = I_{2}(8 \parallel 2) = \frac{8}{5} I_{2} , y_{22} = \frac{I_{2}}{V_{2}} = \frac{I_{2}}{\frac{8}{5}I_{2}} = 0.625 S
By current division,
-I_{1} = \frac{8}{8 + 2} I_{2} = \frac{4}{5} I_{2} , y_{12} = \frac{I_{1}}{V_{2}} = \frac{- \frac{4}{5} I_{2}}{\frac{8}{5} I_{2}} = – 0.5 S
■ METHOD 2 Alternatively, comparing Fig. 19.14 with Fig. 19.13(a),
y_{12} = – \frac{1}{2} S = y_{21}
y_{11} + y_{12} = \frac{1}{4} ⇒ y_{11} = \frac{1}{4} – y_{12} = 0.75 S
y_{22} + y_{12} = \frac{1}{8} ⇒ y_{22} = \frac{1}{8} – y_{12} = 0.625 S
as obtained previously.
