Question 19.9: The ABCD parameters of the two-port network in Fig. 19.34 ar...
The ABCD parameters of the two-port network in Fig. 19.34 are
\left[\begin{array}{cc} 4 & 20 \Omega \\ 0.1 S & 2 \end{array}\right]
The output port is connected to a variable load for maximum power transfer. Find R_L and the maximum power transferred.
What we need is to find the Thevenin equivalent (\pmb{Z_{Th}} \text{ and } \pmb{V_{Th}}) at the load or output port. We find \pmb{Z_{Th}} using the circuit in Fig. 19.35(a). Our goal is to get \pmb{Z_{Th}} = \pmb{V_2/I_2}. Substituting the given ABCD parameters into Eq. (19.22), we obtain
(19.22): \pmb{V _1}= \pmb{A V _2- B I _2} \\\pmb{I _1}= \pmb{C V _2- D I _2}
\pmb{V _1}=4 \pmb{V _2}-20 \pmb{I _2} (19.9.1)
\pmb{I _1}=0.1 \pmb{V _2}-2 \pmb{I _2} (19.9.2)
At the input port, \pmb{V_1} = −10\pmb{I_1}. Substituting this into Eq. (19.9.1) gives
-10 \pmb{I _1}=4 \pmb{V _2}-20 \pmb{I _2}
or
\pmb{I _1}=-0.4 \pmb{V _2}+2 \pmb{I _2} (19.9.3)
Setting the right-hand sides of Eqs. (19.9.2) and (19.9.3) equal,
0.1 \pmb{V _2}-2 \pmb{I _2}=-0.4 \pmb{V _2}+2 \pmb{I _2} \quad \Rightarrow \quad 0.5 \pmb{V _2}=4 \pmb{I _2}
Hence,
\pmb{Z _{ Th }}=\pmb{\frac{ V _2}{ I _2}}=\frac{4}{0.5}=8 \Omega
To find \pmb{V_{Th}}, we use the circuit in Fig. 19.35(b). At the output port \pmb{I_2} = 0 and at the input port \pmb{V_1} = 50 − 10\pmb{I_1}. Substituting these into Eqs. (19.9.1) and (19.9.2),
50-10 \pmb{I _1}=4 \pmb{V _2} (19.9.4)
\pmb{I _1}=0.1 \pmb{V _2} (19.9.5)
Substituting Eq. (19.9.5) into Eq. (19.9.4),
50- \pmb{V _2}=4 \pmb{V _2} \quad \Rightarrow \quad \pmb{V _2}=10
Thus,
\pmb{V _{ Th }}= \pmb{V _2}=10 V
The equivalent circuit is shown in Fig. 19.35(c). For maximum power transfer,
R_L= \pmb{Z _{ Th }}=8 \Omega
From Eq. (4.24), the maximum power is
(4.24): p_{\max }=\frac{V_{ Th }^2}{4 R_{ Th }}
P=I^2 R_L=\left(\frac{ \pmb{V _{ Th }}}{2 R_L}\right)^2 R_L=\frac{ \pmb{V ^2_{ Th }}}{4 R_L}=\frac{100}{4 \times 8}=3.125 W
