Question 19.8: Find the transmission parameters for the two-port network in...

To determine A and C, we leave the output port open as in Fig. 19.33(a) so that \pmb{I_2} = 0 and place a voltage source \pmb{V_1} at the input port. We have

\pmb{V _1}=(10+20) \pmb{I _1}=30 \pmb{I _1} \quad \text { and } \quad \pmb{V _2}=20 \pmb{I _1}-3 \pmb{I _1}=17 \pmb{I _1}

Thus,

\pmb{A} = \pmb{\frac{ V _1}{ V _2}}=\frac{30 \pmb{I _1}}{17 \pmb{I _1}}=1.765, \quad \pmb{C} = \pmb{\frac{ I _1}{ V _2}}=\frac{ \pmb{I _1}}{17 \pmb{I _1}}=0.0588  S

To obtain B and D, we short-circuit the output port so that \pmb{V_2} = 0 as shown in Fig. 19.33(b) and place a voltage source \pmb{V_1} at the input port. At node a in the circuit of Fig. 19.33(b), KCL gives

\frac{\pmb{ V _1- V _a}}{10}-\frac{ \pmb{V _a}}{20}+ \pmb{I _2}=0                     (19.8.1)

But \pmb{V _a}=3 \pmb{I _1} \text { and } \pmb{I _1}=\left( \pmb{V _1- V _a}\right) / 10 . Combining these gives

\pmb{V _a}=3 \pmb{I _1} \quad \pmb{V _1}=13 \pmb{I _1}                    (19.8.2)

Substituting \pmb{V_a} = 3\pmb{I_1} into Eq. (19.8.1) and replacing the first term with \pmb{I_1},

\pmb{I _1}-\frac{3 \pmb{I _1}}{20}+ \pmb{I _2}=0 \quad \Rightarrow \quad \frac{17}{20} \pmb{I _1}=- \pmb{I _2}

Therefore,

\pmb{D} =-\pmb{\frac{ I _1}{ I _2}}=\frac{20}{17}=1.176, \quad \pmb{B} =-\pmb{\frac{ V _1}{ I _2}}=\frac{-13 \pmb{I _1}}{(-17 / 20) \pmb{I _1}}=15.29  \Omega