Find the transmission parameters for the two-port network in Fig. 19.32.

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Accepted Answer

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

[latex] \pmb{V _1}=(10+20) \pmb{I _1}=30 \pmb{I _1} \quad ext { and } \quad \pmb{V _2}=20 \pmb{I _1}-3 \pmb{I _1}=17 \pmb{I _1}[/latex]

Thus,

[latex] \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[/latex]

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

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

But [latex] \pmb{V _a}=3 \pmb{I _1} ext { and } \pmb{I _1}=\left( \pmb{V _1- V _a} ight) / 10 [/latex]. Combining these gives

[latex] \pmb{V _a}=3 \pmb{I _1} \quad \pmb{V _1}=13 \pmb{I _1}[/latex] (19.8.2)

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

[latex] \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}[/latex]

Therefore,

[latex] \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[/latex]

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

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