Question 19.12: Evaluate V2∕Vs in the circuit in Fig. 19.42.

This may be regarded as two two-ports in series. For N_b,

\pmb{z}_{12b} = \pmb{z}_{21b} = 10 = \pmb{z}_{11b} = \pmb{z}_{22b}

Thus,

[ \pmb{z} ]=\left[ \pmb{z} _a\right]+\left[ \pmb{z} _b\right]=\left[\begin{array}{cc} 12 & 8 \\ 8 & 20 \end{array}\right]+\left[\begin{array}{cc} 10 & 10 \\ 10 & 10 \end{array}\right]=\left[\begin{array}{cc} 22 & 18 \\ 18 & 30 \end{array}\right]

But

\pmb{V _1}= \pmb{z _{11} I _1+ z _{12} I _2}=22 \pmb{I _1}+18 \pmb{I _2}                 (19.12.1)

\pmb{V _2}= \pmb{z _{21} I _1+ z _{22} I _2}=18 \pmb{I _1}+30 \pmb{I _2}                 (19.12.2)

Also, at the input port

\pmb{V _1}= \pmb{V _s}-5 \pmb{I _1}                            (19.12.3)

and at the output port

\pmb{V _2}=-20 \pmb{I _2} \quad \Rightarrow \quad \pmb{I _2}=-\frac{ \pmb{V _2}}{20}                     (19.12.4)

Substituting Eqs. (19.12.3) and (19.12.4) into Eq. (19.12.1) gives

\pmb{V _s}-5 \pmb{I _1}=22 \pmb{I _1}-\frac{18}{20} \pmb{V _2} \quad \Rightarrow \quad \pmb{V _s}=27 \pmb{I _1}-0.9 \pmb{V _2}                            (19.12.5)

while substituting Eq. (19.12.4) into Eq. (19.12.2) yields

\pmb{V _2}=18 \pmb{I _1}-\frac{30}{20} \pmb{V _2} \quad \Rightarrow \quad \pmb{I _1}=\frac{2.5}{18} \pmb{V _2}                     (19.12.6)

Substituting Eq. (19.12.6) into Eq. (19.12.5), we get

\pmb{V _s}=27 \times \frac{2.5}{18} \pmb{V _2}-0.9 \pmb{V _2}=2.85 \pmb{V _2}

And so,

\pmb{\frac{ V _2}{ V _s}}=\frac{1}{2.85}=0.3509