Question 19.14: Find the transmission parameters for the circuit in Fig. 19....

We can regard the given circuit in Fig. 19.46 as a cascade connection of two T networks as shown in Fig. 19.47(a). We can show that a T network, shown in Fig. 19.47(b), has the following transmission parameters [see Prob. 19.52(b)]:

\pmb{A} =1+\frac{R_1}{R_2}, \quad \pmb{B} =R_3+\frac{R_1\left(R_2+R_3\right)}{R_2} \\ \space \\ \pmb{C} =\frac{1}{R_2}, \quad \pmb{D} =1+\frac{R_3}{R_2}

Applying this to the cascaded networks N_a \text{ and } N_b in Fig. 19.47(a), we get

\pmb{A_a} = 1 + 4 = 5, \quad \pmb{B_a} = 8 + 4 × 9 = 44  Ω \\ \space \\ \pmb{C_a} = 1  S, \quad \pmb{D_a} = 1 + 8 = 9

or in matrix form,

\left[ \pmb{T _a}\right]=\left[\begin{array}{cc} 5 & 44  \Omega \\ 1  S & 9 \end{array}\right]

and

\pmb{A _b}=1, \quad \pmb{B _b}=6  \Omega, \quad \pmb{C _b}=0.5  S , \quad \pmb{D _b}=1+\frac{6}{2}=4

i.e.,

\left[ \pmb{T _b}\right]=\left[\begin{array}{cc} 1 & 6  \Omega \\ 0.5  S & 4 \end{array}\right]

Thus, for the total network in Fig. 19.46,

[\pmb{ T} ]=\left[ \pmb{T _a}\right]\left[ \pmb{T _b}\right]=\left[\begin{array}{cc} 5 & 44 \\ 1 & 9 \end{array}\right]\left[\begin{array}{cc} 1 & 6 \\ 0.5 & 4 \end{array}\right] \\ \space \\=\left[\begin{array}{cc} 5 \times 1+44 \times 0.5 & 5 \times 6+44 \times 4 \\ 1 \times 1+9 \times 0.5 & 1 \times 6+9 \times 4 \end{array}\right] \\ \space \\=\left[\begin{array}{cc} 27 & 206  \Omega \\ 5.5  S & 42 \end{array}\right]

Notice that

\Delta_{T_a}=\Delta_{T_b}=\Delta_T=1

showing that the network is reciprocal.