Question 17.10: The networks Na and Nb in Figure 17–15 are connected in casc......
The networks N_{a} and N_{b} in Figure 17–15 are connected in cascade and have two-port matrices
[\pmb{t}_{a}] =\begin{bmatrix} 3 &25\\0.2& 2 \end{bmatrix} and [\pmb{z}_{b}] =\begin{bmatrix}200& 0\\-2000& 20 \end{bmatrix} Ω
Find the current gain of this cascade connection.
To get a two-port description of this connection we use the t-parameters since [\pmb{t}] =[\pmb{t}_{a}][\pmb{t}_{b}] in a cascade connection. We are given the t-matrix of N_{a} and the z-matrix of N_{b}. Given the matrix [\pmb{z}_{b}], the z-parameters of N_{b} are seen to be z_{11b} = 200 Ω, z_{12b} = 0, z_{21b} = -2000 Ω, z_{2b} = 20 Ω, and Δ_{zb} = z_{11b}z_{22b} – z_{12b}z_{21b} = 4000. To get the matrix [\pmb{t}_{b}], we use the [\pmb{z}] to [\pmb{t}] conversion shown in Table 17–2.
[\pmb{t}_{b}]=\begin{bmatrix}\frac{ z_{11b}}{z_{21b}}&\frac{Δ_{zb}}{z_{21b}}\\\frac{1}{z_{21b}}&\frac{z_{22b}}{z_{21b}} \end{bmatrix}=\begin{bmatrix}-0.1&-2\\-5 × 10^{-4}& -0.01\end{bmatrix}
Now using [\pmb{t}]= [\pmb{t}_{a}][\pmb{t}_{b}] yields the transmission matrix of the cascade connection as
[\pmb{t}]= [\pmb{t}_{a}][\pmb{t}_{b}]=\begin{bmatrix}3 &25\\0.2 &2\end{bmatrix}\begin{bmatrix}-0.1&-2\\-5 × 10^{-4}& -0.01\end{bmatrix}=\begin{bmatrix}-0.313& -6.25\\-0.021 &-0.420\end{bmatrix}
Given the matrix [t], the t-parameters of the connection are seen to be A = –0.313, B = –6.25 Ω, C = –0.021 S, and D = –0.420. Using the appropriate relationship in Table 17–3 gives us the current gain of the connection.
T_{I} = \frac{-1}{D} = \frac{-1}{-0.420} = 2.381
T A B L E 17–2 TWO-PORT PARAMETER CONVERSION TABLE
| DESIRED PARAMETERS | GIVEN PARAMETERS | |||
| [Z] | [Y] | [H] | [T] | |
| [z] | \begin{bmatrix} z_{11}& z_{12}\\z_{21}& z_{22} \end{bmatrix} | \begin{bmatrix} \frac{y_{22}}{Δ_{y}}&\frac{-y_{12}}{Δ_{y}}\\\frac{-y_{21}}{Δ_{y}}&\frac{y_{11}}{Δ_{y}} \end{bmatrix} | \begin{bmatrix} \frac{Δ_{h}}{h_{22}}&\frac{h_{12}}{h_{22}}\\\frac{-h_{21}}{h_{22}}&\frac{1}{h_{22}}\end{bmatrix} | \begin{bmatrix} \frac{A}{C}&\frac{Δ_{t}}{C}\\\frac{1}{C}&\frac{D}{C}\end{bmatrix} |
| [y] | \begin{bmatrix} \frac{z_{22}}{Δ_{z}}&\frac{-z_{12}}{Δ_{z}}\\\frac{-z_{21}}{Δ_{z}}&\frac{z_{11}}{Δ_{z}}\end{bmatrix} | \begin{bmatrix} y_{11}&y_{12}\\y_{21}& y_{22} \end{bmatrix} | \begin{bmatrix} \frac{1}{h_{11}}&\frac{-h_{12}}{h_{11}}\\\frac{h_{21}}{h_{11}}&\frac{Δ_{h}}{h_{11}} \end{bmatrix} | \begin{bmatrix} \frac{D}{B}&\frac{-Δ_{t}}{B}\\\frac{-1}{B}&\frac{A}{B} \end{bmatrix} |
| [h] | \begin{bmatrix} \frac{Δ_{z}}{z_{22}}&\frac{z_{12}}{z_{22}}\\\frac{-z_{21}}{z_{22}}&\frac{1}{z_{22}} \end{bmatrix} | \begin{bmatrix}\frac{1}{y_{11}}& \frac{-y_{12}}{y_{11}}\\\frac{y_{21}}{y_{11}}&\frac{Δ_{y}}{y_{11}} \end{bmatrix} | \begin{bmatrix} h_{11}& h_{12}\\h_{21}& h_{22} \end{bmatrix} | \begin{bmatrix} \frac{B}{D}&\frac{Δ_{t}}{D}\\\frac{-1}{D}&\frac{C}{D} \end{bmatrix} |
| [t] | \begin{bmatrix}\frac{z_{11}}{z_{21}}&\frac{Δ_{z}}{z_{21}}\\\frac{1}{z_{21}}&\frac{z_{22}}{z_{21}} \end{bmatrix} | \begin{bmatrix}\frac{-y_{22}}{y_{21}}&\frac{-1}{y_{21}}\\\frac{-Δ_{y}}{y_{21}}&\frac{-y_{11}}{y_{21}} \end{bmatrix} | \begin{bmatrix} \frac{-Δ_{h}}{h_{21}}&\frac{-h_{11}}{h_{21}}\\\frac{-h_{22}}{h_{21}}&\frac{-1}{h_{21}} \end{bmatrix} | \begin{bmatrix} A& B\\C& D \end{bmatrix} |
T A B L E 17–3 VOLTAGE GAIN AND CURRENT GAIN IN TERMS OF TWO-PORT PARAMETERS
| DEFINITION | GIVEN PARAMETERS | |||
| [Z] | [Y] | [H] | [T] | |
| T_{V}=\frac{V_{2}}{V_{1}}|_{I_{2}=0} | T_{V}=\frac{z_{21}}{z_{11}} | T_{V}=\frac{-y_{21}}{y_{22}} | T_{V}=\frac{-h_{21}}{Δ_{h}} | T_{V}=\frac{1}{A} |
| T_{I}=\frac{I_{2}}{I_{1}}|_{V_{2}=0} | T_{I}=\frac{-z_{21}}{z_{22}} | T_{I}=\frac{y_{21}}{y_{11}} | T_{I}= h_{21} | T_{I}=\frac{-1}{D} |