The passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port $1$.
Given
$V_{1}=A_{1}V_{2}+B_{1}I_{2}$
$I_{1}=C_{1}V_{2}+D_{1}I_{2}$
$V_{2}=A_{2}V_{3}+B_{2}I_{3}$
$I_{2}=C_{2}V_{3}+D_{2}I_{3}$
$A_{1}, B_{1}, C_{1}, D_{1}, A_{2}, B_{2}, C_{2},$ and $D_{2}$ are the generalized circuit constants. If the Thevenin equivalent circuit at port $3$ consists of a voltage source $V_{T}$ and an impedance $Z_{T}$, connected in series, then
- $V_{T}=\frac{V_{1}}{A_{1}A_{2}}, Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}+B_{1}C_{2}}$
- $V_{T}=\frac{V_{1}}{A_{1}A_{2}+B_{1}C_{2}}, Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}}$
- $V_{T}=\frac{V_{1}}{A_{1}+A_{2}}, Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}+A_{2}}$
- $V_{T}=\frac{V_{1}}{A_{1}A_{2}+B_{1}C_{2}}, Z_{T}=\frac{A_{1}B_{2}+B_{1}D_{2}}{A_{1}A_{2}+B_{1}C_{2}}$