GATE 2024 EE – Question 48
Two passive two-port networks $\mathbf P$ and $\mathbf Q$ are connected as shown in the figure. The impedance matrix of network $\mathbf P$ is $Z_{\mathbf P}=\begin{bmatrix}40\ \Omega&60\ \Omega\\80\ \Omega&100\ \Omega\end{bmatrix}$. The admittance matrix of network $\mathbf Q$ is $Y_{\mathbf Q}=\begin{bmatrix}5\text{ S}&-2.5\text{ S}\\-2.5\text{ S}&1\text{ S}\end{bmatrix}$. Let the ABCD matrix of the two-port network $\mathbf R$ in the figure be $\begin{bmatrix}\alpha&\beta\\\gamma&\delta\end{bmatrix}$. The value of $\beta$ in Ω is _____________ (rounded off to 2 decimal places).
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Correct answer: -19.90 to -19.70
Explanation
The networks are in cascade, so $ABCD_R=ABCD_P\cdot ABCD_Q$. For P (from Z): $A=\dfrac{z_{11}}{z_{21}}=0.5$, $B=\dfrac{\Delta z}{z_{21}}=\dfrac{-800}{80}=-10$, $C=\dfrac1{z_{21}}=0.0125$, $D=\dfrac{z_{22}}{z_{21}}=1.25$. For Q (from Y): $A=-\dfrac{y_{22}}{y_{21}}=0.4$, $B=-\dfrac1{y_{21}}=0.4$, $C=-\dfrac{\Delta y}{y_{21}}=-0.5$, $D=-\dfrac{y_{11}}{y_{21}}=2$. Then $\beta=A_PB_Q+B_PD_Q=0.5(0.4)+(-10)(2)=-19.80$.