GATE 2024 EE Q.48: solution and answer

GATE 2024 · EEQ.48 · 2 marks · NAT

Two passive two-port networks $P$ and $Q$ are connected as shown in the figure. The impedance matrix of network $P$ is $Z_{P} = [40 Ω 80 Ω 60 Ω 100 Ω]$ . The admittance matrix of network $Q$ is $Y_{Q} = [5 S - 2.5 S - 2.5 S 1 S]$ . Let the ABCD matrix of the two-port network $R$ in the figure be $[α γ β δ]$ . The value of $β$ in $Ω$ is ________ (rounded off to 2 decimal places).

GATE 2024 EE question 48: Two boxes connected in cascade. — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

ANSWER

$β = - 19.80 Ω$ .

Official answer: -19.90 to -19.70Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Cascade rule. R is P followed by Q in tandem, so $T_{R} = T_{P} T_{Q}$ . Use the convention $V_{1} = A V_{2} - B I_{2}, I_{1} = C V_{2} - D I_{2}$ . With it, the output pair $(V_{2}, - I_{2})$ of P is exactly the input pair $(V_{1}, I_{1})$ of Q, so the ABCD matrices multiply in the order the signal passes through them.
Z to ABCD for P. From $V_{2} = z_{21} I_{1} + z_{22} I_{2}$ we get $I_{1} = \frac{1}{z _{21}} V_{2} - \frac{z _{22}}{z _{21}} I_{2}$ . Substituting into $V_{1} = z_{11} I_{1} + z_{12} I_{2}$ gives $V_{1} = \frac{z _{11}}{z _{21}} V_{2} - \frac{Δ z}{z _{21}} I_{2}$ . So $A = \frac{z _{11}}{z _{21}}, B = \frac{Δ z}{z _{21}}, C = \frac{1}{z _{21}}, D = \frac{z _{22}}{z _{21}}$ , where $Δ z = z_{11} z_{22} - z_{12} z_{21} = 40 (100) - 60 (80) = - 800 Ω^{2}$ .
$T_{P} = \frac{40}{80} \frac{1}{80} \frac{- 800}{80} \frac{100}{80} = [0.5 0.0125 S - 10 Ω 1.25]$ Check: $A D - B C = 0.625 + 0.125 = 0.75 = z_{12} / z_{21}$ , as it should be for a non-reciprocal network.
Y to ABCD for Q. From $I_{2} = y_{21} V_{1} + y_{22} V_{2}$ we get $V_{1} = - \frac{y _{22}}{y _{21}} V_{2} + \frac{1}{y _{21}} I_{2}$ . Substituting into $I_{1} = y_{11} V_{1} + y_{12} V_{2}$ gives $I_{1} = - \frac{Δ y}{y _{21}} V_{2} + \frac{y _{11}}{y _{21}} I_{2}$ . So $A = - \frac{y _{22}}{y _{21}}, B = - \frac{1}{y _{21}}, C = - \frac{Δ y}{y _{21}}, D = - \frac{y _{11}}{y _{21}}$ , where $Δ y = y_{11} y_{22} - y_{12} y_{21} = 5 (1) - (- 2.5) (- 2.5) = - 1.25 S^{2}$ .
$T_{Q} = - \frac{1}{- 2.5} - \frac{- 1.25}{- 2.5} - \frac{1}{- 2.5} - \frac{5}{- 2.5} = [0.4 - 0.5 S 0.4 Ω 2]$ Check: $A D - B C = 0.8 + 0.2 = 1$ , as it should be because $y_{12} = y_{21}$ (Q is reciprocal).
Only the (1,2) entry is needed. Row 1 of $T_{P}$ times column 2 of $T_{Q}$ : $β = A_{P} B_{Q} + B_{P} D_{Q}$ .
$β = (0.5) (0.4) + (- 10) (2) = 0.2 - 20 = - 19.80 Ω$

KEY IDEA

For networks in cascade, convert each one to its ABCD (transmission) matrix and multiply them in the order the signal passes through: $T_{R} = T_{P} T_{Q}$ . Then compute only the entry you need, here $β = A_{P} B_{Q} + B_{P} D_{Q}$ . You get the Z→ABCD and Y→ABCD formulas by solving the port-2 equation for the port-1 variable. Checking $A D - B C$ ( $= z_{12} / z_{21}$ for P and $= 1$ for reciprocal Q) is a quick test of each conversion.

WATCH OUT FOR

(1) Multiplying in the wrong order: $T_{Q} T_{P}$ gives $β = A_{Q} B_{P} + B_{Q} D_{P} = - 4 + 0.5 = - 3.5 Ω$ . (2) Swapping $z_{12}$ and $z_{21}$ for the non-reciprocal P, i.e. dividing by 60 instead of 80, which gives about $- 26.4 Ω$ . (3) Dropping the minus signs in the Y→ABCD formulas. Because $y_{21}$ is negative here, the signs of every entry of $T_{Q}$ then come out wrong.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2024 EE Q.48: solution and answer

Answer and solution

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