GATE 2020 EC Q.15: solution and answer

GATE 2020 · ECQ.15 · 1 mark · NAT

In the given circuit, the two-port network has the impedance matrix $[Z] = [406060120]$ . The value of $Z_{L}$ for which maximum power is transferred to the load is ________ $Ω$ .

GATE 2020 EC question 15: A two-port network box labelled [Z] sits in the middle. — Figure from the GATE 2020 EC paper. It is part of the question.

Answer and solution

ANSWER

$Z_{L} = 48 Ω$ . This is the Thevenin impedance seen at port 2, $Z_{22} - \frac{Z _{12} Z _{21}}{Z _{11} + 10} = 120 - 72 = 48 Ω$ .

Official answer: 48Source: Official answer key, GATE 2020 EC (IIT Delhi)

Z-parameter equations with both port currents flowing into the network, as drawn: $V_{1} = 40 I_{1} + 60 I_{2}$ and $V_{2} = 60 I_{1} + 120 I_{2}$ .
At the input port, KVL around the source loop gives $V_{1} = 120 - 10 I_{1}$ .
By the maximum power transfer theorem, $Z_{L}$ must equal the complex conjugate of the Thevenin impedance seen at port 2. To find $Z_{t h}$ , set the independent source to zero, which leaves only the $10 Ω$ resistor across port 1, and drive port 2 with a test current $I_{2}$ .
With the source zeroed, $V_{1} = - 10 I_{1}$ . Substituting into the first equation: $- 10 I_{1} = 40 I_{1} + 60 I_{2} \Rightarrow - 50 I_{1} = 60 I_{2} \Rightarrow I_{1} = - 1.2 I_{2}$ .
Substituting into the second equation: $V_{2} = 60 (- 1.2 I_{2}) + 120 I_{2} = - 72 I_{2} + 120 I_{2} = 48 I_{2}$ , so $Z_{t h} = \frac{V _{2}}{I _{2}} = 48 Ω$ .
Check with the standard result for a two-port whose port 1 is terminated by a source resistance $R_{s}$ : $Z_{t h} = Z_{22} - \frac{Z _{12} Z _{21}}{Z _{11} + R _{s}} = 120 - \frac{60 \times 60}{40 + 10} = 120 - 72 = 48 Ω$
$Z_{t h}$ is purely resistive, so $Z_{L} = Z_{t h}^{*} = 48 Ω$ . (For completeness: with port 2 open, $I_{2} = 0$ and $I_{1} = \frac{120}{10 + 40} = 2.4 A$ , so $V_{t h} = 60 \times 2.4 = 144 V$ and $P_{ma x} = \frac{14 4 ^{2}}{4 \times 48} = 108 W$ .)

KEY IDEA

For maximum power, the load must be the conjugate of the Thevenin impedance seen at the output port. For a two-port described by Z-parameters, with a source resistance $R_{s}$ on port 1, that impedance is $Z_{t h} = Z_{22} - \frac{Z _{12} Z _{21}}{Z _{11} + R _{s}}$ . You get it by zeroing the source, writing $V_{1} = - R_{s} I_{1}$ , and eliminating $I_{1}$ .

WATCH OUT FOR

The usual errors are leaving out the 10 Ω source resistance, which gives $120 - \frac{3600}{40} = 30 Ω$ , or taking $Z_{22} = 120 Ω$ on its own as the output impedance. Both ignore how the port-1 termination couples back through $Z_{12} Z_{21}$ . A sign slip in $V_{1} = - 10 I_{1}$ is another trap: with $I_{1}$ defined as flowing into the network, the zeroed source loop forces $V_{1} = - 10 I_{1}$ , not $+ 10 I_{1}$ .

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2020 EC Q.15: solution and answer

Answer and solution

More Two-port questions