ELECTROMAGNETICS · TRANSMISSION LINES

GATE 2020 EC, question 46: Transmission lines — solution and answer

GATE 2020 · ECQ.46 · 2 marks · NAT Checked against the official key

For a 2-port network consisting of an ideal lossless transformer, the parameter $S_{21}$ (rounded off to two decimal places) for a reference impedance of $10 Ω$ , is ________.

GATE 2020 EC question 46, as printed: A dashed rectangle labelled "2 Port Network" at the top holds an ideal transformer drawn as two coupled coils… — As printed in the GATE 2020 EC paper — the figure is part of the question.

Answer and worked solution

ANSWER

$S_{21} = \frac{2 n}{1 + n ^{2}} = \frac{2 ( 2 )}{1 + 4} = 0.80$ for the 2:1 ideal transformer with $Z_{0} = 10 Ω$ at both ports. This matches the official key, 0.8.

Official key: 0.8 to 0.8 · Official answer key printed with the GATE 2020 EC paper (IIT Delhi)

Transformer relations. For an ideal transformer with $N_{1} : N_{2} = n : 1$ , where $n = 2$ : $V_{1} = n V_{2}$ and $I_{1} = - I_{2} / n$ , taking $I_{2}$ as flowing into port 2. A load $Z$ on the secondary appears at the primary as $n^{2} Z$ .
Set up $S_{11}$ and $S_{21}$ . Terminate port 2 in the reference impedance $Z_{0} = 10 Ω$ , so no wave enters port 2 ( $a_{2} = 0$ ). The impedance seen at port 1 is then $Z_{in, 1} = n^{2} Z_{0} = 4 \times 10 = 40 Ω$ .
Reflection coefficient at port 1, using $Γ = \frac{Z _{in} - Z _{0}}{Z _{in} + Z _{0}}$ : $S_{11} = \frac{40 - 10}{40 + 10} = 0.6$
Port-1 voltage from the incident wave. The total voltage at port 1 is the incident wave plus the reflected wave: $V_{1} = V_{1}^{+} + V_{1}^{-} = V_{1}^{+} (1 + S_{11}) = 1.6 V_{1}^{+}$ .
Voltage transfer through the transformer. $V_{2} = V_{1} / n = 1.6 V_{1}^{+} /2 = 0.8 V_{1}^{+}$ . Port 2 is terminated in $Z_{0}$ , so no wave travels into it ( $V_{2}^{+} = 0$ ) and the whole of $V_{2}$ is the outgoing wave: $V_{2} = V_{2}^{-}$ . (The drawing has no polarity dots. We take the usual reading, with both windings wound the same way, so $V_{2}$ is in phase with $V_{1}$ . Reversing one winding would only flip the sign of $S_{21}$ . Its magnitude stays the same.)
Definition of $S_{21}$ . Both ports use the same $Z_{0}$ , so the $Z_{0}$ normalisation cancels: $S_{21} = \frac{b _{2}}{a _{1}}_{a_{2} = 0} = \frac{V _{2}^{-}}{V _{1}^{+}} = 0.8$
General formula (check). $S_{11} = \frac{n ^{2} - 1}{n ^{2} + 1}$ and $S_{21} = \frac{1 + S _{11}}{n} = \frac{2 n}{1 + n ^{2}}$ . With $n = 2$ : $S_{21} = \frac{4}{5} = 0.80$ .
Lossless check. An ideal transformer absorbs no power, so with port 2 matched we need $∣ S_{11} ∣^{2} + ∣ S_{21} ∣^{2} = 1$ : $0.36 + 0.64 = 1$ . The condition holds, so the result is consistent. $S_{21} = 0.80$

KEY IDEA

Match port 2 with $Z_{0}$ . The transformer then shows port 1 an impedance of $n^{2} Z_{0}$ , which gives $S_{11} = \frac{n ^{2} - 1}{n ^{2} + 1}$ . The transformer steps the total port-1 voltage $V_{1}^{+} (1 + S_{11})$ down by $n$ , and that becomes the outgoing wave at the matched port 2, so $S_{21} = \frac{1 + S _{11}}{n} = \frac{2 n}{1 + n ^{2}}$ . The answer depends only on the turns ratio. $Z_{0}$ cancels out, so the given 10 $Ω$ does not change the result.

COMMON MISTAKE

Taking $S_{21}$ as the plain voltage ratio $V_{2} / V_{1} = 1/ n = 0.5$ . That ignores the reflection at port 1: $S_{21}$ is referred to the INCIDENT wave $V_{1}^{+}$ , not to the total voltage $V_{1} = V_{1}^{+} (1 + S_{11})$ . Another slip is using $n$ instead of $n^{2}$ for the impedance transformation, which gives $Z_{in} = 20 Ω$ and wrong values for both $S_{11}$ and $S_{21}$ . Checking $∣ S_{11} ∣^{2} + ∣ S_{21} ∣^{2} = 1$ catches both errors.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 13 Transmission lines questions, with notes

GATE 2020 EC, question 46: Transmission lines — solution and answer

Answer and worked solution

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