GATE 2024 EC Q.15: solution and answer

GATE 2024 · ECQ.15 · 1 mark · MCQ

Consider a lossless transmission line terminated with a short circuit as shown in the figure below. As one moves towards the generator from the load, the normalized impedances $z_{in A}$ , $z_{in B}$ , $z_{in C}$ , and $z_{in D}$ (indicated in the figure) are ________.

GATE 2024 EC question 15: A two-wire lossless transmission line drawn horizontally (two thick parallel lines). — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
$z_{in A} = + 1 j Ω$ , $z_{in B} = \infty$ , $z_{in C} = - 1 j Ω$ , $z_{in D} = 0$
(B)
$z_{in A} = \infty$ , $z_{in B} = + 0.4 j Ω$ , $z_{in C} = 0$ , $z_{in D} = + 0.4 j Ω$
(C)
$z_{in A} = - 1 j Ω$ , $z_{in B} = 0$ , $z_{in C} = + 1 j Ω$ , $z_{in D} = \infty$
(D)
$z_{in A} = + 0.4 j Ω$ , $z_{in B} = \infty$ , $z_{in C} = - 0.4 j Ω$ , $z_{in D} = 0$

Answer and solution

ANSWER

(A): $z_{in A} = + j 1$ , $z_{in B} = \infty$ , $z_{in C} = - j 1$ , $z_{in D} = 0$ .

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

For a lossless line of length $d$ ending in a load $Z_{L}$ , the input impedance is $Z_{in} = Z_{0} \frac{Z _{L} + j Z _{0} tan β d}{Z _{0} + j Z _{L} tan β d}, β = \frac{2 π}{λ} .$ For a short circuit, $Z_{L} = 0$ , so this becomes $Z_{in} = j Z_{0} tan β d$ . The normalized impedance is therefore $z_{in} = \frac{Z _{in}}{Z _{0}} = j tan β d,$ where $d$ is the distance from the load (at $z = 0$ ) back towards the generator.
Point A, $d = λ /8$ : $β d = \frac{2 π}{λ} \cdot \frac{λ}{8} = \frac{π}{4}$ , so $z_{in A} = j tan \frac{π}{4} = + j 1$ (inductive).
Point B, $d = λ /4$ : $β d = \frac{π}{2}$ , and $tan β d \to \infty$ , so $z_{in B} \to \infty$ . A quarter-wave line turns the short into an open circuit.
Point C, $d = 3 λ /8$ : $β d = \frac{3 π}{4}$ , so $z_{in C} = j tan \frac{3 π}{4} = - j 1$ (capacitive).
Point D, $d = λ /2$ : $β d = π$ , so $z_{in D} = j tan π = 0$ . A half-wave line repeats its load, so the input is a short circuit again.
The sequence $+ j 1, \infty, - j 1, 0$ matches option (A). The paper prints $Ω$ after the values, but a normalized impedance has no units, so read those entries as $+ j 1$ and $- j 1$ . On a Smith chart: start at the short-circuit point ( $Γ = - 1$ ) and move clockwise (towards the generator) around the outer $∣Γ∣ = 1$ circle. Each $λ /8$ rotates $Γ$ by $2 β d = π /2$ , so the points go short, $+ j 1$ , open, $- j 1$ , short.

KEY IDEA

Looking into a short-circuited lossless line from a distance $d$ , the normalized impedance is $z_{in} = j tan β d$ . It is purely reactive and repeats every $λ /2$ . At $λ /8$ it is $+ j 1$ , at $λ /4$ it is an open circuit, at $3 λ /8$ it is $- j 1$ , and at $λ /2$ it is a short again.

WATCH OUT FOR

There are three common slips. The first is measuring $d$ from the generator end (point D), which reverses the sequence and leads to option (C). The second is using $β d = 2 π d / λ$ as if it were the Smith-chart rotation angle ( $2 β d$ ), which puts the open circuit at $λ /8$ instead of $λ /4$ . The third is mixing up the sign: a short-circuited stub shorter than $λ /4$ is inductive ( $+ j$ ), not capacitive.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 13 Transmission lines questions, with notes

GATE 2024 EC Q.15: solution and answer

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