The GATE Grind

GATE 2024 EC – Question 15

Electromagnetics · Transmission Lines · 1 mark · Multiple choice

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 $z_{inA}$, $z_{inB}$, $z_{inC}$, and $z_{inD}$ (indicated in the figure) are ________.

Diagram for GATE 2024 EC question 15
  1. $z_{inA}=+1j\ \Omega,\ z_{inB}=\infty,\ z_{inC}=-1j\ \Omega,\ z_{inD}=0$
  2. $z_{inA}=\infty,\ z_{inB}=+0.4j\ \Omega,\ z_{inC}=0,\ z_{inD}=+0.4j\ \Omega$
  3. $z_{inA}=-1j\ \Omega,\ z_{inB}=0,\ z_{inC}=+1j\ \Omega,\ z_{inD}=\infty$
  4. $z_{inA}=+0.4j\ \Omega,\ z_{inB}=\infty,\ z_{inC}=-0.4j\ \Omega,\ z_{inD}=0$

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Show answer and explanation

Correct answer: (A) $z_{inA}=+1j\ \Omega,\ z_{inB}=\infty,\ z_{inC}=-1j\ \Omega,\ z_{inD}=0$

Explanation

For a short-circuited lossless line, $z_{in}=j\tan(\beta d)$ at distance $d$ from the load. The points are at $d=\lambda/8,\ \lambda/4,\ 3\lambda/8,\ \lambda/2$, so $\beta d=45^\circ,\ 90^\circ,\ 135^\circ,\ 180^\circ$. Then $z_{inA}=j\tan45^\circ=+j1$, $z_{inB}=\infty$, $z_{inC}=j\tan135^\circ=-j1$ and $z_{inD}=0$.