ISRO EC 2023 Q.12: solution and answer

Official answer: BSource: ISRO's revised and final answer key of 16 February 2024

1. Return loss is $RL=-20{\log }_{10}|\Gamma |$. The antenna has $RL=10\text{dB}$, so $|{\Gamma }_{ant}|=10^{-10/20}\approx 0.316$. The attenuator's own input mismatch is $RL=40\text{dB}$, so $|{\Gamma }_{att}|=10^{-40/20}=0.01$.

2. A wave sent into the combined port goes through the $6\text{dB}$ attenuator, is reflected by the antenna, and goes back through the attenuator. It is attenuated twice, so the extra loss is $2\times 6=12\text{dB}$.

3. The reflection coming from the antenna, seen at the combined input, therefore has return loss $RL=10+12=22\text{dB}$. In amplitude, $|\Gamma |=0.316\times 10^{-12/20}\approx 0.316\times 0.251\approx 0.079$.

4. The attenuator's own reflection, $0.01$ (40 dB), is much smaller. Depending on its phase it changes the total to somewhere between $0.079-0.01=0.069$ and $0.079+0.01=0.089$.

5. These limits correspond to return losses of about $-20\log (0.089)\approx 21\text{dB}$ and $-20\log (0.069)\approx 23\text{dB}$. The combined return loss is therefore approximately $22\text{dB}$.

6. The other options do not fit. $40\text{dB}$ is the attenuator's own return loss. $16\text{dB}$ would result from counting the attenuator only once ($10+6$). $10\text{dB}$ would ignore the attenuator completely.