GATE EC 2026 Set 1 — Question 34

For a lossless passive two-port network, the sum of reflected power and transmitted power is 1.
$|S_{11}|^2 + |S_{21}|^2 = 1$
Given that $|S_{11}|$ and $|S_{21}|$ intersect at -3 dB. This means at the intersection frequency, $|S_{11}|_{dB} = |S_{21}|_{dB} = -3 \text{ dB}$.
Converting -3 dB to linear scale:
$10 \log_{10} (|S_{11}|^2) = -3 \implies |S_{11}|^2 = 10^{-3/10} \approx 0.5$
$10 \log_{10} (|S_{21}|^2) = -3 \implies |S_{21}|^2 = 10^{-3/10} \approx 0.5$
Check for lossless condition: $0.5 + 0.5 = 1$. This is consistent.For a lossy passive two-port network, some power is dissipated within the network. The sum of reflected power and transmitted power will be less than 1.
$|S_{11}|^2 + |S_{21}|^2 < 1$
Given that $|S_{11}|$ and $|S_{21}|$ intersect at -4 dB. This means at the intersection frequency, $|S_{11}|_{dB} = |S_{21}|_{dB} = -4 \text{ dB}$.
Converting -4 dB to linear scale:
$10 \log_{10} (|S_{11}|^2) = -4 \implies |S_{11}|^2 = 10^{-4/10} \approx 0.3981$
$10 \log_{10} (|S_{21}|^2) = -4 \implies |S_{21}|^2 = 10^{-4/10} \approx 0.3981$The power dissipated ($P_{diss}$) in the lossy network is the difference between the input power and the sum of reflected and transmitted power. Assuming normalized input power $P_{in} = 1$,
$P_{diss} = 1 - |S_{11}|^2 - |S_{21}|^2$
At the intersection frequency for the lossy network:
$P_{diss} = 1 - 0.3981 - 0.3981 = 1 - 0.7962 = 0.2038$Percentage of power dissipated = $P_{diss} \times 100\% = 0.2038 \times 100\% = 20.38\%$.Rounding off to two decimal places, the percentage of power dissipated is approximately $20.38\%$. This value falls within the specified range of 20.00 to 21.00.