GATE 2026 EC Q.34: solution and answer

Official answer: 20.00 to 21.00Source: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Send unit power into port 1 with port 2 terminated in a matched load. Then $|S_{11}{|}^2$ is the fraction of power reflected and $|S_{21}{|}^2$ is the fraction transmitted. Whatever is neither reflected nor transmitted is dissipated inside the network: $$P_{\text{diss}}=1-|S_{11}{|}^2-|S_{21}{|}^2$$

2. Check this against the lossless case. A lossless network has a unitary S-matrix, so $|S_{11}{|}^2+|S_{21}{|}^2=1$. Where the two curves cross, $|S_{11}|=|S_{21}|$, so each squared magnitude is $\frac{1}{2}$, and $10{\log }_{10}(0.5)\approx -3\text{dB}$. That agrees with the first statement in the question.

3. In the lossy network the curves cross at $-4$ dB, so $|S_{11}{|}_{\text{dB}}=|S_{21}{|}_{\text{dB}}=-4\text{dB}$. Since $20{\log }_{10}|S|=10{\log }_{10}|S{|}^2=-4$, $$|S_{11}{|}^2=|S_{21}{|}^2=10^{-4/10}=10^{-0.4}=0.39811$$

4. Reflected plus transmitted power: $|S_{11}{|}^2+|S_{21}{|}^2=2\times 0.39811=0.79621$.

5. Dissipated power: $$P_{\text{diss}}=1-0.79621=0.20379\Rightarrow 20.38\%$$