Question 3-66J4

A perfect (no loss) coaxial cable means all input power is either delivered to the load or reflected; no power is dissipated within the cable. The phrase "7 dB of reflected power" in this context signifies that the power delivered to the load ($P_{OUT}$) is 7 dB lower than the input power ($P_{IN}$). To calculate the output power, we convert this 7 dB reduction into a power ratio: $P_{OUT} = P_{IN} \times 10^{\text{(-dB loss}/10)}$ $P_{OUT} = 5 \text{ W} \times 10^{\text{(-7}/10)}$ $P_{OUT} = 5 \text{ W} \times 10^{-0.7}$ Since $10^{-0.7}$ is approximately $0.1995$ (or roughly $0.2$): $P_{OUT} \approx 5 \text{ W} \times 0.2 = 1 \text{ W}$. Therefore, the output power of the transmission line is approximately 1 watt. Options B, C, and D are incorrect as they do not reflect a 7 dB reduction in power from the input.