To solve this, we need to determine the Carrier-to-Noise ratio, $[C/N]$, during rain attenuation. Given the problem, we know the clear-sky $[C/N]$ is 20 dB, and the rain attenuation exceeds 1.9 dB for 0.1% of the time. We're also given the effective noise temperature of the receiving system is 400 K, and the ambient temperature, $T_a$, is 280 K.

Here are the key points of the approach:

Step 1: Convert the given dB values to linear values

• The clear-sky $[C/N]$ is given as 20 dB, which we convert to a linear ratio: $\left[ \frac{C}{N} \right]_{\text{clear-sky}} = 10^{\frac{20}{10}} = 100$

• The rain attenuation is 1.9 dB, which reduces the carrier power. The rain attenuation in linear terms is: $\text{Attenuation} = 10^{\frac{1.9}{10}} \approx 1.55$

Step 2: Adjust the $[C/N]$ ratio for rain attenuation

• Rain attenuation reduces the carrier power, so the new $[C/N]$ ratio in linear terms becomes: $\left[ \frac{C}{N} \right]_{\text{rain}} = \frac{100}{1.55} \approx 64.52$

Step 3: Convert the new $[C/N]$ ratio back to dB

• Now, convert this value back to dB: $\left[ \frac{C}{N} \right]_{\text{rain}} = 10 \log_{10}(64.52) \approx 18.1 \, \text{dB}$

Thus, the value of $[C/N]$ below which it falls for 0.1% of the time is approximately 18.1 dB.

Would you like more detailed steps or clarifications? Here are some related questions to deepen your understanding:

1. What is the significance of the effective noise temperature in satellite communication?
2. How does rain attenuation specifically affect the signal quality in satellite links?
3. Why do we use decibels (dB) to express $[C/N]$, attenuation, and other ratios in communications?
4. What are typical values of rain attenuation for different climates or regions?
5. How can signal processing techniques mitigate the effects of rain attenuation?

Tip: When working with dB values, remember that a 10 dB increase means multiplying the power ratio by 10, while a 3 dB change corresponds to roughly doubling or halving the power.