What is the attenuation through 1,600-feet of 0.750 cable at 600 MHz and 120° F, given a 1.31 dB/100 feet rate at 68° F?

To determine the attenuation through 1,600 feet of 0.750 cable at 600 MHz and 120° F, one must first calculate the total attenuation at the given temperature and distance based on the provided attenuation rate.

Starting with the attenuation rate of 1.31 dB per 100 feet at 68° F, you first need to adjust this rate since the operating temperature is higher at 120° F. Generally, the attenuation of cables increases with rising temperatures. However, the problem does not provide a specific adjustment factor for attenuation at higher temperatures.

To find the total attenuation over 1,600 feet, you compute how many 100-foot segments are in that distance. There are 16 segments in 1,600 feet (1,600 feet ÷ 100 feet = 16). You then multiply the number of segments by the attenuation rate:

1.31 dB/100 feet x 16 segments = 20.96 dB.

Since the temperature at 120° F is not specifically factored into the attenuation used here, we can account for that by adding an adjustment (which is not explicitly stated in the problem). This adjustment factor typically leads to a slight increase in dB.

Normally