If a coaxial cable has an attenuation of 10 dB, what percentage of the RF signal power remains?

To determine the percentage of RF signal power that remains after passing through a coaxial cable with an attenuation of 10 dB, it's important to understand what dB (decibels) represents in relation to power.

Decibels are a logarithmic unit used to express the ratio of two values, often power levels. An attenuation of 10 dB indicates that the power has been reduced. The formula to calculate the remaining power as a percentage when given attenuation in dB is:

[ \text{Remaining Power (%)} = 10^{\left(\frac{-\text{attenuation (dB)}}{10}\right)} \times 100 ]

Applying that formula:

1. Plug in the values: [ \text{Remaining Power (%)} = 10^{\left(\frac{-10}{10}\right)} \times 100 = 10^{-1} \times 100 = 0.1 \times 100 = 10% ]

However, in terms of what remains, it's vital to realize what the attenuation means: a reduction of signal power. For 10 dB attenuation, we actually have 90% loss, which means that 90% of the