A signal travel from point A to point B . At point A the signal power is 100W and at point B , the power is 90W.

Question

What is the attenuation in decibels?

Please anyone try to solve this numerical

Answer 1

Attenuation in decibels = $10\;log_{10}\;\frac{P_t}{P_r}$ where ${P_t}$ is the transmitted power and ${P_r}$ is the received power.

In the given case attenuation in dB = $10\;log\;\frac{100}{90} = 10\;(log(10)-log(9))=0.46 dB \approx 0.5dB$

Source: Computer Networks 3/e by Andrew S. Tanenbaum