Question

A signal travels from point A to point B. At point A, the signal power is 100 W. At point B, the power is 90 W. What is the attenuation in decibels?

Answer 1

Answer:

$Attenuation = 0.458\ db$

Explanation:

Given

Power at point A = 100W

Power at point B = 90W

Required

Determine the attenuation in decibels

Attenuation is calculated using the following formula

$Attenuation = 10Log_{10}\frac{P_s}{P_d}$

Where $P_s = Power\ Inpu$t and $P_d = Power\ outpu$t

$P_s = 100W$

$P_d = 90W$

Substitute these values in the given formula

$Attenuation = 10Log_{10}\frac{P_s}{P_d}$

$Attenuation = 10Log_{10}\frac{100}{90}$

$Attenuation = 10 * 0.04575749056$

$Attenuation = 0.4575749056$

$Attenuation = 0.458\ db$ (Approximated)