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

`Attenuation = 0.458\ db`

Step-by-step 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)(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)(P_s)/(P_d)`

`Attenuation = 10Log_(10)(100)/(90)`

`Attenuation = 10 * 0.04575749056`

`Attenuation = 0.4575749056`

`Attenuation = 0.458\ db` (Approximated)