Question

A signal travels from point a to point
b. at point a, the signal power is 100 watts. at point b, the power is 90 watts. what is the attenuation in db?

Answer 1

Attenuation of a signal = 10 * log( Input power / output power )

Note-Logarithm is to the base 10.

Here, Power at point A is the input power
          Power at point B is the out power.

There fore , Attenuation in dB = 10 * log (100 / 90 ) 
                                                = 0.457575 dB

Answer 2

Answer:

The answer is  0.457574 db

Step-by-step explanation:

From the question ,

        there are two points A and B

          power at A = 100 watts

          power at B = 90 watts

so here the signal travels from A to B  , therefore  A is the input signal and B is the output signal

To find the attenuation in db = 10 × ㏒( input power / output power)

                     ( log is base to 10)

By substituting the value, we get

                                               =  10 × ㏒ ( 100/90 )

                                               = 10  × ㏒ ( 1.11)

                                               = 10  × 0.0457574

                     attenuation       = 0.457574 db  is the required answer