Question

One CD player is said to have a signal-to-noise ratio of 82 dB, whereas for a second CD player it is 98 dB. What is the ratio of intensities of the signal and the background noise for each device?

Answer 1

1st CD PLAYER-
`(v_2)/(v_1) = 10^(4.1)`

2nd CD PLAYER-
`(v_2)/(v_1) = 10^(4.9)`

Step-by-step explanation:

for voltage ratio

`log(v_2)/(v_1) =(dB)/(20)`

For 1st cd player

`log(v_2)/(v_1) =(82)/(20)`

`(v_2)/(v_1) = 10^(4.1)`

`(v_2)/(v_1)  = 12589.25`

FOR 2ND PLAYER

`log(v_2)/(v_1) =(98)/(20)`

`(v_2)/(v_1) = 10^(4.9)`

`(v_2)/(v_1) = 79432.852`