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

Answer:

1st CD PLAYER- $\frac{v_2}{v_1} = 10^{4.1}$

2nd CD PLAYER- $\frac{v_2}{v_1} = 10^{4.9}$

Explanation:

for voltage ratio

$log\frac{v_2}{v_1} =\frac{dB}{20}$

For 1st cd player

$log\frac{v_2}{v_1} =\frac{82}{20}$

$\frac{v_2}{v_1} = 10^{4.1}$

$\frac{v_2}{v_1}  = 12589.25$

FOR 2ND PLAYER

$log\frac{v_2}{v_1} =\frac{98}{20}$

$\frac{v_2}{v_1} = 10^{4.9}$

$\frac{v_2}{v_1} = 79432.852$