NCEES Computer Engineering Sample problem #140

[sonyisda1]

The problem states the signal-to-noise ratio is 20dB. The solution states 20dB = 100 and uses 100 in the equation for (S/N).

Can anyone explain how they got from 20dB to 100?

thanks!

 

[DK PE]

The problem states the signal-to-noise ratio is 20dB. The solution states 20dB = 100 and uses 100 in the equation for (S/N).
Can anyone explain how they got from 20dB to 100?

thanks!

I don't have this reference but it seems likely they are just converting linear to dB. For example, a signal exiting a 20 dB attenuator is 100 times smaller (in Watts) than the input signal. Just a guess since I don't have book but seems reasonable. Equation for example is power (dB) = 10 log (P1 /P2).

20 dB = 10 log (100)

 

[sonyisda1]

yes that explains it!