Hello,

Have always had trouble in "understanding" db the right way.

...e.g., if I see a loss at some freq. of 3 db, does this imply that I am "losing"

about 30 % of the signal, or,... ?

No. A loss of 3 dB (-3 dB, a loss is a negative) is a loss of 1/2 the signal so that 1/2 remains.

If you have a calculator, you calculate 10^(s/10) to see what the new signal is, where "s" is the loss or gain. A 3 dB loss is -3 dB so one calculates 10^(-3/10)=10^(-.3)=.50 (approx.) or 50% remains (a 50% loss) .

A 6 dB lost would be 10^(-6/10)=10^(-.6)=.25 or 25% of the signal remains (a 75% loss)

A 10 dB loss would be 10^(-10/10)=10^(-1)=.1 or 10% signal remains (a 90% loss)

A 3 dB gain would give 10^(3/10)=10^(.3)=2 (approx.) or twice the signal; while a 6 dB gain will give 4 times the signal and a 10 dB gain gives 10 times the signal (but 20 dB is not 20 times the signal, doing the calculation will see that it is 100 times the signal).

An advantage of using dB is one can add the dB of a series of gains and losses quickly to get a final loss or gain; for example a 10 dB gain followed by a 6 dB loss (-6 dB) will give 10+(-6)=4 dB so there will be 10^(4/10)=2.51or 2.51 times the original signal.

To change a loss or gain to dB one does 10*log10 (new/original), where log10 indicates the logarithm base 10, which is probably what your calculator has (do not use "ln" which is a logarithm in a different base). So for example, if a signal had a 1/3 loss (.33), then the new signal is .67 as much, so new/original=.67 and the calculation is 10*log10(.67)=-1.74 dB or 1.74 dB loss.

Also, read troymail's link.