Not totally sure, since it can depend on what the dB is a loss of. THis is easier to explain in the other direction but just flip the formula around to work backwards.

For voltage, say we have a 2% loss. That translates to a value of 0.98 versus 1.00.

Since dB = -20 log (Voltage out/Voltage ref), (log is log 10)

then we can get to dB as -20 log 0.98 = 0.175dB

Reversing this, say our loss in dB was 0.1dB

0.1 = -20 log (Voltage out/Voltage ref) so Voltage out/Voltage ref = 10^(0.1/-20) = 0.9885531

since this is relative to 1.0, the percentage loss is (1-0.9885531) * 100 = 1.1447 %

Hope that helps

Sash

For voltage, say we have a 2% loss. That translates to a value of 0.98 versus 1.00.

Since dB = -20 log (Voltage out/Voltage ref), (log is log 10)

then we can get to dB as -20 log 0.98 = 0.175dB

Reversing this, say our loss in dB was 0.1dB

0.1 = -20 log (Voltage out/Voltage ref) so Voltage out/Voltage ref = 10^(0.1/-20) = 0.9885531

since this is relative to 1.0, the percentage loss is (1-0.9885531) * 100 = 1.1447 %

Hope that helps

Sash