dansawyeror wrote:
There are three terms, idb, pdb, and vdb. They are not the same. idb and
pdb have the same value: 10*log10 (i/i0), while vdb is 20*log10(v/v0).


Not quite sure what all these are. But to express a voltage or current
ratio in dB, you use:

dB = 20 * log(V2/V1)
dB = 20 * log(I2/I1)

and to express a power ratio in dB, you use:

dB = 10 * log(P2/P1)

If the voltages or currents are measured at the same impedance as the
power, the numerical dB values will be the same. For example, let's
suppose we have a 100 ohm resistor with one amp flowing through it. Then
there's 100 volts across it, and it's dissipating 100 watts.

Double the current to 2 amps. The dB increase in current is

20 * log(2/1) = 6.02 dB

The voltage has also doubled to 200 volts, so the dB increase in voltage is

20 * log(200/100) = 6.02 dB

The power dissipation is now 400 watts, so the dB increase in power is

10 * log(400/100) = 6.02 dB

Cool, huh?

Roy Lewallen, W7EL