Volt/meter to Decibel: Understanding Attenuation Loss from 112dB to -36dB

[Ionito]

I have a table containing the magnitude (dB) of an electric field referred to 1 volt/meter.
The values range from 112 to -36dB. How can I properly interpret the attenuation loss in each case in this table? In other words, for a reference of 1 volt/meter and no signal loss, what is the expect value in dB?

 

[Bob S]

Try

dB = 20Log(V2/V1), or

V2/V1 = 10^^(dB/20)

A change of a factor of two is

20Log(2) = 6.02 dB

Bob S

 

[skeptic2]

If the reference is 1 volt/meter then one would expect that to be equal to 0 dB. However 112 dB referenced to 1 volt/meter seems unreasonably high. Are you sure the reference wasn't to 1 microvolt/meter?

 

[Ionito]

Hi skeptic2, completely sure about 1V/m, that is why I did not understand.

 

[sophiecentaur]

What is the context of the table?
If it is simply tabulating a mathematical relationship there need be no problem. If it it supposed to relate to some physical situation then it seems to have some rather high values: nearly 0.5 MV/m! Hardly a "signal" strength!

 

[Ionito]

I made some research and figure out the following: 1V/m= 0 dB\muV/m.
Therefore, the values I am seeing in the table correspond conversions of this reference. If the signal is not attenuated, I have 1V=1*10⁶*\muV and 20log(V2/V1) in this case is 120. A value of 112dB means that the attenuation is 8dB. I believe this is the explanation.

 

[sophiecentaur]

But that definition of the dB is not right.
1V/m is 0dBV/m
or
1V/m is 120dBmuV/m
That's an identity, surely.

That assumes the same impedance. dB is actually a log of power ratio.