Re: dBs

Cursitor Doom <cd999666@notformail.com> wrote:

On Sun, 26 May 2024 19:25:41 +0200, Jeroen Belleman wrote:
 

On 5/26/24 19:09, Cursitor Doom wrote:

I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything
else in fact, with the sole exception of "fatuous conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
 
CD.

 
That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

 
I was under the impression that one couldn't simply just add dBs to dBms?
 

To use logarithms sensibly, you have to start with a dimensionless number.
Vanilla decibels express power ratios. Watts divided by watts is
dimensionless.

Decorated decibels express absolute power values under various measurement
conditions. This works because the denominator is a constant power fixed by
convention. There are many kinds: dBm, dBW, dBa, dBc, dBm0, dBrnC0
(“dibrinco”), and so forth. (*)

When you subtract two decorated decibel values of the same type, you’re
implicitly dividing the underlying ratios.

Since the denominators are equal, they cancel, leaving the decibel ratio of
the numerators.

This of course is a dimensionless power ratio, expressed in vanilla
decibels.

Doing this backwards, if you want to apply a gain to a decorated-decibel
value, you add the gain in vanilla decibels.

If you add two decorated decibel values, the result is nothing useful,
because you get two copies of the denominator instead of one.



Cheers

Phil Hobbs



--
Dr Philip C D Hobbs  Principal Consultant  ElectroOptical Innovations LLC /
Hobbs ElectroOptics  Optics, Electro-optics, Photonics, Analog Electronics