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From: Jeroen Belleman <jeroen@nospam.please>
Newsgroups: sci.electronics.design
Subject: Re: dBs
Date: Sun, 26 May 2024 23:42:15 +0200
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On 5/26/24 22:58, Cursitor Doom wrote:
> On Sun, 26 May 2024 21:48:21 +0200, Jeroen Belleman wrote:
> 
>> On 5/26/24 19:58, Cursitor Doom 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?
>>
>> You can. That's what decibels were invented for.
>>
>> Let's spell it out then. You know 0 dBm is 1 mW. So -13 dBm is
>> 10^(-13/10) times 1 mW, or 50 uW.
>>
>> A 20dB attenuator divides power by a factor of 10^(20/10), that is, a
>> factor of 100. So before the attenuator, you had 5 mW.
>>
>> 5mW is 10*log(5) is +7 dBm.
>>
>> Jeroen Belleman
> 
> Oh I know you're figures are correct, Jeroen. But to check them I had to
> use look-up tables off the net:
> 
> -13dBm = 0.05mW
> 20dB = 100X
> 0.05X100 = 5mW
> 5mW = =7dBm
> 
> Sometimes you can just straight add-up dBs and other times you can't and I
> can never remember when it's appropriate and when it's not. To be safe, I
> revert to the method I showed above. It's longer, but at least I know I
> can rely on the result. Whoever invent dBs "to make things simpler" needs
> to have their grave desecrated and their name effaced from history IMO.
> 

It's not that bad! Adding dB values works fine in the context of
chains of gain and attenuation. It's easier to add decibels than
to multiply gain values. One gets used to it. I can usually do
it in my head. Anyone working in RF or in control systems gets
proficient at it very quickly.

I agree that there are situations where using decibels can get
confusing. For example, in light detectors, optical power gets
converted into current, so a 20dB change in optical power would
result in only 10dB change in the electrical signal power. But
let's not go there just yet.

Jeroen Belleman