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In my instrumentation class we have been given the following formula to get the power in decibels:

\$10Log(Power(mW)/mW)\$

I'd like to know if taking the power in miliwatts is something widely done or has any importance?

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3 Answers 3

up vote 9 down vote accepted

The important thing to remember when dealing with dB is that it's just a ratio value, if you don't reference it to something it means nothing. For example, when referencing your signal against 1mW, you'll get an answer like 10dBm, while when working in W, you will get an answer like 10dbW. You can use whatever you prefer as long as you are consistent. That being said, values measured in dBm are a lot more common than values in dBW.

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For a selection of reference points that are often used, see en.wikipedia.org/wiki/Decibel, "Suffixes and References". –  WhatRoughBeast May 15 at 12:55

Since Bels (or deciBels) are just expressing ratio they themselves are unit-less. However it is often useful to use them to express the strength of one signal in ratio to a known signal. So for that to be meaning full one must know the units of the known signal. dBm (dbmW) is often used in RF signal chains. But even there negative values are common, many receivers work down to -120dBm input. So really it is just convention. Same as using dB rather than B.

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I'm not 100% sure but you can change mW with W but they should be both in W or mW or MW(megaWatts). With this formula the result will be in mW. If you use watts(w) the result will be in watts and if you use megaWatts the result will me in megaWatts.

Here is how the formula will look with watts

10Log(Power(W)/W)

but you can't do like this: 10Log(Power(W)/mW)

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yeah, I know that. You can do it in whatever units you want, but the numerical value of the decibels won't be the same if you do it in mW or in W for example, so why mW? –  silvrfück May 15 at 10:49
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Milliwatts are commonly used as a power reference because, with the power levels existing in a large percentage of electronic circuits, the result is a positive number which is easier to handle than negative numbers. –  Barry May 15 at 12:41
    
You say "With this formula the result will be in mW". Which formula are you referring to? If you are referring to the Op's formula then no, the result will be in dB and will be a measure of power gain. –  Andy aka May 15 at 13:05

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