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I'm doing a mobile/wireless networking subject and the physics aspect is giving me some trouble. I'm mainly confused about the conversion of dB, dBm and dBW and how to calculate the gain/loss from an input.

The problem I'm trying to solve is what the resulting power (in milliwatts) is at certain points in a circuit (x and y). I've tried to describe the scenario as best I can (let me know if I need to clear anything or post a picture)

In a circuit a 100mW input receives a 10 dB Gain and continues to point 'x' where it receives a 2 dB Loss. Then it receives another 2 dB Loss on the way to point 'y', where it receives another 2 db Loss.

To simplify it I guess it would look something like

100mW --> +10 dB --> -2 dB --> Point 'x' --> -2 dB --> -2 dB --> Point 'y'

Any help would be much appreciated, thanks in advance

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  • $\begingroup$ rapidtables.com/electric/decibel.htm $\endgroup$
    – DavePhD
    Commented Apr 25, 2014 at 3:38
  • $\begingroup$ Thanks for the link, I was able to come up with an answer of ~251 mW by repeatedly using the Power 2 = Power 1 x 10^ (Gain or Loss / 10) formula. If anyone could double check that would be great $\endgroup$
    – NickV
    Commented Apr 25, 2014 at 4:42

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Decibels measure a ratio of power on a logarithmic scale. The decibel measure of a ratio between powers $P_1$ and $P_2$ is given by $10\log_{10}(P_1/P_2)$.

dBm is an expression of power relative to 1 mW. dBW is similarly the ratio between some amount of power and 1 W.

Remember, dB measure a ratio. dBm or dBW actually specify an amount of power.

100mW --> +10 dB --> -2 dB --> Point 'x' --> -2 dB --> -2 dB --> Point 'y'

100 mW is +20 dBm, because $10\log_{10}(\dfrac{100\ \mathrm{mW}}{1\ \mathrm{mW}})$ is 20.

20 + 10 is 30.

30 - 2 is 28.

28 - 2 is 26.

26 - 2 is 24.

So after all these gains and losses you have +24 dBm.

24 dBm is 251.19 mW because $10^{\frac{24}{10}}$ is 251.19.

I was able to come up with an answer of ~251 mW by repeatedly using the Power 2 = Power 1 x 10^ (Gain or Loss / 10) formula.

It looks like you got the right answer, but I'm not sure why you had to repeatedly use a complicated formula. After you convert the input power to dBm, it's just additions and subtractions to work out the output power. Then convert back to mW if you want the answer in mW. That's the advantage of working on a log scale: multiplications and divisions turn into additions and subtractions.

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  • $\begingroup$ I was struggling to understand whether to use dBm or just dB so I just stuck with dB because it was easiest for me to understand. But now I see the difference and using logs seems a lot simpler and should be a big help in future. Thank you! $\endgroup$
    – NickV
    Commented Apr 25, 2014 at 5:40
  • $\begingroup$ Thanks The Photon - this is a nice simple answer for a similar optical problem I am working on! $\endgroup$ Commented Nov 5, 2014 at 23:39

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