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Understanding decibels a bit more

Andrew J

I get most of this - I expect it'll be clearer and faster with practice, but I get the principal that it is logarithmic to powers of 10.  I also understand the underlying principal of decibel gain/loss related to a ratio:

 

db = 10log10(Pout/Pin) for Power

db = 20log10(Vout/Vin) for Volts RMS (or, generically, a ratio of amplitude)

 

So for power, a +3db gain is a doubling of power; for volts, a +6db gain is a doubling of volts (or amplitude.)

I then come across a statement like this:

 

"dBm: This is the value of a signal as referenced to 1 milliwatt. It is an absolute value. A 0 dBm signal equals 1 milliwatt. A 1 watt signal is equal to +30 dBm. If you had a 100-watt transmitter, that would equate to a +50.0 dBm level. If you had a 2 microvolt signal, that would equate to a -100 dBm signal. If you have a 0.5 microvolt (uV) signal, that would equate to -112 dBm."

 

I can follow the first part, no problem.  If 0dBm is 0.001W, then for 1W, dBm = 10log10(1/0.001) = 30; 100W, dBm = 10log10(100/0.001) = 50.  Generically 1W is 10^3 from 1mW and log10(1000) is 3 so 10x3 = 30; ditto 100W ~ 10^5 so log10(100000) is 5 so 10x5 = 50.

 

The second part of the statement though feels like it's making an assumption, one that maybe is so common to those in the know that it doesn't need stating??  That doesn't include me.

 

To relate a voltage value to a power value I need to know the load in Ohms or Amps don't I?  Take as an example, because it works with that logic (nearly), an assumed 50Ohm load, then from the formula P=V^2/R:

 

2uV @ 50Ohms:  P = (2x10^-6)^2 / 50 = (2^-6)^2 / 50 = (4^-12) / 50 = 8^-14.

dBm = 10log10(8^-14 / 0.001) = -100 (actually, pretty much -101)

 

0.5uV @ 50Ohms (not showing working out!): P = 5^-15

dBm = 10log10(5^-15 / 0.001) = 113 (although the statement above has 112)

 

I thought working with decibels was intended to be quick so I think I'm missing a trick here.  I couldn't work out in my head that 0.5uV is -112dBm (or -113dBm) so I feel like I'm doing something wrong.

  • Hi Andrew,

      wrote:

     

    The second part of the statement though feels like it's making an assumption, one that maybe is so common to those in the know that it doesn't need stating??  That doesn't include me.

     

     

    You're totally correct, it is making an assumption. It ought to be saying "a 50 ohm load with 2 microvolt rms across it would be consuming -100dBm of power" or something similar, to be more accurately worded. Usually it's assumed though that the load is 50 ohm (but doesn't need to be. Although it's unusual for it not to be 50 ohm when there's a discussion with dBm featured).

    Also, you're correct on the next part too, 0.5 uV would be -113 dBm, and the -112 dBm statement was incorrect.

    It's annoying, messed up calculations on the Internet sadly, making it harder to learn : (

     

    EDIT: Some of it could be down to people using paper charts or slide rules maybe! : ) I don't know if people had tricks to convert some popular values pre-calculator, but some of it maybe was based off charts etc. Quite a lot of electronic stuff used to be nomograph based too.

  • in reply to shabaz

    The way it is written makes it seem like there is a quick trick to knowing the answer - is that the case?  I've also seen "...range of -110dBm to +16.5dBm (0.7uV to 1.5V RMS)" written like it's obvious.  For me I can only do it by following through the formula above.

     

    At least for one of the very few times in my life I got the maths right image

  • in reply to Andrew J

    Hi Andrew,

     

    There's no quick trick, or at least I'm not aware of it! : )

    I only remember some dBm approximations just out of familiarity, e.g. 0 dBm being 1mW, or 20 dBm being quite high (WiFi power output level aka 100 mW), -90 to -110 dBm being typical received radio power strengths (and knowing this is of the order of a microvolt or so), -145 dBm or lower being GPS received strength, and so on. Converting between that and volts rms is impossible mentally for me at least.

     

    The reason people maybe write it along with the microvolt value, is because some radio receivers specify their sensitivity in microvolts. I can't (directly) measure that anyway : )

    What is worth doing is putting your formula into an Excel file maybe (if you're not already doing it), and slowly keep adding sheets, building it up with all the stuff you come across. I've done that for years, my file is not understandable by anyone, but it works for me! I've even got wire gauge tables in there, and (just looking at it now) for some reason I cannot recall, there's a sheet with hexagon area, and diameter of a circle which would enclose the hexagon.. I have no idea why I ever needed that.

  • in reply to shabaz

    Next time I take the car for a service I’ll let them know I think the brake pads are -3dB worn, see what they say! image

     

    Building up a spreadsheet is a great idea - I do loads of calculations in them so should keep them.  I reckon your hex/circle areas are from when you used to be the bad boy of the local quilting scene judging by the questions my wife asks me.

     

    Thanks Shabaz, helpful as ever.

  • in reply to Andrew J

    I wonder if the squeal from those brake pads is +3dB(A) louder ... or +10dB(A) louder ...

     

    - Gough