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Confusion with dB equation - 10 or 20?

  1. Oct 7, 2012 #1
    how do you know when to use dB= 20log(ratio of input/output quantities) or dB=10log(ratio of output and inout). voltage seems to be 20, power seems to be 10... but how about when doing fourier transforms and you want to find out 3dB of maximum, for example?

  2. jcsd
  3. Oct 7, 2012 #2


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    Power dB is ALWAYS 10log and voltage dB is ALWAYS 20log. I think that covers it doesn't it? Do I misunderstand?
  4. Oct 7, 2012 #3
    i meant in other situations. like if you arent measuring the voltage or the power. but instead some other quantity. how do i know what one to use?
  5. Oct 8, 2012 #4
    Brief answer: decibel is ALWAYS "10x".

    Here is why (aka very long answer): decibel is one of those unfortunate units that they tell you about at the university but spend maybe
    only a few minutes explaning what it really means.

    The unit is actually called 'Bel' (named after Alexander Graham Bell, the man who is generally recognized as the one who invented a
    telephone). It's not technically a unit (as say 1 meter) but rather a ratio of 2 numbers (actually "the order of ratio of 2 numbers"). It
    shows the order by which the 2 numbers differ. So 100 000 and 10 differ by 4 order or 4B (4 Bel) (because log (100 000/10) = 4 ).
    The "log" calculates (from definition) the order of ratio of those 2 numbers.

    Now, say you want to use "bels" in electronics. You have a weak input signal somewhere around 1uV (or 1uW) or any other "micro" level you
    can think of, and after processing you have - say - levels somewhere around 10V or 10W or 10 whatever_units. That's a difference of 7
    orders or 7 Bel. That gives you a range of values between 0 and 7 (if you pronounced the input level as 0 bel) or generally n .. n+7 bel.
    It's a very short range so if you compared 2 values within the circuit you would need to use many decimal points to provide a result that
    is exact enough. Say you'd need to say, that the ratio is 3.58bel. It's a number that's supposedly more difficult to remember than for
    instance 42 or 42.3. That's where the 'deci' comes in.

    "Deci" is a standard (metric) prefix meaning "(one) tenth" much the same way as kilo is a
    prefix meaning thousand (i.e. 1000) and micro meaning "one milionth" (10^-6). So 1 decibel is "one tenth of a bel". To get decibels from
    bels you just multiply bels by 10 (again, exactly the same way as you get hertz (Hz) from kilohertz (kHz) by multiplying by 1000 and volts
    from microvolts by multiplying by 10^-6).

    Now, back to our measurement of 3.58B. If we multiply it by 10 we get 35.8 decibels (35.8dB) which is supposedly easier to remember that

    Using "tenths of a unit" instead of the full unit expands our range of 7 units or orders (or 7 bel) for a typical electronics circuit to 70
    new units (or tenths of the original unit or 70dB). That gives us wider range of many different numbers (easier to remember) and less
    numbers behind a decimal point for a given precision.

    The question is, if dB is always 10x B (bels) where does the "20"come from? Well, it appears when you calculate power transfer (let's call
    it "T") and derive the values of input and output powers from measurements of voltage at both ends of the circuit. P=U x I=U x (U/R) = U^2
    / R

    T=10 log (P_out/P_in)=10 log ( (U_out^2/R) / (U_in^2/R) ) = 10 log (U_out^2 / U_in^2) = 10 log ( U_out/U_in )^2 = 10 x 2 x log (U_out/U_in)
    = 20 log (U_out/U_in)

    It's always 10x (because of the "deci-" prefix), even when it appears as (20x) (actually 2x10x) as demonstrated above (the "2" comes from
    the squared argument "behind" the logarithm, so moving it in front of the logarithm saves us computing the squares, although nothing
    prevents you from calculating the squares or the square of the ratio and then just multiplying the resulting logarithm by 10).

    Hopefully, I explained it to you well enough. I wish somebody explained it to me like this years ago when I had the same question(s).
    Eventually someone did (actually touched on the explanation and I was able to derive the rest), I passed many exams, thought I had a good
    understanding of why using logarithms and bels/decibels makes sense but even now, years later I still find from time to time some usage,
    benefits and properties that I wasn't aware of or had a poor understanding of.

    Wikipedia has another explanation of decibel.

    Good luck.
  6. Oct 8, 2012 #5
    I just want to clarify that when I talk about "orders" I'm talking about "orders of magnitude" (i.e. the "E" in the scientific notation of a number: M x 10^E)
  7. Oct 10, 2012 #6
    If you measure other values that must be squared to obtain a power, then it's 20*log as well. Especially for the pressure of an acoustic wave.
  8. Oct 13, 2012 #7

    I bet there are values where one must calculate 5th root of a cubic of the value but people generally don't call it "6*log" but rather "10*log of ratio of 2 values each ^(3/5)".
  9. Oct 21, 2012 #8


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    This is an old chestnut and the rule is easy to remember if you bear in mind that a Bel is fundamentally Log(base 10) of the ratio of POWERS. Clearly, this implies that a DeciBel is one tenth of a Bel so 10dB represents a Power ratio of 10. There is no need to specify anything about the impedance involved in ratios of Power, of course.
    As a totally side issue, you can ask what the voltage ratio would be for two powers into arbitrary resistances. There is NO simple answer to that question. The question you can answer is "what is the ratio of voltages for a given ratio of powers into the same resistance?"
    The answer to that is that the ratio of the Powers is the square of the ratio of the Volts into the same resistance.
    Doing the sums will tell you that the 'factor of two' comes from this square factor. So a ratio of powers of 10 will be 10dB and a ratio of volts of 10 on the same resistance will produce a ratio of powers of 100 - which is 20dB. 'Voltage dBs' are not a really valid concept. If they were, then a step-up transformer could be regarded as having Power Gain (!!!???).

    People will say I am being too fussy but 'Volts dBs' are a snare and delusion and people who insist on 'using them' are heading for a fall. I am lucky in having spent most of my 'technical life' dealing with RF and matched systems so it's been mostly 50Ω all the time and 20dB has usually implied a voltage ratio of 10. But that's all. An OP Amp, [Edit: with feedback to give it ] a voltage gain of 1 probably has a gain (in yer proper actual dB) of 80dB or more. No wonder students get confused when they're exposed to that sort of thing.
    Last edited: Oct 21, 2012
  10. Oct 22, 2012 #9
    But if both voltages are measured at the same location then 20*log is perfectly valid.
    This is typically the case in a filter where you measure an output voltage or an attenuation at different frequencies.
    There, -3dB means voltage*0.71 without any sort of doubt, and it is useful since voltages are often measured.

    Other example: if you have several stages of op amps, each with a vanishingly small and inconsistent output impedance, a power ratio would be unusable, but a gain in dB is useful.

    So the 20*log convention is well-defined (if not easy), useful, and must be known.
  11. Oct 22, 2012 #10


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    I agree that the 'sums' pan out - because logarithms work. But the UNITS are wrong - in as far as a ratio has units. There is nothing wrong with using standard form to describe ratios of Voltages and you are on totally safe ground then (in fact it plugs straight into circuit calculations with no antilogging needed). You have to admit that the word 'volts' needs to be added as a qualifier somewhere, on every occasion, in order to resolve any ambiguity so why not just leave dBs for the purpose they were invented for- namely, Power.
    I don't know of anyone who has never encountered a problem due to the confusion in a long life with dBs, even if it was just because someone else got it wrong.
    It's along the lines of that awful thing with pounds mass and pounds weight and they both should be stamped out and never allowed to be used by civilised Engineers.
    My prejudices are showing here, I'm afraid.

    Voltage gain in standard form is just as useful and just as calculable.
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