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Logarithmic scales are not uncommon: magnitudes in astronomy, f-stops in photography, the Richter magnitude scale, etc.

But decibels are weird.

I am not even talking yet about the fact that we use a different base for power and amplitudes.

Somehow we are not content with the fact that decibels express a ratio, and that be the end of it, but we pretend it's like any other unit, and we can even use it in expressions. The arithmetic rules though seem very, uhm, creative. For example, a +5 dBm signal goes into a +20 dB amplifier:

$$ 5\ \mathrm{dBm} + 20\ \mathrm{dB} = 25\ \mathrm{dBm} $$

Sometimes, decibel is used as a unit for the underlying quantity:

$$ P = 25\ \mathrm{dBm} = 0.31\ \mathrm{W}$$

Or, I have two noise sources, each of 60 dBA:

$$ 60\ \mathrm{dBA} + 60\ \mathrm{dBA} = 63\ \mathrm{dBA} $$

Or, losses in wave guides are expressed in dB per unit length, e.g. 0.4 dB/km for a typical telecom fiber.

To me, it seems like these notations are mutually exclusive.

Is there a mathematically consistent system that treats the decibel as some sort of unit, that can be composed with itself and perhaps other SI units? What would be the arithmetic in this system? Does the different definition for power and root-power quantities also play into this?

I am not asking for help understanding what decibels are.1 I have a background in RF electronics, and a solid grip on why the decibel came to be (historically) and why we keep using it (it is undeniably useful).


1 Or perhaps I am... What would be the mathematical nature of the dB symbol in the expressions above? An operator? It clearly is not a unit in the usual sense.

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  • $\begingroup$ Related: Are units of angle really dimensionless? $\endgroup$
    – benrg
    Mar 21, 2023 at 0:06
  • $\begingroup$ Perhaps of note, I once read an article from 2010 in Metrologia, the journal for metrology, which set forth a lamenting argument that the axiomization of units is incomplete because we don't have a good set of axioms that work with dB which are consistent with the mathematical operators we are used to. $\endgroup$
    – Cort Ammon
    Mar 21, 2023 at 1:37
  • $\begingroup$ @CortAmmon If you still have that article around, I would love to give it a read. $\endgroup$
    – polwel
    Mar 21, 2023 at 5:50

2 Answers 2

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The main property of a logarithmic scale is that multiplication in the original scale gets converted into addition in the logarithmic scale. This property holds for decibels (dB) as well.

The example of the amplifier you give summarizes this pretty well. In the original scale, a signal in Watts $P$ is amplified by a dimensionless factor $A$, so the new signal $P'=AP$. Then in terms of the logarithmic quantities $L=\log P$, $a=\log A$, $L'=\log P$ (the base of the log doesn't really matter), we have $L' = a + L$.

Now, there is something a bit sneaky I slipped into the above paragraph. Namely, while $A$ is a dimensionless quantity so $a=\log A$ is well defined, $P$ and $P'$ have dimensions of Watts and so $\log P$ and $\log P'$ are not well defined. Therefore, for power, we need to introduce a reference scale $P_{\rm ref}$, and really $L=\log (P/P_{\rm ref})$ and $L'=\log (P'/P_{\rm ref})$. Since $A$ is dimensionless, it is ok to measure $a$ in ${\rm dB}$, but since $P$ and $P'$ are not dimensionless, we need to include the reference scale, and this choice of reference scale is reflected in the choice of unit, we use ${\rm dBm}$ instead of ${\rm dB}$ to indicate that we are using a reference power of $1\ {\rm mW}$.

Now we come to a key point. The whole idea of units to begin with is to let us keep track in changes in arbitrary reference scales. For example, if we convert from using a meter to a kilometer as a reference length, we know to divide all lengths by $1000$ to keep our equations consistent. Well, the key property of logs is that this rescaling will be converted to an addition. So changing the reference scale from say $1\ {\rm mW}$ to $1\ {W}$ will involve subtracting $\log 1000$ from all quantities with units of ${\rm dBm}$. In that sense, logarithmic units are an abuse of notation; they don't obey the same scaling rules as normal units. However, their transformation properties can be derived from the normal ones in a simple way.

It is perfectly fine mathematically to combine logarithmic quantities with different reference scales, so long as we keep track of all the reference scales involved. For example, we could consider Ohm's law $$ V = IR $$ taking a log of both sides and introducing reference scales, $$ \log\left(\frac{V}{V_{\rm ref}}\right) = \log\left(\frac{I}{I_{\rm ref}}\right) + \log\left(\frac{R}{R_{\rm ref}}\right) $$ where we should have $V_{\rm ref}=I_{\rm ref} R_{\rm ref}$ for consistency. If you defined units $\rm dBV$, $\rm dBA$, and $\rm dB\Omega$, then you could write expressions like $$ 15\ {\rm dBV} = 10\ {\rm dBA} + 5\ {\rm dB\Omega} $$ and the logarithmic units would work out. If the base of the log was 10, then (up to factors of 20 or whatever) this would correspond to the equation $$ 10^{15} {\rm V} = 10^{10}\ {\rm A} \times 10^5{\rm \Omega} $$ Now, I can't think of any practical situation where expressing Ohm's law in the logarithmic form would elucidate any physics, but in principle you could do it.

In practice, when one uses logarithmic scales, I think it is most common to multiply a dimensionful quantity by a dimensionless quantity, or two dimensionless quantities together. Therefore, you would naturally expect to see expressions like $10\ {\rm dBm} + 5\ {\rm dB} = 15\ {\rm dBm}$ or $10\ {\rm dB} + 5\ {\rm dB} = 15\ {\rm dB}$ "in the wild".

I believe I have now explained the principles behind your first two examples.

Your third example is of a somewhat different character. In fact ${\rm dBA}$ is not really a logarithmic unit, but a weighted logarithmic unit. Instead of taking the log of a natural physical quantity (the frequency), one is taking the $\log (R_A(f))$ for some complicated function $R_A(f)$ that is meant to model the human ear's frequency response. The introduction of this complicated $R_A(f)$ function means that the "algebra" associated the unit is complicated. I am not sure, but suspect that the "addition" symbol you defined in $60\ {\rm dBA} + 60\ {\rm dBA} = 63\ {\rm dBA}$ is defined so that the signal amplitudes add "normally" in terms of ${\rm dBm}$; if so, then to define the addition symbol acting on quantities with units ${\rm dBA}$, one needs to understand the properties of the complicated function $R_A(f)$. Thus, using ${\rm dBA}$ as a unit is in a sense an even worse abuse of notation than using ${\rm dB}$ as a unit, since one is using the notation to hide the complicated function $R_A(f)$. However, so long as one is clear on the conventions being used, there is no problem defining notation in this way.

Finally, you bring up the example of ${\rm dB/km}$. I actually believe this is a relatively easy extension of the idea of ${\rm dB}$. One just has to remember that under a change of scale of the signal amplitude, a ${\rm dB}$ changes by adding the log of the scale change, while under a change of scale of length, a ${\rm km}$ changes by multiplying the scale change.

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  • $\begingroup$ In your first sentence did you mean to say "addition"? $\endgroup$
    – RC_23
    Mar 20, 2023 at 23:16
  • $\begingroup$ @RC_23 Fixed thanks $\endgroup$
    – Andrew
    Mar 20, 2023 at 23:38
  • $\begingroup$ I only used dBA in the second example for variety. The example would equally stand with two electrical noise sources that contribute to the same signal. Their linear power is added. Just replace dBA by dBm. In any case, your answer is very helpful! $\endgroup$
    – polwel
    Mar 21, 2023 at 5:48
  • $\begingroup$ Normally dB (deci-Bel) is defined as power ratios, and the multiplier is $10$ (deci) so that $10log(P_1/P_0)$ is how many dB-s $P_1$ is more than $P_0$; "log" is base $10$. For voltages or currents whose square is proportional to power the "multiplier" is 20 because $10log v^2= 20log v$, so $15dBV$ means $v=10^{15/20}=5.6$ relative to $1V$. If $A$ means power then twice as much power in dB-s is $3dB$ more because $10log2 \approx 3$. If $A$ means current or voltage then twice that value is $6dB$ more because $10log2^2=6$. $\endgroup$
    – hyportnex
    Mar 21, 2023 at 13:39
  • $\begingroup$ I have now had time to read your answer in more detail. I have indeed thought about the idea of just mapping addition of dB to multiplication of the linear scales. Unfortunately this entirely clashes with the different bases of amplitudes and powers. Assume I have 10 V × 10 A = 100 W. Mapping this to logarithmic scales would falsely suggest that 20 dBV + 20dBA = 40 dBW. $\endgroup$
    – polwel
    Mar 21, 2023 at 15:06
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What would be the arithmetic in this system?

Let $k:=10^{1/10}$. An amplitude $a$ (nondimensionalized so in the case of sound $a=1$ achieves the reference sound pressure $20\operatorname{μPa}$) is $f(a)$ decibels with $f(a):=\log_ka$; $d$ decibels is amplitude $f^{-1}(d)=k^d$. (Equivalently $f(a)=10\log_{10}a,\,f^{-1}(d)=10^{d/10}$.)

Given an $n$-ary function $g$ mapping $n$ amplitudes $a_1,\,\cdots,\,a_n$ to an amplitude, we can compute decibels from decibels viz.$$f(g(f^{-1}(d_1),\,\cdots,\,f^{-1}(d_n)))=\log_kg(k^{d_1},\,\cdots,\,k^{d_n}).$$We say this calculation with decibels is isomorphic to the one with amplitudes. For example, amplitude addition $g(a_1,\,a_2):=a_1+a_2$ can be restated as $\log_k(k^{a_1}+k^{a_2})$, which reproduces the calculation you deemed "creative".

To incorporate quantities other than amplitudes, a relation of the form$$h(a_1,\,\cdots,\,a_n,\,x_1,\,\cdots,\,x_m)=0$$can be restated as$$h(f^{-1}(d_1),\,\cdots,\,f^{-1}(d_n),\,x_1,\,\cdots,\,x_m)=0.$$A relation of the form$$x_m=j(a_1,\,\,\cdots,\,a_n,\,x_1,\,\cdots,\,x_{m-1})$$is the special case $h=j-x_m$.

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