Understanding decibels
Why decibels use logarithms, when it is 10·log and when 20·log, what 3 dB really means, and the difference between dB, dBm and dBV.
Why a logarithmic unit exists at all
Decibels solve two problems at once, and both are practical rather than theoretical.
The first is range. A radio receiver might work with signals from a watt down to a femtowatt — fifteen orders of magnitude. Written out, those numbers are unusable. On a logarithmic scale the same span is about 150 units, and every figure fits on a scrap of paper.
The second is that gains multiply. A chain of an amplifier, a cable and a filter multiplies its stages together, and multiplying a string of numbers is tedious and error-prone. Logarithms turn multiplication into addition, so a signal chain becomes a sum you can do in your head. That single property is why the whole of RF engineering is conducted in decibels.
10·log or 20·log?
This is the part that confuses people, and the rule is simple once you see where it comes from.
dB = 10 · log₁₀(P₂ / P₁) · dB = 20 · log₁₀(V₂ / V₁)
Use 10·log for power — watts, milliwatts, optical intensity. Use 20·log for amplitude — volts, amps, sound pressure.
The 20 is not a different definition. Power goes with the square of voltage (P = V²/R), and log(x²) = 2·log(x), so the square pulls a factor of two out front. Both formulas measure the same physical change; they just start from different quantities. Doubling a voltage into a fixed load quadruples the power, and both expressions correctly report 6 dB.
The catch is the phrase "into a fixed load". The 20·log form assumes the impedance is the same at both points. Compare voltages across different impedances and the shortcut breaks — you have to go back to power. In audio and RF, where impedances are standardised, this rarely bites; in general circuit work it can.
The numbers worth memorising
Four values cover most mental arithmetic in decibels:
- 3 dB — a factor of 2 in power (exactly 3.0103)
- 6 dB — a factor of 2 in voltage, or 4 in power
- 10 dB — a factor of 10 in power
- 20 dB — a factor of 10 in voltage, or 100 in power
From those you can decompose almost anything. 33 dB is 30 + 3, so a thousand times power then doubled: 2,000×. 17 dB is 20 − 3, so 100 divided by 2: 50×. This is quicker than reaching for a calculator, and it builds the intuition that makes a wrong answer look wrong.
The 3 dB figure is why filter cutoffs are quoted where they are. The −3 dB point is where the circuit passes half the power it does in the passband — and in voltage terms that is 1/√2, about 0.707, which is why that number turns up throughout filter design.
dB is a ratio; dBm is a level
This distinction causes more confusion than the logarithms do.
A plain dB figure is dimensionless. It describes a relationship between two quantities — "the amplifier has 20 dB of gain", "the cable loses 3 dB". On its own it cannot tell you how much signal there is, only how much something changed it.
A suffixed figure is absolute, because the suffix names a reference:
- dBm — relative to 1 milliwatt. 0 dBm is 1 mW, 30 dBm is 1 W.
- dBW — relative to 1 watt. 0 dBW is 30 dBm.
- dBV — relative to 1 volt RMS.
- dBu — relative to 0.7746 V, the voltage that gives 1 mW into 600 Ω.
- dBSPL — relative to 20 µPa, roughly the threshold of hearing.
So "the signal is −80 dBm" is a complete statement. "The signal is −80 dB" is not, unless the reference was established elsewhere. When you meet a bare dB where an absolute level is clearly meant, the reference has been dropped, and it is worth finding out which one before trusting the number.
What you can and cannot add
You can add a dB to a dBm: a −80 dBm signal through a 20 dB amplifier becomes −60 dBm. Gains and losses in dB accumulate along a chain by simple addition, and that is exactly what alink budget is — transmit power in dBm, plus and minus a column of gains and losses in dB, giving a received level in dBm.
You cannot meaningfully add two dBm figures. Adding −10 dBm and −10 dBm does not give −20 dBm, and it does not give −5 dBm either. Two equal powers combine to twice the power, which is 3 dB more: −7 dBm. To sum absolute powers you must convert out of decibels, add the watts, and convert back — which the dBm converter will do for you.
Does dBm depend on impedance?
Not for the power. dBm measures power, and 0 dBm is 1 mW regardless of what it is delivered into.
Impedance only matters when you convert to a voltage, because the same power produces different voltages across different resistances. In RF work 50 Ω is assumed almost universally, making 0 dBm about 223 mV RMS. Cable television uses 75 Ω, and legacy audio 600 Ω, and the voltage equivalent changes accordingly. If a datasheet quotes a level in dBm and you need volts, find the impedance first.
Where decibels mislead
Two traps are worth naming.
Small dB numbers hide large differences. A 3 dB improvement sounds marginal and is a doubling of power. A 10 dB difference between two antennas is a factor of ten. Because the scale is compressed, the eye consistently underrates what a few decibels are worth — which is why a link that fails by 6 dB feels like a near miss when it is actually short by a factor of four.
Averaging decibels is usually wrong. The mean of 0 dB and 20 dB is not 10 dB in any physically useful sense, because you have averaged exponents rather than quantities. If you need an average of several levels, convert to linear units, average those, and convert back.