Transmission Levels, dB & dBm

The decibel is not a unit of power, voltage, or loudness. It is a level: the logarithm of the ratio between a quantity and a reference quantity of the same kind. Everything that goes wrong with decibels goes wrong in one of two ways — the reference was never stated, or the same physical change was written as 3 dB by one engineer and 6 dB by another because one measured power and the other measured voltage.

Both failures have exact fixes in the standards. ITU-R V.574 and ITU-T G.100.1 pin down what each suffixed symbol means, which factor applies to which quantity, and — the part almost nobody outside telephony learns — how relative levels (dBr, dBm0) let a transmission plan state what a signal should be independently of where you happened to measure it.

Why anyone bothers with a logarithm #

Two reasons, and only the second one is really about arithmetic.

The dynamic range is absurd. A Wi-Fi access point radiates on the order of 0.1 W and the receiver at the far end of the room decodes something around 10^-11 W. That is ten orders of magnitude in one link, and a satellite downlink is worse. Writing 0.000000000012 W in a budget spreadsheet is how you lose a factor of ten.

Gains and losses cascade multiplicatively, and logarithms turn that into addition. A chain of an amplifier, a cable, a connector, and a path loss multiplies four ratios. In decibels it is a column of signed numbers you can add in your head and audit line by line. That is the entire ergonomic case for the unit, and it is a strong one.

flowchart LR
  TX["PA out<br/>+30 dBm"] --> C1["jumper<br/>-0.5 dB"]
  C1 --> A1["antenna<br/>+14 dBi"]
  A1 --> P["path loss<br/>-100 dB"]
  P --> A2["antenna<br/>+14 dBi"]
  A2 --> C2["feeder<br/>-3 dB"]
  C2 --> RX["at LNA input<br/>-45.5 dBm"]

There is a third, physical reason that is easy to miss: attenuation in a uniform medium is exponential in distance, $P(l) = P_0 e^{-2\alpha l}$, so the loss in decibels is linear in length. That is why cable and fibre specs are quoted as dB per 100 m or dB per km at all — the number is a constant of the medium, not of the run.

The definition, and where the factor of two comes from #

$$L_P = 10\lg\frac{P}{P_\text{ref}}\ \ \text{dB} \qquad\qquad L_F = 20\lg\frac{F}{F_\text{ref}}\ \ \text{dB}$$

The second form is not a separate convention. It is the first form with an assumption folded in. Take two powers dissipated by currents $I_1, I_2$ in resistances $R_1, R_2$ and expand:

$$10\lg\frac{P_1}{P_2} \;=\; 10\lg\frac{I_1^{2}R_1}{I_2^{2}R_2} \;=\; 20\lg\frac{I_1}{I_2} \;+\; 10\lg\frac{R_1}{R_2}$$

The factor 20 appears because the field quantity is squared on its way to power. The impedance term is what you throw away when you write 20 lg. V.574-5 §1.2 states the condition bluntly: the logarithm of the field-quantity ratio is multiplied by 20 "assuming that the impedances are equal", and "use of the decibel when the impedances are not equal is not appropriate unless adequate information is given concerning the impedances involved."

The generalisation covers everything else. If $P_1/P_2 \propto (X_1/X_2)^{\alpha}$ for some other quantity $X$, then the level is $10\alpha \lg(X_1/X_2)$ (V.574-5 §1.3). Power flux density, spectral density, and $E_b/N_0$ all fall out of that rule; sound pressure gets 20 because pressure is a root-power quantity.

The neper, and why old courses look different #

The neper takes the natural logarithm, and it is defined on field quantities: $1\ \text{Np}$ is a factor of $e$ in a field quantity, hence $e^2$ in power. So a power ratio in nepers carries a factor of one half:

$$L_P = \tfrac{1}{2}\ln\frac{P_x}{P_0}\ \ \text{Np} \qquad\qquad L_U = \ln\frac{U_x}{U_0}\ \ \text{Np}$$

$$1\ \text{Np} = (20\lg e)\ \text{dB} = 8.686\ \text{dB}, \qquad 1\ \text{dB} = 0.1151\ \text{Np}$$

This is not archaeology if you read Soviet- or Ukrainian-tradition material: the lecture course this territory is built on defines absolute and relative transmission levels in nepers with exactly that $\tfrac12\ln$ for power, against the GOST reference triad $P_0 = 1\ \text{mW}$, $U_0 = 0.775\ \text{V}$, $I_0 = 1.29\ \text{mA}$. Those three numbers are one number: $0.775\ \text{V}$ across $600\ \Omega$ draws $1.29\ \text{mA}$ and dissipates $1\ \text{mW}$. The same 600 Ω telephone system that gives us dBu, written in the other logarithm.

V.574-5 §3 is explicit that for international exchange of measurements the decibel is the only permitted logarithmic expression, and warns about a genuine ambiguity: in radiometry the neper is sometimes used for power ratios without the ½, which would make 1 Np equal 4.34 dB instead of 8.686 dB. That usage is prohibited in telecommunications.

The landmarks #

Four numbers do most of the mental work. Two of them are not the round numbers people quote.

Read the amplitude column only when the two points share an impedance.

Cascading: what you may add, and what you may not #

Gains and losses in decibels add along a chain, because the underlying ratios multiply. This is safe and it is the reason the unit exists. The link budget is nothing but that column:

Note what happened at line 3: a dB (pure ratio) was added to a dBm (absolute level) and the result is still dBm. That is legal and it is the only kind of addition that is. The general rule:

• $\text{dBm} + \text{dB} = \text{dBm}$ — apply a gain or loss to an absolute level.
• $\text{dBm} - \text{dBm} = \text{dB}$ — a ratio between two absolute levels (this is what an SNR is).
• $\text{dBm} + \text{dBm}$ — meaningless. Two absolute levels never add this way, in either sense of "add".

Adding two powers #

Combining two signals or two noise sources is an addition of powers, and powers do not add in the log domain. You must leave it:

$$L_\Sigma = 10\lg\!\left(10^{L_1/10} + 10^{L_2/10}\right)\ \ \text{dB}$$

Equivalently, and more useful at a whiteboard: take the larger of the two and add an increment that depends only on their difference.

Two equal, uncorrelated noise sources give +3.01 dB. A contributor 10 dB below the dominant one moves the total by 0.4 dB, and one 20 dB below is invisible — which is the practical justification for ignoring minor noise terms in a budget, and the reason chasing a 0.5 dB improvement in a term that is already 15 dB down is wasted work.

Absolute references #

The bare symbol dB means a ratio. A suffix pins a reference and makes the number absolute. V.574-5 §4 lays out the notation rule: dB followed by information in parentheses names a reference value, and the familiar short forms (dBm for dB(mW)) are grandfathered abbreviations of exactly that.

The conversions between the voltage-referenced units are pure arithmetic and worth committing to memory:

$$L_\text{dBu} = L_\text{dBV} + 2.2185$$

That constant assumes the exact reference $\sqrt{0.6} = 0.774597\ldots$ V. The standards print the rounded 0.775 V, which gives 2.2140 instead — so a table quoting 0.775 V next to a 2.2185 conversion is quietly mixing the two. The discrepancy is 4.5 thousandths of a dB and matters only in that it tells you which reference a document is actually using.

$$L_\text{dBm}\big|_{75\,\Omega} = L_\text{dBmV} - 48.75 \qquad\qquad L_\text{dBm}\big|_{50\,\Omega} = L_{\text{dB}\mu\text{V}} - 106.99$$

Those last two are the ones that carry a hidden impedance. dBmV only equals a power level in a 75 Ω system; dBµV only maps to dBm by −107 in a 50 Ω system. Move the same reading to the other impedance and both constants change.

Relative levels: the part telephony got right #

Here is the problem the rest of the industry keeps re-solving badly. A signal travelling through a network has a different absolute power at every point, because there are pads, amplifiers, and codecs between them. A specification that says "dial tone shall be −19 dBm" is worthless unless it also says where. Writing the measurement point into every requirement does not scale.

The fix is to give every point in a signal path a relative level in dBr, and express requirements at a notional reference point.

Which collapses the whole business into one equation you can apply in either direction:

$$\boxed{\ \text{dBm} = \text{dBm0} + \text{dBr}\ }\qquad\qquad \text{dBm0} = \text{dBm} - \text{dBr}$$

The same algebra applies to noise, weighted or not — which is exactly why signal-to-noise ratios can be computed from levels that were never measured at the same place. North American practice calls the same idea a transmission level point (TLP); a point where the nominal test signal is 0 dBm is a zero TLP, and dBm0 means "dBm measured at a zero TLP". Same concept, different vocabulary.

Real transmission plans put the reference somewhere convenient rather than somewhere physical. In the ITU-T plan the transmission reference point T is a virtual two-wire point; the virtual switching point V on the sending side of a four-wire international circuit sits at −3.5 dBr, connected to T through a 3.5 dB hybrid. The conventional load used to compute noise on multi-channel carrier systems is an absolute mean power level of −15 dBm at T (V.574-5 §6.2.2).

Digital full scale: dBFS and dBov are not the same zero #

Once the signal is samples, the natural reference is the largest number the format can hold. Two standards chose different definitions of "largest", and they differ by exactly 3.01 dB.

• dBov (ITU-T G.100.1 §8.7) references the power $P_0 = 1.0$ of a normalised signal at the overload point $x_\text{over} = 1.0$ — a square wave alternating between the extremes. A full-scale sine therefore sits at −3.01 dBov, because a sinusoid's mean square is half its peak square.
• dBFS in the audio world references a full-scale sine wave as 0 dBFS. AES17 defines 0 dB FS as the rms amplitude of a full-scale sine, and states the consequence directly: because full scale is defined on a sine, a square-wave test signal can read as much as +3.01 dB FS. So the same waveform reads 0 dBFS and −3.01 dBov.

Neither is wrong; they are answering different questions (what the arithmetic can hold vs what the converter can reproduce). Mixing them costs you 3 dB of headroom in whichever direction hurts, and 3 dB is half your power.

The number of dB per bit falls out of the amplitude rule: one extra bit doubles the representable amplitude, so it buys 6.02 dB of range (see quantization error for why the usable SNR figure is $6.02N + 1.76$ dB rather than $6.02N$).

Headroom, sensitivity, and other places the numbers show up #

Audio headroom is a subtraction of levels. In the telephone network the nominal active speech level is −11 dBm0, the mean channel load for FDM planning is −15 dBm0 (with pauses included), and the peak clipping level for a sinusoid is +3 dBm0 — 14 dB of headroom over nominal speech. G.100.1 §9.3 justifies that number from the amplitude statistics: speech peaks exceeded 1% of the time sit about 12 dB above the rms value, and subjective tests put 12 dB as the acceptable lower bound before clipping is objectionable. The margin is not a round number someone picked; it is a measured distribution plus a listening test.

Receiver sensitivity is a level, and the floor under it is thermal. Thermal noise power in a bandwidth $B$ is $kTB$; at the standard reference temperature of 290 K this is

$$10\lg(kT \cdot 10^{3}) = -173.98 \approx -174\ \text{dBm/Hz}$$

so a 20 MHz channel has a noise floor of $-174 + 10\lg(2\times10^{7}) = -101$ dBm before the receiver's own noise figure is added. Every sensitivity figure in a datasheet is that floor plus noise figure plus the required SNR — see noise sources and link budgets.

Cable loss is quoted per unit length because it is linear in length — and it is emphatically not linear in frequency. Times Microwave publish LMR-400's attenuation as a closed form in dB per 100 ft, which is the mechanism written down:

$$A = 0.122290\,\sqrt{F_\text{MHz}} \;+\; 0.000260\,F_\text{MHz}$$

The $\sqrt f$ term is conductor loss through the skin effect; the linear term is dielectric loss, and it only starts to matter in the gigahertz. Their published table:

Two traps sit in that table. 100 ft is 30.48 m, so a "per 100" figure read in the wrong units is out by 3.28× — a 30 m feeder at 2.5 GHz loses 6.7 dB, not 6.8 dB (that is the per-100-ft figure) and not 22 dB. And because loss goes as roughly $\sqrt f$, the 900 MHz number tells you almost nothing about the 5.8 GHz run: the same cable that costs you 12.8 dB/100 m on GSM costs 35.5 dB/100 m on 5 GHz Wi-Fi. See transmission media.

The errors, collected #