RF Logarithmic Units Reference Table: dB, dBm, dBW, dBi, and dBd

RF logarithmic units reference table for dB, dBm, dBW, dBi, and dBd, with conversion formulas, power ratios, and worked RF examples.

November 1, 2026 admin 8 minute read

dB expresses a ratio; dBm and dBW express power levels relative to fixed reference powers; dBi and dBd express antenna gain relative to different reference antennas. They are not interchangeable. The reference determines what a logarithmic value means and which calculations are valid. NIST SP 811, Section 8.7, establishes the power- and amplitude-ratio relationships and emphasizes identifying the reference level.

RF Logarithmic Units Reference Table

The table uses conventional RF notation. All logarithms are base 10. Symbols with unit subscripts, such as $(P_{\mathrm{dBm}})$ , represent numerical values in the stated units.

Notation	Quantity and reference	Linear to logarithmic	Logarithmic to linear	Required context
dB, power ratio	Ratio of two positive powers; no fixed absolute reference	(D=10\log_{10} $(P_2/P_1)$ )	$(P_2/P_1=10^{D/10})$	Identify numerator and denominator; use matching power units.
dB, amplitude ratio	Ratio of two positive amplitude magnitudes	(D=20\log_{10} $(A_2/A_1)$ )	$(A_2/A_1=10^{D/20})$	Interpreting this as a power ratio requires the same proportionality between amplitude squared and power.
dBm	Power level relative to 1 mW	(P_{\mathrm{dBm}}=10\log_{10} $(P_{\mathrm{mW}})$ )	$(P_{\mathrm{mW}}=10^{P_{\mathrm{dBm}}/10})$	State the port or reference plane and whether power is average, peak, or measured over a specified interval.
dBW	Power level relative to 1 W	(P_{\mathrm{dBW}}=10\log_{10} $(P_{\mathrm{W}})$ )	$(P_{\mathrm{W}}=10^{P_{\mathrm{dBW}}/10})$	Use the same power definition and reference plane when comparing values.
dBi	Antenna gain relative to an isotropic radiator	(G_{\mathrm{dBi}}=10\log_{10} $(g_i)$ )	$(g_i=10^{G_{\mathrm{dBi}}/10})$	Identify frequency, direction, polarization, and the manufacturer's gain definition.
dBd	Antenna gain relative to a half-wave dipole reference	(G_{\mathrm{dBd}}=10\log_{10} $(g_d)$ )	$(g_d=10^{G_{\mathrm{dBd}}/10})$	Identify the dipole reference and use consistent gain conditions.

Here, $(P_1)$ and $(P_2)$ are powers expressed in the same linear unit; $(A_1)$ and $(A_2)$ are amplitude magnitudes of the same kind; and $(g_i)$ and $(g_d)$ are dimensionless antenna-gain ratios relative to their respective references. NIST supports the 10-log and 20-log relationships; Rohde & Schwarz documents conventional RF power-level notation; Cisco explains the isotropic and dipole antenna references.

A notation distinction matters when citing NIST: SP 811, Section 8.7, Note 3, does not endorse attaching reference identifiers to the dB symbol in forms such as dBm. Its explicit-reference notation would express the quantity as a power level in dB relative to 1 mW. This page uses dBm, dBW, dBi, and dBd because they are established RF industry conventions, while retaining their references explicitly.

Converting Power Levels Without Changing Their Meaning

For power (P>0):

P_{\mathrm{dBm}} =10\log_{10}\left(\frac{P}{1\ \mathrm{mW}}\right)

P_{\mathrm{dBW}} =10\log_{10}\left(\frac{P}{1\ \mathrm{W}}\right)

Because 1 W equals 1,000 mW:

P_{\mathrm{dBm}}=P_{\mathrm{dBW}}+30

P_{\mathrm{W}}=10^{(P_{\mathrm{dBm}}-30)/10}

These conversions change the representation, not the underlying power. They do not require a frequency or an impedance input. Impedance becomes necessary when converting between power and voltage.

Common dBm, dBW, and linear power values

Power level, dBm	Power level, dBW	Power, mW	Power, W
−60	−90	0.000001	0.000000001
−30	−60	0.001	0.000001
−10	−40	0.1	0.0001
0	−30	1	0.001
10	−20	10	0.01
20	−10	100	0.1
30	0	1,000	1
40	10	10,000	10

The table follows directly from the fixed-reference power formulas. Negative dBm means a positive power below 1 mW, not negative power. Zero dBm means 1 mW; zero watts has no finite logarithmic level because (\log_{10}(0)) is undefined. As positive power approaches zero, its level approaches negative infinity.

Reading Gain and Loss in dB

For a signal traveling from input to output, define signed power gain as:

G_{\mathrm{dB}} =10\log_{10}\left(\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\right)

Positive gain means output power exceeds input power. Negative gain means attenuation. A component described as having “2 dB loss” normally gives the loss as a positive magnitude:

L_{\mathrm{dB}} =10\log_{10}\left(\frac{P_{\mathrm{in}}}{P_{\mathrm{out}}}\right)

Therefore, add signed gains but subtract positive loss magnitudes:

P_{\mathrm{out,dBm}} =P_{\mathrm{in,dBm}}+G_{\mathrm{dB}}-L_{\mathrm{dB}}

Do not enter a loss as −2 dB and then subtract it again. Rohde & Schwarz distinguishes signed transfer gain from attenuation reported as a positive loss magnitude in Section 2.5 of application note 1MA98.

Common logarithmic ratios

Signed change, dB	Power ratio $(P_2/P_1)$	Amplitude ratio $(A_2/A_1)$
−20	0.0100	0.1000
−10	0.1000	0.3162
−6	0.2512	0.5012
−3	0.5012	0.7079
0	1.0000	1.0000
+3	1.9953	1.4125
+6	3.9811	1.9953
+10	10.0000	3.1623
+20	100.0000	10.0000

Values are rounded from $(10^{D/10})$ and $(10^{D/20})$ . Thus, “3 dB doubles power” is a useful approximation; an exact doubling is approximately 3.0103 dB. A 6 dB increase approximately doubles amplitude, not power.

Adding component gains and losses assumes that those values describe the actual operating conditions and compatible reference planes. A small-signal amplifier gain is not necessarily valid after compression, and a nominal cable loss does not account automatically for every connector or mismatch in an installation.

When to Use 10 log and When to Use 20 log

Use $(10\log_{10})$ for a power ratio. Use $(20\log_{10})$ for an amplitude ratio, such as voltage magnitude. The voltage-ratio expression also equals the power-ratio expression only when the relevant resistances are equal.

For RMS voltages across positive, purely resistive loads:

P_1=\frac{V_1^2}{R_1}, \qquad P_2=\frac{V_2^2}{R_2}

Therefore:

10\log_{10}\left(\frac{P_2}{P_1}\right) = 20\log_{10}\left(\frac{V_2}{V_1}\right) + 10\log_{10}\left(\frac{R_1}{R_2}\right)

The resistance term disappears only when $(R_1=R_2)$ . Rohde & Schwarz derives this condition in Section 2.3 of application note 1MA98.

For example, 0 dBm is 0.001 W. Across a 50 Ω resistive load:

V_{\mathrm{RMS}} =\sqrt{PR} =\sqrt{0.001\times50} \approx0.2236\ \mathrm{V}

That is the RMS voltage across the load—not automatically the open-circuit source voltage, peak voltage, or peak-to-peak voltage. For complex impedances or mismatched RF systems, use the applicable real-power and traveling-wave definitions rather than treating the impedance magnitude as a resistance.

Converting dBi and dBd

The conventional antenna-gain conversion uses the approximately 2.15 dB maximum gain of an ideal, lossless half-wave dipole in free space relative to an isotropic radiator:

G_{\mathrm{dBi}}\approx G_{\mathrm{dBd}}+2.15

G_{\mathrm{dBd}}\approx G_{\mathrm{dBi}}-2.15

Manufacturer documents may use slightly different approximations. Cisco's antenna reference guide uses 2.14 dB and notes that some documentation rounds the value to 2.2 dB. Keep the stated convention consistent when reproducing a manufacturer's calculation; the offset is a reference conversion, not an antenna measurement tolerance.

Using the 2.15 dB convention:

0 dBd corresponds to approximately 2.15 dBi.
6 dBd corresponds to approximately 8.15 dBi.
8 dBi corresponds to approximately 5.85 dBd.

The conversion does not add physical gain. It expresses the same antenna gain against another reference. A passive antenna concentrates radiation in some directions at the expense of others; its gain is not an increase in total transmitter output power.

Do not assume every physical antenna called a “dipole” has exactly 2.15 dBi gain. Actual gain depends on construction and operating conditions. For example, Cisco lists its AIR-ANT2524DB-R family at 2 dBi in the 2.4 GHz band and 4 dBi in the 5 GHz band. Those are product ratings, not alternative definitions of dBi.

Applying the Units to an EIRP Calculation

For a single transmit path, equivalent isotropically radiated power, or EIRP, combines transmitter power, intervening losses, and antenna gain:

\mathrm{EIRP}_{\mathrm{dBm}} =P_{\mathrm{TX,dBm}} -L_{\mathrm{feed,dB}} +G_{\mathrm{antenna,dBi}}

The reference planes must connect: transmitter power is taken at the transmitter output, feed loss covers the path from that output to the antenna input, and antenna gain uses the corresponding antenna-input reference. Cisco documents this power-plus-gain-minus-loss relationship.

Consider an illustrative planning case, not a measured installation:

Input	Assumed value	Meaning
Transmitter output	20 dBm	100 mW at the transmitter connector
Feed-path loss	2 dB	Total loss between transmitter connector and antenna input
Antenna gain	8 dBi	Gain in the evaluated direction at the operating frequency

The logarithmic calculation is:

\mathrm{EIRP}=20-2+8=26\ \mathrm{dBm}

The corresponding linear calculation is:

P_{\mathrm{antenna}} =0.1\ \mathrm{W}\times10^{-2/10} \approx0.0631\ \mathrm{W}

\mathrm{EIRP} =0.0631\ \mathrm{W}\times10^{8/10} \approx0.398\ \mathrm{W}

The approximately 63.1 mW is power at the antenna input. The approximately 398 mW is isotropic-equivalent power in the evaluated direction, not additional conducted power created by the antenna.

Count each loss once. If an antenna-system rating already includes an attached cable, do not subtract that cable loss again. This example establishes unit arithmetic only: it does not determine coverage, link availability, or compliance with a jurisdiction-specific radiated-power limit.

Subtract Power Levels for Ratios; Add Linear Powers for Totals

Subtracting two dBm levels produces a difference in dB:

D_{\mathrm{dB}} =P_{\mathrm{2,dBm}}-P_{\mathrm{1,dBm}}

For example, −60 dBm is 10 dB above −70 dBm, corresponding to ten times the power. The result is a ratio, not another absolute power level.

Adding independent signal powers is a different operation. Convert each level to linear power, add, and convert the total back:

P_{\mathrm{total,dBm}} = 10\log_{10} \left(\sum_{i=1}^{n}10^{P_{i,\mathrm{dBm}}/10} \right)

Here, each exponential gives a numerical power in milliwatts. Two uncorrelated signals of 20 dBm each contribute 100 mW each, giving:

P_{\mathrm{total}} =200\ \mathrm{mW}

P_{\mathrm{total,dBm}} =10\log_{10}(200) \approx23.01\ \mathrm{dBm}

The result is not 40 dBm. The simple power sum applies when cross terms average to zero, as for uncorrelated signals. Coherent signals require their amplitude and phase relationships; any physical combiner losses must also be accounted for. Rohde & Schwarz addresses these distinctions in Sections 4.1 and 4.3 of application note 1MA98.

Frequency, Bandwidth, and Reference-Plane Checks

Frequency is not an input to a dBm-to-watt conversion or to the conventional dBi-to-dBd reference offset. It is still essential when choosing the gain and loss values to convert. Cisco's reference guide, for example, specifies different antenna gains and cable losses for different operating bands. Use the value applicable to the actual frequency, antenna configuration, and direction—not simply the largest number on the data sheet.

Bandwidth also changes the meaning of a reported level. dBm is a power level; dBm/Hz describes power spectral density relative to 1 mW/Hz. For a density that is flat across an effective noise bandwidth (B):

P_{\mathrm{dBm}} =N_{\mathrm{dBm/Hz}} +10\log_{10}\left(\frac{B}{1\ \mathrm{Hz}}\right)

Here, $(N_{\mathrm{dBm/Hz}})$ is the numerical density level and (B) is in hertz. A nonflat spectrum must be integrated in linear power-density units rather than multiplied by bandwidth as though it were flat.

Before comparing two RF values, retain enough context to identify:

The port or reference plane: transmitter output, antenna input, receiver input, or a radiated equivalent.
The power statistic: average, peak, or another explicitly defined measurement.
The frequency and, where relevant, channel or measurement bandwidth.
The antenna reference and whether the reported gain applies to the required direction and polarization.

A correct logarithmic conversion preserves the original quantity and its conditions. It cannot make power levels measured at different boundaries or under different operating conditions directly comparable.