RF Power: Watts, dB, dBm, and Measurement

01 / 12

What exactly is RF power?

Two transmitters operate at 2.45 GHz. Does that mean they transmit the same power?

No. Frequency is the repetition rate: both signals complete 2.45 billion cycles per second. Voltage amplitude describes an electrical difference at a point. Energy accumulates in joules. Power describes how quickly energy is transferred.

Over a stated interval, average power is the transferred energy divided by elapsed time. One watt means one joule transferred each second.

$P_{\mathrm{avg}}=\frac{\Delta E}{\Delta t}\qquad1\,\mathrm W=\frac{1\,\mathrm J}{1\,\mathrm s}$ P_avg is average power [W]; ΔE is transferred energy [J]; Δt is the measurement interval [s].

Think about itIf 10 mW is delivered continuously for two seconds, how much energy crosses the boundary?

Answer

$10\,\mathrm{mW}\times2\,\mathrm s=20\,\mathrm{mJ}$ . Frequency never enters this calculation. The energy keeps accumulating even though the power stays constant.

The boundary matters. A cable connector can carry conducted power toward an antenna, but that number is not automatically the power that the antenna accepts, radiates, or directs toward a receiver.

01Conducted connector powerat a named cable or connector plane
02Accepted antenna powernet power entering the antenna
03Radiated powerafter antenna dissipation
04ERP or EIRPa directional equivalent using a named reference antenna

Common misconceptionHigher frequency means higher power, and RF power always means radiated power.

Neither is true. A tiny receiver signal and a powerful transmitter can share the same frequency. RF energy can remain guided in a PCB trace or coax. Always name the power quantity and the boundary where it applies.

Go deeperGenerated, available, incident, reflected, and delivered power

A source may be described by its generated output or by available power: the maximum it could deliver under the formal matching condition. At a transmission-line reference plane, power can travel toward a load as incident power and back toward the source as reflected power. Net delivered or absorbed power is incident minus reflected power at that plane. The next module develops the impedance and matching ideas needed to use those terms fully.

At an antenna, accepted power is the net power entering its terminals. Radiated power is lower if the antenna dissipates some power. EIRP and ERP also include directional gain and use different reference antennas, so neither is a synonym for conducted connector power.

Engineering consequence

A trustworthy statement sounds like “10 mW average conducted power at the output connector,” not merely “10 mW RF.” Next we need to see how an alternating voltage can deliver that energy at all.

02 / 12

How a sinusoid delivers energy

A sine-wave voltage averages to zero. Why can it still heat a resistor?

Put the recurring 2.45 GHz sine across an ideal 50 Ω resistor. When voltage is positive, current is positive. Half a cycle later they are both negative. Their directions reverse together, so their product remains positive: energy continues moving into the resistor.

$p(t) = v(t)i(t) \qquad P_{\mathrm{avg}} = (\frac{1}{T}) \int _{0}^{T} p(t)d t$ p(t) is instantaneous power [W];v(t) is instantaneous voltage [V];i(t) is instantaneous current [A];T is one carrier period [s].

Think about itWhen both voltage and current are negative, is resistor power negative too?

Answer

No. A negative value multiplied by a negative value is positive. The resistor keeps absorbing energy during both halves of the voltage cycle.

Interactive power-in-time model

If voltage averages to zero, where does the heat come from?

Predict the sign of power when both voltage and current are negative. Then pause or scrub the slowed two-cycle view and compare all three traces at the same instant.

Position in the two-cycle window

Carrier phase: 90°
Voltage: +1.000 V
Current: +20.00 mA
Instantaneous power: 20.00 mW
Energy since plot start: 1.020 pJ

Assumptions: zero-offset 2.45 GHz sine, ideal 50 Ω resistor, and no reflected power. The animation is slowed and its horizontal axis is normalized; it is not real-time RF. One actual carrier cycle transfers 4.0816 pJ on average.

For this signal, $v(t)=1\sin(2\pi ft)\,\mathrm V$ and $i(t)=20\sin(2\pi ft)\,\mathrm{mA}$ . Their product is $p(t)=20\sin^2(2\pi ft)\,\mathrm{mW}$ . Instantaneous power moves from 0 to 20 mW; its average is exactly 10 mW.

The power trace has two humps during every voltage cycle. Squaring removes the sign, so the positive and negative half-cycles have the same power shape. Power therefore repeats at twice the 2.45 GHz carrier frequency, even though the carrier frequency itself did not change.

Common misconceptionA sine voltage that averages to zero delivers zero average power.

Average voltage and average power are different operations. Power depends on the point-by-point product of voltage and current. Zero-mean voltage can produce positive average heating.

Go deeperWhy the average is exactly half the peak

The identity $\sin^2\theta=\frac{1-\cos2\theta}{2}$ separates the power waveform into a constant half and an oscillating term whose full-cycle average is zero. That is why 20 mW peak instantaneous resistor power becomes 10 mW average power. At 2.45 GHz, one carrier cycle lasts 408.163 ps and transfers 4.0816 pJ on average.

An ideal resistor only absorbs energy. Reactive loads can temporarily store and return energy, allowing instantaneous power to become negative. That behavior belongs with impedance and reactance in the next module.

03 / 12

RMS: the amplitude that predicts heating

What DC voltage would heat the same resistor by the same amount?

That is the practical meaning of root mean square. Square the changing voltage so both polarities contribute, average the squared values across a complete interval, then take the square root. The result is the DC voltage that produces the same average heating in the same resistor.

$V_{\mathrm{rms}} = \sqrt{(\frac{1}{T}) \int _{0}^{T} v^{2}(t)d t}$ V_rms is RMS voltage [V];v(t) is instantaneous voltage [V]; the interval T covers a complete periodic waveform.

For a zero-offset sine wave only, $V_{\mathrm{rms}}=\frac{V_{\mathrm{pk}}}{\sqrt2}$ and $V_{\mathrm{pp}}=2V_{\mathrm{pk}}$ . The recurring 1.000 Vpk sine is therefore 0.7071 Vrms.

For the stated ideal resistive load, RMS voltage and current lead directly to average power.

$P_{\mathrm{avg}} = V_{\mathrm{rms}}I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}^{2}}{R} = I_{\mathrm{rms}}^{2}R$ R is resistance [Ω]. These forms use RMS values and the ideal resistive-load assumption stated here.

Think about itA sine, square wave, and triangle wave all reach ±1 V. Do they deliver equal average power to the same resistor?

Answer

No. Equal peaks do not mean equal mean-square values. How long each waveform spends near its peaks changes its RMS value and its heating.

Interactive RMS heating comparator

Do equal peaks produce equal heating?

Make a prediction, then compare three zero-offset waveforms with the same peak voltage across the same ideal 50 Ω resistor.

Common peak voltage1.00 Vpk

Sine: 0.7071 Vrms10.000 mW average
Square: 1.0000 Vrms20.000 mW average
Triangle: 0.5774 Vrms6.667 mW average

Why: squaring makes both voltage polarities contribute positively. The square wave remains at its peaks, the triangle spends more time near zero, and the sine lies between. These ideal waveforms are periodic, symmetric, and have no DC offset.

Equal peak voltage across one ideal 50 Ω resistor
Zero-offset waveform	Peak	RMS	Average power
Sine	1.000 V	0.7071 V	10.000 mW
Symmetric square	1.000 V	1.0000 V	20.000 mW
Symmetric triangle	1.000 V	0.5774 V	6.667 mW

Common misconceptionEqual peak voltages imply equal power; RMS is average voltage; and Vpk/√2 works for every waveform.

RMS is not the arithmetic average; every symmetric waveform above has 0 V average. The 1/√2 factor belongs to a zero-offset sine, not to square or triangle waves. RMS normally qualifies voltage or current. Power is usually described by its averaging or peak definition—not “RMS power”; for example, average power calculated from RMS voltage.

Engineering consequence

A detector calibrated for one waveform can misread another when crest factor changes. Thermal and true-RMS methods follow mean-square heating; simpler voltage-responding detectors may depend on waveform shape and operating range.

Go deeperRMS with a DC offset

The integral definition includes every part of the waveform. If a signal contains a DC component and a zero-mean AC component, their mean squares add: $V_{\mathrm{total,rms}}^2=V_{\mathrm{DC}}^2+V_{\mathrm{AC,rms}}^2$ . The simple sine conversion must not be applied blindly to an offset or distorted waveform.

04 / 12

Power needs a load and a reference plane

Does “1 Vrms” tell us how much power is being delivered?

Not by itself. The same RMS voltage produces different current and heating in different resistances. For an ideal resistor, divide the squared RMS voltage by the stated resistance.

Voltage-to-power conversions require a stated resistive load
Known quantity	Load	Result
1.000 Vrms	50 Ω	20.000 mW = +13.0103 dBm
1.000 Vrms	75 Ω	13.333 mW = +11.2494 dBm
+10 dBm = 10 mW	stated 50 Ω condition	0.7071 Vrms

Fifty ohms is a common RF interface condition, not a universal property of RF. Seventy-five ohms is common in video and cable systems, while antenna terminals, device pins, resonators, and matching networks can present many other impedances.

A reference plane is the exact location where a level is stated or measured. Move to the far end of a lossy cable and the power level is lower. A source connector, the far end of a cable, an amplifier input, and an antenna terminal are different planes even when they are part of one short bench setup.

P0Generator connectorP1Cable outputP2Load terminals

Think about itA generator is set for 1 Vrms into 50 Ω, but a high-impedance oscilloscope shows about 2 Vrms. Is the generator necessarily wrong?

Answer

No. In a common source model, the displayed voltage is the value expected across the specified 50 Ω load. A high-impedance input draws much less current, so the measured open-circuit-like voltage can approach twice the loaded value. The source model, termination, bandwidth, probe, and reference plane must all be checked.

Common misconceptionEvery RF system is 50 Ω, and any dBm value can therefore be converted directly to voltage.

dBm itself is an absolute power level referenced to 1 mW; its definition does not require 50 Ω, and real RF impedances are not universally 50 Ω. Impedance becomes necessary only when that power is converted to voltage or current. Without the load condition, there is no unique voltage answer.

Go deeperWhy source settings and scope readings can differ

A simple RF generator can be modeled as a voltage source with a series resistance. With a matching load, part of the source voltage appears across that load. Remove the load and the terminal voltage rises. Real generators, cables, probes, and oscilloscopes have frequency limits and imperfect impedance, so “twice” is an idealized expectation—not a calibration rule.

Watts now have physical meaning, but RF systems often span factors of millions or billions. A logarithmic language makes those ratios easier to see and combine.

05 / 12

Why RF engineers use decibels

What is the easiest way to follow a signal through many multiplying gains and losses?

Imagine a weak receiver input, a cable loss, two amplifier gains, and a filter loss. In watts, every stage multiplies the previous value. A logarithm turns those multiplications into additions, which makes both huge dynamic ranges and long chains easier to reason about.

For two power values, their ratio expressed in decibels is:

$G_{\mathrm{dB}} = 10 \log _{10}(\frac{P_{2}}{P_{1}}) \qquad \frac{P_{2}}{P_{1}} = 10^{\frac{G_{\mathrm{dB}}}{10}}$ P₁ and P₂ are powers in the same linear unit; G_dB is their level difference [dB]. Positive means an increase; negative means a decrease.

Think about itDoes a 3 dB loss subtract 3 mW from every signal?

Answer

No. It applies a ratio. A 3 dB loss multiplies power by about 0.501, so 10 mW becomes about 5.01 mW while 100 mW becomes about 50.1 mW. Exact half-power loss is 3.0103 dB.

Useful decibel landmarks for power ratios
Level difference	Exact power ratio	Plain-language reading
+3.0103 dB	2	exactly twice the power
−3.0103 dB	0.5	exactly half the power
+10 dB	10	ten times the power
−10 dB	0.1	one tenth of the power
+20 dB	100	one hundred times the power
−20 dB	0.01	one hundredth of the power

Engineers often round 3.0103 dB to 3 dB. That is an excellent estimate, but the wording should be “approximately double” or “approximately half.” A 3.0000 dB gain is a ratio of 1.9953, not exactly 2.

Common misconceptiondB is an absolute power unit, and a dB loss subtracts a fixed number of milliwatts.

Bare dB expresses a ratio or a difference between levels. It has no fixed watt value. A decibel change multiplies linear power by the same ratio whatever the starting level.

Go deeperWhy cascade arithmetic becomes addition

If one stage multiplies power by r₁ and the next by r₂, the chain ratio is r₁r₂. Because $\log(r_1r_2)=\log r_1+\log r_2$ , their decibel values add. This is a property of logarithms, not a special behavior invented for RF.

06 / 12

dBm and dBW: absolute levels with fixed references

If 0 dBm contains a zero, does it mean there is no signal?

No. A relative decibel becomes an absolute power level when its reference is fixed. dBm compares a power with 1 mW. dBW compares it with 1 W. A level of 0 means equality with that reference—not zero physical power.

$L_{P,\mathrm{dBm}}=10\log_{10}\left(\frac{P}{1\,\mathrm{mW}}\right)$ $P\ [\mathrm{mW}]=10^{L_{P,\mathrm{dBm}}/10}$ $L_{P,\mathrm{dBW}}=10\log_{10}\left(\frac{P}{1\,\mathrm W}\right)$ $L_{P,\mathrm{dBm}}=L_{P,\mathrm{dBW}}+30$ L_P is logarithmic power level;P is the physical power expressed in the same unit as its named reference.

Think about itWhich is physically larger: −10 dBm or −30 dBm? Are either of them negative power?

Answer

−10 dBm is larger: it is 100 µW, while −30 dBm is 1 µW. Both are positive physical powers. A negative logarithmic level only means the power is below the 1 mW reference.

Interactive dB and power-level translator

Is this number a ratio, or a level?

Keep the two ideas separate. The first control compares two powers. The second places one power on a scale with the fixed 1 mW and 1 W references.

01 / Relative power ratio

dB has no fixed watt reference

Power change+3 dB

Linear power ratioP₂ / P₁ = 1.995

Output power is 1.995 times input power. A 3 dB change is approximately ×2; exact doubling is +3.0103 dB.

02 / Absolute power level

dBm and dBW name their references

Power level+10 dBm

Power: 10 mW
Watts: 1.000e-2 W
dBm: +10 dBm
dBW: −20 dBW

Assumption: these are power quantities. dBm is referenced to 1 mW and dBW to 1 W; neither definition assumes 50 Ω. A load is needed only when converting power to voltage or current.

Power ladder: equal 10 dB steps are tenfold power steps
Level in dBm	Linear power	Same level in dBW
−30 dBm	1 µW	−60 dBW
−20 dBm	10 µW	−50 dBW
−10 dBm	100 µW	−40 dBW
0 dBm	1 mW	−30 dBW
+10 dBm	10 mW	−20 dBW
+20 dBm	100 mW	−10 dBW
+30 dBm	1 W	0 dBW
+40 dBm	10 W	+10 dBW

The recurring sine is 10 mW, so its power level is +10 dBm or −20 dBW. Those two numbers describe the same physical power using references 30 dB apart.

Common misconceptiondB and dBm are interchangeable; 0 dBm is no signal; negative dBm is negative power.

Bare dB needs two powers or levels to compare. dBm names the fixed 1 mW reference. 0 dBm is exactly 1 mW, and every finite dBm value represents a positive physical power.

Go deeperA level is a logarithm of a dimensionless ratio

The logarithm does not act on a unit-bearing number by itself. Formally it acts on the ratio of a quantity to a reference quantity of the same kind. That is why careful notation names both the measured quantity and its reference. ITU telecommunications usage retains the practical dBm and dBW symbols; the reference concept must still remain explicit.

07 / 12

dBc: relative to a carrier

A spur is measured at −40 dBc. What does that tell us—and what is still missing?

The suffix “c” names the reference: the carrier. A component at −40 dBc has one ten-thousandth of the carrier power under the stated measurement conditions. Its absolute power remains unknown until the carrier level is known at the same reference plane.

$L_{\mathrm{component,dBc}} = 10 \log _{10}(\frac{P_{\mathrm{component}}}{P_{\mathrm{carrier}}})$ $L_{\mathrm{component,dBm}} = L_{\mathrm{carrier,dBm}} + L_{\mathrm{component,dBc}}$ The second line applies when both levels use the same reference plane and compatible bandwidth, detector, and averaging conditions.

Think about itA +10 dBm carrier has a spur at −40 dBc. What is the spur in dBm and watts?

Answer

Add the relative level to the carrier level: $+10\,\mathrm{dBm}-40\,\mathrm{dB}=-30\,\mathrm{dBm}$ . That is 1 µW.

Interactive dBc spectrum explorer

What does −40 dBc leave unsaid?

Predict the spur's absolute power, then move the carrier. The vertical separation stays fixed in dBc, while both absolute dBm levels move together.

Carrier level+10 dBmSpur relative to carrier-40 dBc

Carrier level: +10 dBm10 mW
Relative spur: -40 dBcre carrier at this plane
Absolute spur: −30 dBm1 µW

Assumptions: carrier and discrete spur are measured at the same reference plane with compatible bandwidth and detector settings. The frequency offset is illustrative. dBc/Hz is a bandwidth-normalized density and is not interchangeable with dBc.

Common misconception−40 dBc means −40 dBm.

dBc is a separation from a named carrier; dBm is a level relative to 1 mW. A −40 dBc component is −30 dBm beside a +10 dBm carrier, but −60 dBm beside a −20 dBm carrier.

What a dBc result should travel with

Name the carrier, absolute carrier level, reference plane, component being measured, integration or resolution bandwidth, detector, and averaging method when they affect the result.

Go deeperdBc is not dBc/Hz

A discrete spur can be reported as an integrated level relative to the carrier. Noise-like energy depends on the bandwidth in which it is measured, so it may instead be normalized per hertz. A dBc/Hz density and a dBc integrated result answer different questions. Likewise, one analyzer marker bin is not automatically the total power in a wide component or channel.

08 / 12

Why voltage ratios use 20 log

If voltage doubles, does power merely double too?

Across the same ideal resistance, power is proportional to RMS voltage squared. Doubling voltage therefore makes four times the power. That square is where the familiar factor of 20 comes from—it is not a separate definition to memorize.

$\frac{P_{2}}{P_{1}} = (\frac{V_{2}}{V_{1}})^{2}(\frac{R_{1}}{R_{2}})$ $10 \log _{10}(\frac{P_{2}}{P_{1}}) = 20 \log _{10}|\frac{V_{2}}{V_{1}}| + 10 \log _{10}(\frac{R_{1}}{R_{2}})$ $\text{When} R_{1} = R_{2}: G_{\mathrm{dB}} = 20 \log _{10}|\frac{V_{2}}{V_{1}}|$ V₁ and V₂ are RMS voltage magnitudes; R₁ and R₂ are the stated ideal resistances.

Think about itAt equal resistance, what power and dB change follow a twofold voltage increase?

Answer

Power becomes four times larger, so the level difference is $10\log_{10}4=20\log_{10}2=+6.0206\,\mathrm{dB}$ .

Equal-resistance voltage and power landmarks
Change	Voltage ratio	Power ratio	Level difference
Double voltage	2	4	+6.0206 dB
Double power	√2 = 1.4142	2	+3.0103 dB
Half voltage	0.5	0.25	−6.0206 dB

Common misconceptionVoltage ratios always use 20 log, regardless of impedance.

The shortcut represents a power-level difference only when the squared voltage ratio maps to the power ratio under the stated impedance conditions. With unequal resistances, keep the additional 10 log₁₀(R₁/R₂) term. Complex impedance and travelling-wave quantities need their own careful definitions in the next module.

Engineering consequence

Never turn a voltage gain into a power gain until the input and output quantities, impedance conditions, and measurement planes are clear.

09 / 12

Gain and loss through an RF chain

What reaches the load after an attenuator, amplifier, and cable?

Keep absolute levels and relative changes distinct. The source starts at an absolute level in dBm. Each component contributes a relative change in dB. Here, gains are positive and losses are positive magnitudes that we explicitly subtract.

$L_{\mathrm{out,dBm}} = L_{\mathrm{in,dBm}} + \sum G_{\mathrm{dB}} - \sum L_{\mathrm{dB}}$ G contains gain magnitudes [dB]; L contains loss magnitudes [dB]. Do not silently mix this convention with a list of already-signed changes.

Think about itStarting at +10 dBm, what remains after −3 dB, +13 dB, and −2 dB?

Answer

Add the signed level changes: $+10-3+13-2=+18\,\mathrm{dBm}$ . That is about 63.1 mW. The ideal blocks change power without changing the 2.45 GHz carrier frequency.

Interactive RF power-budget builder

What reaches the load after every gain and loss?

Predict the final level, then change one block. Each marker is a distinct reference plane; the level belongs to that exact point in the chain.

P0Source+10 dBm10 mWStartSource output
P13 dB attenuator+7 dBm5.012 mW−3 dBAfter attenuator
P2Linear amplifier+20 dBm100 mW+13 dBAmplifier output
P3Cable+18 dBm63.096 mW−2 dBAt ideal load

Source level +10 dBmAttenuator loss 3 dBAmplifier gain +13 dBCable loss 2 dB

Power at P3+18 dBm = 63.096 mW+10 − 3 + 13 − 2 dB

Can I add these?

Two independent contributions arrive at one plane

Choose the operation before checking the total.

Contribution A 0 dBmContribution B 0 dBm

Independent-power total2 mW = +3.010 dBm

Assumptions: all changes apply at the stated frequency, loss controls are positive magnitudes that are subtracted, terminations are suitable, and the amplifier remains linear. The separate-power sum applies to independent or uncorrelated contributions—not coherent same-frequency signals whose phase and network behavior matter.

The recurring illustrative 2.45 GHz power chain
Reference plane	Previous block	Level	Linear power
P0 · source output	—	+10 dBm	10.000 mW
P1 · after attenuator	−3 dB	+7 dBm	5.0119 mW
P2 · amplifier output	+13 dB	+20 dBm	100.000 mW
P3 · after cable / load plane	−2 dB	+18 dBm	63.0957 mW

This arithmetic assumes each quoted gain or loss applies at 2.45 GHz, the ports have the intended terminations, and the amplifier remains linear. A real amplifier cannot increase output without limit. As it approaches compression, its gain falls and distortion grows.

Common misconceptionA 20 dB amplifier always outputs +20 dBm, and nominal gain remains valid in compression.

A 20 dB gain describes an output-to-input ratio under stated conditions. A −30 dBm input would ideally become −10 dBm; a 0 dBm input would ideally become +20 dBm. Either result can be wrong if the device is outside its frequency, bias, temperature, termination, or linear power range.

Go deeperWhy a data-sheet gain is conditional

Gain varies with frequency, input level, output loading, bias, and temperature. Cables and attenuators also have frequency-dependent loss. A practical budget records the conditions, tolerances, and reference planes rather than treating every nominal number as exact.

10 / 12

How separate powers combine

Two independent contributions are each 0 dBm. Is their total still 0 dBm, 3 dBm, or 0 + 0 = 0?

Cascaded dB values can be added because they describe ratios along one path. Separate contributions are different: convert each level to linear power, add those powers, then convert the total back to a logarithmic level.

Think about itWhat is the total of two independent 0 dBm contributions at the same plane?

Answer

Each is 1 mW. The linear total is 2 mW, which is +3.0103 dBm. The answer is not 0 dBm, and writing “0 dBm + 0 dBm” as ordinary addition hides the required conversion.

$P_{\mathrm{total}} = \sum P_{i}$ $\frac{P_{\mathrm{total}}}{1 \mathrm{mW}} = \sum 10^{\frac{L_{\mathrm{i,dBm}}}{10}}$ $L_{\mathrm{total,dBm}}=10\log_{10}\left(\frac{P_{\mathrm{total}}}{1\,\mathrm{mW}}\right)$ Each exponential term is a dimensionless power ratio whose numerical value equals that contribution in milliwatts. Contributions must refer to the same plane and compatible bandwidth, detector, and averaging conditions.

Try the “Can I add these?” panel in the budget builder above. Two 0 dBm contributions become 2 mW. If one contribution is 10 dB below the other, it adds only one tenth as much power and raises the total level by about 0.414 dB.

There is one crucial boundary: independent or mutually uncorrelated powers add this way. Coherent signals at the same frequency retain a stable phase relationship. Their voltages or travelling-wave amplitudes combine first, so the result can reinforce, cancel, or lie between those cases.

Common misconceptionSeparate dBm values add directly, and equal powers always combine to +3 dB.

Independent equal powers give a +3.0103 dB total increase. Two coherent equal-frequency signals need amplitude, relative phase, reference plane, and network behavior. Their two scalar dBm values alone cannot determine the combined result.

Go deeperWhere phase enters a coherent sum

Under the restricted model of two coherent sinusoids combined across the same ideal resistance, the result contains a cross term: $P_{\mathrm{total}}=P_1+P_2+2\sqrt{P_1P_2}\cos\phi$ . Here, P₁ and P₂ are the powers each voltage would deliver alone to that same resistance. At 0° the fields reinforce; at 180° equal signals cancel ideally. A real RF combiner also has port impedance, isolation, loss, and phase behavior, so this expression is intuition—not a substitute for network analysis.

11 / 12

Average, burst, peak, and measured power

Why can three instruments report different power numbers for the same RF burst?

A single number cannot describe every time scale in a changing RF waveform. First decide whether the question concerns one instant, one carrier cycle, the active part of a burst, a full repetition interval, or the highest point of the modulation envelope.

Different power definitions answer different questions
Quantity	Definition	What it reveals
Instantaneous power	p(t) = v(t)i(t)	At one instant; can change within every carrier cycle
Carrier-cycle average	Average over one or more RF cycles	The 10 mW value for the recurring continuous sine
Active-state or burst average	Average inside a defined on-time gate	Ignores the intentional idle interval
Long-term average	Average across active and idle time	Sets heating over the chosen full interval
Envelope power	Carrier-cycle average followed over modulation time	Tracks a changing RF envelope
Peak envelope power · PEP	Maximum envelope power	Not the same as peak instantaneous v(t)i(t)
PAPR	Peak envelope power / a named average	Meaning changes if the averaging interval changes

For an ideal rectangular on/off burst with active power P_on and duty cycle D, the long-term average across complete repetition periods is especially simple.

$P_{\mathrm{long}} = D \times P_{\mathrm{on}}$ $L_{\mathrm{long,dBm}} = L_{\mathrm{on,dBm}} + 10 \log _{10}D$ D is the active fraction from 0 to 1. The relationship assumes zero power while idle and a constant active-state envelope.

Think about itA signal is +18 dBm while active and on for 10% of each interval. What is its long-term average?

Answer

Ten percent means $D=0.1$ , a −10 dB time-average factor. The long-term average is therefore +8 dBm, or 6.3096 mW. The active-state power remains +18 dBm, or 63.0957 mW.

Interactive burst and measurement explorer

How can one burst have several correct power numbers?

Predict what a full-frame average will report, then change the active level and duty cycle. The carrier is idealized; the envelope timing is slowed and normalized.

Active-state level +18 dBmDuty cycle 10%

Active-state average: +18 dBm63.096 mW
Peak envelope power: +18 dBmconstant on-state envelope
Long-term average: +8 dBm6.31 mW
PEP / full-frame average: 10.0 dBon-state envelope PAPR: 0 dB

Ask an instrument

Averaging window decides the answer

An ungated true-average meter covering complete frames ideally reports the long-term average. A correctly gated measurement can instead report the active-state average.

Assumptions: ideal rectangular on/off gating, constant envelope while active, zero power while idle, and an averaging interval containing complete repetition periods. PEP is carrier-cycle-averaged envelope power—not the 20 mW instantaneous peak inside the recurring sine. This is illustrative, not a Bluetooth or Wi-Fi standard waveform.

For this constant-envelope on-state, peak envelope power equals the active-state average: +18 dBm. Relative to the full-frame average it is 10 dB higher, but the on-stateenvelope PAPR is 0 dB. The averaging interval must be named whenever PAPR is quoted.

Do not confuse crest factor with envelope PAPR

A continuous sine voltage has crest factor $\frac{V_{\mathrm{pk}}}{V_{\mathrm{rms}}}=\sqrt2$ . Its carrier-cycle- averaged envelope power is constant, so an unmodulated continuous wave has 0 dB envelope PAPR. Instantaneous resistor power still varies from 0 to twice its average.

What changes an RF power reading
Instrument view	Useful strength	Conditions that change the answer
Scalar power meter	Direct average-power accuracy; gated or peak sensors add timing detail	Sensor type, video bandwidth, dynamic range, zero/calibration, gate, averaging interval
Spectrum analyzer	Shows how measured power is distributed with frequency	RBW, detector, span, sweep or acquisition time, reference level, channel integration
Time-domain instrument	Shows voltage or envelope changing with time	Termination, analog bandwidth, sample rate, record length, trigger, computation

Common misconceptionOne spectrum marker is the total channel power, and every detector reports the same power.

A marker represents the analyzer result at one frequency bin under a particular RBW and detector. A wide signal requires integration across its occupied bandwidth. Sensor response, gate, bandwidth, averaging, crest factor, and instrument limits can also change what is reported.

Go deeperDetector choice and instrument headroom

Thermal sensors compare heating and naturally report true average power, but they are usually slower. Diode sensors can be faster and more sensitive; outside their square-law or calibrated range, waveform crest factor can influence the result. Wideband envelope sensors are needed to preserve fast peaks. On an analyzer, a reference level set too low can cause compression or damage; set too high, it can add attenuation and bury a weak signal closer to the noise floor.

12 / 12

Putting it together — one 2.45 GHz power journey

Can one connected story reconcile voltage, watts, decibels, a spur, and a burst measurement?

Start at the source plane with the same zero-offset 2.45 GHz sine across an ideal 50 Ω resistor. Its frequency describes repetition. Its RMS voltage and the load determine average power. The logarithmic levels describe that same power relative to fixed references.

The connected power model at each stated condition
Concept or plane	Value	Meaning
Carrier frequency	2.45 GHz	Repetition rate; not a power
Peak voltage	1.000 Vpk	Zero-offset sine across the stated load
RMS voltage	0.7071 Vrms	Equivalent resistor-heating voltage
RMS current	14.14 mArms	For the ideal 50 Ω resistor
Average power	10.000 mW	+10 dBm = −20 dBW
Instantaneous resistor power	0 to 20 mW	Repeats at twice the carrier frequency
Final chain plane	+18 dBm	63.0957 mW after −3, +13, and −2 dB
Spur at final plane	−22 dBm	−40 dBc = 6.3096 µW relative to the +18 dBm carrier
10% ideal burst average	+8 dBm	6.3096 mW across complete repetition periods

The carrier leaves P0 at +10 dBm. The 3 dB attenuator brings it to +7 dBm, the linear 13 dB amplifier brings it to +20 dBm, and the 2 dB cable loss leaves +18 dBm at P3. A −40 dBc spur at that final plane is −22 dBm, or 6.3096 µW. Ideal 10% on/off gating makes the carrier's long-term average +8 dBm while its active-state level and PEP remain +18 dBm.

The connected mental model

Power is energy transfer rate. Instantaneous voltage and current create instantaneous power. RMS connects a waveform to resistor heating. The load and reference plane make a power statement physical. dB compares levels; dBm and dBW name fixed references; dBc names a carrier. Cascade ratios add, separate powers sum linearly, and measurement settings decide which time and frequency content reaches the result.

A trustworthy RF power statement names the conditions

Frequency: where the gain, loss, sensor, and termination apply.
Quantity and value: average, burst average, envelope, PEP, density, or another defined power.
Reference: watts, dBm, dBW, or a named relative carrier for dBc.
Reference plane: the exact connector, cable end, device port, or antenna terminal.
Waveform and state: CW, modulated, active burst, duty cycle, or idle-inclusive interval.
Load or termination: especially when converting between voltage, current, and power.
Measurement method: bandwidth, detector or sensor, gate, averaging interval, and peak definition.

Think about itIf the 3 dB attenuator is replaced by a 6 dB attenuator, what changes downstream?

Answer

Every later linear-stage level falls by 3 dB. The final carrier becomes +15 dBm, or 31.6228 mW. At the same resistance, RMS voltage is multiplied by $10^{-3/20}=0.70795$ , approximately 1/√2—not by one half. The carrier frequency remains 2.45 GHz.

Ungraded review

Check your understanding

Answer each question in your own words, then reveal the model answer.

01A zero-offset sine is 1.000 Vpk across an ideal 50 Ω load. What are its RMS voltage, average power, and dBm level?
Model answer
0.7071 Vrms, 10.000 mW, and +10 dBm.
02A +10 dBm carrier passes through −3 dB, +13 dB, and −2 dB stages. What reaches the final plane, and where is a −40 dBc spur?
Model answer
The carrier reaches +18 dBm, or 63.10 mW. The spur is −22 dBm, or about 6.31 µW.
03The +18 dBm carrier is active for 10% of each repetition interval. What changes when long-term average power is reported?
Model answer
The long-term average falls by 10 dB to +8 dBm. The active-state level and ideal peak envelope power remain +18 dBm.

References and further reading

The lesson synthesizes the sources below. Antenna-Theory informed the intuitive teaching approach; standards and measurement references guided notation, definitions, and caveats.

Standards and notation

RF power measurement

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