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The average power of a laser pulse train is its integrated radiant energy per observation time. If an interval $T$ contains $N$ complete pulses, with no uncounted edge contributions or continuous background,Thus the relevant pulse repetition frequency is the actual emitted-pulse rate in the interval. Historical calorimetric calibrations inferred average power and pulse energy from total received energy, injection time and pulse rate.^[ref]

An original counting example exposes a possible cancellation of errors. A nominal $100\mathrm{kHz}$ trigger produces 200 triggers in $2\mathrm{ms}$. Suppose only 180 complete optical pulses arrive, each carrying $0.20\mu \mathrm{J}$. Their total is $36\mu \mathrm{J}$, so the pulse contribution is $18\mathrm{mW}$, corresponding to an actual 90kHz pulse rate. Add a continuous optical background of $2\mathrm{mW}$, and a detector integrating everything reads $20\mathrm{mW}$.

Dividing that total by the nominal trigger rate would happen to give $0.20\mu \mathrm{J}$, despite both the missing pulses and the unremoved background. Removing the background but still using the nominal rate would instead infer $0.18\mu \mathrm{J}$, ten percent below the actual emitted-pulse energy. This missing-pulse bias in average-power inference cannot be diagnosed from the apparently reasonable quotient alone. A pulse-resolving monitor and a suitable background subtraction distinguish the two contributions.

For bursts or variable repetition, count the pulses in the same interval as the energy measurement. A slow integrating detector can establish their mean energy after these corrections but does not resolve pulse-to-pulse energy fluctuation or optical peak power. Laser pulse energy separates these observables.

A finite optical record can bound later radiant energy if an independently justified exponential decay envelope applies at every later time. Let the recorded window end at $b$, and suppose the true nonnegative radiant power satisfiesIntegration gives the exponential-envelope tail-energy boundEquality holds when the entire tail follows the envelope. This is a bound on an unobserved contribution, not automatically a standard uncertainty or a probabilistic interval.

For an original numerical example, suppose the energy inside the recorded window is known to lie in $[48,52]\mathrm{nJ}$, while a justified envelope has $A=0.20\mathrm{W}$ and $\tau =20\mathrm{ns}$. The missing contribution is at most $4\mathrm{nJ}$, so the total laser pulse energy lies in $[48,56]\mathrm{nJ}$. The optical energy outside a time gate fraction, considering only this late contribution, is at most $4/(48+4)\approx 7.69\%$. Any unobserved early energy requires its own bound.

One way to justify the envelope mathematically is a differential inequality for an exponential tail envelope. Suppose $P$ is absolutely continuous on each finite interval after $b$, $P(b)\le A$, and $P^{\prime }(t)\le -P(t)/\tau$ almost everywhere. Multiplying by the integrating factor $e^{(t-b)/\tau }$ givesso $e^{(t-b)/\tau }P(t)\le P(b)\le A$. The decay law must apply throughout the unobserved region; it is a physical assumption to validate, rather than a consequence of fitting the last few samples.

The distinction is testable by a counterexample. Append a later nonnegative pulse whose support lies entirely beyond the recorded interval. Its energy can be arbitrarily large while every recorded sample is unchanged. Such a pulse violates the global envelope but cannot be excluded by a good local exponential fit alone. Similarly, an envelope $A{e}^{-(t-b)/\tau }+P_0$ with a positive constant background $P_0$ has an infinite integral over an infinite time interval. A finite observation horizon or an adequately defined and validated background subtraction is then essential.

Uncertain envelope parameters require a bound valid for their allowed combinations. If the allowed set is the rectangle $0\le A\le A_{\max }$ and $0<\tau \le {\tau }_{\max }$, then $U_{\mathrm{tail}}\le A_{\max }{\tau }_{\max }$. For $A_{\max }=0.25\mathrm{W}$ and ${\tau }_{\max }=25\mathrm{ns}$, this is $6.25\mathrm{nJ}$ and the preceding total interval becomes $[48,58.25]\mathrm{nJ}$. More restrictive joint parameter constraints can tighten this result; inserting separate best-fit values does not establish a guaranteed bound.

Finally, an electrical tail need not be an optical tail. In an original idealized linear time-invariant system, let a photodetector have unit-area impulse response $h(t)={\tau }_d^{-1}{e}^{-t/{\tau }_d}$ for $t\ge 0$, and zero otherwise. An impulsive optical input $U\delta (t)$, with $\delta$ the Dirac delta function, gives the calibrated output $s(t)=Uh(t)$ by convolution. Its electrical tail after $L$ has area $U{e}^{-L/{\tau }_d}$ even though all incident optical energy arrived at time zero. This optical tail and detector-response ambiguity requires a separate response measurement. NIST's historical procedure characterized the photodetector, cable and oscilloscope together before convolving measured responses with trial waveforms. Appendix C describes typical optical excitation of about $120\mathrm{ps}$, a photodetector rise time near $5\mathrm{ns}$, and a $1\mathrm{GHz}$ oscilloscope with $0.4\mathrm{ns}$ rise time. The optical excitation was much shorter than the detector response, while the oscilloscope was fast enough that the detector dominated the measured waveform. Response data were collected at each gain setting and several signal levels.^[ref]

The encircled radiant energy of a Gaussian beam is the radiant energy inside a centred circular receiver at a specified transverse plane. For fixed Gaussian beam radius $w$ and total laser pulse energy $U$,The radiant exposure of a Gaussian beam contains the radial integration and aperture-capture plot. Its $w$ is the irradiance $e^{-2}$ radius, as specified in the primary beam account.^[ref]

An original annular example distinguishes energy near an axis from energy between two radii. Subtracting two encircled values givesChoose $U=5\mu \mathrm{J}$, $w=0.50\mathrm{mm}$, $R_1=0.25\mathrm{mm}$ and $R_2=0.50\mathrm{mm}$. The annulus receives $5(e^{-1/2}-e^{-2})\approx 2.356\mu \mathrm{J}$, or 47.12 percent of the full beam energy. A bright central point does not imply that most energy lies in an arbitrarily small disk around it.

Indeed, the radial radiant-energy density of a Gaussian beam is $dQ/dR=2\pi RH(R)$, where $H$ is local radiant exposure. For this profile it is largest at $R=w/2$, even though $H$ is largest at zero radius. Increasing a centred aperture and comparing integrated detector readings can test the cumulative prediction. Detector linearity, aperture alignment and the finite collection boundary must be controlled; a displaced aperture is not described by the centred formula.

An energy-normalized pulse profile separates an optical pulse's temporal shape from its total laser pulse energy. For nonnegative radiant power $P(t)$ and finite $U=\int P(t)dt>0$, define $p(t)=P(t)/U$. Then $p$ has units inverse time and integrates to one. The radiant energy in a time interval is $U$ times the integral of $p$ over that interval.^[ref]

For an original two-lobe example, let $p=1/(8\mathrm{ns})$ during $0<t<2\mathrm{ns}$ and $p=3/(8\mathrm{ns})$ during $5<t<7\mathrm{ns}$, with zero values elsewhere. The two integrals are $1/4$ and $3/4$. Choosing $U=4\mathrm{nJ}$ gives powers $0.5$ and $1.5\mathrm{W}$, and energies $1$ and $3\mathrm{nJ}$. Multiplying $U$ by two doubles both powers without changing $p$.

The temporal energy centroid of an optical pulse is $\int tp(t)dt$ when this moment exists. Here it is $(1/4)(1)+(3/4)(6)=4.75\mathrm{ns}$, in the gap between lobes. An energy centroid need not be a location of maximum power.

Mathematically, $p$ is a probability density function, but its weighting here is energy over time. It is not automatically a measured photon-arrival distribution, particularly when photon energies vary. A zero-energy trace cannot be normalized this way, and a noisy baseline-corrected trace with negative excursions is not automatically a nonnegative density. The Gaussian pulse energy and peak power article gives one normalized shape family.

Laser pulse energy is the radiant energy carried by one specified laser pulse across a measurement plane. For an isolated pulse whose total radiant power is $P(t)\ge 0$ and whose background is zero,Here $P$ is already integrated over the transverse beam area; it is not local irradiance. The unit of $U$ is the joule. Radiant exposure instead divides received energy by a receiving area, so a beam's $U$ alone does not specify the exposure at its centre. The radiant exposure of a Gaussian beam makes that spatial distinction quantitative.

The Gaussian pulse energy and peak power relation gives $P_{\mathrm{pk}}\simeq 0.939437U/\tau$ when $\tau$ is the power full width at half maximum. The pulse energy shape factor records pulse-shape dependence, while the Gaussian pulse field-envelope width is $\sqrt{2}$ times the power width.

For an original example, stipulate $U=0.50\mu \mathrm{J}$ and a Gaussian pulse power full width at half maximum $\tau =10\mathrm{ns}$. Then $P_{\mathrm{pk}}=46.97\mathrm{W}$. Stretching the same model pulse to $40\mathrm{ns}$ while keeping $U$ unchanged reduces its peak power to $11.74\mathrm{W}$. Both curves below integrate to the same energy; their peak heights and widths differ.

A finite detection time gate collects the Gaussian pulse time-gate energy fraction, which depends on its width and alignment. A centred gate as wide as the power full width at half maximum receives only $76.10\%$ of $U$, because Gaussian tails extend beyond it. For a $\tau$-wide gate, a Gaussian gate centring error of $\tau /2$ reduces capture to $49.07\%$.

The average power of a laser pulse train follows by counting complete pulses. If a long observation of duration $T$ contains $N$ pulses of energies $U_1,\dots ,U_N$, with negligible edge contributions and no continuous background,Here $f_{\mathrm{rep}}$ is the pulse repetition frequency, not the inverse pulse width. An average power readout constrains $\overline{U}$ when $f_{\mathrm{rep}}$ is known; it does not resolve pulse-to-pulse energy fluctuation. Identical, well-separated $0.50\mu \mathrm{J}$ pulses at $1.0\mathrm{MHz}$ give $0.50\mathrm{W}$ average power, even though the stipulated $10\mathrm{ns}$ pulses peak near $47\mathrm{W}$.

If a steady optical background is present, state whether the reported energy includes it. For pulse energy above a measured background $P_{\mathrm{bg}}$, the pulse energy baseline correction uses ${\int }_{\mathrm{gate}}[P(t)-P_{\mathrm{bg}}]dt$. Failing to subtract a constant background adds $P_{\mathrm{bg}}{T}_g$ to the inferred energy. This optical-background model is distinct from an offset introduced later by detector electronics.

Historical NIST equipment illustrates what the readouts measure. Its TC-24 reference absorbed the beam in a glass plate, sensed a plate temperature difference with thermocouples, and integrated the thermopile's electric potential difference during beam injection. Before/after readings enabled background subtraction and compensated thermal drift. Counting pulses converted total injected energy to mean pulse energy. The semiconductor pulse photodetector used an integrating amplifier; calibration related the peak output electric potential difference to pulse energy. That electrical peak did not by itself establish optical peak power.^[ref] Pulse energy measurement with a photodiode connects this kind of measurement to collected electric charge; validating the detector response under the actual pulse conditions remains part of that interpretation.

Optical energy outside a time gate is the radiant energy excluded by a specified ideal temporal acceptance set $G$. For a nonnegative radiant power profile of finite positive total energy,The time-integral definition is the radiometric one.^[ref] This observable needs a gate boundary but no fitted separation into a main field and an optical pulse pedestal.

For an original three-pulse example, let an optical prepulse, main pulse and optical postpulse carry $5$, $12$ and $3\mathrm{nJ}$ respectively, with disjoint supports. A gate containing only the main pulse admits $12\mathrm{nJ}$ of $20\mathrm{nJ}$, so $f_{\mathrm{out}}=0.40$. Widening that gate to include the postpulse reduces the fraction to $0.25$. The optical waveform has not changed; the admitted observable has.

For a translated interval $G_{\delta }=[a+\delta ,b+\delta ]$ with continuous boundary values,Unequal power at the boundaries produces first-order sensitivity to a timing shift. Equal boundary values eliminate this derivative but need not eliminate higher-order loss; quadratic energy loss from centred Gaussian gate timing error gives an existing example.

The total-energy denominator must cover the excluded light too. If the acquisition window misses an unknown tail, pulse-energy interval from incomplete tail measurements supplies conditional bounds rather than an exactly known fraction. A detection time gate on an electrical detector trace also includes the detector impulse response and is not automatically an ideal gate on incoming optical power. A characterized optical response and a stated detector baseline subtraction are needed to compare the two.

A pulse-energy interval from incomplete tail measurements combines measured energy with bounds on optical power outside the acquisition window. It describes a set of compatible total energies, not automatically a standard uncertainty or confidence interval. The pulse-energy bounds from tail duration and pedestal height article treats a particular peak-and-width model; a window-based construction need not assume that pulse shape.

Suppose true radiant energy inside a measured window $G$ lies in a justified interval $[L,H]$, with $0\le L\le H$. Partition the unobserved times into regions of durations ${\Delta }_j$ with independently established nonnegative radiant power caps $M_j$. ThenFor an original measurement, let the observed-window energy lie between $48$ and $52\mathrm{nJ}$. An unobserved early region lasting $2\mathrm{ns}$ at most $0.50\mathrm{W}$ contributes at most $1\mathrm{nJ}$; a late region lasting $20\mathrm{ns}$ at most $0.20\mathrm{W}$ contributes at most $4\mathrm{nJ}$. The total interval is therefore $48$ to $57\mathrm{nJ}$. For these assumptions the fraction of optical energy outside a time gate equal to $G$ is at most $5/(48+5)\approx 9.43\%$.

A finite support cutoff is sufficient but not necessary for a finite bound. Bounding unobserved optical energy with an exponential tail model uses a validated cap $P(t)\le P_b{e}^{-(t-b)/\tau }$ after the window endpoint $b$, which integrates to $P_b\tau$. Without a justified duration or integrable envelope, one below-threshold reading cannot bound all later energy.

Establish the caps with characterized detector sensitivity, baseline and temporal coverage; do not identify detector ringing directly with optical tails. NIST's response analysis explicitly models the electrical trace as a convolution of the optical waveform with detector response.^[ref] The interval above follows from chosen optical bounds, not from an uncorrected finite-bandwidth detector trace.

Radiant exposure measures the radiant energy received per unit surface area during a specified interval. At a fixed surface point $\mathbf{x}$, it is the time integral of incident irradiance $E_{\mathrm{e}}$:The area belongs to the receiving surface: its orientation and location matter. An irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$ specifies a rate, while radiant exposure in $\mathrm{J}{\mathrm{m}}^{-2}$ specifies the accumulated arrival of energy.^[ref]

For a uniform rectangular pulse, $H_{\mathrm{e}}=E_0\tau$, with constant irradiance $E_0$ and duration $\tau$. Stipulate one pulse at $20\mathrm{W}{\mathrm{m}}^{-2}$ for $1\mathrm{s}$, and another at $80\mathrm{W}{\mathrm{m}}^{-2}$ for $0.25\mathrm{s}$. Both deliver $20\mathrm{J}{\mathrm{m}}^{-2}$, but their peak irradiances differ by a factor of four. The two radiant exposures agree only after each complete pulse has arrived. A shorter observation window can intercept different fractions of them, and a nonlinear receiver can respond differently even to their complete pulses.

Integrating over a receiving patch gives its incident radiant energy:Here ${\overline{H}}_A$ is the spatially averaged radiant exposure, rather than necessarily the local maximum. Uniform illumination is the special case where this spatially averaged radiant exposure equals the local value everywhere on the patch. In the rectangular-pulse example, a uniformly illuminated $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ patch receives $20\times 10^{-4}=2\mathrm{mJ}$.

For an original spatial example, consider a circular Gaussian beam at a fixed transverse plane, with constant Gaussian beam radius $w$ throughout the pulse. Let $P(t)$ be its total radiant power and $U=\int P(t)dt$ its laser pulse energy. Integrating its transverse irradiance in time gives the radiant exposure of a Gaussian beam:The normalization follows by integrating $H(r)2\pi rdr$ from $0$ to $\infty$. The disk of radius $w$ contains only $U(1-e^{-2})$, so its actual spatially averaged radiant exposure is $U(1-e^{-2})/(\pi w^2)$, not $U/(\pi w^2)$. With stipulated $U=1\mu \mathrm{J}$ and $w=1\mathrm{mm}$, the peak is $0.6366\mathrm{J}{\mathrm{m}}^{-2}$, while the average inside that disk is $0.2752\mathrm{J}{\mathrm{m}}^{-2}$. State the Gaussian beam radius convention and the receiving area with any reported value.

An energy readout alone therefore does not specify a surface's radiant exposure. In an ideal linear model for pulse energy measurement with a photodiode, dark-subtracted photocurrent $i_{\mathrm{ph}}(t)=R{P}_{\det }(t)$ has constant spectral responsivity $R$ at the chosen wavelength. Integrating the electric current gives signal electric charge $Q_{\mathrm{sig}}=R{U}_{\det }$. If a known patch of area $A$ is uniformly illuminated and all of its incident energy is collected, thenAs a separate stipulated example, $Q_{\mathrm{sig}}=0.40\mathrm{nC}$ and $R=0.40\mathrm{A}{\mathrm{W}}^{-1}$ correspond to $U_{\det }=1.0\mathrm{nJ}$. Uniform illumination over $A=1.0\mathrm{m}{\mathrm{m}}^2$ gives $H_{\mathrm{e}}=1.0\mathrm{mJ}{\mathrm{m}}^{-2}$. Background subtraction, collection geometry, detector linearity and an integration window covering the full response are part of this measurement model. Continuous-wave responsivity alone does not establish linearity for a short pulse: silicon-photodiode experiments have observed pulse-dependent saturation.^[ref]

A historical NIST measurement calibration separates these operations experimentally. A modulator produces pulses from a continuous-wave laser; a calibrated beam splitter sends one beam to a slow thermal reference and another to the pulse detector. The reference records the accumulated energy, while the pulse repetition rate and injection duration determine the mean energy per pulse. The detector's integrating electronics provide a pulse-energy readout. Converting such a readout to radiant exposure additionally requires the illuminated area and its spatial distribution.^[ref]

The beam-use term fluence denotes the same time integral for a directed beam normal to the chosen surface. For light arriving from many directions, specify the angular geometry: planar irradiance and spherical irradiance are different quantities, so a value defined using one cannot simply be substituted for the other.^[ref]

A radiometric energy balance accounts for radiant energy crossing the specified measurement boundary during a specified interval. For a passive optic, suppose the relevant transient has finished, no electromagnetic energy remains stored, and no additional energy source is present. Incident radiation is partitioned into all outgoing radiation and energy transferred to matter. Resolving outgoing paths givesThe channels must be disjoint: a scattering term cannot include energy already counted by a reflection or transmission collector. The definition of radiant energy concerns radiation emitted, transferred or received, so the location and collection geometry belong in each term.^[ref]

For an original example, inject $10\mu \mathrm{J}$ and measure $6\mu \mathrm{J}$ in the transmitted path plus $2\mu \mathrm{J}$ in the specular reflection. Calling the missing $2\mu \mathrm{J}$ absorbed is premature. If a separate collector finds $1.5\mu \mathrm{J}$ of diffuse outgoing radiation outside those channels, the inferred absorbed energy is only $0.5\mu \mathrm{J}$. The uncollected scattering in a radiometric energy balance would otherwise be misidentified as optical absorption.

A physical check compares calibrated incident and outgoing energy readouts while controlling apertures, backgrounds and the observation window. The scattered component may need a different collection geometry from the specular beam. Energy absorbed by matter can remain stored or later leave as heat or radiation; a more general control-volume balance must explicitly include storage, other heat transfer and work. It must not count absorbed energy twice by also treating the same material-energy increase as an independent loss.

Spatially averaged radiant exposure is the received radiant energy divided by the specified receiving area. For a patch $A$ and local radiant exposure $H(\mathbf{x})$,Both local and averaged values have units $\mathrm{J}{\mathrm{m}}^{-2}$, but the average need not equal the local maximum. The radiant exposure definition uses incident energy and the receiving surface's area.^[ref]

For an original unequal-area example, let one $3\mathrm{m}{\mathrm{m}}^2$ region receive a uniform $2\mathrm{J}{\mathrm{m}}^{-2}$, and a neighbouring $1\mathrm{m}{\mathrm{m}}^2$ region receive $10\mathrm{J}{\mathrm{m}}^{-2}$. They receive $6$ and $10\mu \mathrm{J}$ respectively. The average over the combined patch is thereforeThe unweighted average of the two listed exposure values would be $6\mathrm{J}{\mathrm{m}}^{-2}$, which is wrong for these unequal areas. The area weighting of radiant exposure gives the larger region three times the influence. The actual peak remains $10\mathrm{J}{\mathrm{m}}^{-2}$ in the stipulated piecewise-uniform model.

A receiver integrating all incident energy over the complete patch can determine $Q_A$; division by its known collecting area then gives the average. That measurement alone cannot identify the peak or reconstruct the spatial distribution. Subdivision with a characterized spatial response supplies additional information. For a finite aperture intercepting a Gaussian beam, use energy actually entering the aperture, as calculated in radiant exposure of a Gaussian beam, rather than the total energy of the untruncated beam.