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The Gaussian gate centring error is $d=t_c-t_0$, the time displacement between a gate centre and a Gaussian pulse centre. The Gaussian pulse time-gate energy fraction gives the full displaced-gate formula. Equal positive and negative errors collect equal energy in that symmetric model, so an energy deficit alone does not reveal the direction of the misalignment.

A focused expansion explains sensitivity close to the optimum. Let $p$ be the centred energy-normalized pulse profile, let $a=T_g/2$, and define $F(d)={\int }_{d-a}^{d+a}p(t)dt$. For the Gaussian standard deviation $\sigma$,Thus the leading quadratic energy loss from centred Gaussian gate timing error is second order, rather than linear in $d$.

For an original example, take a gate as wide as the power FWHM $\tau$. Its centred fraction is $0.760968$. An error $d=0.10\tau$ gives $0.748042$ from the full formula; the quadratic approximation gives $0.747945$. For $\tau =10\mathrm{ns}$ this is a $1\mathrm{ns}$ displacement, with gate width and total pulse energy unchanged.

A delay scan of a gated readout can locate the maximum, provided the source and gate widths remain stable. Pulse timing jitter produces a distribution of offsets rather than a single deterministic $d$. Also distinguish a gate on optical power from one on a detector's electrical waveform: NIST measured detector impulse responses and used convolution to assess pulse reproduction.^[ref]

For an isolated Gaussian pulse, the relation between laser pulse energy $U$ and peak power $P_{\mathrm{pk}}$ depends on the pulse-width convention. Here $P(t)$ is the total incident radiant power, the background is zero, and $\tau >0$ denotes the power full width at half maximum. Integrating $P$ over the entire pulse defines its positive energy $U>0$.

Pulse energy, peak power and average power resolve different information. Consider a Gaussian pulse withThe chosen pulse duration $\tau$ is the full width at half maximum of radiant power: $P(\pm \tau /2)=P_{\mathrm{pk}}/2$. It refers to radiant power, rather than the wave amplitude of the electric field. The Gaussian integral gives the relation between Gaussian pulse energy and peak power:The pulse energy shape factor is $\kappa =U/(P_{\mathrm{pk}}\tau )$. It equals $1.064467$ for this power profile; a rectangular pulse of duration $\tau$ has $\kappa =1$. Knowing only a width and an integrated energy therefore requires a pulse-shape assumption before inferring peak power.

An equivalent form makes the normalization transparent. Write the pulse centre as $t_0$, and let $\sigma$ denote the standard deviation of its energy-normalized pulse profile:The ratio $p(t)=P(t)/U$ is a probability density function in the mathematical sense: it is nonnegative, has units ${\mathrm{s}}^{-1}$, and integrates to one. Here it weights the pulse's energy over time. Its normal distribution profile has mean $t_0$ and variance ${\sigma }^2$. This gives $P_{\mathrm{pk}}=U/(\sqrt{2\pi }\sigma )$ without a separate area estimate.

For a stipulated $\sigma =2\mathrm{ns}$ and $U=1\mathrm{nJ}$, the power full width at half maximum is $4.70964\mathrm{ns}$ and the peak power is $0.19947\mathrm{W}$. Confusing the standard deviation with the full width at half maximum would insert the wrong duration into the conversion.

The Gaussian pulse field-envelope width is a second distinct convention. If a fixed spatial field profile gives $P(t)=K|A(t){|}^2$ for a constant $K>0$, the wave amplitude envelope obeysConsequently its amplitude full width at half maximum is ${\tau }_E=\sqrt{2}\tau$. Inserting ${\tau }_E$ into the formula for a power-width input underestimates the true peak power by a factor $1/\sqrt{2}$. State which measured waveform defines the width.

An electrical photodetector trace can also be broadened by its impulse response. The optical pulse-width inference therefore needs a characterized detector response.^[ref]

The Gaussian pulse time-gate energy fraction is the share of an isolated pulse's laser pulse energy admitted by a finite time window. The result below acts on incident optical radiant power. Let $\tau$ be the power full width at half maximum, and first place the pulse centre and the detection time gate centre at zero.

A detection time gate need not collect the whole pulse. Integrating the same profile only over $-T_g/2\le t\le T_g/2$ gives the Gaussian pulse time-gate energy fractionwhere $\mathrm{erf}$ is the error function. A gate as wide as the power full width at half maximum collects only $\mathrm{erf}(\sqrt{\ln 2})\simeq 0.760968$, or 76.10% of the pulse energy. The pulse has nonzero Gaussian tails outside that interval. A real readout's impulse response can extend the required integration window further.

Alignment adds a parameter. Let the pulse centre be $t_0$, the gate centre be $t_c$, and write the Gaussian gate centring error as $d=t_c-t_0$. Put $\sigma =\tau /(2\sqrt{2\ln 2})$. Integrating the normalized Gaussian pulse profile over the gate givesAt $d=0$ the two error function terms reduce to the centred formula above. With $T_g=\tau$ and $d=\tau /2$, the captured fraction drops to $0.490734$. The pulse centre then lies on a gate edge. Merely keeping the gate width fixed does not keep its collected energy fixed.

The gate works best when centred. With the centred energy-normalized pulse profile $p(s)=P(t_0+s)/U$ and $a=T_g/2$, differentiation of $F(d)={\int }_{d-a}^{d+a}p(s)ds$ givesA Gaussian function is an even function about its centre and decreases with distance from that centre. Thus $F^{\prime }(d)\le 0$ for $d>0$, and symmetry gives the opposite sign for $d<0$. The same argument proves that a centred interval maximizes mass for an even unimodal density; unimodality supplies the required monotonicity.

A centred gate of width $2\tau$ captures $0.981468$ of the energy. Solving $\mathrm{erf}(\sqrt{\ln 2}{T}_g/\tau )=0.99$ gives $T_g\simeq 2.18771\tau$ for 99% capture in this ideal model. The requirement changes if the centring error is nonzero.

Pulse timing jitter can therefore create variation in gated energy even when every full pulse has the same $U$. A wider gate reduces this loss but also changes admitted background and the chance of including neighbouring pulses. Measure the timing and detector response associated with the chosen gate.

For a detector modeled as a linear time-invariant system, its recorded waveform is a convolution of the optical input with its impulse response. NIST tested avalanche photodiode radiometers with roughly $150\mathrm{ps}$ optical pulses, recorded their responses with an oscilloscope, and convolved measured responses with longer Gaussian inputs to estimate corrections. The detector and amplifier were characterized together.^[ref] A gate applied to that electrical waveform must use the readout model rather than assuming the optical fraction is unchanged.

Quadratic energy loss from centred Gaussian gate timing error is the leading decrease in collected energy when a symmetric gate is displaced slightly from the centre of a Gaussian pulse. At the optimum the first-order change vanishes, but the second-order change is negative. The Gaussian pulse time-gate energy fraction gives the exact displaced-gate integral; here the local loss and its effect on jitter are derived explicitly.

Let $p(t)=e^{-t^2/(2{\sigma }^2)}/(\sqrt{2\pi }\sigma )$ be the centred energy-normalized pulse profile, let $a=T_g/2>0$, and setThe Leibniz integral rule gives $F^{\prime }(d)=p(d+a)-p(d-a)$. Since $p$ is even and $p^{\prime }(t)=-tp(t)/{\sigma }^2$,The Taylor expansion therefore yieldsThis is an absolute fraction of the full pulse energy. The relative loss compared with the centred gate is instead $[F(0)-F(d)]/F(0)$.

For a gate as wide as the power FWHM $\tau =2\sqrt{2\ln 2}\sigma$, $F(0)=0.760968$ and $C{\tau }^2=1.302337$. An offset $d=0.05\tau$ gives leading loss $0.00325584$, or about $0.428$ percent of the centred gate's collected energy. Neither the full pulse energy nor the gate width has changed.

A zero mean offset does not eliminate this loss. For a sufficiently narrow random offset $D$ with mean ${\mu }_D$ and variance ${\sigma }_D^2$, the leading expectation isIf $D$ has a centred normal distribution, the quadratic model also gives $\mathrm{SD}(F(D))\simeq \sqrt{2}C{\sigma }_D^2$. At ${\sigma }_D=0.05\tau$, the leading mean loss is $0.00325584$ and this standard deviation is $0.00460446$. These formulas neglect fourth-order effects; they are local approximations, not bounds for large timing excursions.

An apparatus check scans the gate delay about its optimum and compares the measured decrease with the model, while monitoring gate width and the pulse. A gate on the detector's electrical output need not act directly on optical power: the detector impulse response and readout convolution must be included. NIST's impulse-response measurements illustrate that distinction.^[ref]

The temporal energy centroid of an optical pulse is its energy-weighted mean time. For nonnegative radiant power $P(t)$ with finite positive energy $U$ and finite absolute first moment,The normalization uses radiant energy as integrated radiant flux.^[ref] The centroid is a moment of the energy-normalized pulse profile; it need not coincide with the largest power value, a threshold crossing or the middle of the FWHM interval.

For an original two-component example, let ninety percent of the energy belong to a symmetric narrow component centred at zero and ten percent to a symmetric component centred at $20\mathrm{ns}$. Linearity of the integral gives $t_E=0.9(0)+0.1(20)=2\mathrm{ns}$. Weak delayed energy can move the centroid even when the dominant peak remains near zero. Multiplying the whole power trace by a positive constant leaves $t_E$ unchanged; translating it by $t_0$ shifts its centroid by $t_0$.

A detector response can shift the measured centroid too. Suppose a normalized impulse response $h$ and normalized optical profile $p$ have finite absolute first moments. For their convolution $q=p\ast h$, direct integration with $t=s+u$ givesThus a causal exponential detector response of time constant $2\mathrm{ns}$ shifts the recorded centroid by $2\mathrm{ns}$ in this model. NIST's pulsed-radiometer characterization explicitly measures detector impulse responses and uses convolution to assess reproduced waveforms.^[ref] The centroid identity is the independent moment calculation above.

Baseline drift can cause a separate shift. Suppose a true pulse has $U=1\mathrm{nJ}$ and $t_E=0$, but a positive residual baseline of $1\mathrm{mW}$ remains across the analysis window from $-10$ to $90\mathrm{ns}$. Its added energy is $0.10\mathrm{nJ}$, centred at $40\mathrm{ns}$. The resulting centroid isA waveform measurement should therefore record the time window, baseline estimate and detector timing response with its reported centroid. A negative background-subtracted trace also needs care: it is not automatically a nonnegative energy distribution.