= Gaussian pulse time-gate energy fraction
{c}
{title2=$U_g/U=\operatorname{erf}(\sqrt{\ln2}\,T_g/\tau)$}
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 $-T_g/2\leq t\leq T_g/2$ gives the <Gaussian pulse time-gate energy fraction>
$$
\frac{U_g}{U}
=\operatorname{erf}\!\left(\sqrt{\ln2}\,\frac{T_g}{\tau}\right),
$$
where $\operatorname{erf}$ is the <error function>. A gate as wide as the power <full width at half maximum> collects only $\operatorname{erf}(\sqrt{\ln2})\simeq0.760968$, or \b[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 $\sigma=\tau/(2\sqrt{2\ln2})$. Integrating the normalized <Gaussian pulse> profile over the gate gives
$$
\boxed{\frac{U_g}{U}=
\frac12\left[
\operatorname{erf}\!\left(\frac{d+T_g/2}{\sqrt2\sigma}\right)
-\operatorname{erf}\!\left(\frac{d-T_g/2}{\sqrt2\sigma}\right)
\right].}
$$
At $d=0$ 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 $F(d)=\int_{d-a}^{d+a}p(s)\,ds$ gives
$$
F'(d)=p(d+a)-p(d-a).
$$
A <Gaussian function> is an <even function> about its centre and decreases with distance from that centre. Thus $F'(d)\leq0$ 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 $\operatorname{erf}(\sqrt{\ln2}\,T_g/\tau)=0.99$ gives \b[$T_g\simeq2.18771\tau$ for 99% capture] in this ideal model. The requirement changes if the centring error is nonzero.
\Image[/gaussian-pulse-gate-alignment.png]
{title=Original Gaussian pulse gate model: width and timing offset jointly determine captured energy.}
<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.<Calibration Service for Low-level Pulsed-Laser Radiometers at 1.06 µm: Pulse Energy and Peak Power by Rodney W. Leonhardt>[§§6.1–6.2, printed pp. 13–15: short-pulse excitation, sampling readout and numerical convolution]{ref} A gate applied to that electrical waveform must use the readout model rather than assuming the optical fraction is unchanged.
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