= Quadratic energy loss from centred Gaussian gate timing error
\b[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 set
$$
F(d)=\int_{d-a}^{d+a}p(t)\,dt.
$$
The <Leibniz integral rule> gives $F'(d)=p(d+a)-p(d-a)$. Since $p$ is even and $p'(t)=-tp(t)/\sigma^2$,
$$
F'(0)=0,\qquad F''(0)=2p'(a)=-\frac{2ap(a)}{\sigma^2}.
$$
The <Taylor expansion> therefore yields
$$
\boxed{F(0)-F(d)=Cd^2+O(d^4),\qquad C=\frac{ap(a)}{\sigma^2}>0.}
$$
This 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 $\tau=2\sqrt{2\ln2}\,\sigma$, $F(0)=0.760968$ 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> is
$$
\mathbb E[F(D)]\simeq F(0)-C(\mu_D^2+\sigma_D^2).
$$
If $D$ has a centred <normal distribution>, the quadratic model also gives $\operatorname{SD}(F(D))\simeq\sqrt2 C\sigma_D^2$. 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.
\Image[/gaussian-gate-quadratic-loss.png]
{title=Exact Gaussian gate capture and its local quadratic loss: a zero first derivative does not remove timing sensitivity.}
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.<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, short-pulse excitation and convolution analysis]{ref}
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