= Calibrated two-wire charge centroid
{title2=$\widehat x=pq_R/(q_L+q_R)$}
Can two readout values locate a hit between wires? \b[Only with a specified response model.] Place two readout wires at $0$ and $p$, put $u=x/p\in[0,1]$, and stipulate calibrated, noiseless <electric charges>
$$
q_L=Q(1-u),\qquad q_R=Qu,\qquad Q>0.
$$
These are assumptions for a toy sensor response, not the natural response law of adjacent <anode> wires. Under these assumptions the <centroid>
$$
\widehat x=\frac{0q_L+pq_R}{q_L+q_R}=pu=x
$$
recovers position exactly. The apparent gain in precision comes from adding calibrated analogue information. Taking the mean of two fired wire addresses without their <electric charges> would instead give the fixed midpoint $p/2$.
For a chosen noise model, replace the two <electric charges> by $q_L+n_L$ and $q_R+n_R$, with <independent random variables> $n_L,n_R$ of zero <expected value> and common <standard deviation> $s$. Assume $s/Q\ll1$ and no threshold censoring. First-order expansion of the ratio gives
$$
\delta x\simeq\frac{p}{Q}\bigl[(1-u)n_R-un_L\bigr],\qquad
\sigma_{\mathrm{lin}}(u)=\frac{ps}{Q}\sqrt{(1-u)^2+u^2}.
$$
For stipulated $p=2\,\mathrm{mm}$ and $s/Q=0.02$, this approximation gives $0.0283\,\mathrm{mm}$ at the midpoint and $0.0400\,\mathrm{mm}$ at either endpoint. The endpoint values describe uncensored signed channel readouts, including small negative values after baseline subtraction. They do not assume that both channels would pass a positive threshold. A ratio with a fluctuating denominator also need not have exactly zero bias; this is a local small-noise approximation.
Three original counterexamples show why calibration and selection matter:
* If the right channel retains a relative gain $g=1.10$, then $\widehat x/p=gu/(1-u+gu)$. At $u=1/2$, the bias is $p/42=0.0476\,\mathrm{mm}$ for the chosen pitch, larger than the midpoint noise estimate above.
* Suppose symmetric electronic leakage transforms the normalized right signal to $v=\epsilon+(1-2\epsilon)u$. Treating $v$ as $u$ gives a position bias $p\epsilon(1-2u)$. A known $\epsilon\ne1/2$ permits inversion; $\epsilon=1/2$ destroys position information entirely. This is a stipulated crosstalk model, not a measurement.
* Requiring both noiseless channels to exceed $0.1Q$ accepts only $0.1<u<0.9$: $80\%$ of uniformly distributed hits. Computing a position spread after that cut describes a selected sample. It does not establish improved <spatial resolution> for all incident hits.
Likewise, two equal hits at $u=1/4$ and $u=3/4$ have exactly the same total two-channel response as one hit at $u=1/2$ with twice the <electric charge>. A <centroid> can hide multiplicity even when each charge measurement is exact.
<Diffusion> enabled the lecture's Orsay interpolation, without establishing this model's linear response.<Electronic Imaging of Ionizing Radiation with Limited Avalanches in Gases by Georges Charpak>[p. 30]{ref}
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