Calibrated two-wire charge centroid ($\hat{x}=p{q}_R/(q_L+q_R)$)

Can two readout values locate a hit between wires? 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 chargesThese are assumptions for a toy sensor response, not the natural response law of adjacent anode wires. Under these assumptions the centroidrecovers 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 $s/Q\ll 1$ and no threshold censoring. First-order expansion of the ratio givesFor 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:

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.^[ref]