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Going off of the resolution, it looks to me like the actual label range iswould be 0-32.0, which with 11 significant bits gives an LSB of $\frac{32}{2048} = \frac{1}{64} = 0.015625$.
$1.5 * 64 = 96$
Going off of the resolution, it looks to me like the actual range is 0-32.0, which with 11 significant bits gives an LSB of $\frac{32}{2048} = \frac{1}{64} = 0.015625$.
$1.5 * 64 = 96$
Going off of the resolution, it looks to me like the actual label range would be 0-32.0, which with 11 significant bits gives an LSB of $\frac{32}{2048} = \frac{1}{64} = 0.015625$.
Going off of the resolution, it looks to me like the actual range is 0-32.0, which with 11 significant bits gives an LSB of $\frac{32}{2048} = \frac{1}{64} = 0.015625$.
