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As pichenettes indicated, in this case a peak at the middle of the output indicates 0 lag. The peak's offset from the middle point is your time lag. EDIT: It concerns me that the correlation is so n …

If you want to get an estimate of the noise, do the cross-correlation, and then multiply your transmitted sound by your cross-correlation peak and then subtract it from your received sounds. Make sur …

You divide the peak of the correlation by the energy of $A$. $B$ is a scaled version of $A$, with some noise added to it. Thus, the equation for the cross-correlation peak is- $ A * B' = A * (kA' + …

Using the rising edge is better than cross-correlation when there is the physical equivalent of an inductor in the system, which will introduce inertia into the system. That inertia will cause the ou …

An individual inner product does produce a scalar, but often when a cross correlation is calculated multiple individual cross correlations (i.e. dot products) are calculated at different time offsets. …

I assume by "shift" you mean "shift in time". No, the first correlation peak should not be shifted in time, you should just get multiple peaks, assuming the time difference of arrival is not much sho …

It's not cheating, and it's also not adding any new information. What you are doing is the same thing that any upsampling LPF is doing- adding zeros and then reconstructing the waveform with the alre …

The time lag is equal to the index difference times the sample period. The sample period is: $$ T_s = \frac{1}{f_s} = \frac{1}{16000} = 62.5\mu s $$ Thus, the time lag from the first peak to the seco …

Cross correlation should work. I think the problem is the waveform that you are using. A square wave has bad auto-correlation properties. If it is a periodic square wave it will have multiple peaks …