Doubt on Signal Processing Step

Question

I am using a Radar Sensor. So applying Signal Processing on the Radar Sensor Data. PrevSample and PastSample are Radar IQ samples for a present and previous Sweep detecting tiny movement. In the source code I have some doubts that . Here I am adding the portion of Python Code . Can anyone please explain the following signal processing, and the significance of that, please describe what each term indicate

import numpy as np

Envelope   = np.abs(PresentSample)
Delta = PresentSample * np.conj(PreviousSample)
PhaseWeights = np.imag(Delta)
Weights = np.abs(PhaseWeights) * Envelope
DeltaDist = np.dot(Weights, np.angle(Delta))
DeltaDist *= 2.5 / (2.0 * pi * sum(Weights))

Present and prev samples are 1D array consisting of iq samples. For each processing . Is Delta is cross correlation ? Please give explanation for rest of the codes Why multiply the DetltaDist ?

DeltaDist is the tiny movement, which is used to store in a Buffer. I am asking the signal processing logic of this code/algorithm. I know the code and the functions conj, imag, etc... I just want to know for example why we take conjugate product/ why we take imaginary part like that

Answer 1

It's not clear what your code is doing, but my understanding is:

Delta = PresentSample * np.conj(PreviousSample)

gets the difference between the previous sample and the present one. This might look like:

$$ \Delta = e^{j\omega n} \cdot e^{-j\omega (n-1)} = e^{j \omega} $$

provided the signal you're interested in is like a complex exponential signal.

Then

PhaseWeights = np.imag(Delta)

is taking the imaginary part of

$$ e^{j\omega} = \cos(\omega) + j\sin(\omega). $$

If $\omega$ is small enough, then $$ \Im (\Delta) = \sin(\omega) \approx \omega $$ because $\sin(x) \approx x$ for small $x$.

It's not clear to me why this is a good thing.