Now to put difference quantization in practice we still need to define what the term $U_{n}$ is. The idea here is to let the signal $U_{n}$ be the prediction of $X_{n}$ from past reconstructions (like in our linear predictor set up). Here we define $U_{n}$ as follows $$U_{n}=\sum_{i=1}^{m}a_{i}\hat{X}_{n-i}$$ Pasted image 20240312150852.png

The key feature here is that $U_{n}$ can be generated at both the encoder and decoder since it depends on past reconstruction values.

Expressed end-to-end

$$\begin{align*} \hat{X}_{n}&=\hat{e}_{n}+U_{n}\\ &=Q(e_{n})+U_{n}\\ &=Q(X_{n}-U_{n})+U_{n}\\ &=Q\left( X_{n}-\sum_{i=1}^{m}a_{i}\hat{X}_{n-i} \right)+\sum_{i=1}^{m}a_{i}\hat{X}_{n-i}\\ \end{align*}$$ # Intuition This architecture is clever in the way it cancels out the $U_{n}$ terms from both $X_{n}$ and $\hat{X}_{n}$. It essentially isolates the issue to be strictly concerned with the quantizer alone. Whereas in the previous iteration we had some stuff going on with the pulse response that kept residues of past predictions in the mix. Another way to make sense of this is to look at the source of $P$ here. $P$ is derived from $\hat{X}_{n-i}, i>1$ only, this means we can keep the $U_{n}$ consistent on both sides, a matter that was not the case when we had $P$ being generated both from $X_{n}$ and $\hat{X}_{n}$ on either side.

Equivalent structure

Pasted image 20240312153732.png We see here that the sending side keeps track of what the receiver receives which allows the $U_{n}$ to stay consistent on both sides.