What is the Performance Gain?

Assume $X_{1},\dots,X_{n},\dots$ is WSS. Recall the difference quantization principle $$E[(X_{n}-\hat{X}_{n})^{2}]=E[(e_{n}-\hat{e}_{n})^{2}]=E[(e_{n}-Q(e_{n}))^{2}]$$We see here that our quantization errors do not accumulate. We can assess the performance gain over our baseline scalar quantization.

System SNR

$$\text{SNR}_{\text{sys}}=10\log_{10}G_{\text{clp}}+\text{SNR}_{Q} \ \ \ [\text{dB}]$$

Approximations

1. $$G_{\text{clp}}\approx G_{\text{olp}}=\frac{E[X_{n}^{2}]}{E\left[ \left( X_{n}-\sum^{m}_{i=1}a_{i}X_{n-i} \right)^{2} \right]}$$
2. $G_{Q}$ is mainly determined by $Q$ and not $e_{n}$

Hence we have $$\text{SNR}_{\text{sys}}\approx \text{SNR}_{Q}+10\log_{10}G_{\text{olp}}$$ Here we see the overall performance gain over scalar quantization is $$10\log_{10}G_{\text{olp}}$$ # Maximum Gain With optimal coefficients $a_{1},\dots,a_{m}$ the max gain is $$G_{\text{clp}}\approx G_{\text{olp}}=\frac{E[X_{n}^{2}]}{E\left[ \left( X_{n}-\sum^{m}_{i=1}a_{i}X_{n-i} \right)^{2} \right]}=\frac{r_{0}}{r_{0}-\sum_{i=1}^{m}a_{i}r_{i}}$$