Difference Quantization

Pasted image 20240312145037.png In this system we Set UnU_{n} to be an arbitrary RV with a finite second moment (i.e. E[Un2]<E[U_{n}^{2}]<\infty). Since en=XnUne_{n}=X_{n}-U_{n} and e^n=X^nUn\hat{e}_{n}=\hat{X}_{n}-U_{n} then we have that E[(XnX^n)2]=E[(XnUn(X^nUn))2]=E[(ene^n)2]=E[(enQ(en))2]\begin{align*} E[(X_{n}-\hat{X}_{n})^{2}]&=E[(X_{n}-U_{n}-(\hat{X}_{n}-U_{n}))^{2}]\\ &=E[(e_{n}-\hat{e}_{n})^{2}]\\ &=E[(e_{n}-Q(e_{n}))^{2}] \end{align*}Here we see the MSE of the reconstruction does not accumulate the error and is merely the MSE of the quantizer.

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