Question

The quantization regions of a 3-level quantizer, with input variable X are defined as follows:
$R_1:[-5<X<-1]$
$R_2:[-1<X<1]$
$R_3:[1<X<5]$
For each region, the quantization level is chosen at the centre of region. The samples of a message signal, whose amplitude is uniformly distributed in the range [-3, 2], are applied to the above said quantizer. The quantization noise power due to rounding-off of the message samples lie in the region $R_3$ alone will be

• A. $\frac{4}{3}$
• B. $\frac{8}{15}$
• C. $\frac{7}{15}$
• D. $\frac{4}{15}$

Answer 1

The correct option is C $\frac{7}{15}$

The quantization level for region $R_3$ will be,
$x_{q3}=\frac{1+5}{2}=3$
So, the quantization noise due to rounding off of the samples lie in the region $R_3$ alone is,
$N_{Q3}=E\left[\right(X-x_{q3}{)}^2]={\int }_{<R_3>}(x-3{)}^2{f}_x\left(x\right)dx$
$f_x\left(x\right)=\{\begin{array}{cc}\frac{1}{5};& -3\le x\le 2\\ 0;& otherwise\end{array}$
So, $N_{Q3}=\frac{1}{5}{\int }_1^2(x^2-6x+9)dx=\frac{1}{5}{[\frac{x^3}{3}-3x^2+9x]}_1^2$
$=\frac{1}{5}[(\frac{8}{3}-12+18)-(\frac{1}{3}-3+9)]=\frac{7}{15}$