Question

Consider a non uniform quantizer as shown below



The input to the quantizer is a message signal whose PDF is uniform distributed in the range $[-3,2]$ Volts is applied to this quatizer, then the quantization noise power will be _____ $W$

• A. 1.13

Answer 1

The correct option is A 1.13
$N_o=E\left[\right(X-x_o\left)\right]={\int }_{-\infty }^{\infty }(x-x_q{)}^2{f}_X(x)dx$
Now, since the input is in the range of $[-3\text{to}2]$ the PDF of the input signal can be drawn as



$N_o={\int }_{-3}^0\frac{1}{5}(x+1{)}^{2}dx+{\int }_0^2\frac{1}{5}(x-2{)}^{2}dx$
$={\begin{array}{c}\frac{1}{5}\frac{(x+1{)}^3}{3}\end{array}|}_{-3}^0+{\begin{array}{c}\frac{1}{5}\frac{(x-2{)}^3}{3}\end{array}|}_0^2$
$=\frac{1}{15}(1+8)+\frac{1}{15}(+8)=\frac{17}{15}W=1.13W$