Question

Consider a source $m\left(t\right)$, whose amplitude statistics are as follows:

$f_m\left(m\right)=\{\begin{array}{cc}1/4;& -1\le m\le 1\\ 1/12;& -4\le m\le -1\\ 1/12;& 1\le m\le 4\\ 0;& \text{Otherwise}\end{array}$

The message is passed through a three bit quantize having uniform step size of 1 V. Reconstruction levels of the quantizer are mid points of the decision boundaries. Then the value of signal to quantization noise is equal to

• A. 9.77 dB
• B. 25.28 dB
• C. 16.43 dB
• D. 6.85 dB

Answer 1

The correct option is C 16.43 dB

To create an optimum qunatize of 3-bits all the masssage symbols should be equiprobable thus, the area under each quantized value must be same


Thus, for a 3-bit quantizer we need 8 levels. From the graph we can observe that the quantization level can be choosen as $\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2}$ and $\pm \frac{7}{2}$.
Thus signal power, ${\sigma }_m^2={\int }_{-4}^4{m}^2{f}_m\left(m\right)dm$
$=2[{\left[\frac{1}{4}\frac{m^3}{3}\right]}_0^1+{[\frac{1}{12}.\frac{m^3}{3}]}_1^4]=\frac{11}{3}\text{Watts}$
Quantized noise power,
${\sigma }_q^2=2[{\int }_0^1{(m-\frac{1}{2})}^2\times \frac{1}{4}dm+{\int }_1^2{(m-\frac{3}{2})}^2\times \frac{1}{12}dm+{\int }_2^3{(m-\frac{5}{2})}^2\times \frac{1}{12}dm+{\int }_1^4{(m-\frac{7}{2})}^2\times \frac{1}{12}dm]$
$=2[\frac{1}{4}{\int }_{-1/2}^{1/2}{\lambda }^{2}d\lambda +\frac{1}{12}{\int }_{-1/2}^{1/2}{\lambda }^{2}d\lambda +\frac{1}{12}{\int }_{-1/2}^{1/2}{\lambda }^{2}d\lambda +\frac{1}{2}{\int }_{-1/2}^{1/2}{\lambda }^{2}d\lambda +\frac{1}{12}{\int }_{-1/2}^{1/2}{\lambda }^{2}d\lambda ]$
$={\int }_{-1/2}^{1/2}{\lambda }^{2}d\lambda =2{\int }_0^{1/2}{\lambda }^{2}d\lambda =\frac{1}{12}$
$(SNR{)}_q=\frac{{\sigma }_m^2}{{\sigma }_n^2}=\frac{11/3}{1/12}=44$
$(SNR{)}_{q\left(dB\right)}=10lo{g}_{1}0(44)=16.43dB$