Question

The probability density function of the amplitude of a stationary random signal X(t) is shown below.

$\text{The samples of this random signal are applied to an 8-level uniform quantizer. The decision boundaries of the quantizer are set at}x=0,\pm 1,\pm 2,\pm 3,\pm 4\text{and the quantization levels are set at the center of the two adjacent boundaries. The signal power-to-quantization noise power ratio at the output of the quantizer is\_\_\_\_\_\_\_\_\_\_ dB.}$

• A. 17.16

Answer 1

The correct option is A 17.16
$\text{Step size,}\Delta =1$
$\text{Quantization noise power,}{N}_Q=\frac{{\Delta }^2}{12}=\frac{i}{12}$
$\text{Signal power,}S={\int }_{-\infty }^{\infty }{x}^2{f}_X\left(x\right)dx$
$=2[{\int }_0^1\left(\frac{20}{100}\right)x^{2}dx+{\int }_1^4\left(\frac{10}{100}\right)x^{2}dx]$

$=\frac{20}{100}[2{\left(\frac{x^3}{3}\right)}_0^1+{\left(\frac{x^3}{3}\right)}_1^4]$
$=\frac{20}{300}\left[\right(2\times 1)+(64-1\left)\right]=\frac{13}{3}$
$\text{So,(SQNR)}=\frac{S}{N_Q}=\frac{\left(\frac{13}{3}\right)}{\left(\frac{1}{12}\right)}=52$
${\text{In decilogs, [SQNR]}}_{\text{dB}}=10{\log }_{10}\left(\text{(SQNR)}\right)=10{\log }_{10}\left(52\right)\text{dB}$
$\approx 17.16\text{dB}$