Question

The probability density function of the amplitude of a message signal X(t) is shown below:

If the samples of this message signal are applied to 64-level mid-rise quantizer, then the resultant signal-to-quantization noise ratio will be

• A. 36.12 dB
• B. 34.32 dB
• C. 27.1 dB
• D. 33.11 dB

Answer 1

The correct option is D 33.11 dB

$\text{Signal power,}S={\int }_{-\infty }^{\infty }{x}^{2}fX\left(x\right)dx=2{\int }_0^2{x}^2\left(\frac{2-x}{4}\right)dx$

$=\frac{1}{2}{\int }_0^2(2x^2-x^3)dx=\frac{1}{2}{[\frac{2x^3}{3}-\frac{x^4}{4}]}_0^2=\frac{1}{2}[\frac{16}{3}-\frac{16}{4}]=\frac{2}{3}$

$\text{Quantization noise power,}{N}_Q=\frac{{△}^2}{12}$

$△=\frac{2-(-2)}{64}=\frac{1}{16}$

$\text{So,}{N}_Q=\frac{{\left(\frac{1}{16}\right)}^2}{12}=\frac{1}{3072}$

$SQNR)=\frac{S}{N_Q}=\frac{\left(\frac{2}{3}\right)}{\left(\frac{1}{3072}\right)}=\frac{2\times 3072}{3}=2048$

$ln\text{decilogs,}\left[SQNR\right]=10{\log }_{10}\left(SQNR\right)=33.11\text{dB}$