Question

$\text{The samples of a message signal X(t) are applied to a 16 - level uniform mid-rise quantizer. The peak and rms values of the meassage signal are}{x}_{\text{max}}\text{and}{x}_{\text{rms}}\text{respectively. The dynamic range of the quantizer is from}-x_{\text{max}}\text{to}{x}_{\text{max}}\text{. The quantization error is uniformly distrbuted over}[-\frac{\Delta }{2},\frac{\Delta }{2}],\text{where}\Delta \text{is the uniform step - size. If the signal - to - quantization noise ratio at the output of the quantizer is 25 dB, then the value of the crest or peak factor of the message signal is\_\_\_\_.}$

• A. 1.56

Answer 1

The correct option is A 1.56
$\Delta =\frac{x_{max}-(-x_{max})}{L}=\frac{2x_{max}}{16}=\frac{x_{max}}{8}$
Quantization noise power,
$N_Q=\frac{{\Delta }^2}{12}=\frac{x_{max}^2}{768}$
$\text{Signal power,}S=x_{rms}^2$
Signal - to - quantization noise power,
$\text{SQNR}=\frac{S}{N_Q}=\frac{768x_{rms}^2}{x_{max}^2}$
$10{\log }_{10}\left(SQNR\right)=20{\log }_{10}\left(\frac{x_{rms}}{x_{max}}\right)+10{\log }_{10}\left(768\right)=25dB$
${\log }_{10}\left(\frac{x_{max}}{x_{rms}}\right)=\frac{10{\log }_{10}\left(768\right)-25}{20}=0.1927$
Crest or peak factor,
$\frac{x_{max}}{x_{rms}}={10}^{\left(0.1927\right)}=1.56$