Question

Consider a source $m\left(t\right)$, whose amplitude statistics is defined as
$f_m\left(m\right)=\{\begin{array}{cc}\frac{1}{12}& -4\le m\le -1\\ \frac{1}{4},& -1\le m\le 1\\ \frac{1}{12},& 1\le m\le 4\\ 0,& otherwise\end{array}$
An optimum 3 bit quantizer is designed to sent the message signal over a baseband channel. The signal to quantization noise will be equal to

• A. 16.43 dB
• B. 12.51 dB
• C. 9.2 dB
• D. 20.16 dB

Answer 1

The correct option is A 16.43 dB

3-bit quantizer mean 8-levels.

The eight levels can be chosen as the endpoint of the amplitude level for optimum quantization.
Now, the signal power is equal to
$\begin{array}{cc}{\sigma }_m^2& ={\int }_{-4}^4{m}^2{f}_m\left(m\right)dm=2{\int }_0^4{m}^2{f}_m\left(m\right)dm\\ & =2[{\left(\frac{m^3}{12}\right)}_0^1+{\left(\frac{m^3}{36}\right)}_1^4]=\frac{11}{3}W\\ \text{Now, the quantization noise is equal to}& \\ {\sigma }_q^2& =2[\frac{1}{4}{\int }_0^1{(m-\frac{1}{2})}^{2}dm+\frac{1}{12}{\int }_1^2{(m-\frac{3}{2})}^{2}dm+\frac{1}{12}{\int }_1^3{(m-\frac{5}{2})}^{2}dm+\frac{1}{12}{\int }_1^4{(m-\frac{7}{2})}^{2}dm]\\ {\sigma }_q^2& =\frac{2}{3}[{\begin{array}{c}\frac{1}{4}{(m-\frac{1}{2})}^3\end{array}|}_0^1+{\begin{array}{c}\frac{1}{12}{(m-\frac{3}{2})}^3\end{array}|}_1^2+{\begin{array}{c}\frac{1}{12}{(m-\frac{5}{2})}^3\end{array}|}_2^3+{\begin{array}{c}\frac{1}{12}{(m-\frac{7}{2})}^3\end{array}|}_3^4+]\\ {\sigma }_4^2& =\frac{1}{12}W\\ \therefore (SNR{)}_q& =\frac{{\sigma }_m^2}{{\sigma }_q^2}=\frac{11/3}{1/12}=44\\ \therefore (SN{R}_q{)}_{dB}& =16.43dB\end{array}$