Question

The probability density function of the amplitude of a stationary random signal. The samples of this signal are applied to a 2-level quantizer whose input (X)- output (Y) characteristics are given by.

$X\left(t\right)is{f}_X\left(x\right)=e^{-x}u\left(x\right).$

If the values of $x_0,q_{1}and{q}_2$ are optimized in such a way that the quantization noise power is possible minimum for the given message signal, the value of $x_0$ will be

$Y=\{\begin{array}{cc}{q}_1;& forX\le x_0\\ q_2;& forX>x_0\end{array}$

• A.
  
  $\frac{1}{2}$
• B.
  $\frac{1}{e}$
• C.
  e
• D.
  2

Answer 1

The correct option is A


$\frac{1}{2}$
The conditions to be satisfied by the quantization levels and decision boundaries for optimum quantization are,

The quantization values should be the centroid of the respective quantization regions.

The decision boundaries of the quantizatior, region should be the mid-points of the corresponding quantized values.

Mathematically, the above two conditions can be summarized as,

$q_1={\int }_0^{x_0}x{f}_x\left(x\right)dx{q}_2={\int }_{x_0}^{\infty }x{f}_x\left(x\right)dx{x}_o=\frac{q_1+q_2}{2}$

Given that, $f_x\left(x\right)=e^{-x}u\left(x\right)So,q_1={\int }_0^{x_0}x{e}^{-x}dx=[-x{e}^{-x}-e^{-x}{]}_0^{x_0}=(1-x_0{e}^{x0}-e^{-x0})q_2={\int }^{\infty }x{e}^{-x}dx=[-x{e}^{-x}+e^{-x}]={\int }_{x0}^{\infty }(x_0{e}^{-x_0}+e^{-x_0})]x_0=\frac{q_1+q_2}{2}=\frac{1}{2}$