The amplitude of stationary random signal X(t) is uniformly distributed in the range [0.5V]. The samples of this signal are applied to a a quantizer. Due to a system fault, the quantizer is producing samples at the output with a constant amplitude of 1V. the signal power -to-quantization noise power ratio at the outputof the faulty quantizer is _______dB.

2.84

Open in App

Solution

The correct option is A 2.84

Quantinization noise power,
$N_{Q} = E [(X - x_{q})^{2}]$
$= \int_{- \infty}^{\infty} (X - x_{q})^{2} f_{x} (x) d x = \int_{0}^{5} (x - 1)^{2} (\frac{1}{5}) d x$
$= \frac{1}{5} \int_{0}^{5} (x^{2} - 2 x + 1) d x = \frac{1}{5} [\frac{x^{3}}{3} - x^{2} + x]_{0}^{5}$
$= \frac{1}{15} [125 - 75 + 15] = \frac{65}{15} = \frac{13}{3} W$
Signal power, $S = E [x]^{2} = \int_{- \infty}^{\infty} x^{2} f_{x} (x) d x = \frac{1}{5} \int_{0}^{5} x^{2} d x = \frac{1}{5} [\frac{x^{3}}{3}]_{0}^{5} = \frac{25}{3} W$
$S Q N R = \frac{S}{N_{Q}} = \frac{25}{13}$
IN decilogs, $[S Q N R]_{d B} = 10 l o g_{10} (S Q N R) = 10 l o g_{10} (\frac{25}{13}) d B \approx 2.84 d B)$

Suggest Corrections

Similar questions

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 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.

R_{1} : [- 5 < X < - 1]

R_{2} : [- 1 < X < 1]

R_{3} : [1 < X < 5]

R_{3}

Join BYJU'S Learning Program