Question

A 6-bit PCM system is uniformly quantized with a bit rate of 4.8 kbps. The signal-to-quantization noise ratio that would result when its input is a sine wave with peak amplitude equal to 10 V will be:

• A. 31.5 dB
• B. 37.8 dB
• C. 30 dB
• D. 27 dB

Answer 1

The correct option is B 37.8 dB

n = 6; $A_m=10$

Signal Power = $\frac{A_m^2}{2}=50$

$QuantizationNoisePower=\frac{{\Delta }^2}{12}={\left(\frac{2A_m}{2^n}\right)}^2\times \frac{1}{12}$

$=\frac{4\left(100\right)}{2^{12}}\times \frac{1}{12}$

$\frac{S}{NQ}=\frac{50}{400}\times 2^{12}\times 12=6144$

$\frac{S}{NQ}\left(dB\right)=10lo{g}_{10}\left(68.44\right)=37.8dB$