Topic: Digital Circuits
What is the quantization noise power for N bit uniform quantizer with full scale input voltage 2B?
- A. $\frac{B^{2}}{3(2^{2N})}$
- B. $\frac{2B}{\sqrt{3}(2^{N})}$
- C. $\frac{B^{2}}{12(2^{2N})}$
- D. $\frac{B}{\sqrt{3}(2^{N})}$
Answer: A. $\frac{B^{2}}{3(2^{2N})}$
Explanation: For an N-bit uniform quantizer, the number of quantization levels is $L = 2^N$. The step size $\Delta$ for a full-scale voltage range of $2B$ is $\Delta = \frac{2B}{2^N}$.
The quantization noise power (variance) for a uniform distribution of quantization error is given by $P_q = \frac{\Delta^2}{12}$.
Substituting $\Delta$:
$P_q = \frac{(\frac{2B}{2^N})^2}{12} = \frac{4B^2 / 2^{2N}}{12} = \frac{B^2}{3(2^{2N})}$.