Topic: Communication Systems
Two ideal quantizers A and B have following specifications:
A: 5 bit quantizer with input dynamic range of -1 V to +1 V with $Q_{1}$ as quantization noise power.
B: 8 bit quantizer with input dynamic range of -0.5 V to +0.5 V with $Q_{2}$ as quantization noise power.
Then $Q_{1}/Q_{2}$ will be
- A. 16
- B. 256
- C. 64
- D. 128
Answer: B. 256
Explanation: Quantization noise power is given by $Q = \frac{s^{2}}{12}$, where step size $s = \frac{V_{p-p}}{2^{n}}$.
For Quantizer A: Dynamic range is from -1 V to +1 V, so $V_{p-p} = 2 \text{ V}$. Since it is a 5-bit quantizer, $n=5$. Step size $s_{1} = \frac{2}{2^{5}} = \frac{1}{2^{4}}$. The quantization noise power is $Q_{1} = \frac{(1/2^{4})^{2}}{12}$.
For Quantizer B: Dynamic range is from -0.5 V to +0.5 V, so $V_{p-p} = 1 \text{ V}$. Since it is an 8-bit quantizer, $n=8$. Step size $s_{2} = \frac{1}{2^{8}}$. The quantization noise power is $Q_{2} = \frac{(1/2^{8})^{2}}{12}$.
The ratio $\frac{Q_{1}}{Q_{2}} = \frac{(1/2^{4})^{2}}{(1/2^{8})^{2}} = \frac{2^{16}}{2^{8}} = 2^{8} = 256$.