Lecture 2 - Part 1

Contents

• Quantisation
• Signal-to-Quantisation-Noise-Ratio (SQNR)
• SQNR Problems and Solutions
  • Problem 1: Uniform Quantizer SQNR
  • Problem 2: SQNR with Different Bit Depth
  • Problem 3: Quantization Noise Power

Quantisation

Signal-to-Quantisation-Noise-Ratio (SQNR)

Assume that an ADC has dynamic range $\pm V_0$ volts and uses $N$ bits. The quantization step size is

$$\Delta = \frac{2V_0}{2^N} = 2^{1-N} V_0.$$

The quantization noise power (uniform quantizer assumption) is

$$\text{NoisePower} = \frac{\Delta^2}{12}.$$

The RMS noise (standard deviation) is

$$\sigma_e = \frac{\Delta}{\sqrt{12}}.$$

The RMS value of a full-scale sine wave with peak $V_0$ is

$$\text{RMS}_{\text{signal}} = \frac{V_0}{\sqrt{2}}.$$

Therefore the linear SNR (signal RMS divided by noise RMS) is

$$\text{SQNR} = \frac{\text{RMS}_{\text{signal}}}{\sigma_e} = \frac{V_0/\sqrt{2}}{\Delta/\sqrt{12}} = \frac{V_0\sqrt{12}}{\Delta\sqrt{2}} = \frac{V_0}{\Delta}\sqrt{6}.$$

Substituting $\Delta = \dfrac{2V_0}{2^N}$ yields

$$\text{SQNR} = \frac{V_0}{2V_0/2^N}\sqrt{6} = 2^{N-1}\sqrt{6} = 2^N \sqrt{\tfrac{3}{2}}.$$

In decibels:

$$\text{SQNR}_{\mathrm{dB}} = 20\log_{10}\!\left(2^N \sqrt{\tfrac{3}{2}}\right) = N\cdot 20\log_{10}(2) + 10\log_{10}\!\left(\tfrac{3}{2}\right).$$

Numerically this is typically written as:

$$\text{SQNR}_{\mathrm{dB}} \approx 6.02\,N + 1.7609.$$ So every extra bit in an ADC gives approximately a 6dB increase in the SNR of the ADC output SQNR.

SQNR Problems and Solutions

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Problem 1: Uniform Quantizer SQNR

A sinusoidal signal $x(t) = A \sin(2 \pi f t)$ with amplitude $A = 1$ is uniformly quantized using a 3-bit quantizer. Assume the signal is full-scale.

Questions:

1. Calculate the quantization step size $\Delta$.
2. Determine the maximum SQNR in dB.

Solution:

1. Quantization step size:
   For an $n$-bit uniform quantizer covering full scale $[-A, A]$:

$$\Delta = \frac{2A}{2^n} = \frac{2 \cdot 1}{2^3} = \frac{2}{8} = 0.25$$

2. Maximum SQNR for sinusoidal input:

$$\text{SQNR}_{\text{dB}} = 6.02 n + 1.76$$

Here $n = 3$:

$$\text{SQNR}_{\text{dB}} = 6.02 \cdot 3 + 1.76 = 19.82 \text{ dB}$$

Answer: $\Delta = 0.25, \quad \text{SQNR} \approx 19.8 \text{ dB}$

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Problem 2: SQNR with Different Bit Depth

A full-scale sinusoidal signal is quantized using an 8-bit uniform quantizer.

Questions:

1. What is the SQNR in dB?
2. If the bit depth is increased to 12 bits, by how much does the SQNR improve?

Solution:

1. 8-bit SQNR:

$$\text{SQNR}_{\text{dB}} = 6.02 \cdot 8 + 1.76 = 50.92 \text{ dB}$$

2. 12-bit SQNR:

$$\text{SQNR}_{\text{dB}} = 6.02 \cdot 12 + 1.76 = 74.0 \text{ dB}$$

SQNR improvement:

$$74 - 50.92 \approx 23.08 \text{ dB}$$

Answer: 8-bit SQNR ≈ 50.9 dB, improvement with 12-bit ≈ 23 dB.

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Problem 3: Quantization Noise Power

A full-scale sinusoidal signal with amplitude $A = 2$ is quantized using a 4-bit uniform quantizer.

Questions:

1. Find the quantization step size $\Delta$.
2. Calculate the quantization noise power $\sigma_q^2$.
3. Determine SQNR (linear, not dB).

Solution:

1. Step size:

$$\Delta = \frac{2A}{2^n} = \frac{2 \cdot 2}{16} = 0.25$$

2. Quantization noise power (uniform quantization):

$$\sigma_q^2 = \frac{\Delta^2}{12} = \frac{0.25^2}{12} = \frac{0.0625}{12} \approx 0.00521$$

3. Signal power (sinusoid $A \sin(\cdot)$):

$$P_s = \frac{A^2}{2} = \frac{2^2}{2} = 2$$

4. SQNR (linear):

$$\text{SQNR} = \frac{P_s}{\sigma_q^2} = \frac{2}{0.00521} \approx 384$$

Answer:

$$\Delta = 0.25, \quad \sigma_q^2 \approx 0.00521, \quad \text{SQNR} \approx 384$$