Skip to main content

Quantization; SQNR for linear quantization

Short answer

Quantization approximates samples by the nearest of L levels with step Δ = 2mp/L. Error is uniformly distributed in ±Δ/2, giving noise power Δ²/12. For a full-scale sine, SQNR = 1.5 L² or 1.76 + 6.02n dB; each extra bit adds 6 dB.

Define quantization. Derive the expression for evaluating signal to quantization noise ratio (SQNR) for linear quantization.

Quantization

Quantization is the process of approximating each continuous-amplitude sample of a signal by the nearest of a finite set of allowed levels. With LL levels spaced Δ\Delta apart it turns a sampled (discrete-time, continuous-amplitude) signal into a discrete-amplitude signal that can be coded with n=log⁡2Ln = \log_2 L bits. The difference between the true sample and the quantized value is the quantization error (quantization noise).

SQNR for linear (uniform) quantization

Let the message m(t)m(t) lie in the range −mp-m_p to +mp+m_p and let the quantizer have LL equally spaced levels.

  1. Step size
Δ=2mpL\Delta = \frac{2m_p}{L}
  1. Error range: each sample is replaced by the nearest quantization level (the centre of its step), so the error qq lies between −Δ/2-\Delta/2 and +Δ/2+\Delta/2.
  2. Error pdf: for a busy signal with many levels, qq is assumed uniformly distributed, fQ(q)=1/Δf_Q(q) = 1/\Delta for ∣q∣≤Δ/2|q| \le \Delta/2.
  3. Quantization noise power (mean square error):
Nq=q2‾=∫−Δ/2Δ/2q21Δ dq=1Δ[q33]−Δ/2Δ/2=Δ212=(2mp/L)212=mp23L2\begin{aligned} N_q = \overline{q^2} &= \int_{-\Delta/2}^{\Delta/2} q^2 \frac{1}{\Delta}\,dq = \frac{1}{\Delta}\left[\frac{q^3}{3}\right]_{-\Delta/2}^{\Delta/2} \\ &= \frac{\Delta^2}{12} = \frac{(2m_p/L)^2}{12} = \frac{m_p^2}{3L^2} \end{aligned}
  1. SQNR: with signal power S=m2(t)‾S = \overline{m^2(t)},
SQNR=SNq=3L2 m2(t)‾mp2\text{SQNR} = \frac{S}{N_q} = 3L^2\,\frac{\overline{m^2(t)}}{m_p^2}
  1. Sinusoidal message m(t)=mpcos⁡ωmtm(t) = m_p\cos\omega_m t (full load): S=mp2/2S = m_p^2/2, so
SQNR=mp2/2mp2/(3L2)=32L2=1.5×22nSQNRdB=10log⁡101.5+20nlog⁡102≈1.76+6.02n dB\begin{aligned} \text{SQNR} &= \frac{m_p^2/2}{m_p^2/(3L^2)} = \frac{3}{2}L^2 = 1.5 \times 2^{2n} \\ \text{SQNR}_{dB} &= 10\log_{10}1.5 + 20n\log_{10}2 \approx 1.76 + 6.02n\ \text{dB} \end{aligned}

Key results

  • SQNR rises as L2L^2, i.e. by about 6 dB for each extra bit.
  • For a signal whose power equals mp2m_p^2 (e.g. a square-like signal) the result is 3L23L^2, i.e. 4.77+6.02n4.77 + 6.02n dB.
  • Example: 8-bit PCM with a full-scale sine gives 1.76+48.16≈49.91.76 + 48.16 \approx 49.9 dB.
  • Weak signals give a much lower SQNR because NqN_q is fixed while SS falls; this is why non-uniform quantization (companding) is used for speech.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Communication Systems chapter 4 questions →