E1 hierarchy; PCM SQNR at 32 kbps

E1 digital hierarchy

E1 is the ITU-T TDM-PCM carrier. Each frame carries 32 time slots of 8 bits in 125 µs: TS0 for frame sync, TS16 for signalling, and 30 voice slots. Its rate is $R_b = 256 \times 8000 = 2.048$ Mb/s. Higher levels multiplex 4 lower streams each, adding framing and stuffing bits.

E1 uses A-law companding and the HDB3 line code.

Numerical

Given: $f_m = 4$ kHz, amplitude $\pm 1.1$ V, $R_b = 32$ kb/s. Assumption: Nyquist-rate sampling, a uniform quantiser covering $\pm 1.1$ V, and a full-scale sinusoidal test signal.

Sampling rate:

$$f_s = 2 f_m = 2 \times 4 = 8\ \text{kHz}$$

Bits per sample:

$$n = \frac{R_b}{f_s} = \frac{32\,000}{8000} = 4\ \text{bits}, \qquad L = 2^4 = 16\ \text{levels}$$

SQNR:

$$\begin{aligned} \Delta &= \frac{2 \times 1.1}{16} = 0.1375\ \text{V} \\ N_q &= \frac{\Delta^2}{12} = \frac{0.1375^2}{12} = 1.5755 \times 10^{-3}\ \text{V}^2 \\ P_s &= \frac{(1.1)^2}{2} = 0.605\ \text{V}^2 \\ \text{SQNR} &= \frac{0.605}{1.5755 \times 10^{-3}} = 384 = \tfrac{3}{2} L^2 \\ \text{SQNR (dB)} &= 10 \log_{10} 384 = 25.84\ \text{dB} \end{aligned}$$

This agrees with the rule $\text{SQNR} = 1.76 + 6.02n = 1.76 + 24.08 = 25.84$ dB.

Answer: $n = 4$ bits per sample; SQNR = 384 ≈ 25.84 dB.