E1 hierarchy; PCM of 4.2 MHz TV signal with 512 levels

E1 hierarchy of TDM-PCM telephony

E1 is the ITU-T (CEPT) primary digital multiplex used in Europe, Asia and Nepal.

• Each voice signal (300–3400 Hz) is sampled at 8 kHz and A-law coded into 8 bits: 64 kbps per channel.
• One frame lasts 125 µs and has 32 time slots × 8 bits = 256 bits.
• TS0: frame alignment word; TS16: signalling (multiframe of 16 frames); TS1–15, TS17–31: 30 voice channels.
• Bit rate: $32 \times 8 \times 8000 = 2.048$ Mbps.

|TS0|TS1|...|TS15|TS16|TS17|...|TS31|  = 125 us
 sync  voice 1-15  sig   voice 16-30

Higher orders multiplex 4 tributaries each, adding justification and framing bits:

Numerical

Given: $W = 4.2$ MHz, $L = 512$, sampling at the Nyquist rate.

i) Code word length

$$n = \log_2 512 = 9\ \text{bits}$$

ii) Transmission bandwidth (minimum, Nyquist)

$$\begin{aligned} f_s &= 2W = 8.4\ \text{MHz} \\ B_T &= \frac{R_b}{2} = \frac{nf_s}{2} = nW = 9 \times 4.2 = 37.8\ \text{MHz} \end{aligned}$$

iii) Bit rate

$$R_b = nf_s = 9 \times 8.4 \times 10^6 = 75.6\ \text{Mbps}$$

iv) SQNR (full-load sinusoid)

$$\begin{aligned} \text{SQNR} &= 1.5L^2 = 1.5 \times 512^2 = 393\,216 \\ \text{SQNR}_{dB} &= 1.76 + 6.02 \times 9 = 55.95\ \text{dB} \end{aligned}$$

(Using the form $3L^2$ for a signal with power $m_p^2$: $786\,432$, i.e. 58.96 dB.)

Answer: $n = 9$ bits; $B_T = 37.8$ MHz; $R_b = 75.6$ Mbps; SQNR ≈ 55.95 dB.