---
title: "E1 hierarchy; PCM of 4.2 MHz TV signal with 512 levels"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:e1-pcm-tv-4-2mhz-512"
source: https://www.bctengineer.com/q/semester_6/communication_systems/e1-pcm-tv-4-2mhz-512
site: BCT Engineering
---

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

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2074 Chaitra (CS II), 4+6 marks
- Video: [Communication Engineering - Multiplexing Hierarchy for Digital Communication](https://www.youtube.com/watch?v=DrPbb3cDvFE) — Dr. Sapna Katiyar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/e1-pcm-tv-4-2mhz-512
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Explain E1 hierarchy of TDM-PCM Telephony. A television signal having a bandwidth of 4.2 MHz is transmitted using binary PCM system. Given that the number of quantization level is 512. Determine:
i) Code word length
ii) Transmission bandwidth
iii) Bit rate
iv) SQNR

## Short answer

E1 hierarchy: 30 voice channels at 64 kbps each, 32 time slots/frame, 2.048 Mbps. Higher orders multiplex 4× lower: E2=8.448 Mbps, E3=34.368 Mbps, E4=139.264 Mbps. For TV signal: W=4.2 MHz, L=512. Code word length n=log₂512=9 bits. Nyquist fs=2W=8.4 MHz. Bit rate Rb=nfs=75.6 Mbps. Transmission bandwidth BT=nW=37.8 MHz. SQNR=1.76+6.02×9=55.95 dB (full-load sinusoid).

## Answer

#### 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.

```text
|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:

| Level | Inputs | Voice channels | Rate (Mbps) |
|---|---|---|---|
| E1 | 30 voice | 30 | 2.048 |
| E2 | 4 × E1 | 120 | 8.448 |
| E3 | 4 × E2 | 480 | 34.368 |
| E4 | 4 × E3 | 1920 | 139.264 |

#### 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How many voice channels does E1 carry, and how are the other slots used?
   A. 32 voice channels
   B. 30 voice channels; TS0 for frame alignment and TS16 for signalling
   C. 16 voice channels
   D. 24 voice channels plus 8 for sync
   Answer: B. Of 32 slots, TS0 is sync and TS16 signalling, leaving 30 voice slots. (Key point: E1 frame)

2. With L = 512, the code word length is:
   A. 8 bits
   B. 9 bits
   C. 10 bits
   D. 512 bits
   Answer: B. n = log2 512 = 9. (Key point: Code word length)

3. For W = 4.2 MHz at the Nyquist rate and 9 bits, Rb is:
   A. 37.8 Mbps
   B. 75.6 Mbps
   C. 8.4 Mbps
   D. 151.2 Mbps
   Answer: B. fs = 8.4 MHz and Rb = 9 x 8.4 = 75.6 Mbps. (Key point: Bit rate)

4. The minimum transmission bandwidth and SQNR (full-load sine) are:
   A. 37.8 MHz and about 55.95 dB
   B. 75.6 MHz and 49.9 dB
   C. 4.2 MHz and 55.95 dB
   D. 37.8 MHz and 58.96 dB
   Answer: A. BT = Rb/2 = 37.8 MHz, and SQNR = 1.76 + 6.02 x 9 = 55.95 dB. (Key point: Bandwidth and SQNR)

## Related in Communication Systems

- [PAM, PWM, PPM; PCM of 4.8 MHz TV signal with 512 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apam-pwm-ppm-tv-4-8mhz) — Communication Systems, chapter 4
- [PCM bit rate for 4.5 MHz TV signal with 1024 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-tv-4-5mhz-1024) — Communication Systems, chapter 4
- [T1 and E1 TDM-PCM hierarchy with frame diagrams](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3At1-e1-frame-diagrams) — Communication Systems, chapter 5
- [Companding; T1 hierarchy of TDM-PCM telephony](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompanding-t1) — Communication Systems, chapter 5
- [E1 digital hierarchy; SQNR in uniformly quantized PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-hierarchy-sqnr-uniform-pcm) — Communication Systems, chapter 5, asked 2×
- [E1 hierarchy; PCM SQNR at 32 kbps](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-pcm-4khz-1-1v-32kbps) — Communication Systems, chapter 5