--- title: "E1 hierarchy; PCM SQNR at 32 kbps" subject: "Communication Systems" chapter: 5 key: "semester_6/communication_systems:e1-pcm-4khz-1-1v-32kbps" source: https://www.bctengineer.com/q/semester_6/communication_systems/e1-pcm-4khz-1-1v-32kbps site: BCT Engineering --- # E1 hierarchy; PCM SQNR at 32 kbps Communication Systems · Chapter 5: Multiplexing Techniques - Asked in: 2069 Chaitra (CS II), 4+3 marks - Video: [SNR in PCM | Quantization noise in PCM | Sgnal to noise ratio in PCM](https://www.youtube.com/watch?v=NHlhFWeFebM) — Techno Tutorials ( e-Learning) - 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-4khz-1-1v-32kbps - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-5 ## Question Explain the E1 digital hierarchy. A speech signal with maximum frequency of 4 kHz and maximum amplitude of ±1.1 V is applied to a PCM system with its bit rate of 32 kbps. Calculate the SQNR and number of bits per sample. ## Short answer E1 is the ITU-T TDM-PCM carrier with 32 time slots (30 voice), 125 µs frame, rate 2.048 Mb/s. Sampling at Nyquist rate fs=8 kHz, bit rate 32 kbps yields n=4 bits/sample. For a full-scale sine, SQNR = 1.76 + 6.02×4 = 25.84 dB (linear 384). ## Answer #### 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. | Level | Made of | Channels | Bit rate | |---|---|---|---| | E1 | 30 voice + 2 | 30 | 2.048 Mb/s | | E2 | 4 × E1 | 120 | 8.448 Mb/s | | E3 | 4 × E2 | 480 | 34.368 Mb/s | | E4 | 4 × E3 | 1920 | 139.264 Mb/s | | E5 | 4 × E4 | 7680 | 564.992 Mb/s | 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A speech signal has fm = 4 kHz. What minimum sampling rate is used? A. 4 kHz B. 8 kHz C. 16 kHz D. 32 kHz Answer: B. Nyquist: fs = 2 fm = 8 kHz. (Key point: Sampling rate) 2. With Rb = 32 kb/s and fs = 8 kHz, how many bits per sample are used? A. 8 B. 2 C. 3 D. 4 Answer: D. n = Rb/fs = 32000/8000 = 4 bits, so L = 16 levels. (Key point: Bits per sample) 3. What is the SQNR for n = 4 bits with a full-scale sinusoid? A. About 25.84 dB B. About 24.08 dB C. About 48 dB D. About 12 dB Answer: A. SQNR = 1.76 + 6.02 x 4 = 25.84 dB (a linear ratio of 384 = 1.5 L^2). (Key point: SQNR value) 4. For a uniform quantiser covering +/-1.1 V with 16 levels, what are the step size and quantisation noise formula? A. Delta = 0.1375 V, Nq = Delta^2/12 B. Delta = 0.275 V, Nq = Delta^2/12 C. Delta = 0.1375 V, Nq = Delta^2/2 D. Delta = 2.2 V, Nq = Delta/12 Answer: A. Delta = 2 x 1.1 / 16 = 0.1375 V, and the noise power is Delta^2/12. (Key point: Quantisation noise) ## Related in Communication Systems - [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× - [Bits per sample and information rate (4.5 kHz, 8 levels)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abits-info-rate-4-5khz-8-levels) — Communication Systems, chapter 6 - [E1 hierarchy; PCM of 4.2 MHz TV signal with 512 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-pcm-tv-4-2mhz-512) — Communication Systems, chapter 4 - [Non-uniform quantization; PCM numerical (4 kHz, ±2.4 V, 64 kHz)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-companding-pcm-4khz-64khz) — Communication Systems, chapter 4 - [PCM SQNR increases 6 dB per bit](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-sqnr-6db-per-bit) — 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 ## Related in other subjects - [T1 and E1 frame structures and bit rates](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3At1-e1-frame-structures) — Telecommunications, chapter 3 - [E1 telephone hierarchy](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Ae1-hierarchy) — Computer Networks and Security, chapter 2 - [Parity bits and 3-bit even parity generator](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aparity-even-generator-3bit) — Digital Logic, chapter 4, asked 2× - [Parity bits and 3-bit odd parity generator](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aparity-odd-generator-3bit) — Digital Logic, chapter 4, asked 2×