--- title: "European TDM system" subject: "Telecommunications" chapter: 3 key: "semester_8/telecommunications:european-tdm-system" source: https://www.bctengineer.com/q/semester_8/telecommunications/european-tdm-system site: BCT Engineering --- # European TDM system Telecommunications · Chapter 3: Signal Multiplexing - Asked in: 2070 Asar, 7 marks - Video: [European E1 TDM Hierarchy](https://www.youtube.com/watch?v=ZS5_Q-Z64Kk) — Aman Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/telecommunications/european-tdm-system - Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-3 ## Question Explain the European TDM system used in telecommunication system. ## Short answer The European TDM system uses the CEPT hierarchy based on E1 at 2.048 Mbps with 32 time slots (TS0 sync, TS16 signalling, 30 voice channels) and A-law companding. Higher levels multiplex four lower-order tributaries with bit interleaving and pulse stuffing: E2 8.448 Mbps (120 ch), E3 34.368 Mbps (480 ch), E4 139.264 Mbps (1920 ch), E5 564.992 Mbps (7680 ch). ## Answer The **European TDM system** is the CEPT (ITU-T G.732/G.751) plesiochronous digital hierarchy, also used in Nepal, Asia and most of the world outside North America and Japan. Its basic building block is the **E1 (2.048 Mbps) 30-channel PCM system**. #### E1 (30-channel PCM) frame - Each voice channel: 8000 samples/s × 8 bits = 64 kbps, coded with **A-law** companding ($A = 87.6$). - One frame = 32 time slots × 8 bits = 256 bits in 125 µs. ```text +-----+-----+-----+------+------+-----+------+ | TS0 | TS1 | ... | TS15 | TS16 | ... | TS31 | +-----+-----+-----+------+------+-----+------+ sync <-voice 1-15-> sig <-voice 16-30-> <------ 32 x 8 = 256 bits in 125 us -------> ``` $$ \begin{aligned} \text{Bits per frame} &= 32 \times 8 = 256\ \text{bits} \\ R_{E1} &= 256 \times 8000 = 2.048\ \text{Mbps} \end{aligned} $$ - **TS0:** frame-alignment word (×0011011) in even frames; alarm and national bits in odd frames. - **TS16:** signalling. For channel-associated signalling, 16 frames form a **multiframe** (2 ms); TS16 of frame 0 carries the multiframe alignment, and frames 1–15 each carry 4 signalling bits for two channels, so all 30 channels are served. TS16 can instead be a 64 kbps common-channel (SS7) link. - **TS1–TS15 and TS17–TS31:** 30 voice channels. - Line code: **HDB3**, at 2.048 Mbps with ±50 ppm tolerance. #### Higher levels Each higher order combines **four** lower-order tributaries by bit interleaving with pulse stuffing (justification), plus frame-alignment and service bits. ```text 30 x 64k --> [MUX] --> E1 2.048 Mbps (30 ch) 4 x E1 --> [MUX] --> E2 8.448 Mbps (120 ch) 4 x E2 --> [MUX] --> E3 34.368 Mbps (480 ch) 4 x E3 --> [MUX] --> E4 139.264 Mbps (1920 ch) 4 x E4 --> [MUX] --> E5 564.992 Mbps (7680 ch) ``` | Level | Made from | Voice channels | Bit rate | Overhead | |---|---|---|---|---| | E0 | 1 voice channel | 1 | 64 kbps | – | | E1 | 30 voice + 2 TS | 30 | 2.048 Mbps | 128 kbps (TS0, TS16) | | E2 | 4 E1 | 120 | 8.448 Mbps | 256 kbps | | E3 | 4 E2 | 480 | 34.368 Mbps | 576 kbps | | E4 | 4 E3 | 1920 | 139.264 Mbps | 1.792 Mbps | | E5 | 4 E4 | 7680 | 564.992 Mbps | 7.936 Mbps | $$ \begin{aligned} E2 &: 4 \times 2.048 = 8.192;\ \ 8.448 - 8.192 = 0.256\ \text{Mbps overhead} \\ E3 &: 4 \times 8.448 = 33.792;\ \ 34.368 - 33.792 = 0.576\ \text{Mbps overhead} \\ E4 &: 4 \times 34.368 = 137.472;\ \ 139.264 - 137.472 = 1.792\ \text{Mbps overhead} \\ E5 &: 4 \times 139.264 = 557.056;\ \ 564.992 - 557.056 = 7.936\ \text{Mbps overhead} \end{aligned} $$ Features: regular ×4 structure; signalling does not steal voice bits; overhead 2/32 at E1. Higher levels are now carried in SDH (STM-1 at 155.52 Mbps can hold 63 E1s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the E1 rate of 2.048 Mbps obtained? A. 24 x 8 + 1 = 193 bits x 8000 B. 32 x 8 = 256 bits x 8000 frames/s C. 30 x 8 = 240 bits x 8000 D. 32 x 8 = 256 bits x 4000 Answer: B. A frame has 32 time slots of 8 bits = 256 bits every 125 µs, so 256 x 8000 = 2.048 Mbps. (Key point: E1 rate) 2. What are TS0 and TS16 used for? A. TS0 carries signalling; TS16 frame alignment B. TS0 carries voice; TS16 carries voice C. TS0 carries frame alignment/alarm; TS16 carries signalling D. Both carry only parity bits Answer: C. TS0 holds the frame alignment word and alarms; TS16 carries signalling (multiframe CAS or a 64 kbps SS7 link). (Key point: TS0 and TS16) 3. How is each higher level of the European hierarchy built? A. Four lower-order tributaries bit-interleaved with pulse stuffing plus overhead B. Seven tributaries by byte interleaving C. Exactly 4 times the rate with no overhead D. Six tributaries with robbed-bit signalling Answer: A. Each level combines four tributaries with stuffing and framing, so E2 = 8.448 Mbps, more than 4 x 2.048 = 8.192. (Key point: Higher-order structure) 4. Which companding law and line code does E1 use? A. mu-law and B8ZS B. mu-law and AMI C. A-law and HDB3 D. A-law and NRZ Answer: C. E1 uses A-law (A = 87.6) companding and the HDB3 line code. 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