--- title: "Short note: Digital transmission hierarchies" subject: "Telecommunications" chapter: 3 key: "semester_8/telecommunications:short-note-digital-hierarchies" source: https://www.bctengineer.com/q/semester_8/telecommunications/short-note-digital-hierarchies site: BCT Engineering --- # Short note: Digital transmission hierarchies Telecommunications · Chapter 3: Signal Multiplexing - Asked in: 2065 Baisakh (old course), 8 marks - Video: [Digital Transmission Hierarchy | T1, T2, T3 & T4 Multiplexing | Data Rate, Chennels, & Applications](https://www.youtube.com/watch?v=huH4sxrkuP8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/telecommunications/short-note-digital-hierarchies - Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-3 ## Question Write a short note on hierarchies in digital transmission system. ## Short answer Digital hierarchies multiplex 64 kbps channels into higher-rate streams. PDH uses plesiochronous clocks and bit stuffing, with rates such as E1 (2.048 Mbps) and T1 (1.544 Mbps), but requires full demultiplexing to access a single channel. SDH/SONET uses synchronous multiplexing and pointers, enabling direct add/drop; typical STM-1 rate is 155.52 Mbps. ## Answer A **digital transmission hierarchy** is a standard set of bit rates in which lower-rate digital signals are multiplexed step by step into higher-rate signals, starting from the 64 kbps PCM voice channel (8000 samples/s × 8 bits). Two generations exist: **PDH** and **SDH/SONET**. #### Plesiochronous Digital Hierarchy (PDH) Tributaries at each level have their own clocks, which are almost but not exactly equal (plesiochronous). Higher-order multiplexers use **bit interleaving with pulse stuffing** to absorb rate differences, so each level's rate is slightly more than the sum of its inputs. | Level | North America (T/DS) | Europe (E) | Japan (J) | |---|---|---|---| | 0 | 64 kbps | 64 kbps | 64 kbps | | 1 | 1.544 Mbps (24 ch) | 2.048 Mbps (30 ch) | 1.544 Mbps (24 ch) | | 2 | 6.312 Mbps (96 ch) | 8.448 Mbps (120 ch) | 6.312 Mbps (96 ch) | | 3 | 44.736 Mbps (672 ch) | 34.368 Mbps (480 ch) | 32.064 Mbps (480 ch) | | 4 | 274.176 Mbps (4032 ch) | 139.264 Mbps (1920 ch) | 97.728 Mbps (1440 ch) | Bit rate examples: $T1 = (24 \times 8 + 1) \times 8000 = 1.544$ Mbps; $E1 = 32 \times 8 \times 8000 = 2.048$ Mbps. **Drawbacks of PDH:** to drop one 2 Mbps stream from a 140 Mbps line, the whole signal must be demultiplexed level by level ("multiplexer mountain"); the three regional standards do not interwork; little overhead for management and protection. #### Synchronous Digital Hierarchy (SDH) / SONET All equipment is locked to a common master clock, and byte-interleaved frames with **pointers** locate each tributary directly. | SDH | SONET | Rate | |---|---|---| | STM-1 | OC-3 | 155.52 Mbps | | STM-4 | OC-12 | 622.08 Mbps | | STM-16 | OC-48 | 2488.32 Mbps | | STM-64 | OC-192 | 9953.28 Mbps | Benefits: a tributary (e.g. one E1 out of the 63 in an STM-1) can be added or dropped directly with an add-drop multiplexer; one world standard; rich overhead for OAM; ring protection; carries PDH signals and data. It is the backbone of present-day optical networks, usually over WDM. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does PDH use pulse stuffing when multiplexing to a higher level? A. Tributary clocks are only nearly equal, so rate differences must be absorbed B. To make the system fully synchronous C. To add pointers for add-drop D. To remove the 64 kbps base rate Answer: A. PDH is plesiochronous: clocks are almost but not exactly equal, so stuffing absorbs the difference. (Key point: PDH stuffing) 2. What is the main drawback of PDH when dropping one 2 Mbps stream from a 140 Mbps line? A. The line must be amplified B. The whole signal must be demultiplexed level by level C. Only North America can do it D. The stream must be converted to analogue Answer: B. PDH has no pointers, so access to a tributary needs full step-by-step demultiplexing. (Key point: Multiplexer mountain) 3. How is the E1 rate of 2.048 Mbps obtained? A. 24 x 8 x 8000 B. (24 x 8 + 1) x 8000 C. 32 x 8 x 8000 D. 30 x 8 x 8000 x 2 Answer: C. E1 has 32 time slots of 8 bits, 8000 frames per second: 32 x 8 x 8000 = 2.048 Mbps. (Key point: E1 bit rate) 4. What allows SDH to add or drop one tributary directly from STM-1? A. Bit interleaving with stuffing B. Separate clocks for each tributary C. Byte-interleaved frames with pointers and a common master clock D. Three regional standards Answer: C. Synchronous operation and pointers locate each tributary, so an add-drop multiplexer can access it directly. 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