--- title: "TDM frame, time slot and 24-channel system" subject: "Telecommunications" chapter: 3 key: "semester_8/telecommunications:tdm-frame-24-channel" source: https://www.bctengineer.com/q/semester_8/telecommunications/tdm-frame-24-channel site: BCT Engineering --- # TDM frame, time slot and 24-channel system Telecommunications · Chapter 3: Signal Multiplexing - Asked in: 2064 Poush (old course), 6+10 marks - Video: [DS1 (T1) Fundamentals](https://www.youtube.com/watch?v=eKMjMFbxx4E) — VeEX Inc. - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/telecommunications/tdm-frame-24-channel - Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-3 ## Question With an example, clarify the concept of frame and time slot in case of TDM (Time Division Multiplexing) system. With respect to first order PCM system, explain the mechanism involved in American 24 voice channel multiplex system. ## Short answer A frame repeats at the sampling rate and contains one time slot per channel plus sync. For 4 channels with 8-bit samples and 1 sync bit, 33 bits/frame give 264 kbps. The T1 system multiplexes 24 µ-law coded voice channels (8 bits each) plus one framing bit in 125 µs frames, achieving 1.544 Mbps with robbed-bit signalling in frames 6 and 12. ## Answer #### Frame and time slot in TDM In TDM several channels share one line in time. A **time slot** is the short interval allotted to one channel, in which its sample (or bits) is sent. A **frame** is one complete cycle in which every channel gets its time slot once, plus any synchronisation bits. The frame repeats at the sampling rate. ```text |<------------- frame (125 us) ------------->| | sync | slot 1 | slot 2 | slot 3 | slot 4 | sync | ... ch A ch B ch C ch D ``` **Example:** 4 voice channels A, B, C, D, each sampled at 8 kHz and coded with 8 bits, plus 1 sync bit per frame. $$ \begin{aligned} \text{Frame time} &= \frac{1}{8000} = 125\ \mu\text{s} \\ \text{Bits per frame} &= 4 \times 8 + 1 = 33\ \text{bits} \\ \text{Line rate} &= 33 \times 8000 = 264\ \text{kbps} \\ \text{Bit duration} &= \frac{1}{264\,000} \approx 3.79\ \mu\text{s} \\ \text{Slot duration} &= 8 \times 3.79 \approx 30.3\ \mu\text{s} \end{aligned} $$ So in each 125 µs frame, A's 8 bits fill slot 1, B's slot 2, and so on; the receiver uses the sync bit to know where slot 1 starts. Each channel gets its own slot in every frame, so it receives 8000 samples per second, enough to rebuild speech up to 4 kHz. Key points: - Slot position identifies the channel (synchronous TDM), so no addresses are needed. - Frame length is fixed by the sampling rate (125 µs for voice), not by the number of channels; more channels mean shorter slots and a higher line rate. - Frame alignment (sync) bits let the receiver find frame boundaries. - Several frames can form a **multiframe** for signalling. #### American 24-voice-channel PCM multiplex (first-order T1/DS1) This is the first-order PCM system of North America (Bell D-type channel bank). **1. Band limiting.** Each of the 24 voice inputs passes through an anti-aliasing filter (passband 300–3400 Hz). **2. Sampling and multiplexing.** An electronic commutator samples the 24 channels one after another, each at 8000 samples/s, producing an interleaved PAM stream of $24 \times 8000 = 192\,000$ samples/s. **3. Companding and coding.** A shared codec compresses each sample with the **µ-law** ($\mu = 255$) characteristic and codes it into an 8-bit word (sign bit + 7 bits for segment and step), giving 256 levels. Companding keeps the signal-to-quantisation-noise ratio nearly constant for weak and strong talkers. **4. Frame formation.** The 24 words (192 bits) are preceded by **one framing bit (F)**. ```text 1 bit 8 bits 8 bits 8 bits +-----+-------+-------+-----+--------+ | F | Ch1 | Ch2 | ... | Ch24 | +-----+-------+-------+-----+--------+ <--------- 193 bits in 125 us --------> F = framing bit ``` $$ \begin{aligned} \text{Bits per frame} &= 24 \times 8 + 1 = 193\ \text{bits} \\ \text{Frame rate} &= 8000\ \text{frames/s}\ (T = 125\ \mu\text{s}) \\ R_{T1} &= 193 \times 8000 = 1.544\ \text{Mbps} \end{aligned} $$ **5. Superframe and framing pattern.** Twelve frames form a **superframe** (D4). The F-bits of odd frames carry the terminal framing pattern 101010 and those of even frames the signal framing pattern 001110, together 100011011100. The receiver searches for this pattern to find frame and superframe boundaries. (The later Extended Superframe, ESF, uses 24 frames with F-bits for framing, a CRC-6 and a 4 kbps data link.) **6. Signalling (robbed bit).** In frames 6 and 12 the least significant bit of every channel is replaced by signalling bits A and B (on-hook/off-hook, dial pulses). Voice thus gets 8 bits in 10 frames and 7 bits in 2 frames; the loss in quality is small. Signalling rate per channel: $$ \frac{8000}{6} \approx 1333\ \text{bps} \ (\text{A and B bits together}) $$ **7. Line coding and transmission.** The 1.544 Mbps stream is converted to **bipolar AMI** (with **B8ZS** to break long runs of zeros) and sent over two twisted pairs (four-wire), with regenerative repeaters about every 6000 ft (1.8 km). **8. Reception.** The receiver regenerates the pulses, recovers the bit clock from the transitions, locks to the framing pattern, extracts the signalling bits, distributes each 8-bit word to its channel, decodes and expands it (µ-law), and passes each channel through a reconstruction LPF to recover speech. ```text TRANSMIT ch1..ch24 -[LPF]-[sampler/commutator]-[u-law codec] -[add F bit, robbed-bit sig.]-[AMI/B8ZS]-> line RECEIVE line -[regenerator]-[clock rec.]-[frame sync] -[codec decode]-[distributor]-[LPF]-> ch1..ch24 ``` **Summary of figures** | Quantity | Value | |---|---| | Channels | 24 | | Sampling rate | 8 kHz | | Bits/sample | 8 (µ-law) | | Bits/frame | 193 | | Frame time | 125 µs | | Line rate | 1.544 Mbps | | Overhead | 8 kbps | | Superframe | 12 frames (1.5 ms) | T1 lines are then multiplexed to T2 (4 × T1, 6.312 Mbps) and T3 (28 × T1, 44.736 Mbps). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a frame in TDM? A. The interval for one channel's sample B. One complete cycle in which every channel gets its slot once, plus sync bits C. A group of 12 channels D. The guard band between channels Answer: B. A slot is one channel's interval; a frame is one full cycle of all slots plus sync, repeating at the sampling rate. (Key point: Frame vs slot) 2. What fixes the frame length of 125 µs? A. The number of channels B. The sampling rate (8 kHz) C. The bit rate of the line D. The coding law Answer: B. Frame length is set by the sampling rate; more channels give shorter slots and a higher line rate, not a longer frame. (Key point: Frame length) 3. How is signalling carried in the T1 D4 system? A. A separate time slot TS16 B. In frames 6 and 12 the LSB of each channel is replaced by signalling bits A and B C. By extra framing bits in every frame D. By a separate wavelength Answer: B. Robbed-bit signalling steals the LSB in frames 6 and 12, so voice has 7 bits in those frames, with little quality loss. (Key point: Robbed-bit signalling) 4. What is the purpose of mu-law companding in the T1 coder? A. To reduce the number of channels B. To keep the signal-to-quantisation-noise ratio nearly constant for weak and strong talkers C. To remove DC D. To generate the framing pattern Answer: B. Compressing the sample before 8-bit coding gives similar SQNR for weak and loud speech. (Key point: Companding in T1) ## Related in Telecommunications - [Analog vs digital multiplexing; 24-channel system](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aanalog-digital-mux-24-channel) — Telecommunications, chapter 3 - [Pulse stuffing and complete TDM PCM system](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Amultiplexing-pulse-stuffing-tdm-pcm) — Telecommunications, chapter 3 - [Multiplexing techniques and TDM standards](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Amultiplexing-techniques-tdm-standards) — Telecommunications, chapter 3 - [Impairments to voice channel transmission](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Avoice-channel-impairments) — Telecommunications, chapter 2 ## Related in other subjects - [TDM of 24 voice signals: PAM bandwidth and PCM sampling rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-24-voice-pam-pcm) — Communication Systems, chapter 5 - [Time division multiplexing explained](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Atdm-explain) — Data Communication, chapter 6, asked 3× - [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 - [TDM vs FDM for voice; E1 and T1 hierarchies](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-fdm-voice-e1-t1) — Communication Systems, chapter 5