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TDM principle and T1 carrier

Short answer

TDM assigns each signal a time slot within a recurring frame; a commutator samples channels sequentially at ≥2fmax. The T1 system frames 24 voice channels (8 bits each) plus one framing bit every 125 µs, yielding 1.544 Mbps. Higher levels: T2 6.312 Mbps (4 T1s), T3 44.736 Mbps (7 T2s), T4 274.176 Mbps (6 T3s).

Explain the working principle of TDM. Describe T1 carrier system showing the frame structure and different multiplexing levels.

Working principle of TDM

In Time Division Multiplexing, several signals share one line by taking turns. Time is divided into frames, and each frame into time slots; each channel gets one slot per frame and uses the full line bandwidth during it.

 ch1 -[LPF]-o                     o-[LPF]- ch1
 ch2 -[LPF]-o \                 / o-[LPF]- ch2
              (rotating)  link (rotating)
 chN -[LPF]-o / switch ------> \  o-[LPF]- chN
            commutator     decommutator
              ^                    ^
              +-- clock / sync ----+
  1. Each signal is band-limited by an LPF.
  2. A commutator samples the channels in turn at fs≥2fmf_s \geq 2 f_m (8 kHz for voice), so one frame lasts 1/fs1/f_s = 125 µs.
  3. The samples are coded (PCM) and sent with framing bits.
  4. A decommutator at the receiver, kept in step by the framing pattern, routes each slot to its own output, where an LPF restores the signal.

T1 carrier system

Frame structure: 24 voice channels × 8 bits + 1 framing bit.

 1 bit  8 bits  8 bits        8 bits
+-----+-------+-------+-----+--------+
|  F  |  Ch1  |  Ch2  | ... |  Ch24  |
+-----+-------+-------+-----+--------+
<--------- 193 bits in 125 us -------->
F = framing bit
Bits per frame=24×8+1=193 bitsFrame rate=8000 frames/s (T=125 μs)RT1=193×8000=1.544 Mbps\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}
  • Framing bits over 12 frames (superframe) give the alignment pattern; robbed-bit signalling uses the LSB of each channel in frames 6 and 12.
  • µ-law coding, AMI/B8ZS line code, repeaters about every 1.8 km.

Multiplexing levels:

LevelMade fromVoice channelsBit rateOverhead
DS01 voice channel164 kbps–
DS1 (T1)24 DS0241.544 Mbps8 kbps (framing)
DS2 (T2)4 DS1966.312 Mbps136 kbps
DS3 (T3)7 DS267244.736 Mbps552 kbps
DS4 (T4)6 DS34032274.176 Mbps5.76 Mbps

T2=4×T1+136T2 = 4 \times T1 + 136 kbps overhead; T3=7×T2+552T3 = 7 \times T2 + 552 kbps; T4=6×T3+5.76T4 = 6 \times T3 + 5.76 Mbps (framing and stuffing bits).

✓ Answer checked by Claude Opus 5.5 (high effort)

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