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TDM frame, time slot and 24-channel 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.

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.

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.

 |<------------- 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.

Frame time=18000=125 μsBits per frame=4×8+1=33 bitsLine rate=33×8000=264 kbpsBit duration=1264 000≈3.79 μsSlot duration=8×3.79≈30.3 μs\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 24×8000=192 00024 \times 8000 = 192\,000 samples/s.

3. Companding and coding. A shared codec compresses each sample with the µ-law (μ=255\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).

 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}

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:

80006≈1333 bps (A and B bits together)\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.

 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

QuantityValue
Channels24
Sampling rate8 kHz
Bits/sample8 (µ-law)
Bits/frame193
Frame time125 µs
Line rate1.544 Mbps
Overhead8 kbps
Superframe12 frames (1.5 ms)

T1 lines are then multiplexed to T2 (4 × T1, 6.312 Mbps) and T3 (28 × T1, 44.736 Mbps).

✓ Answer checked by Claude Opus 5.5 (high effort)

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