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Chapter 3 · 4 hours

Signal Multiplexing

IOE past exam questions

Past questions and answers

25 questions set from this chapter, 3 of them more than once. Most asked first.

  • Asked 2 times
  • 2081 Chaitra · 8 marks
  • 2073 Chaitra · 8 marks

Compare North American TDM and the European E1 TDM hierarchy by clearly indicating data rates and numbers of users with the help of proper diagrams.

Answer

The North American (T/DS) hierarchy (AT&T/ANSI) and the European E hierarchy (CEPT/ITU-T G.732/G.751) are the two plesiochronous digital hierarchies (PDH). Both sample each voice channel at 8 kHz with 8-bit PCM (64 kbps) and use a 125 µs frame, but they differ in frame size, signalling and the multiplexing steps.

North American T1 frame

24 channels × 8 bits + 1 framing bit per frame.

 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}

Signalling is "robbed bit": the least significant bit of each channel in every 6th frame (12-frame superframe D4, or 24-frame ESF).

European E1 frame

32 time slots × 8 bits; TS0 for frame alignment, TS16 for signalling (CAS) and TS1–15, TS17–31 for 30 voice channels.

+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... | TS15 | TS16 | ... | TS31 |
+-----+-----+-----+------+------+-----+------+
 sync  <-voice 1-15->  sig   <-voice 16-30->
<------ 32 x 8 = 256 bits in 125 us ------->
Bits per frame=32×8=256 bitsRE1=256×8000=2.048 Mbps\begin{aligned} \text{Bits per frame} &= 32 \times 8 = 256\ \text{bits} \\ R_{E1} &= 256 \times 8000 = 2.048\ \text{Mbps} \end{aligned}

North American hierarchy

24 x DS0 --> [MUX] --> T1  1.544 Mbps (24 ch)
 4 x T1  --> [M12] --> T2  6.312 Mbps (96 ch)
 7 x T2  --> [M23] --> T3  44.736 Mbps (672 ch)
 6 x T3  --> [M34] --> T4  274.176 Mbps (4032 ch)
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

European hierarchy

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)
LevelMade fromVoice channelsBit rateOverhead
E01 voice channel164 kbps–
E130 voice + 2 TS302.048 Mbps128 kbps (TS0, TS16)
E24 E11208.448 Mbps256 kbps
E34 E248034.368 Mbps576 kbps
E44 E31920139.264 Mbps1.792 Mbps
E54 E47680564.992 Mbps7.936 Mbps

Extra bits at each higher level carry frame alignment, stuffing (justification) and control, as the tributaries are not exactly synchronous.

Comparison

PointNorth American (T)European (E)
Primary rate1.544 Mbps2.048 Mbps
Channels at primary level2430 (+2 overhead slots)
Bits per frame193256
Framing1 bit per frame (F bit)Whole time slot TS0
SignallingRobbed bit (in voice LSB)Separate slot TS16
Companding lawµ-law (μ=255\mu = 255)A-law (A=87.6A = 87.6)
Multiplexing factor4, 7, 6 (irregular)4 at every level
Higher levels6.312, 44.736, 274.176 Mbps8.448, 34.368, 139.264, 564.992 Mbps
Channels per level96, 672, 4032120, 480, 1920, 7680
Line code (primary)AMI / B8ZSHDB3

The E system wastes no voice bits on signalling and has a regular ×4 structure; the T system gives slightly higher efficiency at the primary level (only 8 kbps overhead) but degrades voice LSBs. Interworking needs converters (e.g. µ-law ↔ A-law).

  • Asked 2 times
  • 2080 Chaitra · 4+2+2 marks
  • 2075 Asoj · 4+2+2 marks

What are the strength and weakness of TDM and statistical TDM? How statistical TDM recover weakness of TDM? Explain STDM with its frame format.

Answer

In synchronous TDM every input channel gets a fixed time slot in every frame whether or not it has data. In statistical (asynchronous) TDM slots are given only to inputs that currently have data, so the link capacity is shared on demand.

Strengths and weaknesses of synchronous TDM

Strengths

  • Simple multiplexer and demultiplexer; position of a slot identifies the channel, so no address overhead.
  • Fixed, constant delay; ideal for voice and other constant-bit-rate traffic.
  • Easy synchronisation and well-standardised (T1, E1).

Weaknesses

  • Wasted capacity: idle sources still get slots; for bursty data traffic most slots go empty.
  • Link rate must equal the sum of all input rates, even if most are idle.
  • Inflexible: number of channels and rates fixed.

Strengths and weaknesses of statistical TDM

Strengths

  • Much better link utilisation; the link rate can be less than the sum of input rates.
  • Supports more terminals on the same line for bursty data.
  • Flexible, can handle variable rates.

Weaknesses

  • Each slot must carry an address (and length), adding overhead.
  • Needs buffers; when many inputs are active at once, data is queued, so delay is variable and buffers may overflow.
  • More complex and costly multiplexer; not ideal for real-time voice.

How statistical TDM overcomes the weakness of TDM

The main weakness of TDM is that slots for idle inputs are wasted. A statistical TDM scans the input buffers and fills the frame only with data from active inputs, tagging each piece with its source address. Since on average only a fraction of terminals are active, a link of lower capacity serves the same terminals, and no capacity is wasted on silence.

 Inputs   A: data   B: idle   C: data   D: idle
 Sync TDM frame:   | A | B(empty) | C | D(empty) |
 Stat TDM frame:   | A,data | C,data |

STDM frame format

Statistical TDM usually uses an HDLC-type frame:

+----+------+-----+----------------------+-----+----+
|Flag| Addr | Ctl | sub-frames (data)    | FCS |Flag|
+----+------+-----+----------------------+-----+----+
                  |                      |
         +--------+                      +-----+
         | Addr | Len | Data | Addr | Len | Data |
         +------+-----+------+------+-----+------+
  • Flag: marks the start and end of frame (01111110).
  • Address / Control: link-level fields of the frame.
  • Sub-frames: each holds the source address, optional length field and data of one active input. With one source per frame, only address + data is needed; with many, address, length and data are repeated.
  • FCS: frame check sequence (CRC) for error detection.

Address and length fields are the price paid for the gain in efficiency.

  • Asked 2 times
  • 2078 Bhadra · 1+6 marks
  • 2074 Asoj · 2+6 marks

Define multiplexing. Compare FDM, TDM and WDM with neat diagram.

Answer

Multiplexing is the technique of sending several independent signals together over a single shared transmission medium, so that the high capacity of the medium is used efficiently and the cost per channel falls.

FDM (Frequency Division Multiplexing)

Each signal modulates a different carrier, so the channels sit side by side in frequency with guard bands between them; all channels are transmitted at the same time. Used for analogue telephony carrier, radio/TV broadcasting.

 amplitude
   |  [ch1]  [ch2]  [ch3]  [ch4]
   |  |   |g |   |g |   |g |   |
   +--+---+--+---+--+---+--+---+--> f
      f1     f2     f3     f4
 g = guard band

TDM (Time Division Multiplexing)

Each signal is sampled and given a short time slot in turn; all channels use the full bandwidth but at different times. Used in digital telephony (T1/E1), GSM.

 |<-------------- one frame ------------->|
 | S1 | S2 | S3 | ... | SN | S1 | S2 | ...
  slot slot              slot
 S_i = time slot of channel i

WDM (Wavelength Division Multiplexing)

Used in optical fibre: each signal is carried on a different wavelength (colour) of light; a prism-like multiplexer combines them and a demultiplexer separates them. It is FDM at optical frequencies.

 l1 -->|\                    /|--> l1
 l2 -->| MUX ==== fibre ==== DEMUX|--> l2
 l3 -->|/   (l1+l2+...+lN)   \|--> l3
 lN -->                         --> lN
 l = wavelength (lambda)

Comparison

PointFDMTDMWDM
Shared resourceFrequency bandTimeWavelength of light
Signal typeMainly analogueMainly digitalOptical (digital)
Channel separationGuard bandsGuard times / syncWavelength spacing
MediumCopper, coax, radioCopper, fibre, radioOptical fibre
Key devicesModulators, filtersCommutator, clock, buffersLasers, optical filters, prisms/gratings
SynchronisationNot neededEssentialNot needed between channels
Main impairmentCrosstalk from filter overlap, intermodulationTiming jitter, slipsChannel crosstalk, dispersion
Example12-ch group, FM radioT1 (24 ch), E1 (30 ch)DWDM 40–96 λ backbones
  • 2079 Chaitra · 2+6 marks

What is Time Division Multiplexing? Explain digital carrier standards T1 and E1 with proper elaboration.

Answer

Time Division Multiplexing (TDM) is a technique in which several signals share one transmission line by taking turns in time: each channel is given a short time slot in a repeating frame and uses the whole bandwidth of the line during its slot. For voice, each channel is sampled at 8 kHz and coded into 8-bit PCM words, so a frame lasts 125 µs.

T1 carrier (North American DS1)

  • 24 voice channels, each 8 bits per frame, plus 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: the F bits of successive frames form a fixed pattern (in the 12-frame D4 superframe, 100011011100) used to find frame boundaries.
  • Signalling: robbed-bit; in frames 6 and 12 the LSB of each channel carries signalling (A and B bits), so voice uses 7⅚ bits on average.
  • Companding: µ-law, μ=255\mu = 255.
  • Line code: bipolar AMI, with B8ZS to avoid long zero runs; 1.544 Mbps over twisted pair with repeaters about every 1.8 km.
  • Efficiency: 24×64=1.53624 \times 64 = 1.536 Mbps payload, only 8 kbps overhead.

E1 carrier (European CEPT-1)

  • 32 time slots of 8 bits each.
+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... | TS15 | TS16 | ... | TS31 |
+-----+-----+-----+------+------+-----+------+
 sync  <-voice 1-15->  sig   <-voice 16-30->
<------ 32 x 8 = 256 bits in 125 us ------->
Bits per frame=32×8=256 bitsRE1=256×8000=2.048 Mbps\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 alternate frames, alarms and spare bits in the others.
  • TS16: channel-associated signalling, shared among 30 channels over a 16-frame multiframe (4 bits per channel per multiframe), or used as a 64 kbps common-channel signalling link.
  • Voice: TS1–TS15 and TS17–TS31 = 30 channels.
  • Companding: A-law (A=87.6A = 87.6). Line code: HDB3.

Comparison summary

PointT1E1
Channels2430
Bits/frame193256
Rate1.544 Mbps2.048 Mbps
Framing1 bit/frameTS0
SignallingRobbed bitTS16
Law / line codeµ-law / AMI, B8ZSA-law / HDB3

Higher levels are formed by multiplexing: 4 T1 → T2 (6.312 Mbps), 7 T2 → T3 (44.736 Mbps); 4 E1 → E2 (8.448 Mbps), 4 E2 → E3 (34.368 Mbps).

  • 2081 Bhadra · 5+3 marks

Describe the role of Pulse Stuffing in maintaining synchronization in Time Division Multiplexing (TDM) systems. How does it ensure the accuracy of data transmission in digital carrier systems?

Answer

Pulse stuffing (bit stuffing or justification) is the addition of extra non-information bits to a digital tributary so that its rate exactly matches the slot rate offered by a higher-order TDM multiplexer. It is the way plesiochronous digital hierarchy (PDH) multiplexers handle inputs whose clocks are nominally equal but not exactly the same.

Why it is needed

When several tributaries (e.g. four E1 streams into an E2) come from different places, each has its own clock with a small tolerance (±50 ppm for E1). The multiplexer reads all tributaries at one common rate. If a tributary is a little slower than the read rate the multiplexer buffer runs empty (underflow); if it is faster it overflows. Either case causes slips – lost or repeated bits – and loss of frame alignment.

Role in maintaining synchronisation (positive stuffing)

  1. The multiplexer runs each tributary slot at a rate slightly higher than the maximum possible tributary rate. That is why the higher-order rate is more than the sum of the inputs (8.448 > 4 × 2.048 Mbps).
  2. Each input is written into an elastic buffer with its own clock and read out with the multiplexer clock.
  3. A phase comparator watches the buffer fill. When it is nearly empty, the multiplexer inserts a stuff (dummy) bit in a reserved position instead of a data bit.
  4. Stuffing control (justification control) bits in the frame tell the receiver whether the reserved position holds a stuff bit or data. They are sent several times (e.g. 3 times) and decoded by majority vote, so a single bit error does not cause a wrong decision.
  5. In effect each tributary gets a variable number of data bits per frame, matching its actual rate, while the aggregate stream has a fixed rate.
 tributary in (slightly slow)
 bits:  d d d d d d d d d d d ...
 MUX frame (fixed rate):
 | C | d d d d | C | d d d d | C | S/d d d d |
   C = stuffing control bits (repeated, e.g. 3)
   S/d = stuff bit if C=111, data bit if C=000

How it ensures accurate data transmission

  • At the receiver (demultiplexer): frame alignment locates the control bits; majority logic decides whether a stuff bit is present; stuff bits are removed, and the remaining data bits are written into a buffer.
  • A phase-locked loop smooths the gapped clock and regenerates the original tributary clock, so the tributary leaves the demultiplexer at exactly its original rate with no bits lost or added.
  • Thus no slips occur even though the tributary clocks differ; only a small waiting-time jitter remains, which the PLL keeps within limits.
  • Repetition of control bits protects against wrong de-stuffing, which would otherwise shift all following bits and corrupt the channel.

Other types: negative stuffing (sending an extra data bit in a spare position when the tributary is fast) and positive/zero/negative justification, as used in SDH pointers.

  • 2081 Baisakh · 3+2+3 marks

What is Time Division Multiplexing (TDM)? What are the types of TDM? Explain it. What is pulse stuffing and why it needed in TDM?

Answer

Time Division Multiplexing (TDM) is a technique in which several signals share one channel by occupying it in turn: time is divided into frames, and each frame into time slots, one slot per input. Each input uses the full channel bandwidth during its slot. Digital telephony uses PCM TDM: each voice channel is sampled 8000 times/s, so a frame is 125 µs long.

 |<-------------- one frame ------------->|
 | S1 | S2 | S3 | ... | SN | S1 | S2 | ...
  slot slot              slot
 S_i = time slot of channel i

Types of TDM

  1. Synchronous TDM: every input has a fixed slot in every frame, even if it has no data. The slot position identifies the channel. Simple and good for constant-rate voice (T1, E1), but slots of idle inputs are wasted.
  2. Statistical (asynchronous) TDM: slots are given only to inputs that have data, and each slot carries the source address. The link rate can be less than the sum of input rates; suits bursty data but needs buffers and address overhead.
  3. By the way the frame is built, TDM may also be:
    • Bit interleaved – one bit from each tributary in turn (used in PDH higher-order multiplexers).
    • Word (byte) interleaved – one 8-bit word from each channel in turn (used in T1/E1, SDH).

Example: in E1, 30 voice channels plus 2 control slots are byte interleaved: 32×8×8000=2.04832 \times 8 \times 8000 = 2.048 Mbps.

Pulse stuffing and why it is needed

Pulse stuffing (justification) is the insertion of extra dummy bits into a tributary stream so that its rate exactly matches the slot rate of a higher-order multiplexer.

Need:

  • Tributaries (e.g. four E1 signals into an E2) come from different sources with independent clocks, which differ slightly (plesiochronous).
  • The multiplexer reads them at one common rate. Without correction, buffers would underflow or overflow, causing slips (lost/repeated bits) and frame loss.
  • The multiplexer is therefore run slightly faster than any tributary; whenever a tributary's buffer is nearly empty, a stuff bit is inserted. Stuffing control bits, repeated three times and decoded by majority vote, tell the demultiplexer where stuff bits are, so it can remove them and recover the original clock with a PLL.
  • This is why higher PDH rates exceed the sum of inputs: 8.448>4×2.048=8.1928.448 > 4 \times 2.048 = 8.192 Mbps.
  • 2080 Bhadra · 3+5 marks

Difference between Analog and Digital Multiplexing. Explain the mechanism involved in American 24 voice channel multiplex system.

Answer

Analog versus digital multiplexing

PointAnalog multiplexing (FDM)Digital multiplexing (TDM/PCM)
PrincipleChannels separated in frequencyChannels separated in time slots
SignalContinuous analogueSampled, quantised, coded bits
EquipmentModulators, band-pass filters, oscillatorsSampler, coder (codec), clock, digital logic
NoiseAccumulates over every repeaterRegenerated at each repeater; no accumulation
CrosstalkFrom filter overlap, non-linearityVery low
SynchronisationNot neededNeeded (frame and bit sync)
Cost/sizeBulky filtersCheap, IC-based
Example12-channel group, 60-channel supergroupT1 (24 ch), E1 (30 ch)

American 24-voice-channel PCM multiplex (T1 / DS1)

Mechanism:

  1. Filtering: each voice signal is band-limited to 300–3400 Hz by a low-pass filter.
  2. Sampling: each channel is sampled at 8000 samples/s (> 2 × 3.4 kHz) by a PAM gate controlled by the commutator; samples of the 24 channels are interleaved in time.
  3. Compression and coding: samples are compressed by µ-law (μ=255\mu = 255) and coded into 8-bit words by a shared codec (256 levels).
  4. Framing: the 24 words (192 bits) plus one framing bit form a frame of 193 bits, repeated every 125 µs.
 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}
  1. Superframe and signalling: 12 frames form a superframe (D4); the F-bits form the pattern 100011011100 for frame and superframe alignment. In frames 6 and 12 the LSB of every channel is "robbed" to carry signalling bits A and B (on-hook/off-hook, dialling).
  2. Line coding: the stream is sent in bipolar AMI (later B8ZS) on twisted pair, with regenerative repeaters about every 6000 ft (1.8 km).
  3. Receiver: the decoder finds frame alignment from the F-bits, distributes each 8-bit word to its channel, decodes and expands it, and a low-pass filter recovers the voice.
ch1 -[LPF]-\                         /-[LPF]- ch1
ch2 -[LPF]-- sampler -> encoder ->   - decoder ...
ch24-[LPF]-/  (8 kHz)  (8 bit, u-law) \-[LPF]- ch24
                  + F bit -> AMI line 1.544 Mbps
  • 2080 Baisakh · 2+6 marks

Define multiplexing. Discuss about FDM and TDM with the help of neat block diagrams.

Answer

Multiplexing is the process of combining several independent message signals into one composite signal so that they can be sent over a single transmission medium; the reverse process at the receiver is demultiplexing. It saves cost because one high-capacity link replaces many separate links.

Frequency Division Multiplexing (FDM)

The available bandwidth is divided into frequency bands; each signal modulates its own carrier and occupies one band. All signals are sent at the same time.

 TRANSMITTER
 ch1 -[LPF]-[Mod f1]-[BPF]-+
 ch2 -[LPF]-[Mod f2]-[BPF]-+--(SUM)--> line
 chN -[LPF]-[Mod fN]-[BPF]-+

 RECEIVER
       +-[BPF f1]-[Demod f1]-[LPF]-> ch1
 line -+-[BPF f2]-[Demod f2]-[LPF]-> ch2
       +-[BPF fN]-[Demod fN]-[LPF]-> chN

Working:

  1. Each baseband signal is band-limited by an LPF (voice 0.3–3.4 kHz).
  2. It modulates a sub-carrier f1,f2,…,fNf_1, f_2, \ldots, f_N (SSB-SC in telephony to save bandwidth).
  3. A BPF keeps only the wanted sideband; channels are spaced 4 kHz apart, leaving guard bands.
  4. All outputs are summed and transmitted.
  5. At the receiver, BPFs separate each band, demodulators bring it back to baseband and LPFs recover each message.

Example: CCITT basic group – 12 voice channels in 60–108 kHz.

 amplitude
   |  [ch1]  [ch2]  [ch3]  [ch4]
   |  |   |g |   |g |   |g |   |
   +--+---+--+---+--+---+--+---+--> f
      f1     f2     f3     f4
 g = guard band

Time Division Multiplexing (TDM)

The time axis is divided into frames and each frame into time slots; each signal is sampled and sent in its own slot in turn, using the full bandwidth during that slot.

 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 ----+

Working:

  1. Each signal is band-limited by an LPF.
  2. A commutator (electronic rotating switch) samples channel 1, 2, …, N in turn at the sampling rate fs≥2fmf_s \geq 2f_m (8 kHz for voice).
  3. The interleaved samples (PAM) are usually quantised and coded (PCM) and sent with frame synchronisation bits.
  4. At the receiver a decommutator, synchronised to the transmitter by the clock and framing pattern, sends each sample to the correct output, where an LPF rebuilds the signal.
 |<-------------- one frame ------------->|
 | S1 | S2 | S3 | ... | SN | S1 | S2 | ...
  slot slot              slot
 S_i = time slot of channel i

Example: E1 – 32 slots × 8 bits × 8000 = 2.048 Mbps carrying 30 voice channels.

FDMTDM
Shares frequencyShares time
Analogue signalsMostly digital
Needs filters, guard bandsNeeds synchronisation, guard time
Crosstalk from non-linearityLittle crosstalk
  • 2079 Bhadra · 3+5 marks

What is wavelength division multiplexing? Describe T carrier system showing the frame structure of T1 level and different multiplexing levels with data rates.

Answer

Wavelength Division Multiplexing (WDM)

WDM is the multiplexing of several optical signals on one optical fibre by giving each a different wavelength (colour) of light. It is FDM in the optical domain.

 l1 -->|\                    /|--> l1
 l2 -->| MUX ==== fibre ==== DEMUX|--> l2
 l3 -->|/   (l1+l2+...+lN)   \|--> l3
 lN -->                         --> lN
 l = wavelength (lambda)
  • Each channel has its own laser at wavelength λ1,λ2,…,λN\lambda_1, \lambda_2, \ldots, \lambda_N (typically in the 1550 nm band).
  • A passive multiplexer (prism, diffraction grating or thin-film filter) combines them; at the far end a demultiplexer separates them. Erbium-doped fibre amplifiers boost all wavelengths together.
  • CWDM uses wide spacing (20 nm, up to 18 channels); DWDM uses close spacing (0.8 nm/100 GHz or less, 40–160 channels), giving terabits per second on one fibre.

T carrier system

The T carrier system is the North American digital TDM hierarchy (AT&T). The basic unit is a 64 kbps PCM voice channel (DS0): 8000 samples/s × 8 bits.

T1 frame structure: 24 channels × 8 bits + 1 framing bit = 193 bits per 125 µs frame.

 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 from 12 frames form a superframe pattern; signalling is carried by robbing the LSB of each channel in frames 6 and 12.

Multiplexing levels:

24 x DS0 --> [MUX] --> T1  1.544 Mbps (24 ch)
 4 x T1  --> [M12] --> T2  6.312 Mbps (96 ch)
 7 x T2  --> [M23] --> T3  44.736 Mbps (672 ch)
 6 x T3  --> [M34] --> T4  274.176 Mbps (4032 ch)
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
T2:4×1.544=6.176;  6.312−6.176=0.136 Mbps overheadT3:7×6.312=44.184;  44.736−44.184=0.552 Mbps overheadT4:6×44.736=268.416;  274.176−268.416=5.76 Mbps overhead\begin{aligned} T2 &: 4 \times 1.544 = 6.176;\ \ 6.312 - 6.176 = 0.136\ \text{Mbps overhead} \\ T3 &: 7 \times 6.312 = 44.184;\ \ 44.736 - 44.184 = 0.552\ \text{Mbps overhead} \\ T4 &: 6 \times 44.736 = 268.416;\ \ 274.176 - 268.416 = 5.76\ \text{Mbps overhead} \end{aligned}

The overhead at each higher level carries framing, pulse-stuffing (justification) control and alarm bits.

  • 2076 Chaitra · 7 marks

Define multiplexing and explain the working principle of FDM system with a neat block diagram.

Answer

Multiplexing is the technique of sending several independent signals simultaneously over one common transmission medium, so that the medium's capacity is shared and the cost per channel is reduced.

Principle of FDM

In Frequency Division Multiplexing, the total bandwidth of the medium is divided into non-overlapping frequency bands. Each message signal is shifted by modulation to its own band; all the shifted signals are added and sent at the same time. At the receiver, filters separate the bands and demodulators bring each one back to baseband.

Block diagram

 TRANSMITTER
 ch1 -[LPF]-[Mod f1]-[BPF]-+
 ch2 -[LPF]-[Mod f2]-[BPF]-+--(SUM)--> line
 chN -[LPF]-[Mod fN]-[BPF]-+

 RECEIVER
       +-[BPF f1]-[Demod f1]-[LPF]-> ch1
 line -+-[BPF f2]-[Demod f2]-[LPF]-> ch2
       +-[BPF fN]-[Demod fN]-[LPF]-> chN

Working

  1. Band limiting: each input (e.g. voice) passes through an LPF to limit it to 0.3–3.4 kHz.
  2. Modulation: each signal modulates a different sub-carrier f1,f2,…,fNf_1, f_2, \ldots, f_N. Telephone FDM uses SSB-SC so each voice channel needs only 4 kHz (3.1 kHz speech + guard band).
  3. Band-pass filtering: removes the unwanted sideband and carrier.
  4. Combining: a summing network adds all channels into a composite baseband signal, which may then modulate a final radio carrier.
  5. Receiver: BPFs centred on each band pick out each channel; product demodulators with locally generated carriers (same frequency and phase) recover the baseband, and LPFs remove the remaining high-frequency components.
 amplitude
   |  [ch1]  [ch2]  [ch3]  [ch4]
   |  |   |g |   |g |   |g |   |
   +--+---+--+---+--+---+--+---+--> f
      f1     f2     f3     f4
 g = guard band

Guard bands between channels allow for practical filter roll-off and reduce crosstalk.

Example – FDM telephone hierarchy (CCITT)

LevelMade fromChannelsBand
Basic group12 voice channels1260–108 kHz
Supergroup5 groups60312–552 kHz
Mastergroup5 supergroups300812–2044 kHz
Supermastergroup3 mastergroups9008516–12388 kHz

Other uses: AM/FM radio and TV broadcasting, cable TV, first-generation (AMPS) mobile.

  • 2076 Asoj · 6 marks

Clarify the North American (T1) and European (E1) frame structures along with their multiplexing hierarchy, and for each hierarchical level provide the bit rate calculation.

Answer

Both systems use 8-bit PCM samples taken 8000 times per second, so each frame is 125 µs long and each voice channel is 64 kbps.

North American T1 frame

24 channels × 8 bits + 1 framing (F) bit. Signalling is robbed from the LSB of each channel in frames 6 and 12 of a 12-frame superframe.

 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}

European E1 frame

32 time slots × 8 bits; TS0 = frame alignment, TS16 = signalling, 30 voice slots.

+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... | TS15 | TS16 | ... | TS31 |
+-----+-----+-----+------+------+-----+------+
 sync  <-voice 1-15->  sig   <-voice 16-30->
<------ 32 x 8 = 256 bits in 125 us ------->
Bits per frame=32×8=256 bitsRE1=256×8000=2.048 Mbps\begin{aligned} \text{Bits per frame} &= 32 \times 8 = 256\ \text{bits} \\ R_{E1} &= 256 \times 8000 = 2.048\ \text{Mbps} \end{aligned}

Multiplexing hierarchy and bit rates

North American:

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
T2:4×1.544=6.176;  6.312−6.176=0.136 Mbps overheadT3:7×6.312=44.184;  44.736−44.184=0.552 Mbps overheadT4:6×44.736=268.416;  274.176−268.416=5.76 Mbps overhead\begin{aligned} T2 &: 4 \times 1.544 = 6.176;\ \ 6.312 - 6.176 = 0.136\ \text{Mbps overhead} \\ T3 &: 7 \times 6.312 = 44.184;\ \ 44.736 - 44.184 = 0.552\ \text{Mbps overhead} \\ T4 &: 6 \times 44.736 = 268.416;\ \ 274.176 - 268.416 = 5.76\ \text{Mbps overhead} \end{aligned}

European:

LevelMade fromVoice channelsBit rateOverhead
E01 voice channel164 kbps–
E130 voice + 2 TS302.048 Mbps128 kbps (TS0, TS16)
E24 E11208.448 Mbps256 kbps
E34 E248034.368 Mbps576 kbps
E44 E31920139.264 Mbps1.792 Mbps
E54 E47680564.992 Mbps7.936 Mbps
E2:4×2.048=8.192;  8.448−8.192=0.256 Mbps overheadE3:4×8.448=33.792;  34.368−33.792=0.576 Mbps overheadE4:4×34.368=137.472;  139.264−137.472=1.792 Mbps overheadE5:4×139.264=557.056;  564.992−557.056=7.936 Mbps overhead\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}

The extra bits at each higher level are used for frame alignment, pulse stuffing (justification) control and alarms, because the tributaries are plesiochronous.

  • 2075 Chaitra · 4+4 marks

What are the different multiplexing techniques used in telecommunication? Explain briefly. Also discuss the North American and European standard of TDM with their hierarchies and data rates.

Answer

Multiplexing is combining several signals for transmission over one shared medium.

Multiplexing techniques

  1. Frequency Division Multiplexing (FDM): each signal is placed in its own frequency band by modulation; guard bands separate the bands; all channels are sent together. Used in analogue carrier telephony, radio and TV broadcasting.
  2. Time Division Multiplexing (TDM): each channel is given a time slot in a repeating frame and uses the full bandwidth during its slot. Synchronous TDM (fixed slots, e.g. T1/E1) and statistical TDM (slots on demand, with addresses).
  3. Wavelength Division Multiplexing (WDM): in optical fibre, each signal is carried on a different wavelength of light (CWDM, DWDM).
  4. Code Division Multiplexing (CDM/CDMA): all users share the same band at the same time, each spread by a unique orthogonal code; used in 3G mobile and GPS.
  5. Space Division Multiplexing: separate physical paths (pairs in a cable, fibres, beams).

North American TDM standard

Basic frame: 24 channels × 8 bits + 1 framing bit = 193 bits per 125 µs.

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}
24 x DS0 --> [MUX] --> T1  1.544 Mbps (24 ch)
 4 x T1  --> [M12] --> T2  6.312 Mbps (96 ch)
 7 x T2  --> [M23] --> T3  44.736 Mbps (672 ch)
 6 x T3  --> [M34] --> T4  274.176 Mbps (4032 ch)
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

European TDM standard

Basic frame: 32 time slots × 8 bits (TS0 sync, TS16 signalling, 30 voice).

Bits per frame=32×8=256 bitsRE1=256×8000=2.048 Mbps\begin{aligned} \text{Bits per frame} &= 32 \times 8 = 256\ \text{bits} \\ R_{E1} &= 256 \times 8000 = 2.048\ \text{Mbps} \end{aligned}
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)
LevelMade fromVoice channelsBit rateOverhead
E01 voice channel164 kbps–
E130 voice + 2 TS302.048 Mbps128 kbps (TS0, TS16)
E24 E11208.448 Mbps256 kbps
E34 E248034.368 Mbps576 kbps
E44 E31920139.264 Mbps1.792 Mbps
E54 E47680564.992 Mbps7.936 Mbps

The European system uses a regular factor of 4 at every level and keeps signalling in a separate slot; the American system uses robbed-bit signalling and factors 4, 7 and 6.

  • 2073 Shrawan · 4+4 marks

What is wave length division multiplexing? Explain its light sources characteristics and differentiate between them.

Answer

Wavelength Division Multiplexing (WDM) is the technique of sending several optical signals at the same time over one fibre, each on a different wavelength (colour) of light. It is FDM at optical frequencies and multiplies the capacity of an installed fibre.

 l1 -->|\                    /|--> l1
 l2 -->| MUX ==== fibre ==== DEMUX|--> l2
 l3 -->|/   (l1+l2+...+lN)   \|--> l3
 lN -->                         --> lN
 l = wavelength (lambda)

Each channel drives its own source at λ1,λ2,…\lambda_1, \lambda_2, \ldots; a passive optical multiplexer (grating, prism or thin-film filter) combines them and a demultiplexer separates them. CWDM uses 20 nm spacing; DWDM uses 0.8 nm (100 GHz) or closer, with EDFAs amplifying all channels together.

Light sources and their characteristics

Two semiconductor sources are used: the LED (light-emitting diode) and the LD (injection laser diode). A good optical source must:

  • emit at a wavelength where fibre loss and dispersion are low (850, 1310, 1550 nm);
  • have a narrow spectral width, so dispersion is low and channels can be closely packed in WDM;
  • couple enough power into the small fibre core;
  • be easy to modulate at high speed with a linear light–current characteristic;
  • be small, reliable, long-lived and stable with temperature.

LED: works by spontaneous emission when electrons and holes recombine in a forward-biased p-n junction. Output is incoherent with a wide spectrum (30–60 nm) and wide beam, so coupling to fibre is poor. It is cheap, simple and reliable, and is used with multimode fibre over short distances.

Laser diode: works by stimulated emission in a p-n junction with an optical cavity. Above the threshold current the output rises steeply and is coherent and narrow (1–3 nm; less than 0.1 nm for DFB lasers). It couples much more power into single-mode fibre and can be modulated at many Gbps, so it is the source used for DWDM and long-haul links.

 Optical
 power       LD /
   |           /
   |          /      LED
   |         /   .-------
   |        / .-'
   |    _.-/'
   +--------+-----------> current
          I_th (threshold of LD)

Difference between LED and laser diode

PointLEDLaser diode
EmissionSpontaneousStimulated
LightIncoherentCoherent
Spectral widthWide, 30–60 nmNarrow, 1–3 nm (DFB < 0.1 nm)
Output powerLow (µW to ~1 mW into fibre)High (several mW)
BeamWide angle, poor couplingNarrow, good coupling
Modulation speedUp to ~100–200 MbpsSeveral Gbps
ThresholdNone, linearThreshold current needed
Temperature sensitivityLowHigh, needs control
Cost, lifeCheap, long lifeCostly, shorter life
Fibre / useMultimode, short LAN linksSingle mode, long-haul, DWDM

For WDM, laser diodes (especially DFB lasers with temperature control) are needed because only they give the narrow, stable wavelengths that allow many closely spaced channels.

  • 2072 Chaitra · 2+2+4 marks

What is multiplexing? Why pulse stuffing is needed? Explain TDM of analog and digital sources and then show complete TDM PCM system with data rates.

Answer

Multiplexing is combining several signals into one composite signal for transmission over a single shared channel, and separating them again at the receiver.

Why pulse stuffing is needed

When digital tributaries with independent clocks (e.g. four E1 streams entering an E2 multiplexer) are combined, their bit rates differ slightly. The multiplexer reads them at a fixed, slightly higher rate; without correction its buffers would empty or overflow, causing slips (lost or repeated bits). Pulse stuffing inserts dummy bits whenever a tributary's buffer runs low; stuffing control bits (repeated and decided by majority vote) tell the demultiplexer which bits to remove, so the original stream and clock are recovered exactly. This is why E2=8.448>4×2.048E2 = 8.448 > 4 \times 2.048 Mbps.

TDM of analog sources

Analog signals are band-limited, then sampled in turn by a rotating switch (commutator) to give an interleaved PAM signal; each sample is usually quantised and coded (PCM).

  • Example: 4 voice channels, each limited to 3.4 kHz and sampled at 8 kHz, give 4×8000=32 0004 \times 8000 = 32\,000 samples/s on the line. With 8-bit coding: 32 000×8=25632\,000 \times 8 = 256 kbps.
  • If sources have different bandwidths, the commutator samples the wider-band source more often (sub-commutation / super-commutation), keeping each rate a multiple of the frame rate.

TDM of digital sources

Digital sources are already in bits, so they are interleaved directly, bit by bit or word by word.

  • If all sources have the same rate, each gets one slot per frame: 4 sources of 64 kbps → 256 kbps.
  • If rates differ but are multiples, faster sources get more slots per frame. Example: sources of 64, 64 and 128 kbps → frame with slots A, B, C, C → 256 kbps.
  • If rates are not exact multiples or not synchronous, pulse stuffing is used to bring each to its slot rate.

Complete TDM–PCM system with data rates (T1, 24 channels)

 ch1 -[LPF]-\
 ch2 -[LPF]--[sampler 8 kHz]--[u-law 8-bit coder]
 ch24-[LPF]-/  (commutator)            |
                      [add F bit, AMI/B8ZS]
                                       |
              line 1.544 Mbps, regenerative repeaters
                                       |
 ch1..ch24 <-[LPF]<-[distributor]<-[decoder]<-[frame sync]

Data rates at each stage:

Samples per channel=8000 s−1Channel rate=8000×8=64 kbpsFrame=24×8+1=193 bits per 125 μsLine rate=193×8000=1.544 Mbps\begin{aligned} \text{Samples per channel} &= 8000\ \text{s}^{-1} \\ \text{Channel rate} &= 8000 \times 8 = 64\ \text{kbps} \\ \text{Frame} &= 24 \times 8 + 1 = 193\ \text{bits per } 125\ \mu\text{s} \\ \text{Line rate} &= 193 \times 8000 = 1.544\ \text{Mbps} \end{aligned}

The receiver locks onto the framing bits, splits the 8-bit words to the 24 channels, decodes and expands them, and low-pass filters to recover speech. Four such T1 streams, with pulse stuffing, form a T2 at 6.312 Mbps.

  • 2072 Kartik · 6+4 marks

What is digital carrier system? Discuss the advantages and disadvantages of various multiplexing techniques.

Answer

A digital carrier system is a transmission system that carries many telephone (or data) channels in digital form over one medium using PCM and time division multiplexing, with regenerative repeaters along the line. Each voice channel is sampled 8000 times per second and coded in 8 bits (64 kbps); channels are interleaved into frames of 125 µs.

Main parts

 voice -> [codec: sample, quantise, code] -> [TDM mux]
       -> [line coder] -> ~[regen. repeater]~
       -> [line decoder] -> [demux] -> [codec] -> voice
  • Primary carriers: T1 (24 ch, 1.544 Mbps, North America/Japan) and E1 (30 ch, 2.048 Mbps, Europe/Nepal).
  • Higher orders: T2/T3 (6.312, 44.736 Mbps), E2/E3/E4 (8.448, 34.368, 139.264 Mbps), then SDH (STM-1 155.52 Mbps and above) over fibre and microwave.
  • Uses line codes (AMI, B8ZS, HDB3) and frame/signalling bits.

Advantages: noise does not accumulate (regeneration), easy integration with digital switching, cheap ICs, easy encryption and error monitoring. Disadvantages: needs more bandwidth than analogue, and needs strict synchronisation.

Advantages and disadvantages of multiplexing techniques

1. Frequency Division Multiplexing (FDM)

  • Advantages: simple for analogue signals; no synchronisation; continuous transmission for all users; well suited to radio/TV broadcast.
  • Disadvantages: bulky, costly filters and oscillators; guard bands waste bandwidth; crosstalk and intermodulation from non-linear amplifiers; noise accumulates in analogue repeaters.

2. Synchronous Time Division Multiplexing (TDM)

  • Advantages: digital, IC based and cheap; little crosstalk; regenerative repeaters; easily mixed voice and data; fits digital switching (T1/E1).
  • Disadvantages: needs accurate bit and frame synchronisation; slots wasted when sources are idle; higher bandwidth than analogue FDM for voice.

3. Statistical TDM

  • Advantages: slots only for active sources, so better link utilisation; more terminals per line.
  • Disadvantages: address/length overhead; buffering, variable delay and possible overflow; complex.

4. Wavelength Division Multiplexing (WDM)

  • Advantages: huge capacity on existing fibre; protocol and bit-rate transparent; easy upgrade by adding wavelengths.
  • Disadvantages: costly stable lasers and optical filters; channel crosstalk, dispersion and non-linear effects; only for fibre.

5. Code Division Multiplexing (CDM/CDMA)

  • Advantages: all users share the whole band at all times; resistant to interference and jamming; soft capacity; secure.
  • Disadvantages: needs power control (near–far problem); complex receivers; capacity limited by mutual interference.
TechniqueMain gainMain cost
FDMNo sync, analogueFilters, guard bands
TDMDigital, cheap ICsSync, idle slots
STDMEfficiency for bursty dataOverhead, delay
WDMVery high capacityOptical components
CDMRobust, flexiblePower control, complexity
  • 2071 Shrawan · 2+5 marks

Why multiplexing is needed in telecommunication? Describe wavelength division multiplexing in brief.

Answer

Why multiplexing is needed

Multiplexing lets many signals share one transmission medium. It is needed because:

  • A transmission medium (coax, fibre, microwave) has far more capacity than one voice channel needs; multiplexing uses that capacity fully.
  • It greatly reduces the cost per channel: one cable, fibre or radio link replaces hundreds of separate lines.
  • Fewer physical lines mean less right-of-way, duct space and maintenance.
  • Radio spectrum is limited; multiplexing lets many users share it.
  • It makes it possible to build trunk networks and hierarchies (T1/E1, SDH) between exchanges.

Wavelength Division Multiplexing

WDM carries several optical signals over a single fibre, each on a different wavelength of light. It is the optical equivalent of FDM.

 l1 -->|\                    /|--> l1
 l2 -->| MUX ==== fibre ==== DEMUX|--> l2
 l3 -->|/   (l1+l2+...+lN)   \|--> l3
 lN -->                         --> lN
 l = wavelength (lambda)

Working:

  1. Each input signal modulates its own laser at a different wavelength λ1,λ2,…,λN\lambda_1, \lambda_2, \ldots, \lambda_N (in the 1310/1550 nm windows).
  2. A passive optical multiplexer (diffraction grating, prism, arrayed waveguide grating or thin-film filter) combines all wavelengths into one fibre.
  3. Optical amplifiers (EDFA) can boost all wavelengths together on long routes.
  4. At the far end a demultiplexer separates the wavelengths and photodetectors convert each to an electrical signal.

Types:

  • CWDM (coarse): 20 nm spacing, up to 18 channels, cheap uncooled lasers, metro distances.
  • DWDM (dense): 0.8 nm (100 GHz) or less, 40–160 channels, long-haul backbones, terabits per second per fibre.

Advantages: huge increase in capacity of existing fibre; each wavelength is independent and can carry any format and bit rate; easy upgrades; bidirectional use of one fibre is possible. Limitations: costly, stable lasers and filters; crosstalk between adjacent wavelengths; dispersion and fibre non-linearities at high power.

  • 2071 Chaitra · 10 marks

Compare and contrast among various multiplexing techniques used in telecommunication.

Answer

Multiplexing is the sharing of one transmission medium among several signals. The main techniques used in telecommunication are FDM, TDM (synchronous and statistical), WDM and CDM.

Frequency Division Multiplexing (FDM)

Each signal modulates a different carrier and occupies its own band, with guard bands between bands; all signals are sent at the same time. Classic use: analogue telephone carrier (12-channel group at 60–108 kHz, 60-channel supergroup), radio and TV broadcasting.

 amplitude
   |  [ch1]  [ch2]  [ch3]  [ch4]
   |  |   |g |   |g |   |g |   |
   +--+---+--+---+--+---+--+---+--> f
      f1     f2     f3     f4
 g = guard band

Time Division Multiplexing (TDM)

Each signal gets a time slot in a repeating frame and uses the full bandwidth in its slot. Synchronous TDM gives fixed slots (T1 = 24 ch at 1.544 Mbps, E1 = 30 ch at 2.048 Mbps); statistical TDM gives slots only to active sources and tags them with addresses.

 |<-------------- one frame ------------->|
 | S1 | S2 | S3 | ... | SN | S1 | S2 | ...
  slot slot              slot
 S_i = time slot of channel i

Wavelength Division Multiplexing (WDM)

Several optical carriers of different wavelengths share one fibre; combined and separated by passive optical devices. CWDM and DWDM give from a few to over 100 channels of 10–100 Gbps each.

Code Division Multiplexing (CDM)

All users transmit at the same time in the same band, each multiplied by a unique spreading code; the receiver correlates with the wanted code to extract its signal. Used in CDMA/3G mobile and GPS.

          freq
 FDM:  ch3 |=========|      TDM: | 1 | 2 | 3 | 1 | 2 |
       ch2 |=========|            ------ time ------>
       ch1 |=========|
           +--- time -->     CDM: all users, all time,
                                  all freq, diff. codes

Comparison

PointFDMSync TDMStat TDMWDMCDM
Shared byFrequencyTime slotTime slot on demandWavelengthCode
SignalAnalogueDigitalDigital (data)OpticalDigital
SeparationGuard bands, filtersFrame syncAddress in slotOptical filtersOrthogonal codes
Sync neededNoYesYesNoYes (code sync)
Efficiency with bursty dataPoorPoor (idle slots)GoodPoor per λGood
Main impairmentCrosstalk, intermod.Jitter, slipsBuffer delay, overflowDispersion, λ crosstalkMulti-user interference
HardwareModulators, filtersCodec, clock, logicBuffers, processorLasers, gratings, EDFASpreading/correlators
MediumCopper, coax, radioCopper, fibre, radioData linksFibre onlyRadio
Capacity example12/60/300/900 ch groups24, 30, 672, 1920 chMany terminals40–160 λ × 10–100 GbpsDozens of users per carrier
Typical useOld carrier, broadcastPCM trunks, PDH/SDHTerminal concentratorsOptical backbone3G mobile, GPS

Summary

  • FDM is simplest for analogue signals but suffers from noise build-up and costly filters; it has been largely replaced.
  • TDM fits digital switching and transmission and is the basis of T1/E1, PDH and SDH.
  • Statistical TDM is best for bursty data and leads to packet switching.
  • WDM gives the highest capacity on fibre and sits under SDH/OTN backbones.
  • CDM gives flexible, interference-resistant sharing of radio spectrum. Modern networks combine them, e.g. TDM (SDH) channels carried on WDM wavelengths.
  • 2070 Asar · 7 marks

Explain the European TDM system used in telecommunication system.

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=87.6A = 87.6).
  • One frame = 32 time slots × 8 bits = 256 bits in 125 µs.
+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... | TS15 | TS16 | ... | TS31 |
+-----+-----+-----+------+------+-----+------+
 sync  <-voice 1-15->  sig   <-voice 16-30->
<------ 32 x 8 = 256 bits in 125 us ------->
Bits per frame=32×8=256 bitsRE1=256×8000=2.048 Mbps\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.

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)
LevelMade fromVoice channelsBit rateOverhead
E01 voice channel164 kbps–
E130 voice + 2 TS302.048 Mbps128 kbps (TS0, TS16)
E24 E11208.448 Mbps256 kbps
E34 E248034.368 Mbps576 kbps
E44 E31920139.264 Mbps1.792 Mbps
E54 E47680564.992 Mbps7.936 Mbps
E2:4×2.048=8.192;  8.448−8.192=0.256 Mbps overheadE3:4×8.448=33.792;  34.368−33.792=0.576 Mbps overheadE4:4×34.368=137.472;  139.264−137.472=1.792 Mbps overheadE5:4×139.264=557.056;  564.992−557.056=7.936 Mbps overhead\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).

  • 2070 Chaitra · 3+5 marks

Explain FDM hierarchy. Describe T carrier system showing the frame structure of T1 level and different multiplexing levels with data rates.

Answer

FDM hierarchy

In analogue telephone networks voice channels (0.3–3.4 kHz, 4 kHz slots) are stacked in frequency in steps, by SSB modulation, to form larger and larger groups. The CCITT (ITU-T) hierarchy:

LevelMade fromVoice channelsFrequency bandBandwidth
Voice channel–10–4 kHz4 kHz
Basic group12 channels1260–108 kHz48 kHz
Basic supergroup5 groups60312–552 kHz240 kHz
Basic mastergroup5 supergroups300812–2044 kHz1232 kHz
Supermastergroup3 mastergroups9008516–12388 kHz3872 kHz

(The Bell/AT&T system uses a 600-channel mastergroup of 10 supergroups at 564–3084 kHz, and a 3600-channel jumbo group of 6 mastergroups.)

 12 x voice --[SSB mod]--> GROUP      (12 ch, 60-108 kHz)
  5 x group --[SSB mod]--> SUPERGROUP (60 ch, 312-552 kHz)
  5 x SG    --[SSB mod]--> MASTERGROUP(300 ch)
  3 x MG    --[SSB mod]--> SUPERMASTER(900 ch)

Standard building blocks let equipment from different makers interconnect, and whole groups can be routed together.

T carrier system

The T carrier is the North American digital TDM hierarchy. Each voice channel (DS0) is PCM coded at 8000×8=648000 \times 8 = 64 kbps with µ-law companding.

T1 frame structure:

 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}
  • 12 frames form a superframe (D4); the F-bits form an alignment pattern.
  • In frames 6 and 12 the LSB of every channel is robbed for signalling.
  • Line code: AMI/B8ZS on twisted pair.

Multiplexing levels:

24 x DS0 --> [MUX] --> T1  1.544 Mbps (24 ch)
 4 x T1  --> [M12] --> T2  6.312 Mbps (96 ch)
 7 x T2  --> [M23] --> T3  44.736 Mbps (672 ch)
 6 x T3  --> [M34] --> T4  274.176 Mbps (4032 ch)
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
T2:4×1.544=6.176;  6.312−6.176=0.136 Mbps overheadT3:7×6.312=44.184;  44.736−44.184=0.552 Mbps overheadT4:6×44.736=268.416;  274.176−268.416=5.76 Mbps overhead\begin{aligned} T2 &: 4 \times 1.544 = 6.176;\ \ 6.312 - 6.176 = 0.136\ \text{Mbps overhead} \\ T3 &: 7 \times 6.312 = 44.184;\ \ 44.736 - 44.184 = 0.552\ \text{Mbps overhead} \\ T4 &: 6 \times 44.736 = 268.416;\ \ 274.176 - 268.416 = 5.76\ \text{Mbps overhead} \end{aligned}

The extra overhead at T2, T3 and T4 carries framing, pulse-stuffing control and alarm bits.

  • 2069 Chaitra · 2+5 marks

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

Answer

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).

  • 2064 Poush (old course) · 6+10 marks

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.

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.

 |<------------- 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-channel bank).

1. Band limiting. Each of the 24 voice inputs passes an anti-aliasing LPF (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 followed/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 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 signalling bits, distributes each 8-bit word to its channel, decodes and expands (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).

  • 2065 Magh (old course) · 4+12 marks

Why line coding is required before the PCM signals to be connected to the digital line? Justify that Unipolar Non Return to Zero (NRZ), Return to Zero (RZ) and bipolar Alternate Mark Inversion (AMI) codes can not be used in line coding.

Answer

Why line coding is required

The binary output of a PCM coder (simple on–off pulses) is not suitable for sending directly on a cable. Line coding converts it into a waveform that suits the line and the regenerative repeaters. The line code must give:

  • No DC component: lines are coupled through transformers and capacitors (and carry DC power for repeaters), so the signal must have no DC and little low-frequency energy.
  • Timing (clock) content: regenerative repeaters extract the clock from the transitions of the received signal; there must be enough transitions whatever the data.
  • Small bandwidth: the spectrum should be narrow to reduce attenuation and crosstalk on the cable.
  • Error monitoring: the code should have redundancy so that errors can be detected in service (e.g. bipolar violations).
  • Transparency: any data pattern, including long runs of 0s or 1s, must pass without loss of timing.
  • Simple, low-cost coding and decoding.

Waveforms of the three codes

 data:     1   0   1   1   0   0   0   0
          +---+   +-------+
 NRZ-L    |   |   |       |
 (unipol) +   +---+       +---------------  0

          +-+     +-+ +-+
 RZ       | |     | | | |
 (unipol) + +-----+ +-+ +-----------------  0

          +---+       +---+
 AMI      |   |       |   |
       ---+   +---+   +   +---+------------ 0
                  |   |   (no pulses
                  +---+    for 0s)

1. Unipolar NRZ cannot be used

In unipolar NRZ, '1' = +V for the whole bit, '0' = 0 V.

  • Large DC component: the average value is V/2V/2 for random data (and VV for a run of 1s). Lines use transformers and series capacitors (and carry DC to power repeaters), which block DC. The DC part is lost, the baseline wanders, and the decision threshold shifts, causing errors.
  • No timing for long runs: a long run of 1s or 0s has no transitions at all, so the repeater's clock-recovery circuit loses lock, causing bit slips.
  • No discrete clock line in the spectrum: its power spectrum has no component at the bit rate fbf_b, so simple tuned-circuit clock extraction is not possible.
  • No error monitoring: any pattern is legal, so in-service errors cannot be detected.
  • Most energy is at low frequency, where crosstalk on pairs is high. Hence unipolar NRZ is unsuitable.

2. Unipolar RZ cannot be used

In RZ, '1' = +V for half a bit and returns to 0; '0' = 0 V.

  • Still has DC: the average is V/4V/4 for random data, so the same baseline wander problem.
  • Long runs of 0s give no transitions, so timing is lost (runs of 1s are fine).
  • Bandwidth doubles: pulses are half as wide, so the first spectral null is at 2fb2f_b; more attenuation and crosstalk, and repeaters must be closer.
  • Half the pulse energy is lost for the same peak voltage, lowering noise margin.
  • No error detection. Though RZ does contain a clock component at fbf_b, its DC content and bandwidth rule it out.

3. Bipolar AMI cannot be used (on its own)

In AMI, '0' = 0 V, and each '1' is a pulse of alternating polarity (+, −, +, …). It solves several problems:

  • No DC: positive and negative pulses balance, so the average is zero and the spectrum is zero at DC.
  • Narrow spectrum: most energy near fb/2f_b/2.
  • Error detection: two successive 1s of the same polarity (bipolar violation) show an error.

But it fails one requirement:

  • Long runs of zeros give no pulses at all. PCM data from an idle channel or from data terminals can contain many consecutive zeros. During such a run the repeaters receive nothing, so the timing-recovery circuit drifts and loses bit synchronisation. Every following bit may then be mis-read.
  • Therefore plain AMI is not transparent; early T1 systems had to force a 1 into all-zero words ("zero code suppression"), which corrupts data and limits users to 7 bits (56 kbps) per channel.

Example – data 1 0 1 1 0 0 0 0 0 0 0 0 1: AMI gives + 0 − + 0 0 0 0 0 0 0 0 −; the eight zeros produce 8 bit periods with no pulse, and the repeater clock has nothing to lock to.

Summary

RequirementUnipolar NRZUnipolar RZBipolar AMI
Zero DCNoNoYes
Timing in long 0 runsNoNoNo
Timing in long 1 runsNoYesYes
Bandwidthfbf_b (null)2fb2f_b (null)fbf_b (peak at fb/2f_b/2)
Error monitoringNoNoYes (violations)
Suitable alone?NoNoNo

What is used instead

Modified bipolar codes that keep AMI's benefits but replace long zero strings with patterns containing deliberate bipolar violations:

  • HDB3 (High Density Bipolar 3) in E1/E2/E3: no more than 3 consecutive zeros on the line.
  • B8ZS (Bipolar 8 Zero Substitution) in T1: eight zeros replaced by 000VB0VB. The receiver recognises the violation pattern and restores the zeros, so DC balance, timing and error monitoring are all achieved.
  • 2065 Baisakh (old course) · 2+1+8+5 marks

Why line coding is required in PCM (Pulse Code Modulation) system? Is it possible to implement NRZ (Not Return to Zero), RZ (Return to Zero) and AMI (Alternate Mark Inversion) coding techniques? If not, explain why it is not possible and which coding technique solves this problem?

Answer

Why line coding is required in PCM

The PCM coder gives a plain binary stream. Before it is sent on a cable through regenerative repeaters, it must be converted into a line code that matches the line. The code must have:

  • No DC component: lines are coupled through transformers and capacitors (and carry DC power for repeaters), so the signal must have no DC and little low-frequency energy.
  • Timing (clock) content: regenerative repeaters extract the clock from the transitions of the received signal; there must be enough transitions whatever the data.
  • Small bandwidth: the spectrum should be narrow to reduce attenuation and crosstalk on the cable.
  • Error monitoring: the code should have redundancy so that errors can be detected in service (e.g. bipolar violations).
  • Transparency: any data pattern, including long runs of 0s or 1s, must pass without loss of timing.
  • Simple, low-cost coding and decoding.

Is it possible to use NRZ, RZ and AMI?

No. None of the three meets all the requirements, so none is used on its own for PCM lines.

Why they are not possible

 data:     1   0   1   1   0   0   0   0
          +---+   +-------+
 NRZ-L    |   |   |       |
 (unipol) +   +---+       +---------------  0

          +-+     +-+ +-+
 RZ       | |     | | | |
 (unipol) + +-----+ +-+ +-----------------  0

          +---+       +---+
 AMI      |   |       |   |
       ---+   +---+   +   +---+------------ 0
                  |   |   (no pulses
                  +---+    for 0s)

Unipolar NRZ ('1' = +V for full bit, '0' = 0):

  • Average (DC) value V/2V/2; DC is blocked by line transformers and coupling capacitors, causing baseline wander and wrong decisions.
  • Long runs of 1s or 0s have no transitions, so repeaters cannot recover the clock.
  • No spectral line at fbf_b to extract timing; no error detection.

Unipolar RZ ('1' = +V for half bit):

  • Still has DC (average V/4V/4).
  • Long runs of 0s give no pulses, so timing is lost.
  • Pulses are half-width, so the bandwidth is double (2fb2f_b first null), causing more attenuation and crosstalk.

Bipolar AMI ('1' = alternate +/− pulses, '0' = 0):

  • Good: zero DC, narrow spectrum, simple error detection via bipolar violations.
  • Problem: a long run of 0s sends no pulses, so the repeater clock drifts and synchronisation is lost. PCM traffic (idle channels, data) often contains such runs, so AMI is not transparent.
CodeDC freeTiming in 0-runsBandwidthError check
NRZNoNofbf_bNo
RZNoNo2fb2f_bNo
AMIYesNofbf_bYes

Coding techniques that solve the problem

The solution is zero-substitution bipolar codes: AMI with long strings of zeros replaced by special patterns that contain deliberate bipolar violations (V). The receiver recognises a violation, knows it is a substitution and replaces the pattern by zeros. This keeps zero DC and error monitoring while guaranteeing pulses for timing.

1. HDB3 (High Density Bipolar of order 3) – used on E1, E2, E3 (ITU-T G.703). Rule: any string of four zeros is replaced by

  • 000V if the number of 1s (pulses) since the last substitution is odd;
  • B00V if it is even.

Here V is a pulse of the same polarity as the previous pulse (a violation), and B is a normal bipolar pulse (opposite to the previous pulse). This makes successive V pulses alternate, so there is no DC build-up. At most 3 zeros appear in a row.

Example (the first pulse is taken as +):

Data101100001000000001
AMI+0−+0000−00000000+
HDB3+0−+000+V−000−V+B00+V−
  • After 1 0 1 1 (three 1s, odd) the first 0000 becomes 000V, V = + (same as previous +).
  • After that, one 1 (odd) → 000V, with V = − (same as previous −).
  • Next 0000 follows with no 1 in between (even, zero) → B00V: B = + (opposite of −), V = + (same as B).
  • The V pulses (+, −, +) alternate, so the line stays DC balanced.

2. B8ZS (Bipolar with 8-Zero Substitution) – used on T1. Any string of eight zeros is replaced by 000VB0VB: if the previous pulse is +, the substitute is 000+−0−+; if it is −, it is 000−+0+−. Two violations in fixed positions identify the substitution.

Data1100000000101
AMI+−00000000+0−
B8ZS+−000−V+B0+V−B+0−

3. Other codes: B6ZS and B3ZS (T2, T3), CMI and Manchester (biphase) for some interfaces, 4B3T and 2B1Q for ISDN lines.

Result

HDB3/B8ZS give: no DC component, a guaranteed pulse at least every 4 (or 8) bits for clock recovery, bandwidth similar to AMI, in-service error detection (an unexpected violation means an error), and full transparency for 64 kbps clear channels.

  • 2065 Baisakh (old course) · 8 marks

Write a short note on hierarchies in digital transmission system.

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.

LevelNorth America (T/DS)Europe (E)Japan (J)
064 kbps64 kbps64 kbps
11.544 Mbps (24 ch)2.048 Mbps (30 ch)1.544 Mbps (24 ch)
26.312 Mbps (96 ch)8.448 Mbps (120 ch)6.312 Mbps (96 ch)
344.736 Mbps (672 ch)34.368 Mbps (480 ch)32.064 Mbps (480 ch)
4274.176 Mbps (4032 ch)139.264 Mbps (1920 ch)97.728 Mbps (1440 ch)

Bit rate examples: T1=(24×8+1)×8000=1.544T1 = (24 \times 8 + 1) \times 8000 = 1.544 Mbps; E1=32×8×8000=2.048E1 = 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.

SDHSONETRate
STM-1OC-3155.52 Mbps
STM-4OC-12622.08 Mbps
STM-16OC-482488.32 Mbps
STM-64OC-1929953.28 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.

  • 2064 Poush (old course) · 10 marks

Write a short note on line coding principles in PCM system.

Answer

Line coding is the conversion of the binary PCM bit stream into an electrical waveform (pulse pattern) suitable for transmission over the cable and through regenerative repeaters. A plain on–off signal has DC, loses timing on long runs and gives no error check, so a line code is always used.

Requirements of a good line code

  • No DC component: lines are coupled through transformers and capacitors (and carry DC power for repeaters), so the signal must have no DC and little low-frequency energy.
  • Timing (clock) content: regenerative repeaters extract the clock from the transitions of the received signal; there must be enough transitions whatever the data.
  • Small bandwidth: the spectrum should be narrow to reduce attenuation and crosstalk on the cable.
  • Error monitoring: the code should have redundancy so that errors can be detected in service (e.g. bipolar violations).
  • Transparency: any data pattern, including long runs of 0s or 1s, must pass without loss of timing.
  • Simple, low-cost coding and decoding.

Common line codes

 data:     1   0   1   1   0   0   0   0
          +---+   +-------+
 NRZ-L    |   |   |       |
 (unipol) +   +---+       +---------------  0

          +-+     +-+ +-+
 RZ       | |     | | | |
 (unipol) + +-----+ +-+ +-----------------  0

          +---+       +---+
 AMI      |   |       |   |
       ---+   +---+   +   +---+------------ 0
                  |   |   (no pulses
                  +---+    for 0s)
  1. Unipolar NRZ: 1 = +V, 0 = 0 for the full bit. Simple, but has large DC and no timing in long runs. Not used on lines.
  2. Polar NRZ: 1 = +V, 0 = −V. DC is zero only for balanced data; poor timing.
  3. Unipolar RZ: pulse for half a bit for 1s. Has DC, double bandwidth, timing lost on 0-runs.
  4. Manchester (biphase): a transition in the middle of every bit. Excellent timing and no DC, but needs twice the bandwidth; used in Ethernet, not on PCM trunks.
  5. Bipolar AMI: 0 = no pulse, 1s alternately + and −. No DC, bandwidth about fb/2f_b/2 centred, and bipolar violations reveal errors. Its weakness: long runs of zeros give no pulses, so repeaters lose timing.
  6. HDB3: AMI in which each string of four zeros is replaced by 000V or B00V (chosen so violations alternate). At most three zeros in a row. Standard for E1/E2/E3.
  7. B8ZS: AMI in which eight zeros are replaced by 000VB0VB. Standard for T1; B3ZS/B6ZS for higher T levels.
  8. CMI, 4B3T, 2B1Q: used on some interfaces and ISDN lines.

Example HDB3 coding (first pulse +):

Data101100001000000001
AMI+0−+0000−00000000+
HDB3+0−+000+V−000−V+B00+V−

Choice

CodeDC freeTimingBandwidthError checkUse
NRZNoPoorLowNoInside equipment
ManchesterYesExcellentHighPartlyLAN
AMIYesFails on 0-runsLowYesEarly T1
HDB3YesGoodLowYesE1–E3
B8ZSYesGoodLowYesT1

Thus PCM lines use bipolar codes with zero substitution (HDB3, B8ZS), which keep AMI's low bandwidth and zero DC while guaranteeing clock content.

Questions from Old Question Collection (BEI EX 756) (IOE BEI IV/II Telecommunication (EX 756) papers, 2079 to 2081), Old Question Collection (EX 703) (IOE BEX IV/I Telecommunication (EX 703) papers, 2069 to 2081) and Old Questions (EX 703 and earlier) (IOE EX 703 papers 2069-2075 and older-course BEX IV/II papers 2064-2069). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗