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Chapter 5 · 3 hours

Multiplexing Techniques

IOE past exam questions

Past questions and answers

29 questions set from this chapter, 8 of them more than once. Most asked first.

  • Asked 5 times
  • 2076 Baisakh (CS I) · 4 marks
  • 2076 Bhadra (CS I) · 4 marks
  • 2070 Bhadra (CS I) · 5 marks
  • 2067 Mangsir (CS I) · 5 marks
  • 2065 Kartik (CS I) · 4 marks

Write a short note on FDM in telephony.

Answer

FDM in telephony combines many voice channels onto one wideband link (coaxial cable, microwave) by giving each a separate frequency slot. It was the standard in analog long-distance telephone networks.

Principle

  • Each voice signal is band-limited to 300–3400 Hz and given a 4 kHz slot (the extra 900 Hz is a guard band).
  • Each channel SSB-SC modulates (usually LSB) a different subcarrier; SSB halves the bandwidth compared with DSB.
  • The outputs are added and sent together; at the receiver band-pass filters separate them and synchronous demodulators recover each voice.
ch1->[SSB mod f1]->[BPF]-+
ch2->[SSB mod f2]->[BPF]-+->(sum)-> line
 ...                     |
ch12->[SSB f12]->[BPF]---+

Hierarchy (ITU-T/CCITT)

LevelMade ofChannelsBand
Basic group12 voice1260–108 kHz
Supergroup5 groups60312–552 kHz
Mastergroup5 supergroups300812–2044 kHz
Supermastergroup3 mastergroups9008516–12388 kHz

In the basic group, carriers are 64, 68, ..., 108 kHz with LSB selection.

Requirements: sharp SSB filters (crystal/mechanical), stable carriers derived from one master oscillator, pilot tones for level and frequency control, and linear amplifiers to avoid intermodulation crosstalk.

FDM telephony has largely been replaced by digital TDM (E1/T1, SDH), but the principle remains in radio and cable systems.

  • Asked 4 times
  • 2080 Chaitra (CS I) · 5 marks
  • 2073 Magh (CS I) · 5 marks
  • 2073 Bhadra (CS I) · 5 marks
  • 2068 Bhadra (CS I) · 4 marks

Write a short note on filter and oscillator requirement in FDM.

Answer

In an FDM system the channels are separated only by frequency, so the quality of the filters and carrier oscillators decides how much crosstalk and distortion occur.

voice->[Mod]<-osc fc1  ->[BPF]-+
voice->[Mod]<-osc fc2  ->[BPF]-+-> composite
           (from one master)   |

Filter requirements

  • Sharp cut-off: SSB generation needs filters that pass one sideband and reject the other, which is only a few hundred hertz away (e.g. 300 Hz lower edge of voice). Very steep skirts are needed, so crystal, ceramic or mechanical filters with high Q are used.
  • Flat passband and low ripple across 300–3400 Hz so all speech frequencies have equal gain.
  • Linear phase / low group-delay distortion, especially for data and video channels.
  • High stop-band attenuation (about 60 dB) to keep adjacent-channel crosstalk low.
  • Guard bands (900 Hz per voice channel) relax the filter slope.
  • Stability with temperature and ageing.
  • Multi-stage modulation (group → supergroup → mastergroup) is used partly so that filters work at low frequencies where sharp filters are practical.

Oscillator requirements

  • High frequency stability and accuracy: in SSB-SC any carrier offset between transmitter and receiver shifts every voice frequency by the same amount; only a few hertz of error is tolerable for speech and less for data.
  • All carriers are therefore derived from a single crystal master oscillator by frequency multiplication/division (frequency synthesis), so they are locked to each other.
  • Low phase noise and harmonics, since spurious outputs cause interchannel interference.
  • Synchronization: a pilot carrier is sent so the receiver can lock its oscillators to the transmitter's.
  • Carriers must have stable amplitude for constant modulator gain.
  • Asked 4 times
  • 2079 Chaitra (CS I) · 4+6 marks
  • 2075 Baisakh (CS I) · 2+5 marks
  • 2069 Bhadra (CS I) · 4+6 marks
  • 2067 Shrawan (CS I) · 4+6 marks

Define FDM. Explain FDM hierarchy used in telephony.

Answer

FDM

Frequency Division Multiplexing (FDM) is a technique in which several signals share one channel by being placed in different, non-overlapping frequency bands, each modulating its own carrier. All signals are transmitted simultaneously; guard bands between them prevent interference, and band-pass filters separate them at the receiver.

 |ch1|g|ch2|g|ch3|g| ... |chN|   g = guard band
 f1      f2      f3          fN   --> frequency

FDM hierarchy in telephony

In analog telephony, each voice signal (300–3400 Hz) is allotted 4 kHz and translated by SSB-SC (LSB) modulation. Channels are combined step by step into larger blocks. The ITU-T (CCITT) hierarchy:

12 voice ch --> Group (60-108 kHz)
5 Groups    --> Supergroup (312-552 kHz)
5 Supergrps --> Mastergroup (812-2044 kHz)
3 Mastergrps--> Supermastergroup (8516-12388 kHz)

1. Basic group

  • 12 voice channels.
  • Carriers at 64, 68, 72, ..., 108 kHz; lower sidebands selected.
  • Occupies 60–108 kHz (48 kHz bandwidth).

2. Supergroup

  • 5 groups = 60 channels.
  • Each group modulates carriers 420, 468, 516, 564, 612 kHz (LSB).
  • Occupies 312–552 kHz (240 kHz).

3. Mastergroup

  • 5 supergroups = 300 channels (CCITT).
  • Occupies 812–2044 kHz, with 8 kHz guard bands between supergroups.

4. Supermastergroup

  • 3 mastergroups = 900 channels, occupying 8516–12388 kHz.
LevelCompositionChannelsBand
Group12 voice1260–108 kHz
Supergroup5 groups60312–552 kHz
Mastergroup5 supergroups300812–2044 kHz
Supermastergroup3 mastergroups9008.516–12.388 MHz

AT&T (North American) version: mastergroup = 10 supergroups = 600 channels (564–3084 kHz); jumbo group = 6 mastergroups = 3600 channels (0.564–17.548 MHz).

Why a hierarchy: the same standard equipment (channel banks, group and supergroup translators) is reused; filters at each stage work at practical frequencies; blocks of channels can be routed as a unit.

Requirements: sharp SSB filters, carriers derived from one master oscillator, pilot tones for gain/frequency control, linear wideband amplifiers to limit intermodulation.

  • Asked 2 times
  • 2074 Bhadra (CS I) · 4 marks
  • 2073 Bhadra (CS I) · 5 marks

Write a short note on FDMA in satellite communications.

Answer

FDMA (Frequency Division Multiple Access) in satellite communication lets many earth stations share one satellite transponder by giving each station its own frequency band within the transponder bandwidth. All stations transmit continuously and at the same time.

   Transponder (e.g. 36 MHz)
 |  ES-A  |g|  ES-B  |g|  ES-C  |g| ...
 <------------ frequency ------------>

Types

  • MCPC (Multiple Channels Per Carrier): each earth station FDM-multiplexes many voice channels onto one carrier (FDM/FM/FDMA). Pre-assigned, suited to heavy fixed routes.
  • SCPC (Single Channel Per Carrier): each voice/data channel has its own carrier. Can be demand assigned (DAMA), e.g. INTELSAT SPADE system: about 800 channels at 45 kHz spacing in a 36 MHz transponder, assigned on request.

Features and problems

  • Simple: no network-wide timing or synchronization needed.
  • Intermodulation: the transponder's TWT amplifier is nonlinear; many carriers produce intermodulation products, so the amplifier is operated with input/output back-off (a few dB), which reduces capacity and power.
  • Guard bands between carriers waste spectrum.
  • Power control needed so that a strong carrier does not suppress weak ones (capture effect).
  • Capacity falls sharply as the number of carriers rises.

Comparison: FDMA is the oldest and simplest satellite access method; TDMA uses the transponder more efficiently (single carrier at saturation) but needs precise timing; CDMA allows overlapping use with spread spectrum.

  • Asked 2 times
  • 2073 Bhadra (CS I) · 5 marks
  • 2070 Magh (CS I) · 5 marks

Write a short note on Frequency Division Multiplexing (FDM).

Answer

Frequency Division Multiplexing (FDM) sends several signals at the same time over one channel by giving each signal a separate frequency band. Each message modulates a different carrier; the modulated signals are added and transmitted together.

Tx: m1->[Mod fc1]-+          Rx: +->[BPF1]->[Demod]->m1
    m2->[Mod fc2]-+->(sum)->ch->-+->[BPF2]->[Demod]->m2
    m3->[Mod fc3]-+              +->[BPF3]->[Demod]->m3

Working

  1. Each message is band-limited by a LPF.
  2. It modulates its own subcarrier (often SSB-SC to save bandwidth).
  3. Band-pass filters limit each modulated signal to its slot; guard bands separate adjacent slots.
  4. All are summed and sent (possibly modulating a main carrier).
  5. At the receiver, band-pass filters separate the channels and demodulators recover the messages.

Total bandwidth ≈N×(channel bandwidth)+guard bands\approx N \times (\text{channel bandwidth}) + \text{guard bands}; e.g. 12 voice channels in 4 kHz slots need 48 kHz (telephone group, 60–108 kHz).

Applications: analog telephone trunks (group, supergroup), AM/FM broadcasting, cable TV, FM stereo, satellite FDMA, and OFDM in Wi-Fi/4G as a digital form.

Advantages: no synchronization between channels, simple for analog signals, continuous transmission. Disadvantages: crosstalk from nonlinearity (intermodulation) and imperfect filters, needs many sharp filters and stable oscillators, wasted guard bands, and fading affects a whole channel.

  • Asked 2 times
  • 2072 Magh (CS I) · 2+6 marks
  • 2071 Magh (CS I) · 2+6 marks

What do you understand by FDMA? Explain the use of FDM in telephony hierarchy.

Answer

FDMA

Frequency Division Multiple Access (FDMA) is a multiple-access method in which a shared medium (e.g. a satellite transponder or a cellular band) is divided into frequency sub-bands, and each user/station is assigned its own sub-band for the duration of its use. All users transmit at the same time on different frequencies. Examples: satellite SCPC/MCPC systems, 1G AMPS cellular (30 kHz channels). FDM is the multiplexing of signals at one location; FDMA is the same idea applied to users at different locations.

FDM in telephony hierarchy

Each voice channel (300–3400 Hz) is given a 4 kHz slot and frequency-translated by SSB-SC (LSB) modulation. Channels are built up in standard blocks (ITU-T/CCITT):

12 ch  -> Basic group       60-108 kHz
x5     -> Supergroup  (60)  312-552 kHz
x5     -> Mastergroup (300) 812-2044 kHz
x3     -> Supermaster (900) 8516-12388 kHz
  1. Basic group: 12 voice channels modulate carriers 64, 68, ..., 108 kHz; LSBs selected; band 60–108 kHz (48 kHz).
  2. Supergroup: 5 groups translated by carriers 420, 468, 516, 564, 612 kHz into 312–552 kHz; 60 channels.
  3. Mastergroup: 5 supergroups, 300 channels, 812–2044 kHz (8 kHz guard bands between supergroups).
  4. Supermastergroup: 3 mastergroups, 900 channels, 8516–12388 kHz.
LevelChannelsBandwidth
Group1248 kHz
Supergroup60240 kHz
Mastergroup3001232 kHz
Supermastergroup9003872 kHz

In the AT&T (Bell) system a mastergroup is 10 supergroups (600 channels, 564–3084 kHz) and a jumbo group is 6 mastergroups (3600 channels).

Advantages of the hierarchy: standard modular equipment, filters at convenient frequencies, and whole blocks can be switched or routed together. Carriers come from one master oscillator, and pilot tones keep levels and frequencies correct.

  • Asked 2 times
  • 2072 Asoj (CS I) · 4 marks
  • 2068 Bhadra (CS I) · 4 marks

Write a short note on satellite communication system.

Answer

A satellite communication system uses an artificial satellite in orbit as a repeater in space: it receives signals from an earth station, amplifies them, changes their frequency and retransmits them to other earth stations over a wide area.

          [Satellite: transponder]
          ^ uplink        \ downlink
         /  (6 / 14 GHz)   v (4 / 12 GHz)
 [Earth station A]     [Earth station B]

Main parts

  • Space segment: satellite with transponders (receive antenna, LNA, frequency down-converter, power amplifier such as TWTA/SSPA, transmit antenna), solar panels, batteries, and attitude/orbit control.
  • Ground segment: earth stations with large dish antennas, high-power amplifiers (uplink), low-noise amplifiers (downlink), modems and multiplexers; plus telemetry, tracking and command (TT&C) stations.

Key features

  • Geostationary orbit (GEO) at about 35 786 km above the equator; the satellite appears fixed, so antennas need not track. Three GEO satellites can cover almost the whole Earth.
  • Different uplink and downlink frequencies avoid interference: C-band 6/4 GHz, Ku-band 14/12 GHz, Ka-band 30/20 GHz.
  • Typical transponder bandwidth: 36 MHz.
  • Multiple access: FDMA, TDMA, CDMA.
  • One-way earth–satellite–earth delay about 240–270 ms for GEO.
  • LEO and MEO constellations give lower delay (used for mobile broadband, GPS).

Advantages: wide coverage, broadcast capability, distance-independent cost, quick setup in remote and hilly areas (useful in Nepal). Disadvantages: high launch cost, propagation delay and echo, rain fade at high frequencies, limited lifetime (about 15 years).

Applications: TV broadcasting (DTH), telephony and VSAT networks, navigation (GPS), weather and remote sensing, internet.

  • Asked 2 times
  • 2072 Chaitra (CS II) · 4+4 marks
  • 2071 Chaitra (CS II) · 4+4 marks

Explain E1 digital hierarchy as related to telephony system. Evaluate the expression of SQNR in uniformly quantized PCM system.

Answer

E1 digital hierarchy

E1 is the ITU-T (CEPT) primary PCM-TDM multiplex used in Europe and most of Asia, including Nepal.

E1 frame

  • Voice (300–3400 Hz) sampled at 8 kHz, A-law companded, 8 bits/sample: 64 kbps per channel.
  • Frame duration 125 µs, 32 time slots × 8 bits = 256 bits.
  • TS0: frame synchronisation/alignment; TS16: signalling (multiframe of 16 frames); TS1–15 and TS17–31: 30 voice channels.
  • Rate =256×8000=2.048= 256 \times 8000 = 2.048 Mbps.
|TS0|TS1 ... TS15|TS16|TS17 ... TS31|  125 us
sync   voice 1-15  sig   voice 16-30

Higher orders (PDH) multiplex 4 tributaries each, plus framing and justification bits:

LevelCompositionVoice chRate
E130 + 2 slots302.048 Mbps
E24 × E11208.448 Mbps
E34 × E248034.368 Mbps
E44 × E31920139.264 Mbps

(The North American counterpart is T1: 24 channels, 1.544 Mbps.)

SQNR of uniformly quantized PCM

Message range ±mp\pm m_p, L=2nL = 2^n levels.

  1. Step size: Δ=2mpL\Delta = \dfrac{2m_p}{L}.
  2. Error qq uniform in (−Δ/2,Δ/2)(-\Delta/2, \Delta/2), pdf 1/Δ1/\Delta:
Nq=∫−Δ/2Δ/2q2Δ dq=Δ212=mp23L2N_q = \int_{-\Delta/2}^{\Delta/2}\frac{q^2}{\Delta}\,dq = \frac{\Delta^2}{12} = \frac{m_p^2}{3L^2}
  1. Signal power S=m2‾S = \overline{m^2}:
SQNR=SNq=3L2m2‾mp2\text{SQNR} = \frac{S}{N_q} = 3L^2\frac{\overline{m^2}}{m_p^2}
  1. For a full-load sinusoid, m2‾=mp2/2\overline{m^2} = m_p^2/2:
SQNR=32L2=1.5×22nSQNRdB=1.76+6.02n dB\begin{aligned} \text{SQNR} &= \frac{3}{2}L^2 = 1.5 \times 2^{2n} \\ \text{SQNR}_{dB} &= 1.76 + 6.02n\ \text{dB} \end{aligned}

So SQNR increases by about 6 dB for each added bit. Example: for E1 channels with 8-bit uniform coding, 1.76+48.16=49.91.76 + 48.16 = 49.9 dB for a full-scale tone (companding keeps it high for weak signals too).

  • 2081 Chaitra · 5 marks

Write a short note on T1 and E1 hierarchy used in telephony.

Answer

T1 and E1 are the two standard TDM-PCM digital carrier systems used in telephony. Each one interleaves several 64 kb/s PCM voice channels into one serial stream. T1 is the North American (Bell) standard. E1 is the European (ITU-T) standard, and Nepal Telecom uses it too. Both sample speech at 8 kHz with 8 bits per sample, so one frame lasts 1/8000=1251/8000 = 125 µs.

T1 system

  • 24 voice channels × 8 bits + 1 framing bit = 193 bits per frame.
  • Rb=193×8000=1.544R_b = 193 \times 8000 = 1.544 Mb/s.
  • Signalling is "robbed bit": the LSB of each channel in frames 6 and 12 of a superframe.

E1 system

  • 32 time slots × 8 bits = 256 bits per frame. TS0 carries frame sync, TS16 carries signalling, and the other 30 slots carry voice.
  • Rb=256×8000=2.048R_b = 256 \times 8000 = 2.048 Mb/s.

Hierarchies

Higher levels are made by multiplexing lower-level streams. Extra bits are added for framing and bit stuffing, so each higher rate is a little more than the sum of its inputs.

T1 levelRate (Mb/s)ChannelsE1 levelRate (Mb/s)Channels
DS1 (T1)1.54424E12.04830
DS2 = 4 DS16.31296E2 = 4 E18.448120
DS3 = 7 DS244.736672E3 = 4 E234.368480
DS4 = 6 DS3274.1764032E4 = 4 E3139.2641920

T1 uses µ-law companding (μ=255\mu = 255) and the AMI/B8ZS line code. E1 uses A-law (A=87.6A = 87.6) and the HDB3 line code.

  • 2081 Chaitra · 5 marks

Write a short note on FDM vs TDM.

Answer

Frequency division multiplexing (FDM) sends several signals at the same time over one channel by giving each signal a different frequency band. Time division multiplexing (TDM) sends several signals over one channel by giving each signal a different time slot, one after another.

FDM

  • Each message is band-limited, then modulates its own sub-carrier (usually SSB in telephony).
  • Guard bands between adjacent channels let the receiver separate them with band-pass filters.
  • Example: 12 telephone channels in the 60–108 kHz basic group, and AM/FM broadcasting.

TDM

  • Each message is sampled. A commutator takes one sample from each channel in turn to form a frame.
  • The receiver separates the channels with a synchronised decommutator.
  • Example: a T1 frame carries 24 PCM voice channels at 1.544 Mb/s.
 FDM: frequency axis          TDM: time axis
 | ch1 | ch2 | ch3 |  f       |1|2|3|1|2|3|1|2|3|  t
  all at the same time         one band, shared in time

Comparison

PointFDMTDM
Sharing basisEach signal gets its own frequency bandEach signal gets its own time slot
Time useAll signals sent at the same timeSignals sent one after another
Signal typeMainly analogMainly digital (PAM/PCM samples)
HardwareModulators, sharp band-pass filters, oscillators per channelSampler, commutator, mostly digital circuits
GuardGuard bands between channelsGuard times between slots
Crosstalk causeNon-linearity, poor filtersPulse spreading (ISI), poor sync
SynchronisationNot criticalFrame sync is essential
Fading effectNarrow-band fading hits only some channelsFading during a slot hits one channel
ExampleRadio/TV broadcast, old analog telephone trunksT1/E1 digital telephony, GSM slots

FDM suits analog signals. TDM suits digital signals, and it is used in modern telephone networks because digital hardware is cheap and regenerative repeaters stop noise from building up along the link.

  • 2080 Chaitra · 2+4 marks

What is the significance of constellation diagram? Compare E1 and T1 TDM PCM hierarchy.

Answer

Significance of constellation diagram

A constellation diagram plots every possible symbol of a digital modulation scheme as a point in the I–Q (in-phase vs quadrature) plane. The distance of a point from the origin shows the symbol's amplitude, and its angle shows the symbol's phase.

      Q                      Q
      |                  o   |   o
  o---+---o   I          ----+----  I
      |                  o   |   o
     BPSK                   QPSK

Why it matters:

  • It shows the number of symbols M, and so the bits per symbol (log⁡2M\log_2 M), at a glance.
  • The minimum distance between points sets the noise margin. Points that are farther apart give a lower bit error rate for the same noise.
  • It shows the average and peak power of the scheme: the mean and maximum squared distance from the origin.
  • On a received-signal display, spread-out clouds show noise. A rotated pattern shows phase error, and a squeezed pattern shows amplitude non-linearity. This makes it a practical test tool.
  • It lets us compare schemes such as BPSK, QPSK and 16-QAM by bandwidth efficiency and noise immunity.

Comparison of E1 and T1 hierarchy

PointT1 (North America, Japan)E1 (Europe, ITU-T, Nepal)
Voice channels2430 (+2 overhead slots)
Time slots per frame24 + 1 framing bit32
Bits per frame24×8+1=19324 \times 8 + 1 = 19332×8=25632 \times 8 = 256
Frame duration125 µs125 µs
Bit rate1.544 Mb/s2.048 Mb/s
Framing1 F-bit per frame, pattern over 12/24 framesTS0 carries frame alignment word
SignallingRobbed-bit (LSB of frames 6 and 12)Separate slot TS16 (CAS)
MultiframeSuperframe 12 frames / ESF 24 frames16 frames (2 ms)
Companding lawµ-law, μ=255\mu = 255A-law, A=87.6A = 87.6
Line codeAMI / B8ZSHDB3
Higher levels6.312, 44.736, 274.176 Mb/s8.448, 34.368, 139.264, 564.992 Mb/s

Frame structures:

|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)
|<-------- one frame = 125 us = 256 bits -------->|
+-----+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... |TS15 | TS16 | TS17 | ... | TS31 |
+-----+-----+-----+-----+------+------+-----+------+
 sync  voice       voice signal voice        voice
 each time slot = 8 bits; 32 slots per frame
 voice channels: TS1-TS15 and TS17-TS31 (30)
  • 2079 Chaitra · 3+2+5 marks

Compare TDM and FDM. Show that for voice application. Compare E1 and T1 hierarchies.

Answer

Comparison of TDM and FDM

PointFDMTDM
Sharing basisEach signal gets its own frequency bandEach signal gets its own time slot
Time useAll signals sent at the same timeSignals sent one after another
Signal typeMainly analogMainly digital (PAM/PCM samples)
HardwareModulators, sharp band-pass filters, oscillators per channelSampler, commutator, mostly digital circuits
GuardGuard bands between channelsGuard times between slots
Crosstalk causeNon-linearity, poor filtersPulse spreading (ISI), poor sync
SynchronisationNot criticalFrame sync is essential
Fading effectNarrow-band fading hits only some channelsFading during a slot hits one channel
ExampleRadio/TV broadcast, old analog telephone trunksT1/E1 digital telephony, GSM slots

Voice application

Reading: "show the comparison for voice channels". Take N=24N = 24 telephone channels, each band-limited to 3.4 kHz.

FDM (SSB, 4 kHz slot per channel including guard band):

BFDM=N×4 kHz=24×4=96 kHzB_{FDM} = N \times 4\ \text{kHz} = 24 \times 4 = 96\ \text{kHz}

TDM-PCM (8 kHz sampling, 8 bits/sample, T1 frame):

Bits per frame=24×8+1=193 bitsFrames per second=fs=8000 frames/sRb=193×8000=1 544 000 b/s=1.544 Mb/s\begin{aligned} \text{Bits per frame} &= 24 \times 8 + 1 = 193\ \text{bits} \\ \text{Frames per second} &= f_s = 8000\ \text{frames/s} \\ R_b &= 193 \times 8000 = 1\,544\,000\ \text{b/s} = 1.544\ \text{Mb/s} \end{aligned}

The minimum (Nyquist) bandwidth is Rb/2=772R_b/2 = 772 kHz.

So for voice, FDM needs about 8 times less bandwidth. TDM-PCM is still preferred because:

  • regenerative repeaters remove noise at each hop,
  • digital switching and circuits are cheap,
  • it can carry speech and data together.

FDM was used in older analog trunks. TDM (T1/E1) is used in today's digital telephone networks.

Comparison of E1 and T1 hierarchies

PointT1 (North America, Japan)E1 (Europe, ITU-T, Nepal)
Voice channels2430 (+2 overhead slots)
Time slots per frame24 + 1 framing bit32
Bits per frame24×8+1=19324 \times 8 + 1 = 19332×8=25632 \times 8 = 256
Frame duration125 µs125 µs
Bit rate1.544 Mb/s2.048 Mb/s
Framing1 F-bit per frame, pattern over 12/24 framesTS0 carries frame alignment word
SignallingRobbed-bit (LSB of frames 6 and 12)Separate slot TS16 (CAS)
MultiframeSuperframe 12 frames / ESF 24 frames16 frames (2 ms)
Companding lawµ-law, μ=255\mu = 255A-law, A=87.6A = 87.6
Line codeAMI / B8ZSHDB3
Higher levels6.312, 44.736, 274.176 Mb/s8.448, 34.368, 139.264, 564.992 Mb/s
  • 2078 Chaitra · 3+2 marks

Differentiate between FDMA and TDMA. Draw T1 and E1 telephone hierarchy.

Answer

FDMA vs TDMA

FDMA (frequency division multiple access) gives each user its own carrier frequency band for the whole call. TDMA (time division multiple access) gives all users the same band, and each user transmits in bursts during its own time slot.

PointFDMATDMA
Resource sharedFrequency bandTime slot
TransmissionContinuousBurst (bursty)
SynchronisationNot neededStrict time sync needed
GuardGuard bandsGuard times
EquipmentSeparate RF filters and oscillators per userOne carrier shared, digital processing
IntermodulationProblem in non-linear amplifiers (e.g. satellite TWT)Less, since only one carrier at a time
ExampleAMPS (1G), SCPC satelliteGSM (2G), TDMA satellite

T1 and E1 telephone hierarchy

T1 hierarchy (Bell / North America):

 24 voice ch x 64 kb/s
        |
   [ DS1 / T1 ]  1.544 Mb/s    24 ch
        | x4
   [ DS2 / T2 ]  6.312 Mb/s    96 ch
        | x7
   [ DS3 / T3 ]  44.736 Mb/s   672 ch
        | x6
   [ DS4 / T4 ]  274.176 Mb/s  4032 ch

E1 hierarchy (ITU-T / Europe):

 30 voice ch x 64 kb/s (+ 2 overhead slots)
        |
   [ E1 ]  2.048 Mb/s     30 ch
        | x4
   [ E2 ]  8.448 Mb/s     120 ch
        | x4
   [ E3 ]  34.368 Mb/s    480 ch
        | x4
   [ E4 ]  139.264 Mb/s   1920 ch
        | x4
   [ E5 ]  564.992 Mb/s   7680 ch

Both systems use 125 µs frames. A T1 frame carries 193 bits, giving 1.544 Mb/s. An E1 frame carries 256 bits, giving 2.048 Mb/s.

  • 2077 Chaitra (CS I) · 6+2 marks

Describe FDM in Telephony with its uses and structure. What are the uses of filters in FDM?

Answer

Frequency division multiplexing (FDM) in telephony places many voice channels, each about 4 kHz wide, side by side in frequency on one cable or radio link. Each voice channel is shifted to its own slot by single-sideband (SSB) modulation. SSB is used because it needs only 4 kHz per channel.

Structure (CCITT FDM hierarchy)

Channels are combined in stages. Each stage takes the previous blocks, SSB-modulates them onto new carriers, and stacks them in frequency.

Level (CCITT)Made ofVoice channelsFrequency bandBandwidth
Voice channel—10–4 kHz (0.3–3.4 kHz used)4 kHz
Basic group12 voice channels1260–108 kHz48 kHz
Supergroup5 groups60312–552 kHz240 kHz
Mastergroup5 supergroups300812–2044 kHz1232 kHz
Supermastergroup3 mastergroups9008516–12388 kHz3872 kHz
 Group: 12 channels, carriers 64, 68, ..., 108 kHz
 (LSB of each channel kept)

 |ch12|ch11|ch10| ch9| ch8| ... | ch2| ch1|
 60   64   68   72   76         100  104  108 kHz
 |<----------------- 48 kHz ----------------->|

 Supergroup: 5 groups, carriers 420, 468, 516,
 564, 612 kHz (LSB kept)

 | G5 | G4 | G3 | G2 | G1 |
 312  360  408  456  504  552 kHz
 |<-------- 240 kHz -------->|

Working

  1. Each speech signal is band-limited to 300–3400 Hz by a low-pass/band-pass filter.
  2. It SSB-modulates a channel carrier (64, 68, …, 108 kHz). The lower sideband is kept, giving the 60–108 kHz basic group.
  3. Five groups are shifted to 312–552 kHz to form a supergroup, and five supergroups form a mastergroup.
  4. At the receiver, the same steps are reversed. Band-pass filters pick out each block, and SSB demodulators with locally generated carriers recover the speech.
  5. Pilot tones are sent for carrier synchronisation and level control.

Uses

  • Long-distance analog telephone trunks on coaxial cable and microwave links.
  • Satellite FDMA (SCPC) links.
  • The same idea is used in radio/TV broadcasting, FM stereo, and ADSL on telephone lines.

Uses of filters in FDM

  • Input low-pass filters band-limit each message so that it fits its 4 kHz slot.
  • Band-pass filters after the modulators remove the unwanted sideband and the carrier leakage (SSB generation). They keep each channel inside its own band.
  • Receiver band-pass filters separate each channel or group from the composite signal (demultiplexing).
  • Output low-pass filters after demodulation remove high-frequency products and recover the baseband speech.
  • Sharp filters keep guard bands small and reduce adjacent-channel crosstalk.
  • 2075 Bhadra (CS I) · 2+6 marks

What is frequency division multiplexing (FDM)? Describe the method of FDM in telephony.

Answer

Frequency division multiplexing

FDM is a technique in which several message signals are sent at the same time over one channel by shifting each one, with modulation, to a different non-overlapping frequency band. Guard bands separate the bands, and the receiver uses band-pass filters to separate the channels.

Method of FDM in telephony

Telephone FDM uses SSB-SC modulation, so each voice channel takes only 4 kHz (0.3–3.4 kHz speech plus guard band). Channels are built up in a hierarchy:

Level (CCITT)Made ofVoice channelsFrequency bandBandwidth
Voice channel—10–4 kHz (0.3–3.4 kHz used)4 kHz
Basic group12 voice channels1260–108 kHz48 kHz
Supergroup5 groups60312–552 kHz240 kHz
Mastergroup5 supergroups300812–2044 kHz1232 kHz
Supermastergroup3 mastergroups9008516–12388 kHz3872 kHz

Step 1: Basic group. Twelve voice channels modulate carriers at 64, 68, 72, …, 108 kHz. Band-pass filters keep only the lower sidebands, which fill 60–108 kHz.

Step 2: Supergroup. Five basic groups modulate carriers at 420, 468, 516, 564 and 612 kHz. The lower sidebands fill 312–552 kHz (60 channels).

Step 3: Mastergroup and above. Five supergroups form a 300-channel mastergroup (812–2044 kHz). Three mastergroups form a 900-channel supermastergroup.

 Group: 12 channels, carriers 64, 68, ..., 108 kHz
 (LSB of each channel kept)

 |ch12|ch11|ch10| ch9| ch8| ... | ch2| ch1|
 60   64   68   72   76         100  104  108 kHz
 |<----------------- 48 kHz ----------------->|

 Supergroup: 5 groups, carriers 420, 468, 516,
 564, 612 kHz (LSB kept)

 | G5 | G4 | G3 | G2 | G1 |
 312  360  408  456  504  552 kHz
 |<-------- 240 kHz -------->|

Receiver. The demultiplexer reverses each step. Band-pass filters pick out each supergroup, then each group, then each channel. Each is SSB-demodulated with a local carrier locked to transmitted pilot tones and then low-pass filtered to get the speech back.

Advantages are that many calls share one cable, the hardware is analog and simple, and no sampling is needed. The drawbacks are crosstalk from non-linearity and noise that builds up over long links, which is why FDM was later replaced by TDM-PCM (T1/E1).

  • 2071 Magh (old course) · 8 marks

Prove that signaling rate for time division multiplexing of 24 voice channels is equal to 1.544 Mbps.

Answer

In the T1 (DS1) TDM-PCM system, 24 voice channels are sampled, quantised, encoded and interleaved in time into one frame.

Given standard values

  • Voice band limited to about 3.4 kHz, so the sampling rate used is fs=8f_s = 8 kHz (above the Nyquist rate of 6.8 kHz).
  • Each sample is coded with n=8n = 8 bits (256 levels, µ-law).
  • Number of channels N=24N = 24.
  • One framing bit is added per frame for synchronisation.

Frame structure

|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)

Proof

One frame holds one sample from every channel, so frames repeat at the sampling rate:

Tframe=1fs=18000=125 μsT_{frame} = \frac{1}{f_s} = \frac{1}{8000} = 125\ \mu\text{s}

Bits in one frame:

Information bits=N×n=24×8=192 bitsFraming bit=1 bitTotal=192+1=193 bits/frame\begin{aligned} \text{Information bits} &= N \times n = 24 \times 8 = 192\ \text{bits} \\ \text{Framing bit} &= 1\ \text{bit} \\ \text{Total} &= 192 + 1 = 193\ \text{bits/frame} \end{aligned}

Signalling rate:

Rb=bits per frameTframe=193125×10−6=193×8000=1 544 000 b/s\begin{aligned} R_b &= \frac{\text{bits per frame}}{T_{frame}} = \frac{193}{125 \times 10^{-6}} \\ &= 193 \times 8000 = 1\,544\,000\ \text{b/s} \end{aligned}

Without the framing bit, the rate would be 24×64 kb/s=1.53624 \times 64\ \text{kb/s} = 1.536 Mb/s. The framing bit adds 8 kb/s of overhead.

Each bit lasts Tb=1/1.544 MHz≈0.648T_b = 1/1.544\ \text{MHz} \approx 0.648 µs, and the minimum (Nyquist) transmission bandwidth is Rb/2=772R_b/2 = 772 kHz.

Answer: Signalling rate Rb=193×8000=1.544R_b = 193 \times 8000 = 1.544 Mb/s (proved).

  • 2071 Bhadra (CS I) · 2+4 marks

Describe the principle of frequency division multiplexing (FDM). Briefly explain SCPC and DAMA types of FDMA.

Answer

Principle of FDM

Frequency division multiplexing shares the bandwidth of one channel among many signals by giving each signal its own frequency band at the same time. Each message is band-limited and modulates a different carrier fc1,fc2,…f_{c1}, f_{c2}, \dots. The modulated signals are added and sent together. Guard bands between bands prevent overlap. At the receiver, band-pass filters tuned to each band separate the signals, and demodulators recover the messages.

 Spectrum of composite FDM signal
   | ch1 |g| ch2 |g| ch3 |g| ch4 |
   fc1      fc2     fc3     fc4      f
   (g = guard band)

When the same idea is used to let many earth stations share one satellite transponder, it is called FDMA.

SCPC (Single Channel Per Carrier)

  • Each voice or data channel gets its own carrier inside the transponder band, instead of first being grouped into a multi-channel FDM baseband.
  • Example: a 36 MHz transponder divided into narrow channel slots (e.g. 45 kHz each).
  • Often used with voice activation: the carrier is switched off during silence, which saves transponder power.
  • Suits thin-route traffic (small stations with only a few circuits).
  • In pre-assigned SCPC, channels are fixed to station pairs, so they sit idle when not in use.

DAMA (Demand Assigned Multiple Access)

  • Channels are not fixed. All stations share a common pool of carrier frequencies, and a channel is given to a station only when a call is requested. It is returned to the pool when the call ends.
  • A common signalling channel (central or distributed control) handles the requests.
  • Example: the INTELSAT SPADE system (SCPC, PCM, multiple access, demand assignment) uses about 800 SCPC channels at 45 kHz spacing in a 36 MHz transponder.
  • Advantage: far better use of the transponder when traffic is random and many stations have light traffic.
PointPre-assigned SCPCDAMA
Channel allotmentFixedOn demand
Efficiency for light trafficLowHigh
ControlSimpleNeeds signalling channel
  • 2068 Jestha (old course) · 4+6 marks

Differentiate between FDM and TDM. Draw spectral details and explain various standard groups of FDM telephone hierarchy.

Answer

Difference between FDM and TDM

PointFDMTDM
Sharing basisEach signal gets its own frequency bandEach signal gets its own time slot
Time useAll signals sent at the same timeSignals sent one after another
Signal typeMainly analogMainly digital (PAM/PCM samples)
HardwareModulators, sharp band-pass filters, oscillators per channelSampler, commutator, mostly digital circuits
GuardGuard bands between channelsGuard times between slots
Crosstalk causeNon-linearity, poor filtersPulse spreading (ISI), poor sync
SynchronisationNot criticalFrame sync is essential
Fading effectNarrow-band fading hits only some channelsFading during a slot hits one channel
ExampleRadio/TV broadcast, old analog telephone trunksT1/E1 digital telephony, GSM slots

FDM telephone hierarchy and spectral details

Telephone FDM uses SSB-SC modulation. Each voice channel (300–3400 Hz) gets a 4 kHz slot that includes its guard band. The standard (CCITT) groups are:

Level (CCITT)Made ofVoice channelsFrequency bandBandwidth
Voice channel—10–4 kHz (0.3–3.4 kHz used)4 kHz
Basic group12 voice channels1260–108 kHz48 kHz
Supergroup5 groups60312–552 kHz240 kHz
Mastergroup5 supergroups300812–2044 kHz1232 kHz
Supermastergroup3 mastergroups9008516–12388 kHz3872 kHz

1. Basic group (12 channels, 60–108 kHz). Channel kk modulates a carrier fk=112−4kf_k = 112 - 4k kHz (k=1…12k = 1 \dots 12, i.e. 108, 104, …, 64 kHz). The lower sideband is kept, so channel kk occupies fk−4f_k - 4 to fkf_k kHz.

2. Supergroup (60 channels, 312–552 kHz). Five groups modulate carriers at 420, 468, 516, 564 and 612 kHz. Taking the lower sideband turns each 60–108 kHz group into a 48 kHz block. For example, 420−108=312420 - 108 = 312 to 420−60=360420 - 60 = 360 kHz.

3. Mastergroup (300 channels, 812–2044 kHz). Five supergroups are placed with 8 kHz guard bands between them. (The Bell system mastergroup instead uses 10 supergroups = 600 channels in 564–3084 kHz.)

4. Supermastergroup (900 channels, 8516–12388 kHz). Three mastergroups.

 Group: 12 channels, carriers 64, 68, ..., 108 kHz
 (LSB of each channel kept)

 |ch12|ch11|ch10| ch9| ch8| ... | ch2| ch1|
 60   64   68   72   76         100  104  108 kHz
 |<----------------- 48 kHz ----------------->|

 Supergroup: 5 groups, carriers 420, 468, 516,
 564, 612 kHz (LSB kept)

 | G5 | G4 | G3 | G2 | G1 |
 312  360  408  456  504  552 kHz
 |<-------- 240 kHz -------->|

Each step needs only one set of modulators and filters for a whole block of channels. This keeps equipment simple. The wide guard bands at higher levels make the filters easy to build.

  • 2064 Shrawan (CS I) · 4+4 marks

Discuss a block diagram of an analog communication system. Explain FDM system with neat diagrams, assuming three different message signals are to be transmitted simultaneously.

Answer

Block diagram of an analog communication system

An analog communication system carries a continuously varying message from a source to a user over a channel. The message is first converted to an electrical signal and then modulated onto a carrier.

 [Information source]
         |
 [Input transducer]
         |
 [Transmitter (modulator)]
         |
 [Channel] <---- noise, interference
         |
 [Receiver (demodulator)]
         |
 [Output transducer] --> user
  • Information source and input transducer: produce the message, e.g. speech converted to voltage by a microphone.
  • Transmitter: amplifies, filters and modulates (AM/FM/PM) the message onto a carrier suited to the channel, then power-amplifies it.
  • Channel: a wire, coaxial cable, optical fibre or free space. It attenuates and distorts the signal and adds noise.
  • Receiver: selects the wanted signal, amplifies it, demodulates it and filters it to recover the message.
  • Output transducer: a loudspeaker or display that turns the electrical signal back into its original form.

FDM system for three message signals

Assume three messages m1(t),m2(t),m3(t)m_1(t), m_2(t), m_3(t), each band-limited to W=4W = 4 kHz, and SSB modulation with carriers fc1=100f_{c1} = 100 kHz, fc2=105f_{c2} = 105 kHz and fc3=110f_{c3} = 110 kHz (upper sideband kept, 1 kHz guard band).

 TRANSMITTER
 m1 ->[LPF]->[SSB mod fc1]->[BPF1]--+
 m2 ->[LPF]->[SSB mod fc2]->[BPF2]--+-->(+)--> channel
 m3 ->[LPF]->[SSB mod fc3]->[BPF3]--+

 RECEIVER
          +->[BPF1]->[demod fc1]->[LPF]-> m1
 channel -+->[BPF2]->[demod fc2]->[LPF]-> m2
          +->[BPF3]->[demod fc3]->[LPF]-> m3

Spectrum of the multiplexed signal:

  |  m1  |g|  m2  |g|  m3  |
  100    104 105  109 110   114  kHz
  |<--------- 14 kHz ---------->|

Working:

  1. LPFs band-limit each message to WW.
  2. Each message modulates its own carrier. BPFs keep one sideband, so each signal sits in its own band.
  3. The adder combines them. The composite signal (bandwidth 3W3W + guard bands = 14 kHz here) is sent over one channel.
  4. At the receiver, BPFs centred on each band separate the three signals. Coherent demodulators with fc1,fc2,fc3f_{c1}, f_{c2}, f_{c3} and LPFs recover m1,m2,m3m_1, m_2, m_3.

Guard bands and good filters prevent crosstalk between the channels.

  • 2080 Bhadra (CS II) · 8 marks

With the help of the frame diagrams, discuss T1 and E1 hierarchy of TDM-PCM telephony.

Answer

TDM-PCM telephony carries many 64 kb/s PCM voice channels on one line by interleaving their 8-bit samples in time. The two standards are T1 (North America, Japan) and E1 (Europe, ITU-T, used in Nepal). Both sample speech at 8 kHz, so each frame lasts 125125 µs.

T1 frame

|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)
Bits per frame=24×8+1=193 bitsFrames per second=fs=8000 frames/sRb=193×8000=1 544 000 b/s=1.544 Mb/s\begin{aligned} \text{Bits per frame} &= 24 \times 8 + 1 = 193\ \text{bits} \\ \text{Frames per second} &= f_s = 8000\ \text{frames/s} \\ R_b &= 193 \times 8000 = 1\,544\,000\ \text{b/s} = 1.544\ \text{Mb/s} \end{aligned}
  • 12 frames form a superframe (24 in the extended superframe, ESF). The F-bits of these frames carry the frame alignment pattern (100011011100 for the superframe).
  • Signalling uses the robbed-bit method: in frames 6 and 12, the LSB of every channel carries signalling (A and B bits).
  • µ-law companding (μ=255\mu = 255) and AMI/B8ZS line coding are used.

E1 frame

|<-------- one frame = 125 us = 256 bits -------->|
+-----+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... |TS15 | TS16 | TS17 | ... | TS31 |
+-----+-----+-----+-----+------+------+-----+------+
 sync  voice       voice signal voice        voice
 each time slot = 8 bits; 32 slots per frame
 voice channels: TS1-TS15 and TS17-TS31 (30)
Rb=32×8×8000=2.048 Mb/sR_b = 32 \times 8 \times 8000 = 2.048\ \text{Mb/s}
  • TS0 carries the frame alignment word (0011011) in alternate frames, plus alarm bits.
  • TS16 carries channel-associated signalling. 16 frames form a multiframe of 2 ms, and each frame's TS16 carries the signalling of 2 channels.
  • A-law companding (A=87.6A = 87.6) and HDB3 line coding are used.

T1 hierarchy

 24 voice ch x 64 kb/s
        |
   [ DS1 / T1 ]  1.544 Mb/s    24 ch
        | x4
   [ DS2 / T2 ]  6.312 Mb/s    96 ch
        | x7
   [ DS3 / T3 ]  44.736 Mb/s   672 ch
        | x6
   [ DS4 / T4 ]  274.176 Mb/s  4032 ch

E1 hierarchy

 30 voice ch x 64 kb/s (+ 2 overhead slots)
        |
   [ E1 ]  2.048 Mb/s     30 ch
        | x4
   [ E2 ]  8.448 Mb/s     120 ch
        | x4
   [ E3 ]  34.368 Mb/s    480 ch
        | x4
   [ E4 ]  139.264 Mb/s   1920 ch
        | x4
   [ E5 ]  564.992 Mb/s   7680 ch

At each higher level, the higher rate is a little more than (number of inputs × lower rate). The extra bits are used for framing and bit stuffing (justification), which align input streams whose clocks differ slightly. For example, 4×1.544=6.1764 \times 1.544 = 6.176 Mb/s, but DS2 runs at 6.312 Mb/s.

  • 2079 Bhadra (CS II) · 5+5 marks

Explain T1 digital carrier system with its multiplexing hierarchy. Explain the principles of Adaptive Delta Modulation (ADM).

Answer

T1 digital carrier system

T1 is the North American TDM-PCM carrier that sends 24 voice channels over one twisted-pair line at 1.544 Mb/s.

  • Each voice channel is band-limited to 3.4 kHz and sampled at 8 kHz.
  • Each sample is µ-law compressed (μ=255\mu = 255) and coded into 8 bits, giving 64 kb/s per channel.
  • One frame = one 8-bit sample from each of the 24 channels + 1 framing bit = 193 bits in 125 µs.
|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)
Bits per frame=24×8+1=193 bitsFrames per second=fs=8000 frames/sRb=193×8000=1 544 000 b/s=1.544 Mb/s\begin{aligned} \text{Bits per frame} &= 24 \times 8 + 1 = 193\ \text{bits} \\ \text{Frames per second} &= f_s = 8000\ \text{frames/s} \\ R_b &= 193 \times 8000 = 1\,544\,000\ \text{b/s} = 1.544\ \text{Mb/s} \end{aligned}
  • Signalling uses robbed bits (the LSB in frames 6 and 12 of a 12-frame superframe).
  • The line code is bipolar AMI (B8ZS on newer lines). Repeaters are placed about every 1.8 km.

Multiplexing hierarchy:

 24 voice ch x 64 kb/s
        |
   [ DS1 / T1 ]  1.544 Mb/s    24 ch
        | x4
   [ DS2 / T2 ]  6.312 Mb/s    96 ch
        | x7
   [ DS3 / T3 ]  44.736 Mb/s   672 ch
        | x6
   [ DS4 / T4 ]  274.176 Mb/s  4032 ch

Higher levels add framing and stuffing bits, so DS2 = 6.312 Mb/s rather than 4×1.544=6.1764 \times 1.544 = 6.176 Mb/s.

Adaptive Delta Modulation (ADM)

Linear delta modulation (DM) sends 1 bit per sample, which says whether the signal is above or below its staircase approximation. A fixed step size Δ\Delta causes two problems:

  • Slope overload when the signal changes faster than Δfs\Delta f_s.
  • Granular noise when the signal is nearly flat.

ADM changes the step size to suit the signal slope:

  • If several successive output bits are the same (e.g. 1 1 1), the staircase is lagging behind, so the step size is increased. A common rule is to multiply it by 1.5 or 2.
  • If the bits alternate (1 0 1 0), the signal is flat, so the step size is decreased (e.g. divided by the same factor), down to a minimum Δmin\Delta_{min}.
 m(t) -->(+)--> [Comparator] --+--> output bits b(k)
          ^-                   |
          |                    v
          |            [Step size logic]
          |                    | delta(k)
          +--- [Accumulator] <-+

The receiver uses the same logic on the received bits, so no extra side information is sent.

Advantages: much less slope overload and granular noise, a wider dynamic range, and good speech quality at 16–32 kb/s. An example is CVSD (continuously variable slope DM), used in military and Bluetooth voice.

  • 2076 Chaitra (CS II) · 4+3 marks

Twenty four voice signals are sampled uniformly and then have to be time division multiplexed. The highest frequency component for each voice signal is equal to 3.4 kHz. Now, (i) If the signals are pulse amplitude modulated using Nyquist rate of sampling, what would be the minimum channel bandwidth required. (ii) If the signal are pulse code modulated with an 8 bit encoder, what would be the sampling rate? The bit rate of the system is given as 1.5×10⁶ bits/sec.

Answer

Given: N=24N = 24 voice channels, fm=3.4f_m = 3.4 kHz.

(i) PAM-TDM, Nyquist sampling

Nyquist rate for each channel:

fs=2fm=2×3.4=6.8 kHzf_s = 2 f_m = 2 \times 3.4 = 6.8\ \text{kHz}

Total pulse rate of the TDM signal:

r=Nfs=24×6.8 kHz=163.2 k pulses/sr = N f_s = 24 \times 6.8\ \text{kHz} = 163.2\ \text{k pulses/s}

The minimum channel bandwidth needed for TDM-PAM is half the pulse rate. This is the Nyquist bandwidth, the same as NfmN f_m:

Bmin=Nfs2=24×6.82=81.6 kHz  (=24×3.4 kHz)\begin{aligned} B_{min} &= \frac{N f_s}{2} = \frac{24 \times 6.8}{2} \\ &= 81.6\ \text{kHz} \;(= 24 \times 3.4\ \text{kHz}) \end{aligned}

Answer (i): Minimum channel bandwidth = 81.6 kHz.

(ii) PCM with 8-bit encoder, Rb=1.5×106R_b = 1.5 \times 10^6 b/s

Each frame has one 8-bit sample from each channel:

Rb=N n fsR_b = N \, n \, f_s fs=RbNn=1.5×10624×8=1.5×106192=7812.5 Hz\begin{aligned} f_s &= \frac{R_b}{N n} = \frac{1.5 \times 10^6}{24 \times 8} \\ &= \frac{1.5 \times 10^6}{192} = 7812.5\ \text{Hz} \end{aligned}

Check: 7812.5>68007812.5 > 6800 Hz (the Nyquist rate), so there is no aliasing, with a guard band of 7812.5−6800=1012.57812.5 - 6800 = 1012.5 Hz.

Answer (ii): Sampling rate fs=7812.5f_s = 7812.5 samples/s ≈ 7.81 kHz.

  • 2076 Chaitra (CS II) · 5 marks

Write a short note on the E1 digital hierarchy.

Answer

E1 is the ITU-T (European) standard TDM-PCM digital carrier. It multiplexes 30 voice channels of 64 kb/s plus 2 overhead channels into a 2.048 Mb/s stream. It is used in Europe, Asia and Nepal.

E1 frame

  • Speech is sampled at 8 kHz, so a frame lasts 125 µs.
  • Each frame has 32 time slots × 8 bits = 256 bits.
  • TS0: frame alignment word and alarm bits.
  • TS16: signalling (channel-associated). 16 frames form a 2 ms multiframe.
  • TS1–TS15 and TS17–TS31: 30 voice channels.
|<-------- one frame = 125 us = 256 bits -------->|
+-----+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... |TS15 | TS16 | TS17 | ... | TS31 |
+-----+-----+-----+-----+------+------+-----+------+
 sync  voice       voice signal voice        voice
 each time slot = 8 bits; 32 slots per frame
 voice channels: TS1-TS15 and TS17-TS31 (30)
Rb=32×8×8000=2.048 Mb/sR_b = 32 \times 8 \times 8000 = 2.048\ \text{Mb/s}

E1 digital hierarchy

Four lower-level streams are combined at each step. Extra bits for framing and justification (bit stuffing) are added.

LevelMade ofVoice channelsBit rate
E130 voice + 2302.048 Mb/s
E24 × E11208.448 Mb/s
E34 × E248034.368 Mb/s
E44 × E31920139.264 Mb/s
E54 × E47680564.992 Mb/s

E1 uses A-law companding (A=87.6A = 87.6) and the HDB3 line code. Compared with T1 (24 channels, 1.544 Mb/s), E1 carries more channels and has a separate slot for signalling, so all 8 bits of each voice slot stay available for speech.

  • 2075 Asoj (CS II) · 4 marks

Explain the differences between T1 and E1 digital hierarchy.

Answer

T1 and E1 are TDM-PCM digital carrier systems. Both use 8 kHz sampling, 8-bit samples and 125 µs frames, but they differ in channel count, framing and rates.

PointT1 (North America, Japan)E1 (Europe, ITU-T, Nepal)
Voice channels2430 (+2 overhead slots)
Time slots per frame24 + 1 framing bit32
Bits per frame24×8+1=19324 \times 8 + 1 = 19332×8=25632 \times 8 = 256
Frame duration125 µs125 µs
Bit rate1.544 Mb/s2.048 Mb/s
Framing1 F-bit per frame, pattern over 12/24 framesTS0 carries frame alignment word
SignallingRobbed-bit (LSB of frames 6 and 12)Separate slot TS16 (CAS)
MultiframeSuperframe 12 frames / ESF 24 frames16 frames (2 ms)
Companding lawµ-law, μ=255\mu = 255A-law, A=87.6A = 87.6
Line codeAMI / B8ZSHDB3
Higher levels6.312, 44.736, 274.176 Mb/s8.448, 34.368, 139.264, 564.992 Mb/s
  • 2074 Asoj (CS II) · 3+3 marks

Describe E1 frame and its TDM hierarchy along with signaling rate.

Answer

E1 frame

E1 is the ITU-T TDM-PCM carrier that carries 30 voice channels. Each voice channel is sampled at 8 kHz and coded with 8 bits (A-law), giving 64 kb/s. One frame takes one byte from each of 32 time slots:

|<-------- one frame = 125 us = 256 bits -------->|
+-----+-----+-----+-----+------+------+-----+------+
| TS0 | TS1 | ... |TS15 | TS16 | TS17 | ... | TS31 |
+-----+-----+-----+-----+------+------+-----+------+
 sync  voice       voice signal voice        voice
 each time slot = 8 bits; 32 slots per frame
 voice channels: TS1-TS15 and TS17-TS31 (30)
  • TS0: frame alignment word (0011011) in alternate frames, plus alarm/CRC bits.
  • TS16: channel-associated signalling. 16 frames form a 2 ms multiframe.
  • 30 slots carry speech.

Signalling (bit) rate:

Bits per frame=32×8=256Rb=256×8000=2 048 000 b/s=2.048 Mb/s\begin{aligned} \text{Bits per frame} &= 32 \times 8 = 256 \\ R_b &= 256 \times 8000 = 2\,048\,000\ \text{b/s} = 2.048\ \text{Mb/s} \end{aligned}

Bit duration Tb=1/2.048 MHz≈0.488T_b = 1/2.048\ \text{MHz} \approx 0.488 µs.

E1 TDM hierarchy

 30 voice ch x 64 kb/s (+ 2 overhead slots)
        |
   [ E1 ]  2.048 Mb/s     30 ch
        | x4
   [ E2 ]  8.448 Mb/s     120 ch
        | x4
   [ E3 ]  34.368 Mb/s    480 ch
        | x4
   [ E4 ]  139.264 Mb/s   1920 ch
        | x4
   [ E5 ]  564.992 Mb/s   7680 ch
LevelRateChannels
E12.048 Mb/s30
E28.448 Mb/s120
E334.368 Mb/s480
E4139.264 Mb/s1920

Each level combines 4 streams of the level below. A few extra bits are added for framing and justification, so for example 4×2.048=8.1924 \times 2.048 = 8.192 Mb/s, but E2 = 8.448 Mb/s.

  • 2071 Shrawan (CS II) · 2+4 marks

What do you mean by companding? Explain T1 hierarchy of TDM-PCM telephony.

Answer

Companding

Companding = compressing + expanding. Before uniform quantisation, the signal passes through a compressor that boosts weak amplitudes and compresses strong ones. At the receiver, an expander applies the inverse curve. The result works like non-uniform quantisation: small step sizes for weak signals and large steps for loud ones.

This gives a nearly constant SQNR over a wide range of speech levels, using only 8 bits.

 Tx: m(t) -> [Compressor] -> [Uniform quantizer
                              + encoder] ~~> channel
 Rx: channel ~~> [Decoder] -> [Expander] -> m(t)

Standard laws: µ-law (μ=255\mu = 255, North America/T1) and A-law (A=87.6A = 87.6, Europe/E1).

y=ln⁡(1+μ∣x∣)ln⁡(1+μ)sgn⁡(x),∣x∣≤1y = \frac{\ln(1 + \mu |x|)}{\ln(1 + \mu)} \operatorname{sgn}(x), \quad |x| \le 1

T1 hierarchy of TDM-PCM telephony

T1 (DS1) multiplexes 24 PCM voice channels:

|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)
Bits per frame=24×8+1=193 bitsFrames per second=fs=8000 frames/sRb=193×8000=1 544 000 b/s=1.544 Mb/s\begin{aligned} \text{Bits per frame} &= 24 \times 8 + 1 = 193\ \text{bits} \\ \text{Frames per second} &= f_s = 8000\ \text{frames/s} \\ R_b &= 193 \times 8000 = 1\,544\,000\ \text{b/s} = 1.544\ \text{Mb/s} \end{aligned}

Higher levels are formed by multiplexing T1 streams and adding framing and stuffing bits:

 24 voice ch x 64 kb/s
        |
   [ DS1 / T1 ]  1.544 Mb/s    24 ch
        | x4
   [ DS2 / T2 ]  6.312 Mb/s    96 ch
        | x7
   [ DS3 / T3 ]  44.736 Mb/s   672 ch
        | x6
   [ DS4 / T4 ]  274.176 Mb/s  4032 ch
LevelMade ofChannelsRate
DS1 (T1)24 voice241.544 Mb/s
DS1C2 DS1483.152 Mb/s
DS2 (T2)4 DS1966.312 Mb/s
DS3 (T3)7 DS267244.736 Mb/s
DS4 (T4)6 DS34032274.176 Mb/s
  • 2070 Asar (CS II) · 3+3 marks

Explain functional block diagram of the PCM system. Find the signaling rate of the T1 system and draw its frame diagram.

Answer

Functional block diagram of PCM

Pulse code modulation (PCM) turns an analog signal into a stream of binary code words by sampling, quantising and encoding it.

 TRANSMITTER
 m(t)->[LPF]->[Sampler]->[Quantizer]->[Encoder]->
                                     PCM bits
 CHANNEL with REGENERATIVE REPEATERS
 ->[Equalizer]->[Decision/Timing]->[Regenerate]->
 RECEIVER
 ->[Regenerator]->[Decoder]->[Reconstruction LPF]->m'(t)
  • Anti-aliasing LPF: limits the message to WW Hz.
  • Sampler: takes samples at fs≥2Wf_s \ge 2W (8 kHz for speech).
  • Quantiser: rounds each sample to one of L=2nL = 2^n levels (with companding for speech).
  • Encoder: turns each level into an nn-bit code word, then a line code is applied.
  • Regenerative repeaters: reshape and retime the pulses along the line, so noise does not build up.
  • Receiver: the regenerator cleans the pulses, the decoder turns code words back into PAM samples, and the LPF rebuilds m(t)m(t).

Signalling rate of T1

T1 sends 24 voice channels, each sampled at 8 kHz with 8 bits per sample, plus 1 framing bit per frame.

Bits per frame=24×8+1=193 bitsFrames per second=fs=8000 frames/sRb=193×8000=1 544 000 b/s=1.544 Mb/s\begin{aligned} \text{Bits per frame} &= 24 \times 8 + 1 = 193\ \text{bits} \\ \text{Frames per second} &= f_s = 8000\ \text{frames/s} \\ R_b &= 193 \times 8000 = 1\,544\,000\ \text{b/s} = 1.544\ \text{Mb/s} \end{aligned}

Answer: T1 signalling rate = 1.544 Mb/s (frame of 193 bits every 125 µs).

T1 frame diagram

|<--------- one frame = 125 us = 193 bits --------->|
+---+--------+--------+--------+-------+--------+
| F |  Ch 1  |  Ch 2  |  Ch 3  |  ...  | Ch 24  |
+---+--------+--------+--------+-------+--------+
 1 bit  8 bits  8 bits  8 bits           8 bits
 F = framing bit (pattern spread over 12 frames)
  • 2070 Chaitra (CS II) · 3+4 marks

What are the signalling (bit) rate and bandwidth requirement for the T1 and E1 digital carrier systems? Explain briefly about Differential Pulse Code Modulation (DPCM) encoder.

Answer

Bit rate and bandwidth of T1 and E1

Both systems sample speech at fs=8f_s = 8 kHz with 8 bits per sample.

T1: 24 channels + 1 framing bit:

Rb=(24×8+1)×8000=193×8000=1.544 Mb/sR_b = (24 \times 8 + 1) \times 8000 = 193 \times 8000 = 1.544\ \text{Mb/s}

E1: 32 slots (30 voice + 2 overhead):

Rb=32×8×8000=256×8000=2.048 Mb/sR_b = 32 \times 8 \times 8000 = 256 \times 8000 = 2.048\ \text{Mb/s}

Bandwidth. The minimum (Nyquist) bandwidth for binary transmission is B=Rb/2B = R_b/2. With practical NRZ/AMI pulses, the first-null bandwidth is about RbR_b.

SystemBit rateMin. bandwidth Rb/2R_b/2First-null BW ≈ RbR_b
T11.544 Mb/s772 kHz1.544 MHz
E12.048 Mb/s1.024 MHz2.048 MHz

DPCM encoder

Differential PCM uses the fact that speech samples next to each other are strongly correlated. It does not quantise each sample directly. It quantises the difference between the sample and a prediction made from earlier samples. The difference is smaller than the sample, so fewer bits are needed for the same quality.

 m[n] ->(+)--e[n]-->[Quantizer]--eq[n]--+-->[Encoder]-> bits
        ^-                              |
        |                               v
        |                   m^[n] ---> (+)
        |                     ^         | mq[n]
        +-------- m^[n] --[Predictor]<--+
  • Prediction error: e[n]=m[n]−m^[n]e[n] = m[n] - \hat m[n].
  • Quantised error: eq[n]=e[n]+q[n]e_q[n] = e[n] + q[n].
  • Predictor input: mq[n]=m^[n]+eq[n]=m[n]+q[n]m_q[n] = \hat m[n] + e_q[n] = m[n] + q[n].

The predictor (a linear filter on past values of mqm_q) works on the quantised samples. The receiver has the same predictor, so encoder and decoder stay in step and quantisation errors do not build up.

The SNR improves by the prediction gain Gp=σm2/σe2G_p = \sigma_m^2 / \sigma_e^2. DPCM saves about 1–2 bits per sample compared with PCM, e.g. 32–48 kb/s instead of 64 kb/s for speech.

  • 2069 Chaitra (CS II) · 4+3 marks

Explain the E1 digital hierarchy. A speech signal with maximum frequency of 4 kHz and maximum amplitude of ±1.1 V is applied to a PCM system with its bit rate of 32 kbps. Calculate the SQNR and number of bits per sample.

Answer

E1 digital hierarchy

E1 is the ITU-T TDM-PCM carrier. Each frame carries 32 time slots of 8 bits in 125 µs: TS0 for frame sync, TS16 for signalling, and 30 voice slots. Its rate is Rb=256×8000=2.048R_b = 256 \times 8000 = 2.048 Mb/s. Higher levels multiplex 4 lower streams each, adding framing and stuffing bits.

LevelMade ofChannelsBit rate
E130 voice + 2302.048 Mb/s
E24 × E11208.448 Mb/s
E34 × E248034.368 Mb/s
E44 × E31920139.264 Mb/s
E54 × E47680564.992 Mb/s

E1 uses A-law companding and the HDB3 line code.

Numerical

Given: fm=4f_m = 4 kHz, amplitude ±1.1\pm 1.1 V, Rb=32R_b = 32 kb/s. Assumption: Nyquist-rate sampling, a uniform quantiser covering ±1.1\pm 1.1 V, and a full-scale sinusoidal test signal.

Sampling rate:

fs=2fm=2×4=8 kHzf_s = 2 f_m = 2 \times 4 = 8\ \text{kHz}

Bits per sample:

n=Rbfs=32 0008000=4 bits,L=24=16 levelsn = \frac{R_b}{f_s} = \frac{32\,000}{8000} = 4\ \text{bits}, \qquad L = 2^4 = 16\ \text{levels}

SQNR:

Δ=2×1.116=0.1375 VNq=Δ212=0.1375212=1.5755×10−3 V2Ps=(1.1)22=0.605 V2SQNR=0.6051.5755×10−3=384=32L2SQNR (dB)=10log⁡10384=25.84 dB\begin{aligned} \Delta &= \frac{2 \times 1.1}{16} = 0.1375\ \text{V} \\ N_q &= \frac{\Delta^2}{12} = \frac{0.1375^2}{12} = 1.5755 \times 10^{-3}\ \text{V}^2 \\ P_s &= \frac{(1.1)^2}{2} = 0.605\ \text{V}^2 \\ \text{SQNR} &= \frac{0.605}{1.5755 \times 10^{-3}} = 384 = \tfrac{3}{2} L^2 \\ \text{SQNR (dB)} &= 10 \log_{10} 384 = 25.84\ \text{dB} \end{aligned}

This agrees with the rule SQNR=1.76+6.02n=1.76+24.08=25.84\text{SQNR} = 1.76 + 6.02n = 1.76 + 24.08 = 25.84 dB.

Answer: n=4n = 4 bits per sample; SQNR = 384 ≈ 25.84 dB.

Questions from Old Question Collection (BEI EX 656) (BEI Communication Systems (EX 656) exam papers, 2078 to 2081 Chaitra), Communication System I (EX 652) (BEX Communication System I (EX 652) papers 2064 to 2080, plus two old BCT Communication Systems papers (2068, 2071)) and Communication System II (EX 702) (BEX Communication System II (EX 702) exam papers, 2069 to 2081). Answers are written for this site; check them against your class notes.

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