Top Banner

of 50

Multiplexing in Communication system

Jun 02, 2018

Download

Documents

Bhavin V Kakani
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
  • 8/11/2019 Multiplexing in Communication system

    1/50

    Multiplexing in

    Communication SystemBHAVIN V KAKANI

    IT, NU

  • 8/11/2019 Multiplexing in Communication system

    2/50

    Multiplexing

    Multiplexing is the set of techniques thatallows the simultaneous transmission ofmultiple signals across a single data link.

    A Multiplexer(MUX) is a device thatcombines several signals into a singlesignal.

    A Demultiplexer(DEMUX) is a device thatperforms the inverse operation.

  • 8/11/2019 Multiplexing in Communication system

    3/50

  • 8/11/2019 Multiplexing in Communication system

    4/50

    Categories of Multiplexing

  • 8/11/2019 Multiplexing in Communication system

    5/50

    Frequency-division Multiplexing (FDM)

    FDM is an analog technique that can be applied whenthe bandwidth of a link is greater than the combinedbandwidths of the signals to be transmitted.

  • 8/11/2019 Multiplexing in Communication system

    6/50

    Frequency-division Multiplexing (FDM)

    In FDM signalsgenerated by eachdevice modulatedifferent carrierfrequencies. These

    modulated signals arecombined into asingle compositesignal that can betransported by the

    link.

    FDM is an analog mul tiplexing technique

    that combines signals.

  • 8/11/2019 Multiplexing in Communication system

    7/50

    Modulating (Modulation)

    In analog transmission, the sendingdevice produces a high frequency signal(a sine wave) that acts as a basis for the

    information signal. This base signal iscalled the carrier signal.

    Digital information is then modulated onthe carrier signal by modifying one or

    more of its characteristics (amplitude,frequency, phase). This kind ofmodification is called modulation andthe information signal is called amodulating signal.

  • 8/11/2019 Multiplexing in Communication system

    8/50

    Modulation (Amplitude Shift keyingASK)

    Time-domain description

  • 8/11/2019 Multiplexing in Communication system

    9/50

    Modulation (Amplitude Shift keyingASK)

    Frequency-domain description

  • 8/11/2019 Multiplexing in Communication system

    10/50

    Frequency-division Multiplexing (FDM)

    In FDM signals generated by each device modulate

    different carrier frequencies. These modulated

    signals are combined into a single composite signal

    that can be transported by the link.

    Carrier frequencies are separated by enoughbandwidth to accommodate the modulated signal.

    These bandwidth ranges arte the channels through

    which various signals travel.

    Channels must separated by strips of unusedbandwidth (guard bands) to prevent signal

    overlapping.

  • 8/11/2019 Multiplexing in Communication system

    11/50

  • 8/11/2019 Multiplexing in Communication system

    12/50

  • 8/11/2019 Multiplexing in Communication system

    13/50

    Frequency-division Multiplexing (FDM)

    In FDM, signals are modulated onto separatecarrier frequencies using either AM or FMmodulation.

  • 8/11/2019 Multiplexing in Communication system

    14/50

    Example 1

    Assume that a voice channel occupies a bandwidth of 4

    KHz. We need to combine three voice channels into a link

    with a bandwidth of 12 KHz, from 20 to 32 KHz. Show

    the configuration using the frequency domain without the

    use of guard bands.

    Solution

    Shift (modulate) each of the three voice channels to adifferent bandwidth, as shown in Figure 6.6.

  • 8/11/2019 Multiplexing in Communication system

    15/50

    Example 1

  • 8/11/2019 Multiplexing in Communication system

    16/50

    Example 2

    Five channels, each with a 100-KHz bandwidth, are to be

    multiplexed together. What is the minimum bandwidth of

    the link if there is a need for a guard band of 10 KHz

    between the channels to prevent interference?

    Solution

    For five channels, we need at least four guard bands.

    This means that the required bandwidth is at least

    5 x 100 + 4 x 10 = 540 KHz,

    as shown in Figure 6.7.

  • 8/11/2019 Multiplexing in Communication system

    17/50

    Example 2

  • 8/11/2019 Multiplexing in Communication system

    18/50

    Wave-division Multiplexing (WDM)

    Wave-division multiplexing isconceptually the same as FDM, exceptthat multiplexing and demultiplexing

    involve light signals transmitted throughfiber-optic channels.

    The purpose is to combine multiple lightsources into one single light at the

    multiplexer and do the reverse at thedemultiplexer.

    Combining and splitting of light sourcesare easily handled by a prism.

  • 8/11/2019 Multiplexing in Communication system

    19/50

    The only difference with electrical FDM is that onoptical system is completely passive and thushighly reliable.

    Due to its enormous bandwidth of around 25KGHz, there is a potential of multiplexing manychannels together over long routes.

    Potential application of WDM is in the FTTC (FiberTo The Curb) systems.

    Important application of WDM is the SONETnetworks in which multiple optical fiber lines aremultiplexed.

  • 8/11/2019 Multiplexing in Communication system

    20/50

  • 8/11/2019 Multiplexing in Communication system

    21/50

    Time-division Multiplexing (TDM)

    Time-division multiplexing (TDM) is a digital process thatcan be applied when the data rate capacity of thetransmission medium is greater than the data raterequired by the sending and receiving devices.

  • 8/11/2019 Multiplexing in Communication system

    22/50

    Data Rate

    For T DM, the data rate of the multiplexed signal isalways n times the data rate of the individualsources, where n is the number of sources.

    If 3 sources are multiplexed, then the data rate ofthe TDM signal is 3 times higher than the individualdata rate.

  • 8/11/2019 Multiplexing in Communication system

    23/50

    TDM

    TDM is a digital mul tiplexing technique to

    combine data.

  • 8/11/2019 Multiplexing in Communication system

    24/50

    Time-division Multiplexing (TDM)

    TDM can be implemented in two ways:synchronous TDM and asynchronous TDM.

    In synchronous time-division multiplexing, the

    term synchronous means that the multiplexerallocates exactly the same time slot to eachdevice at all times, whether or not a devicehas anything to transmit.

    Frames

    Time slots are grouped into frames. A frameconsists of a one complete cycle of timeslots, including one or more slots dedicatedto each sending device.

  • 8/11/2019 Multiplexing in Communication system

    25/50

  • 8/11/2019 Multiplexing in Communication system

    26/50

    TDM frames

  • 8/11/2019 Multiplexing in Communication system

    27/50

    Example 5

    Four 1-Kbps connections are multiplexed together. A unit

    is 1 bit. Find (1) the duration of 1 bit before multiplexing,

    (2) the transmission rate of the link, (3) the duration of a

    time slot, and (4) the duration of a frame?

    Solution

    We can answer the questions as follows:

    1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms).2. The rate of the l ink is 4 Kbps.

    3. The duration of each time slot 1/4 ms or 250 s.

    4. The duration of a frame 1 ms.

  • 8/11/2019 Multiplexing in Communication system

    28/50

  • 8/11/2019 Multiplexing in Communication system

    29/50

    I nter leaving

  • 8/11/2019 Multiplexing in Communication system

    30/50

    Example 6

    Four channels are multiplexed using TDM. If each

    channel sends 100 bytes/s and we multiplex 1 byte per

    channel, show the frame traveling on the link, the size of

    the frame, the duration of a frame, the frame rate, and the

    bit rate for the link.

    Solution

    The multiplexer is shown in Figure 6.15.

    E l 6

  • 8/11/2019 Multiplexing in Communication system

    31/50

    Example 6

  • 8/11/2019 Multiplexing in Communication system

    32/50

    Example 7

    A multiplexer combines four 100-Kbps channels using a

    time slot of 2 bits. Show the output with four arbitrary

    inputs. What is the frame rate? What is the frame

    duration? What is the bit rate? What is the bit duration?

    Solution

    Figure 6.16 shows the output for four arbitrary inputs.

    E l 7

  • 8/11/2019 Multiplexing in Communication system

    33/50

    Example 7

  • 8/11/2019 Multiplexing in Communication system

    34/50

    Time-division Multiplexing (TDM)

    Framing Bits

    Because the time slot order in a synchronous

    TDM system does not vary from frame to

    frame, very little overhead information needs

    to be included in each frame. However, one

    or more synchronization bits are usually

    added to the beginning of each frame.

    These bits, called framing bits, allows thedemultiplexer to synchronize with the

    incoming stream so that it can separate the

    time slot accurately.

    F i bit

  • 8/11/2019 Multiplexing in Communication system

    35/50

    Framing bits

  • 8/11/2019 Multiplexing in Communication system

    36/50

    Example

    Suppose that we have four input devices on a synchronous TDM

    link, where the transmissions are interleaved by character. If

    each device is generating 250 characters per second, and each

    frame is carrying 1 character from each device, what is the

    minimum data rate of this link?

  • 8/11/2019 Multiplexing in Communication system

    37/50

    The link must be able to carry250 frames per second.

    If we assume that eachcharacter consists of 8bits, then each frame has4x8 + 1= 33 bits ( 32 bits forthe four characters plus 1framing bit).

    On the other hand, eachdevice is creating2000bps, because 250characters per second x 8bits =2000 bits per second,and the link is carrying8250 bps, because 250frames per second x33 bitsis 8250 bps.

  • 8/11/2019 Multiplexing in Communication system

    38/50

    Example 8

    We have four sources, each creating 250 characters persecond. If the interleaved unit is a character and 1

    synchronizing bit is added to each frame, find (1) the data

    rate of each source, (2) the duration of each character in

    each source, (3) the frame rate, (4) the duration of each

    frame, (5) the number of bits in each frame, and (6) the

    data rate of the link.

    Solution

    See next slide.

  • 8/11/2019 Multiplexing in Communication system

    39/50

    Solution (continued)

    We can answer the questions as follows:

    1. The data rate of each source is 2000 bps = 2 Kbps.

    2. The duration of a character is 1/250 s, or 4 ms.

    3. The link needs to send 250 frames per second.4. The duration of each frame is 1/250 s, or 4 ms.

    5. Each frame is 4 x 8 + 1 = 33 bits.

    6. The data rate of the link is 250 x 33, or 8250 bps.

  • 8/11/2019 Multiplexing in Communication system

    40/50

    Example 9

    Two channels, one with a bit rate of 100 Kbps and

    another with a bit rate of 200 Kbps, are to be multiplexed.How this can be achieved? What is the frame rate? What

    is the frame duration? What is the bit rate of the link?

    Solution

    We can allocate one slot to the first channel and two slots

    to the second channel. Each frame carries 3 bits. The

    frame rate is 100,000 frames per second because itcarries 1 bit from the first channel. The frame duration is

    1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x

    3 bits/frame, or 300 Kbps.

    DS hierarchy

  • 8/11/2019 Multiplexing in Communication system

    41/50

    DS hierarchy

  • 8/11/2019 Multiplexing in Communication system

    42/50

    Table 6.1 DS and T l ines rates

    Service LineRate

    (Mbps)

    Voice

    Channels

    DS-1 T-1 1.544 24

    DS-2 T-2 6.312 96

    DS-3 T-3 44.736 672

    DS-4 T-4 274.176 4032

  • 8/11/2019 Multiplexing in Communication system

    43/50

    Table 6.2 E line rates

    E LineRate

    (Mbps)

    Voice

    Channels

    E-1 2.048 30

    E-2 8.448 120

    E-3 34.368 480

    E-4 139.264 1920

  • 8/11/2019 Multiplexing in Communication system

    44/50

  • 8/11/2019 Multiplexing in Communication system

    45/50

    Asynchronous TDM

    Synchronous TDM does not guarantee that the full

    capacity of a link is used. Because the time slots arepreassigned and fixed, whenever a connected device

    is not transmitting, the corresponding slot is empty.

    Asynchronous time-division multiplexing, or statistical

    time-division multiplexing, is designed to avoid thistype of waste.

    Like synchronous TDM, asynchronous TDM allows a

    number of lower-speed input lines to be multiplexed

    to a single higher-speed line. However, in

    asynchronous TDM the total speed of the input lines

    can be greater than the capacity of the link.

  • 8/11/2019 Multiplexing in Communication system

    46/50

    In an asynchronous system,

    if we have n input lines, the

    frame contains no more than

    m slots, with m less than n.

    The number of time slots in

    an asynchronous TDM

    frame (m) is based onstatistical analysis of the

    number of input lines that

    are likely to be transmitting

    at any given time.

    In this case any slot is

    available to any of the

    attached input lines that has

    data to send.

  • 8/11/2019 Multiplexing in Communication system

    47/50

  • 8/11/2019 Multiplexing in Communication system

    48/50

    Asynchronous TDM

    Addressing and Overhead

    In asynchronous TDM each time slot mustcarry an address telling the demultiplexerhow direct the data. This address, forlocal use only, is attached by themultiplexer and discarded by the

    demultiplexer once it has been read. Asynchronous TDM is efficient only when

    the size of the time slots kept relativelylarge.

  • 8/11/2019 Multiplexing in Communication system

    49/50

    Inverse Multiplexing

    Inverse multiplexing takesthe data stream from onehigh-speed line andbreaks it into portions thatcan be sent across severallower-speed linessimultaneously, with noloss in the collective datarate.

    Figure 6.21 Multiplexing and inverse mul tiplexing

  • 8/11/2019 Multiplexing in Communication system

    50/50