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    Frame Structure of the Digital Signal Hierarchies 2..4 Siemens

    TT2510EU01AL_011

    Contents

    1 Frame Structure of 8, 34, 140 Mbit/s Hierarchies 3

    2 Timing Sequence of the Multiplex Process 11

    3 Transmission of Additional Data Channels with Y-Bits 13

    Frame Structure of the Digital SignalHierarchies 2..4

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    1 Frame Structure of 8, 34, 140 Mbit/sHierarchies

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    Alignment word

    The frame of all hierarchy levels begin with the frame alignment word (FAW), by

    means of which the receiving system (demultiplexer) detects the beginning of theframe and is thus able to interpret the following bit positions correctly. Besides, an in-service-supervision of the incoming signals bit error rate can be performed bycontinuous evaluation of the FAW.

    Signaling bits D, N

    Immediately after the FAS the signaling bits D and N are transmitted. They provideinformation about the state of the opposite transmission direction. Hereby, urgentalarms (failure) are signaled via the D-bit (remote alarm indication RAI), and non-urgent alarms (interference) via the N-bit. If it is possible to renounce the backward

    transmission of non-urgent alarms, the N-bit can be used for the asynchronoustransmission of external data (so-called Y-data channels via V.11 interface).

    Blocks TB

    Here the signals (tributary bits) of channels 1...4 are transmitted bit-by-bit interleaved.

    Blocks JS

    These blocks consist of 4 bits and contain the justification service bits of channels1...4. In order to provide a protection against transmission errors, the justification bitsare transmitted in a redundant way and evaluated on the receiving end by majority

    decision. The 3 JS blocks (5 at 140 Mbit/s) contain the same information in the biterror-free state. If, due to a transmission error, one of the justification service bits (2at 140 Mbit/s) is wrongly detected, the majority decision nevertheless allows thecorrect evaluation of the following justification bit positions. A wrong interpretation ofthe justification bit position would inevitably result in a desynchronization of theaffected subsystem.

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    10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2

    102

    104

    106

    108

    1010

    1012

    1014

    100

    1 year

    1 day

    1 h

    3.10-5

    (3J)

    (2J)

    Average time interval between

    loss of frame alignment due

    to errors, s

    Line error rate Pe

    Fig. 1

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    Probability of a loss of synchronization depending on the bit error rate

    due to

    l Loss of the justification information: (2 J) at 8 Mbit/s

    (3 J) at 140 Mbit/s

    l Loss of the frame alignment: (2) at 8 Mbit/s

    (3) at 140 Mbit/s

    Block JT

    This block contains the justification bit positions (justifying bit or tributary bit) and isintegrated into a TB block. By use respectively non-use of this bit position, thetransmission capacity is matched of the individual channels (as described in theprevious sections).

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    8-Mbit/s-Pulse Frame

    Number of bits per block4x212=848 bits

    1 10 1112 13 212 1 4 5 212 1 4 5 212 5 8 9

    TB (200 bit) TB (208 bit) TB (208 bit) TB (208 bit)JS JS JS

    JT

    Block I II III IV

    ND0000101111

    Service

    bits

    Frame alignment

    signal

    TB Tributary bits

    JS Justification service

    bits

    JT Justifying bit or

    tributory bit

    of the 2-Mbit/s-

    signals Nr. 1 to 4

    bit-by-bit interleaved

    1 4

    Bitrate: either 8448 kbit/s 3 x 10-5

    or 8448 kbit/s 30 ppm

    Number of bits per frame: 848Number of tributary bits per frame: 820...824Frame length: 100.38 sNominal justification bitrate: ca. 4.23 kbit/s

    Fig. 2

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    34-Mbit/s-Pulse Frame

    Number of bits per block4x384=1.536 bits

    1 10 1112 13 384 1 4 5 384 1 4 5 384 1 4 5 8 9 384

    TB (372 bit) TB (380 bit) TB (380 bit)

    JT

    Block I II III IV

    ND0000101111

    Service

    bits

    Frame alignment

    signal

    TB Tributary bits

    JS Justification service

    bits

    JT Justifying bit or

    tributory bit

    8-Mbit/s-signals

    Nr. 1 to 4

    bit-by-bit interleaved

    TB (380 bit) JSJSJS

    Bitrate: either 34368 kbit/s 2 x 10-5

    or 34368 kbit/s 20 ppmNumber of bits per frame: 1536Number of tributary bits per frame: 1508...1512Frame length: 44.69 sNominal justification bitrate: ca. 9.75 kbit/s

    Fig. 3

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    140-Mbit/s-Pulse-Frame

    Y1

    Number of bits per block6x488=2928 bits

    1 12 17 488 1 4 5 488 1 4 5 488 1 4

    JS JS JS

    Block I II III IV

    9 488

    Frame alignment

    signal

    TB Tributary bits

    JS Justi ficat ion service

    bitsJT Justifying bit or

    tributory bit

    1 4

    JS

    V

    1 4

    VI

    TB(472 bit)

    TB(484 bit)

    TB(484 bit)

    1 45 85 488 5 488

    TB(484 bit)

    TB(484 bit)

    JS TB(484 bit)

    JT

    000001011111 Y2ND

    Service

    bits

    13 16

    Bitrate: either 139264 kbit/s 15 x 10-6

    or 139264 kbit/s 15 ppmNumber of bits per frame: 2928Number of tributary bits per frame: 2888...2892Frame length: 21.03 sNominal justification bitrate: ca. 19.93 kbit/s

    Fig. 4

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    In addition to the service bits this frame has two data bits, which can be used for theasynchronous transmission of external data signals with bitrates of up to approx.10 kbit/s.

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    2 Timing Sequence of the Multiplex Process

    0

    T

    1 1 1

    J

    00 0

    Tcontent of the justification

    bit position JT

    justification service bitposition JS

    T: occupied with tributary signal

    J: occupied with justification bit

    1: Justification bit in following JTposition

    0: Tributary bit in following JT

    position

    JS/JT FAS JS JS JS/JT FAS JS JS JS FAS

    Address difference bit

    Frame

    3

    4

    5

    6

    N N+1 N+2

    Fig. 5 Time sequence of the address difference in the elastic store depending on the frame structure

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    The figure shows an example of the time sequence of the address differencebetween write in respect. read out address of the elastic store along several frames.The rising edges (reduction of the distance between addresses) occur in the tributary

    information blocks, i.e. when the store is read out. Every time if no tributaryinformation is transmitted (with JS, FAS) the read out process is interrupted (1 clockat JS, 3 clock at FAS) and the difference between the addresses increasesaccordingly (1 address at JS, 3 addresses at FAS). Despite of these interruptions ofthe read out process, the actual read out timing frequency is higher than the write infrequency.

    The necessity of the justification is determined in frame Nr. N (getting below thedifference of 3 bits). Then all justification service bits (JS) of the affected channel areset to 1 in frame N + 1.

    When the justification bit position (JT) in frame N + 1 is reached, the read-out

    address is stopped during one clock cycle ; the transmitted bit is interpreted asjustification bit and the difference between the addresses increases by one address.

    Example:

    Let us look at a multiplex system 2/8 Mbit/s and at the structure of the 8 Mbit frame.The nominal write in frequency is fE = 2048 kHz. The elastic store is read out duringthe tributary information blocks with one fourth of the system clock:

    fP = 8448 kHz/4 = 2112 kHz

    This read out clock is interrupted by the JS, FAS blocks. The actual read out rateresults from the relation of tributary bits per frame to the overall number of bits:

    kHzbits

    bitskHzf

    M23.2052

    848

    8242112 ==

    The nominal justification bitrate fj can be calculated from the difference between fM

    and fE.

    fj = 2052.23 kHz - 2048 kHz = 4.23 kHz

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    3 Transmission of Additional Data Channelswith Y-Bits

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    With the frame of the 8 and 34 Mbit/s hierarchy the N-alarm bits can be used fortransmission of external data channels. The 140 Mbit/s frame has two speciallydesigned bit positions (so-called Y-bits) for this purpose. The bitrate for one Y-bit

    corresponds to the frame clock.This bitrate cannot be used for the external signal, as for this a synchronization to theframe clock would be necessary. Therefore, the maximum allowable bitrate for theexternal data signal is restricted to approx. one fifth of the Y-bitrate. One bit of thesignal to be transmitted is then sampled several times by the Y-bits (Oversampling).

    The distortion of the transmitted signal results from the relation of the Y-bitrate to thebitrate of the signal to be transmitted.

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    original

    signal

    Y-bits

    reproduced

    signal

    0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 0

    Fig. 6

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