tdm mux

Copyright 1999 by Marko VuskovicC

M. Vuskovic Broadband Communication Networks CS678

Appendix A

DIGITAL SIGNALS, TRANSMISSIONMEDIA AND MULTIPLEXING

Table of contents:

A.1 DIGITAL SIGNALS A.1.1 Digital Signal Encoding A.1.2 Fourier Series A.1.3 Average Power and Bandwidth A.1.4 Relation Between Bandwidth and Data Rate A.1.5 Digitized Voice Signal A.1.6 Sampling Theorem A.1.7 Data Rate of a Digitized Voice Signal A.1.8 Teleconferencing and Music A.1.9 Video Signals

A.2 TRANSMISSION MEDIA A.2.1 Two-Wire Open Lines A.2.2 Twisted-Pair Lines A.2.3 Coaxial Cable A.2.4 Optical Fiber A.3 MULTIPLEXING A.3.1 Frequency-Division Multiplexing A.3.2 Principle of the Frequency Shift A.3.3 Time-Division Multiplexing A.3.4 Frame Alignment in TDM A.3.5 Duplex Multiplexers A.3.6 TDM Hierarchies A.3.7 Digital Signal Service (DS) A.3.8 Integrated Services Digital Network (ISDN) A.3.9 Synchronous Optical Network (SONET) A.3.10 Statistical Multiplexing

A-2


NOTICE: This appendix is focused on the physical OSI layer of computer networks. It covers the most important facts about digital transmission media and communication infrastructure, which are needed to understand other appendices and chapters. Analog transmission media and modems are not covered.

The appendix doesn't require a prior background in computer networks.

A-3


DIGITAL SIGNALS

Digital Signal Encoding

How is binary data (bit patterns like 01101000) represented by voltage (or light)?

NRZ Encoding Non-Return-to-Zero encoding. The simplest and most common encoding: logical 0 represented by high voltage level, logical 1 by low voltage level.

time

Highvoltage

Lowvoltage

0 1 1 0 1 0 0 0

First bit Last bit

time

Highvoltage

Lowvoltage

time

Highvoltage

Lowvoltage

0 0 0 0 0 0 0 0

Can't read data in case of longer sequences of zeros or ones -extremely precise and well synchronized clocks needed at transmiterand receiver side.

How many zeros?

Problem:

A-4


DIGITAL SIGNALS (Cont.)

NRZI Encoding Non-Return-to-Zero Invert on ones. Presence or absence of signal transition: transition low to high, or high to low means bit value 1, no transition means bit value 0.

This is an example of differential encoding. The synchronization problem remains, but the noise immunity improves: the bit value determined by comparing adjacent voltage levels instead of comparing with an absolute value is more reliable. Also, the sense of polarity can't be lost even if leads are reversed. With differential encoding, wires are symmetric (no polarity mark).

time

Highvoltage

Lowvoltage

0 1 1 0 1 0 0 0

First bit Last bit

A-5



time

Highvoltage

Lowvoltage

0 1 1 0 1 0 0 0

First bit Last bit

Manchester Encoding Transition of signal levels at the middle of a bit interval (low to high transition represents bit value 1, high to low transition represents 0.) There is always a transition regardles of the encoded bit pattern. These transitions can therefore be used for synchronization. Manchester encoding is also called: self-synchronizing code, or self-clocking code and is used in Ethernet LANs). Manchester encoding is noise imune (can even perform on-the-fly error recovery if one bit is corrupted), however it is polarity sensitive.

Differential Manchester Encoding

Transitions at the middle of bit intervals always exist but they are used only for clocking. The encoding is based on transitions at the beginning of the bit interval: presence of transition menas 0, absence of transition means 1.

Differential Manchester encoding maintains the advantages of the NRZI and Manchester encoding encoding (noise immunity) and is popular in token ring LANs. Manchester encoding is polarity independent.

time

Highvoltage

Lowvoltage

0 1 1 0 1 0 0 0

First bit Last bit

A-6



time

+V

-V

0 1 1 0 1 0 0 0

First bit Last bit

0

Manchester encoding requires changes of signal levels two times during a bit interval. This dou-bles the required bit rate, which is acceptable in LANs. However, in WANs the distances and the bit rates are much higher and hence the bit rate becomes more important. The following two en-coding schemes solve this problem. They use three-level codes (also called bipolar codes).

AMI Encoding (Alternate Mark Inversion)

0s are represented by a zero voltage level, while 1s are represented by +V or -Vin alternation (first 1 is positive). Consequently, there is no loss of synchronization in thecase of long strings of 1s. However, long strings of 0s still present a problem. Because 1s alternate, there is no DC component (i.e. the mean values of the signal is zero.)

In addition, the pulse alternation provides good error detection. For example, two adjacent voltage levels with the same polarity indicate an error. Similarly, deleting a pulse can be detected because the alternation rule is violated.

NOTE: There is a similar bipolar encoding standard called pseudoternary encoding with 0s and 1s reversed (1s - no voltage, 0s +V or -V in alternation). This encoding is used in ISDN.

A-7



B8ZS Encoding (Bipolar with 8-bit zero substitution)

Based on AMI, B8ZS solves the problem of long sequences of 0s by adding a special bit pattern - technique called bit scrambling. In B8ZS, the bit scrambling is applied whenever there are 8 or more consecutive 0s. The sequence of 8 zeros is replaced with the patterns (000+-0-+) or (000-+0+-), depending on the polarity of the last nonzero bit (before 8 or more zeros.)

time

+V

-V

Positive level proceeds the long sequence of 0s

0

8 bits (previously all 0s)

Violates the alternation rule,meaning either an error, or an

inserted bit pattern(violation bit, V bit)

Balance bit, B bit(used to balance out the dc

component of the inserted bit pattern)

Inserted bit pattern (+,-,0,-,+)

time

+V

-V

Negative level proceeds the long sequence of 0s

0

8 bits (previously all 0s)

Violates the alternation rule,meaning either an error, or an

inserted bit patternInserted bit pattern (-,+,0,+,-)

The receiver can easily recognize the inserted bit patterns and replace them with zeros.It is not very likely that a 5-bit error occurs that exactly matches the scrambling bit pattern. The B8ZS encoding is used in North America. Europe and Japan use a similarscheme called HDB3 (High density bipolar-3 zeros) also based on AMI.

A-8



There are two important concepts used in this section: bandwidth and bit rate. In order to understand these concepts, we need to use some basic facts about Fourier analysis.

Fourier Series

Fourier showed that each periodic signal s(t) = s(t+T), where T is the period, can be represented by an infinite time series of sinusoidal signals:

)sin()cos()( 001

0 tibtiaats ii

i ++=

=

Tf 22 00 ==

dttitsT

b

dttitsT

a

dttsT

a

T

i

T

i

T

)sin()(2

)cos()(2

)(1

00

00

00

=

=

=

where

The coefficients of the time series above are constants, which can be computed as follows:

is angular velocity and

is frequency (measured in Hz).

A-9



t

+A

-A

0

T

++++= )7sin(71)5sin(

51)3sin(

31)sin(4)( 0000 tttt

Ats

Example 1

f0

1

1/3

1/51/7 1/9

Signal in time domain

Signal in frequency domain

SPECTRAL DIAGRAM

First harmonic(fundamental)harmonic)

2fofo 3fo 4fo 5fo 6fo 7fo 8fo 9fo

Third harmonic

A4

A-10



Used from: "Understanding Data Communications and networks"by W. A. Shay, PSW Pub. Co., Boston, 1995

This diagram shows the goodness of approximation in terms of the number of spectral components.

A-11



t

+A

-A

0

T

+++= )5sin()5cos(51)3sin()3cos(

31)sin()cos(4)( 000 ttt

Ats

+= )7sin(71)5sin(

51)3sin(

31)sin(4)4/cos()( 0000 tttt

Ats

=

T

212

,4T

=

,4

=

Example 2

where:

Special case:

thus

giving:

A-12



Average Power and Bandwidth From the Fourier transformation it is clear that if we want to transmit a signal across a

communication channel without distortion, we need to allow all harmonics to be transmit-ted. That would require an infinite bandwidth of the media, which is practically not possible.

If we transmit a signal over a channel with a limited bandwidth, the signal at the receiv-er's side will be distorted. The question is whether the distorted signal can be fully recov-ered?

In case of binary signals the bit values are normally sampled at the middle of the bit in-terval. Suppose we have a bit pattern of alternating 0s and 1s (0101010101...). If we rep-resent the signal with a square wave form as in example 1, it becomes clear that such a signal can be recovered from just the first harmonic. Therefore, a bandwidth of the com-munication channel limited by the first harmonic would be sufficient.

t

+A

-A

0

T

Two bits -one period Sampling pulses

A-13



dttsEb

a=

2)(

dttsT

PT

=0

2)(1

For periodic signals it is more convenient to use the definition of the average power over a period T:

It is usually said that a communication channel must be able to transmit those harmonics which contain more than one half of the signal's energy (half-power bandwidth).

If the instantaneous power of the signal is defined as , the energy of the signal con-tained in a time interval a

A-14



Relation Between Bandwidth and Bit Rate

A bit pattern with the maximal number of transitions is an alternating pattern of zeroes and ones: 1010101010101... Such signal would require a maximal bandwidth. There-fore, the pattern 10101010101... can be used as a worst case scenario for a transmis-sion channel. If the pattern is encoded with NRZ we can see that the required bandwidth (B) is equal to the frequency of the first harmonic, i.e. 1/T. Since one period covers two bits (group 01, or 10), we can derive the following conclusion: data rate of a binary signal (D) is twice of its bandwidth:

Data rate, also called bit rate, is measured in:

bps - bits per second kbps - kilo bits per second (1 kbps = 1000 bps) Mbps - mega bits per second (1 Mbps = 1000 kbps)

Bandwidth of a transmission media is the maximal frequency of a sinusoidal signal whichcan go through the media without a significant attenuation. Bit rate of a transmission mediais the number of bits per second that can go through the media and can be read at theother end.

NOTICE: Communication engineers always use the word "bandwidth" in its strict sense - the width of the frequency spectrum of a signal, or of a transmission media, measured in Hz. The computer network designers and managers very often use the word "bandwidth" for the bit rate measured in bps. This may sometimes create a confusion.

D = 2 B

A-15


Digitized Voice Signal

A voice signal is an analog signal which is more complex than the periodic squarewave signal considered in the previous section. To simplify, we can assume that a voice signal is a mixture of sinusoidal signals in which frequencies are in the continuous range be-tween 300 Hz and 3.4 kHz. In order to transmit a voice signal across a digital transmis-sion channel, it has to be converted from analog form to a digital form. This is normally done in two phases: sampling and quantization.

Digitizedvoice signal

Analogvoicesignal

PAMsignal

PCMsignal

Sampling Quantization

Clock(125 s)

PAM - Pulse Amplitude ModulationPCM - Pulse Code Modulation

Analogvoicesignal

PAMsignal

PCMsignal

t

t

t

125 s 125 s

6.2

10.18.1

6.07.9

13.3

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

8-bit (256-level)quantization

Sampling period


Question: How do we determine the required sampling rate for a voice signal?

A-16


Sampling Theorem

There is a required minimal sampling rate by which an analog signal must be sampled in order to be able to reconstruct the original analog signal from its PAM form. This rate can be defined by the Nyquist sampling theorem:

The amplitude of an analog signal must be sampled at a minimum rate of greater than twice the highest frequency component.

NOTICE: Theoretically, most analog signals have an indefinite bandwidth. However, for practical reasons, we apply the limited bandwidth, such as half-power bandwidth.

Data Rate of a Digitized Voice Signal

A significant part of the energy of a voice signal is between 300 and 3400 Hz. This can be rounded to a bandwidth of 4000 Hz. The required sampling rate is therefore 8 kHz,i.e. the sampling period is 1/8000 = 125 x 10-6 sec = 125 s. For reasonably good voice quality (i.e. telephony) the voice signal is quantized into 256 levels, which requires 8-bit coding. The bit rate of a digitized voice signal is therefore 8000 samples/sec x 8 bits/sample = 64,000 bps = 64 kbps.

t

T/2

s(t)x(t)

Sampling period s(t) or x(t) ?


Given sample values,we need to reconstructthe sinusoidal signal

A-17


(MPEG = Motion Pictures Expert Group)

Teleconferencing and Music

Teleconferencing involves several persons speaking simultaneously. Therefore the requirement for teleconferencing is much higher than in telephony, where we suppose transmission of a single voice. The standard for teleconferencing is a sampling rate of 16 kHz x 8 bit coding which gives 128 kbps. The compression of this signal can reduce the bit rate below 64 kbps.

Music on CD ROM requires an even higher data rate, which is 44.1 kHz x 16 bit = 705.6 kbps. This can be reduced by compression to 128 kbps.

Video Signals


Transactiontype

FormatName

SignalStructure

Raw bit rate[Mbps]

Compressedbit rate[Mbps]

VideoConferencing

TV

High qualityTV

MPEG-1(NTSC)

MPEG-2(NTSC)

HDTV

352 pixel x 240 linex 12 bit x 30 frame/s



30.4

124.4

995.3

A-18


TRANSMISSION MEDIA

Two-Wire Open Lines

The simplest.

n(t)m(t)

r(t) = (s(t)+m(t)+n(t)) - (-s(t)+n(t)) = 2 s(t) + m(t)

n(t)

RT

RT - Termination resistor

s(t)

-s(t)

NoiseSender Receiver

TTL TTL

-

+

Termination resistance is used to terminate the line so that the receiver absorbs theentire signal, otherwise the signal would bounce back towards the sender (reflection),which would cause more noise. This resistor has a value of the characteristic impedance.

Usually differential circuits are used which help reduce noise (common mode rejection).The sender generates a double-ended signal, while the receiver subtracts the voltages at the input lines.

If noise attacks both conductors equally (noise signal n(t)), the common mode rejection will successfully eliminate the noise. However, if only one conductor is attacked (noisesignal m(t)). the common mode rejection doesn't work.

Distance < 50 mbit rate < 20 kbps

A-19


TRANSMISSION MEDIA (Cont.)

Twisted-Pair Lines

Also called: unshielded twisted-pair (UTP)

Coaxial Cable

Skin effect and radiation effect are largely eliminated in coaxial cables,therefore they can be used for higher bit rates. The external (coaxial) conductoracts as shield against electromagnetic interference.

Helps common mode rejection because the noise is not likely to affect only one line.In addition, multi-core twisted cables reduce cross-talk between the twisted pairs.

Problems:

Skin effect - tendency of the current to flow on the outer surface of the wire at higher frequencies. This reduces the effective cross section of the conductor leading to higher attenuation. Radiation - At higher frequencies part of the signal power radiates away from the conductor (becomes like a radio antenna)

m(t)Distance < 100 m at 10 Mbps < 15 km at 2400 bps

Copper conductor

Dielectricinsulator

Outer conductor

(shield)

Outer insulator

and coverDistance < 1 km at 500 Mbps

A-20


TRANSMISSION MEDIA (Cont.)

Optical Fiber

Opticaltransmitter

(LED)

Opticalreceiver

(pohotodiodephototransistor)

Multi-mode stepped index fiber

Multi-mode graded index fiber

Monomode (singlemode) fiber

Optical core(pure glass)

Optical cladding

uniform refractive index

LED

Variable refractive index

Laserdiode

No electrical interferences, radiation, skin effect, and crosstalk.High security - almost impossible to tap.

Distance 1to 10 km at 1000 Mbps

A-21


MULTIPLEXING

DEMUXMUX

Channel 1 Channel 1Channel 2 Channel 2

Channel 3 Channel 3

Channel n Channel n

(twisted pair,coaxial cable,optical fiber,microwave link,satelite link)

Physical link

How to transmit several signals across the same physical link ?

There are three commonly used methods:

FDM - Frequency-division multiplexing

TDM - Time-division multiplexing

STDM - Statistical TDM

Multiplexer Demultiplexer

A-22


MULTIPLEXING (Cont.)

Frequency-Division Multiplexing

Used mainly in the transmission of analog voice signals. Also used in transmission of digital signals across coaxial cables (in broadband mode), or across wireless communication links (microwave and satelite).

Modulator(FSK, PSK)

MixerFilter

f1Modulator

(FSK, PSK)MixerFilter

f2

Modulator(FSK, PSK)

MixerFilter

Transmitter

fn

+

Demodulator(FSK, PSK)

MixerFilter

f1

f1


MixerFilter

f2

f2


MixerFilter

ReceiverSignalsplitter

SignalCombiner

fn

s2(t)

s2(t)

sn(t)

sn(t)

s1(t)

s1(t)

f

Bandwidth of the linkBandwidth ofthe channel

Frequency shift

Communicationlink

Carrierfrequency

A-23



Principle of the Frequency Shift

Signal component:

Carrier:

Mixing (multiplication):

After filtering out the upper side band:

Mixing at the receiver side:

After filtering out the high frequency component:

)sin()( tSts d=

)sin()( ttc jj =

tStm djj )cos()( =

[ ]ttStSttstctm

djdj

djjj

)cos()cos(21

)sin()sin()()()(

+=

===

[ ]ttStSttmtctr

djd

djjjjj

)2sin()sin(21

)cos()sin()()()(

+=

===

tSts dj )sin()( =

ffi - fd fi fi + fd

Lowerside-band

Upperside-band

A-24



Time-Division Multiplexing

Also called: synchronous time-division multiplexing. (It is called "synchronous" because the time slots are fixed and preasigned to channels, this is not "synchronous transmission mode")

MUX

C1 B1 A1

C2 B2 A2

C3 B3 A3

DEMUX

C1 B1 A1

C2 B2 A2

C3 B3 A3

C1C2C3 B1B2B3 A1A2A3

TDM frame A

TDM frame B

TDM frame C

Channel #1Channel #2

Each channel hasdata rate n bps

Each channel hasdata rate n bps

Multiplexed signal hasdata rate 3 x n bps

Data unit(normally 8 bits -byte multiplexing)

Each TDM frame has fixed number of data units,

for each channel one

Flow control and error control are not included in TDM. This is something that mustbe performed on a per-channel basis, at higher level protocols, like data link protocols(see later). Consequently, the multiplexed signal doesn't have a header and trailer which would contain information for flow and error control.

Each frame is given a constant time slot. The inverse of the time slot is called theframe rate. (As shown later, the frame rate is usually 8000 frames/s.)

A-25



Frame Alignment in TDM

The clocks at the transmitter and receiver side can not be generally the same, a form of synchronization is needed. Therefore each TDM frame gets an additional bit (called a framing bit). After demultiplexing at the receiver side, this bit results in an additional, n+1-th channel. The resulting bit pattern must coincide with a predefined pattern (usually an alternating sequence 1010101010...). If this pattern brakes down, the receiver knows it is out of synchronism. Then it enters the search mode in which it picks and extracts a bit including every next m-th bit that follows (m is frame size). Then it compares the resulting bit pattern from the n+t-th channel with the predefined bit pattern. If it doesn't match, it will continue the same process with the next bit in sequence. Once the match occurs, the receiver must make sure that the pattern matching persists over multiple frames, before it concludes that it is back in synchronism.

Aggregated bit rate of the multiplexed signal is:

frame size [bits] x frame rate [frames/s]

Example:

(Suppose three channels, each channel has 8 bits, frame rate is 8000 frames/s}

Frame size: 3 x 8 + 1 = 25 bits Bit rate: 25 x 8000 = 200,000 bps = 200 kbps

n+1 th channel,synchroinization pattern

MUX

C1 B1 A1

C2 B2 A2

C3 B3 A3

C1C2C3 B1B2B3 A1A2A3DE

MUX

C1 B1 A1

C2 B2 A2

C3 B3 A3

Frame

0 1 0

Framing bits

A-26



Duplex Multiplexers

Full-duplex communication requires a combination of a multiplexer and demultiplexer at the same site.

MUX

DEMUX

MUX

DEMUX

Full duplex link

Full-duplexcustomers

Full-duplexcustomers

MUX

DEMUX

DEMUX

MUX

The full-duplex arrangement above can be represented in a simplified manner:

A-27



TDM Hierarchies Synchronous TDM is an approach used to share a single transmission media among several independent communication channels. The capacity of the media (in terms of bit rate) must be equal or greater than the sum of the bit rates of individual channels. In order to combine a larger number of channels, the multiplexing can be performed in several stages. This leads to a hierarchy of multiplexed signals:

MUX

1

2

k

w1w2

wk

x

MUX

1

2

m

x1x2

xm

y

MUX

1

2

n

y1y2yn

z

The following sections discuss three important hierarchies of digital services:

DS - Digital Signal Service

ISDN - Integrated Services Digital Network

SONET/SDH - Synchronous Optical Network / Synchronous Digital Hierarchy

NOTICE: The discussion will focus only on the physical aspects (OSI layer 1)

A-28



Digital Signal Service (DS) DS-1

Basic transmission format of the hierarchy of digital signals adopted in North America and Japan.

DS-1 circuits are also called T1 links, or T1 transmission services.

DS-1 frame:

125 s (=1/8000 frames/s), 24 channels, 193 bits (=24x8+1)

1 2 3 4 5 6 7 8 9 10 11 1312 14 15 16 17 18 19 20 21 22 23 24

8-bit data unit(64 kbps service)

7-bit data unit(56 kbps service)

Signaling bit(used for network

control)

Aggregate bit rate of a T1 link:

(24 x 8 + 1) * 8000 = 1.544 Mbps

framingbit

number ofchannels

size ofdata units framing

bit

framerate

aggregate bit rate of T1

There is also a Europian standard (ITU-T) called E1. It has 30 8-bit voice channels,one framing byte and one signaling byte. The agregate bit rate of E1 is:

(30x8+8+8)x8000 = 2.048 Mbps

A-29



DS-2 and higher services

All DS services with higher bit rates are obtained by multiplexing the basic DS-1

A similar hiearchy exists in Europe based on E1 (E2, E3, E4, E5)

MUX

MUX

MUX

MUX

DS-26.312 Mbps

(772+17)

DS-1C3.152 Mbps

DS-344.376 Mbps

(5523+24) DS-4274.176 Mbps(33282+990)

MUX

DS-11.544 Mbps

(192+1)

64 kpbs(8)

24 channels

DS Carrier Bit Rate Multiplex Voice Service System [Mbps] Factor Channels

DS-1 T-1 1.544 24 24DS-1C T-1C 3.152 2 48DS-2 T-2 6.312 4 (from DS-1) 96DS-3 T-3 44.736 7 672DS-4 T-4 274.176 6 4032

A-30



Custom T-1 Topologies

A T-1 signal doesn't require the standard DS-1 format (22 64 kpbs channels plus two 56 kpbs channels). If a customer leases the entire T-1 link he or she can arrange his or her own topology. The only condition is that the aggregate bit rate is 1.544 Mbps.

A typical example:

Aggregate bit rate:

((256+768+128+6x64)x1000/8000+1)x8000 = 1544000 bps =1.544 Mbps

data (256 kbps)

video (768 kbps)

CAD/CAM (128 kbps)

voice (64 kbps)

voice (64 kbps)

voice (64 kbps)

voice (64 kbps)

voice (64 kbps)

voice (64 kbps)

1.544 MbpsTDM

No signaling bits needed(all channels carry 8 bit units)Framing bit is required though

A-31



Bit Stuffing

Sometimes the sum of bit rates of T-1 components are below the required aggregate bit rate. For example some of the voice (or data) channels can be 56, 48, 40, 32, 24, 16 or 8 kpbs. In order to maintain the correct aggregate bit rate a technique called bit stuffing (or pulse stuffing) is applied. For example, in case of 56 kbps we have 56000/8000 = 7 bits in data unit. These seven bits can be extended with a dummy bit, which makes 8 bits per 125 s, i.e. 64 kbps. Similarly, 48 kbps need two additional bits, 40 kbps need three additional bits etc.

The condition for bit stuffing is that the source bit rate is a multiple of the frame rate. For example 56000/8000 = 7, however 19200/8000 = 2.4, can not be handled by bit stuffing.

64 kbps

56 kbps

48 kbps

No bit stuffing

One bit stuffing

Two bit stuffing

1.544 Mbps

Multiplexing with bit stuffing is called plesiochronous multiplexing, which means nearlysynchronous multiplexing ("plesio" means "nearly" in Greek).

The resulting higher-order signals are called Plesiochronous Digital Hierarchy (PDH).

A-32



Subrate Multiplexing

How can we connect a source in which the bit rate is not a multiple of the frame rate, or which the bit rate is below the frame rate? For example, 9.6 kbps would have 9600/8000 = 1.2 bits per channel, while 2400 bps would have 2400/8000 = 0.3 bits per channel. In such a case we can use a subrate multiplexing technique which spans several frames. For 9600 bps we use 6 bits of a given channel in every 5-th frame. Five other users can share this single channel at 9.6 kbps.

n+1 n+2 n+3 n+4 n+5 n+6 n+7 n+8 n+9 n+10 n+11

5x125 = 625 s

frame1 2 24

channelBit used to indicate which subrate is used by this channel (bit robbing)

125 s

Bit rate: 6 bits / 0.000625 = 9600 bps

Using 6 bits every 5-th frame gives 9600 bps 6 bits every 10-th frame gives 4800 bps 6 bits every 20-th frame gives 2400 bps

A-33


MULTIPLEXING (Cont.)Add-Drop Multiplexers

How can a low rate customer be connected to a T-x link? The signal must be completely demultiplexed folowing the entire hierarchy, then the customer can be added, and the demultiplexed signals must be multiplexed back again up the hierachy. Such complex multiplexers are called drop-and-insert multiplexers, or add-drop multiplexers (ADM)

Suppose we are connecting a 256 kbps duplex customer to a 44.376 Mbps duplex link (T-3).

ADMA BT3 T3

256 kbps

T3

T2

T3

T2

T2

T1

6.312 Mbps

44.736 Mbps 44.736 Mbps

T2

T1

6.312 Mbps

1.544 Mbps 1.544 Mbps

256 kbps

A-34



Integrated Services Digital Network (ISDN)

Also called Narrowband ISDN (N-ISDN) as opposed to Broadband ISDN (B-ISDN), which is something totally different.

Introduced with an idea to replace existing analog telephone network (PSTN - Public Switched Telephone Network), also called POTS (Plain Old Telephone Service). ISDN is entirely digital and can transport digitized voice, video and data over the same link.

The basis of the ISDN are 64 kbps channels, which can be combined in order to allocate required bandwidth for more demanding transactions (video conferencing, fast facsimile, etc.)

TA

TA

ISDN PCadapter

NT1

LAN hub

NT2

Router

ISDN device(video telephone,ISDN terminal)

Non ISDN device

Terminaladapter

Networktermination

type 1

Networktermination

type 2

Full-duplex, two-pair,Unshielded twisted pairPseudoternary encoding

Full-duplex, four-pair,Unshielded twisted pairPseudoternary encoding

TerminaladapterNon ISDN

device

Non ISDN device

100 to 500 m

RS-232C

R

R

TSS U

S

SS

R, S, T and U are reference points

A-35


MULTIPLEXING (Cont.)ISDN Channels

There are three types of channels used in ISDN:

Bearer channel (B channel) - 64 kbps, used to transport the payload (data, voice, video), Data channel (D channel) - 16 kbps or 64 kbps - normally used for signaling (out-of-band signaling), but can also be used to transport data

Hybrid channels (H channel) - 384, 1536, 1920 kbps - used for transactions requiring higher bit rate.

There are two ISDN services: Basic rate interface service (BRI) - multiplexes two B channels and one 16 kbps D channel (2B+D) - total bit rate: 2x64+16 +48 overhead = 192 kbps - for individual users and small offices

Primary rate interface service (PRI) - multiplexes 23 B channels and one 64 kbps D channel (23B+D, N. America and Japan) or 30 B channels and one D channel (30B+D, Europe) - total bit rate: 23x64+64+8 (overhead) = 1.544 Mbps, or 30x64+64+64(overhead) = 2.048 Mbps - PRI is compatible with T1, or E1 lines - Transmission over H channels is available in PRI by combining B channels (384 = 6x64, 1536 = 24x64, 1920 = 30x64)BRI Frame

Comments: Besides the B and D fields, BRI frame has overhead bits (shaded area). They include the framing bit, level balancing bits, etc. The size of overhead is 12 bits. Why does the BRI frame have two B1 channels and two B2 channels? Answer: to make the frame longer. (As seen later, its size matches the payload of ATM cells.) PRI Frame is identical with the DS-1/E1 format (same frame size and same frame rate)

D B1 B2B2B1 D D D

4000 frames/s, 250 s, 4x8+4+12 (overhead bits) = 48 bits

8 bits 8 bits 8 bits 8 bits1 bit 1 bit 1 bit 1 bit

A-36



Inverse Multiplexing

PRI service enables transmission of digital signals at bit rates higher than 64 kbps. This is done by segmenting the high bit rate signal into a number of individual low bit rate signals which are then sent across the ISDN link as individual data streams. At the receiving side the individual channels are again combined into one high bit rate signal. The segmentation and reassembly is done by a pair of inverse multiplexers, which are sitting at the customer's premise.

ISDN

NT

Inversemultiplexer

video video

Inversemultiplexer

NT

Singlehigh bit rate

circuit(384 kbps)

Singlehigh bit rate

circuit(384 kbps)

Multiplelow bit rate

circuits(6 x 64 kbps)

Multiplelow bit rate

circuits(6 x 64 kbps)

Transmitting inverse multiplexer: - sets up required number of 64 kbps channels to destination - segments the high bit rate stream into a number of low bit rate streams

Receiving inverse multiplexer: - accepts the required number of streams - compensates for delays (different channels may go through different paths) - resynchronizes the input channels - reassembles the channels into one high bit rate stream

A-37



Synchronous Optical Network (SONET)

Originally proposed by BellCore in early 1980s, then standardized by ANSI. ITU-T has later standardized a compatible version called Synchronous Digital Hierarchy (SDH). The main characteristics of SONET/SDH is that it is truly synchronous (not plesiochronous), based on a single clock with extremely high accuracy.

SONET/SDH uses optical fibers, and is the main carrier for B-ISDN and ATM. It can also transport DS-x and E-x signals.

STS-N/OC-N Signals

The structure of the SONET/SDH hierarchy is based on the 51.84 Mbps signal called: Synchronous Transport Signal (STS-1).

STS-1 frame (payload and overhead)

810 octets = 6480 bits, 125 s, 8000 frames/s

Aggregate bit rate: 6480 x 8000 = 51.84 Mbps

STS-1 frame has an overhead of 27 octets = 216 bits. The bit rate of the STS-1 payloadis (6480 - 216) x 8000 = 50.112 Mbps, which is greater than 44.736 Mbps (or 34.368 Mbps),meaning that an STS-1 payload can accommodate a DS-3/T-3 (or E3) signal, or a combination of multiple DS-1, DS-1c, DS-2, E-1 and E-2 signals.

The SONET/SDH hierarchy of signals can be obtained by multiplexing the basic STS-1(STS-N = N x STS-1)

STS-N designate electrical signals. The corresponding optical equivalents are designatedas OC-N (Optical Carrier). For example OC-3 is an optical equivalent of STS-3.

SONET SDH Bit Rate [Mbps]

STS-1 51.84STS-3 STM-1 155.52STS-9 STM-3 466.65STS-12 STN-4 622.08STS-18 STM-6 933.12STS-24 STM-8 1244.16STS-36 STM-12 1866.24STS-48 STM-16 2488.32

A-38



SONET/SDH Configuration

A SONET/SDH networks have three basic structural elements: sections, lines and paths.

Section is a single piece of transmission cable between two regenerators. Regenerators are used to recover the attenuated signal and to extend the lenght of lines.

Line consists of several sections, it connects two network nodes that provide access to SONET from other LANs or WANs. The line terminating devices are add-drop multiplexers (ADM) or digital-cross connect (DCS). The later device, besides adding and dropping a payload, it also functions as a SONET hub.

Path is an end-to-end transmission way. It connects two service adapters which map the user payload from a non-SONET to a SONET format.

Section

Path (end-to-end)

Line Line

Section

Path terminating equipment(Terminal multiplexer,

access node,service adapter -

maps the user payloadlike DS-1, DS-2, FDDI, ATM

into a SONET format)

Line terminating equipment(ADM - add-drop-multiplexer

or DCS- digital-cross connect)

Section terminating equipment(regenerators)

A-39



Example of SONET/SDH Topology

FDDI LAN(Fiber Distributed Data Interface100 Mbps, 100 km, 500 stations)

802.5(token ring, 16 Mbps)

802.3(Ethernet, 10 Mbps)

ADMADM

ATM

E-1, E-3

T1, T3

ADM

ADM

DCS

SA SA

SASA

SA

OC-n

OC-n

OC-n

OC-n

OC-n/STS-n

OC-n/STS-n

OC-n/STS-n

OC-n/STS-nOC-n/STS-n

OC-n/STS-n

Survivablerings

Used from: U. Black, ATM Resource Library, Vol IPrentice Hall, 1999

SA - terminal/service adapterDCS - Digital cross-connect systemADM - Add-drop multiplexer

A-40



STS-1 Frame

The structure of the STS-1 frame is organized into a two-dimensional array of octets:

The frame overhead contains alignment bits, frame identifier, error checks, payload pointers,and other data necessary for administration, monitoring and traffic management.

The octets are transmitted row-wise (the top leftmost octet first).

Bit rates:

Aggregate: 90 columns x 9 rows x 8 bits x 8000 frames/s = 51.84 Mbps

Payload: 87 columns x 9 rows x 8 bits x 8000 frames/s = 50.112 Mbps.

3 columns 87 columns

90 columns

SECTIONOVERHEAD

LINEOVERHEAD

SYNCHRONOUS PAYLOAD ENVELOPE(SPE)

Path overhead(part of the SPE)

Payload doesn't have to fill the entire SPEand doesn't have to start from the beginning.In other words payload floates within the envelope.Sometimes the payload spans several frames. Therefore the line overhead has pointers which locate the payload within theSPE.

A-41



Multiplexing STS frames

Lower rate frames are interleaved, making up the higher rate frames. For example, an STS-n frame has the following general format:


270 columns

SECTIONOVERHEAD

LINEOVERHEAD

3xn columns 87xn columns

90xn columns

SECTIONOVERHEAD

LINEOVERHEAD

SYNCHRONOUS PAYLOAD ENVELOPE(SPE)

For example, the very popular STS-3 frame:

Aggregate bit rate: 270 columns x 9 rows x 8 bits x 8000 frames/s = 155.52 MbpsPayload bit rate: 261 columns x 9 rows x 8 bits x 8000 frames/s = 150.336 Mbps

NOTICES: Multiplexing STS signals is much simpler than the multiplexing DS signals. In order to extract a lower-rate STS component it is not necessary to perform a multistep demultiplexing down the entire hierarchy as in the case of DS. The ADM can be done in a single step.

STS-3 is a prefered carrier for ATM cells; one STS-3 frame can accomodate 44 cells.

ATM cells 53 octets

A-42



Virtual Tributaries

Although the SONET/SDH is designed to carry broadband payloads, it is made backward compatible with the existing DS hierarchy. To achieve this, the SPE is subdivided into sections called virtual tributaries (VT).


90 columns

SECTIONOVERHEAD

LINEOVERHEAD

Pathoverhead

VT VT VT VT VT

There are four types of VTs:

VT1.5 - accommodates DS-1 (1.544 Mbps, 3 columns)

VT2 - accommodates E-1 (2.048 Mbps, 4 columns)

VT3 - accommodates DS-1c (3.152 Mbps, 6 columns)

VT6 - accommodates DS-2 (6.312 Mbps, 12 columns)

The frame overhead contains enough data to separate the VTs without demultiplexingthe entire structure of the STS frame.

A-43



Statistical Multiplexing

In synchronous TDM there is a strict ordering of multiplexed channels - each channel has an exact position within the frame. This can result in inefficient transmission of bursty data. For example, a customer may have silent periods during telephone conversation. Consequently, there will be many idle inputs to the multiplexer and empty time slots in the transmitted frames.

MUX

C1 B1 A1

B3

B4

A3

A4

A1A4B1B4 A3C1B3

Empty time slots(wasted bandwidth)

Synchronous TDMFrameidle

idle

A1A4B1B4 A3C1B3STATMUX

C1 B1 A1

B3

B4

A3

A4

Bandwidth available

Asynchronous TDM

idle

idle

The solution is statistical TDM (STDM), also called asynchronous TDM (ATDM) in whichthe idle inputs are ignored and the data items are packed into frames as dense aspossible.

A-44



While synchronous TDM requires that the sum of input data rates must be less or equal to the data rate of the output line, in STDM the sum of the mean values of input rates must be less or equal to the rate of the output line. Consequently, either the rates ofinputs can be increased, or the number of inputs can be extended. In order to accommodatetemporary data surges, when rates in all inputs exceed the average values, the buffers of an appropriate size must be used.

In statistical TDM the position of time slots loses significance. In order be able to demultiplex the frames, addressing is needed. Each data item must be accompanied with an address, i.e. the channel (input) number.

Adding addresses to data results in inefficiency, specially if data items are short incomparison with the address field. Therefore the data items must be grouped in larger fields.This however brings a need to have an additional field, the data length. The added controlfields and data constitute a subframe. The subframe may be more complicated as it will be shown in later chapters.

C1

C5

B1

B5

A1

A5

B3

B4

A3

A4

C55 B33 C11 B55 B44 A33 1 B1 5 A5 4 A4 1 A1

1

2

3

4

5

STATMUX

Address

idle

idle

STATMUX

C1

C5

B1

B5

A1

A5

B3

B4

A3

A4

A1A4A5B1B4 A3C1 B5B3C5

idle

idle

Available bandwidth usedto fit one more channel

Address Length Data

tdm mux

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