BROADBAND TRANSMISSION

STANDARDS

DSL, ADSL and other flavors

2000 Bijan Mobasseri 2

Background

For the longest time it was taught that analog phone lines are bandlimited to 4 KHz

Nyquist’s first criterion then states that maximum theoretical data rate on such lines is 8000 pulses/sec.

So what happened?

2000 Bijan Mobasseri 3

Historical trends

Modems started out at 300 bits/sec using binary FSK modulation

Over time speed grew to 1200, 2400, 9600 and finally 56K bits/sec.

This increase was made possible by using high density digital modulations

And yes, this was all done within the confines of 4 KHz bandwidth

56K, V.90, was the first standard taking advantage of digital backbone at 64 Kbps

2000 Bijan Mobasseri 4

26 Mb/sec over ordinary phone lines?

DSL in its various flavors, allows for up to 26 MB/sec over twisted pair. How is that possible?

The answer is that it is the switch that has limited us not the phone lines

Solution: avoid the switch by extracting digital signals before they get to the switch and re-route them to a broadband network

2000 Bijan Mobasseri 5

Original objectives

Cover a serving distance of 18000 feet on a single twisted pair telephone subscriber loop at T1 rate of 1.544 Mb/s. Motivation: video-on-demand (meets H.261 MPEG rate)

Other features• POTS availability: 300 Hz-4KHz• Upstream control channel: 16-64Kb/s in 10 KHz-50 KHz• Downstream: 1.544 Mb/s in 100-500 KHz

2000 Bijan Mobasseri 6

DSL vs. voice band modems

COCO Local loopNarrowbandnetwork

internet

Old Way

upstreamdownstream

2000 Bijan Mobasseri 7

DSL vs. voice band modems

internet

Broadband splitDSL DSLsplitLocal loop<1 mile

26 Mb/s 3-26Narrowbandnetwork

CO

CO

Broadband

POTS/ISDN VDSL

120 KHz 300 KHz 30 MHz

Splitter allows for the coexistence of POTS and DSL on the same line

2000 Bijan Mobasseri 8

Spectrum picture

POTS

Upstream Downstream

ADSL

1.5 Mb/s8 Mb/s

4 20 140 552 1104

G.Lite

2000 Bijan Mobasseri 9

Why the “A” in ADSL?

Asymmetric DSL rates are• Downstream: 1.544 Mb/s for distances less than 18000 feet (to CO) up to

6.144 Mb/s up to 12000 feet• Upstream: 16 to 640 Kb/s

Asymmetry is due to the bundling of the twisted pair telephone wires. Symmetric rates interfere with each other

Symmetric rates must be much slower and cover shorter distances

2000 Bijan Mobasseri 10

DSL flavors

ADSL• Modulation: CAP or DMT• Data rate: 8 Mb/sec downstream (in 240KHz-2 MHz) and up to 1 Mb/sec

upstream in the 25--200 KHz simultaneously with POTS

HDSL• Four wire access for achieving symmetrical rates at T1 (1.544 Mb/sec) or

E1(2.048 Mb/sec).• Modulation: CAP64(passband) or 2B1Q( baseband)• Spectrum in copper loop:0-300 KHz or 0-425 KHz• Fractional T1 rates supported (i.e. nx64 Kb/sec)• No simultaneous voice data, HDSL2 will do that

2000 Bijan Mobasseri 11

More DSL flavors

SDSL• Single-pair (two wire) implementation of 2B1Q or CAP modem• Echo cancellation and adaptive equalization• Symmetric data rates of 384 Kb/s, 768 Kb/s, 1 Mb/s or 2 Mb/s. • SDSL is expected to eclipse HDSL due to its single pair, symmetrical data

rate property

VDSL• Supports much shorter distances• Very high-speed DSL: 6.5-51.8 Mb/s downstream. • 1.6-6.5 Mb/s upstream for asymmetrical services and 6.5-25.9 Mb/s for

symmetrical services(short <1000 ft, two-wire loops)• Modulation: M-ary CAP, DMT

DSL Modems

Discrete Multi Tone (DMT)or

Orthogonal Frequency Division Multiplexing(OFDM)

2000 Bijan Mobasseri 13

The general problem:multipath

Transmitted signal arrives at the receiver through a number of paths

2000 Bijan Mobasseri 14

Effect on pulse transmission

Multiple copies of the signal interfere with each other(ISI). Let T be bit duration. Pulses are delayed differently forming a delay spread

Largest delay is called maximum delay spread

max

T

A received symbol can be influenced by

max/T previous symbols

We want max delay spread to be less thanT, i.e. bit duration

2000 Bijan Mobasseri 15

What are the choices?Single carrier approach

Either reduce delay spread or increase T Increasing T means reducing bitrate. Don’t like that. Example (DVB)

• Transmission rate: R=1/T=7.4 Msymbols/sec

• Maximum channel delay: max=224 sec

For a single carrier modulation max/T=1600

The complexity of removing this much interference is enormous

2000 Bijan Mobasseri 16

Multi-carrier approach

Split data at rate R into N parallel stream of rate Rmc=1/Tmc=R/N

Each slower data stream is modulated by a different carrier frequency and transmitted in the same band

ISI is reduced by a factor of N max/Tmc= max/NT

For DVB, we have N=8192 streams resulting in

max/Tmc=0.2

2000 Bijan Mobasseri 17

What are we talking about?Multicarrier modulation

Modulation method used in DSL is of a frequency division multiplexing flavor.

Channel is divided into many subchannels This approach bundles bits and transmits them over

different frequency bands to counter channel characteristics

frequency

Bits per hertz

frequency frequency

line gain Bits per hertz

2000 Bijan Mobasseri 18

Channel allocation

DMT in current ADSL creates 256 downstream subbands, 4 kHz each

Each channel can be modulated with QAM at up to 15 bits/sec/hz.

Theoretical transmission rate is then 15.36 Mb/sec over a zero length line. How?

15(b/s/Hz) x 4000(Hz/channel) x 256(channels)=15.36 Mb/sec

2000 Bijan Mobasseri 19

Bit allocation

Instead of using equalizers to correct channel response, DMT spreads data over all channels according to the SNR in each one.

frequency

Bits per hertz

frequency frequency

line gain Bits per hertz

Sweet spot (up to 15 b/s/hertz)

Ideal channel

2000 Bijan Mobasseri 20

Orthogonal Frequency Division Multiplexing:first step

Take a bitstream running and divide it into N parallel channels each at a reduced rate

11001010110101

1 0 1 ...

0 1 1 ….

1 1 1….

1 0 0...

N

On each line, groupBits into a symbol.The symbols are much wider than the original bitstream thus defeatingpossible ISI

2000 Bijan Mobasseri 21

Modulation in OFDM

Each symbol is mapped to a QAM constellation. Channels are modulated by orthogonal frequencies

1 0 1 11 0 1 0

0 0 0 1

1 1 1 0

16 lines16 frequencies

f1

f2

f3

f16

2000 Bijan Mobasseri 22

OFDM signal model

The OFDM signal can be written as the sum of N pulses modulating N orthogonal carriers

Fixed k means summing symbols from different channels, each modulating a different frequency

v(t) = xk,nn=0

N−1

∑k∑ φn t−kT( ), φn t( ) =e

j2π n−

N−12

⎛ ⎝

⎞ ⎠ t

T

Symbol coming from a M-ary QAM const. M=sqrt(N)

2000 Bijan Mobasseri 23

OFDM epoch

Let’s look at “epoch” k=0. This is taking N symbols one from each channel and adding them up.

1 0 1 11 0 1 0

0 0 0 1

1 1 1 0

f1

f2

f3

f16

k=0 k=1

v(t) = x0,n exp j2πntNTs

⎧ ⎨ ⎩

⎫ ⎬ ⎭ n=0

N−1

∑ 0<t<T

T

Assigned toOne of these

2000 Bijan Mobasseri 24

Generating OFDM by IFFT

Sample v(t) at t=kTs

OFDM can be generated by an inverse FFT

vk =v kTs( ) = xk,n expj2πnk

N{ }n=0

N−1∑

k=0,1,...,NX 0,0 X 0,1 X 0,N-1

IFFT

V 0 V 1 V N-1

2000 Bijan Mobasseri 25

Variable bit loading

Input data at Mfs bits /sec grouped into blocks of M bits at a block (or symbol) rate of fs then ...

Serial to parallel

conversion-M bits at

a time

Mfs bits /sec

m1

m2

m3

mn bitsM = mi

i=1

n

∑

forexample

ifM =8bits/ symbol

m1 =1,m2 =2,m3 =3,

m4 =1,m5 =1

2000 Bijan Mobasseri 26

Bringing in the multitones

Each group of bits then modulate a separate carrier: the mnth group modulates a carrier with carrier frequency fc,n

There are Nc carriers spaced ∆f Hz apart. The total number of carriers where Nc=n.

For the previous slide, we need 5 carriers because we broke up 8 bits into 5 groups of bits

fcn

fc1

1 bit

2 bits

1 bit

3 bits

2000 Bijan Mobasseri 27

Selecting modulations

M = mii=1

n

∑

forexample

ifM =8bits/ symbol

m1 =1,m2 =2,m3 =3,

m4 =1,m5 =1

3 bits2 bits

When a block contains 2 bits, they represent 4 states. The 4 states come from a 4-QAM modulation. Same goes for 3 bits