4.modn

CHAPTER 11

IF bandpass (carrier)modulation

11.3.1 Binary amplitude shift keying (and on-off keying)

In binary amplitude shift keyed (BASK) systems the two digitalsymbols 0 and 1 are represented by pulses of a sinusoidalcarrier (frequency, fc) with amplitudes A0 and A1.In practice, one of the amplitudes, A0, is inv ariably chosen to bezero resulting in on-off keyed (OOK) IF modulation:

f (t) = A1 (t/To) cos 2pi fc t , for a digital 10 , for a digital 0

where To is the symbol duration and is the rectangular pulsefunction.

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2 IF bandpass (carrier) modulation

Figure 11.1 Onoff keying (OOK) modulator, waveforms, spectra and phasor states.

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Binary amplitude shift keying (and on-off keying) 3

Decision instant voltage, f (kTo), at the receiver OOK matchedfilter (correlator) output:

f (kTo) = kE1 , digital 1 or 0 , digital 0 (11.2)E1 is the normalised energy in symbol 1. The normalised o/pnoise power, 2 is:

2 = k2E1 N0/2 (11.3)where N0 (V2/Hz) is the normalised one sided noise powerspectral density at the matched filter or correlator input. Theprobability of symbol error is thus:

Pe =12

1 erf

12

E1N0

12

(11.4)

This can be expressed in terms of the time averaged energy persymbol, E = 12 (E1 + E0) where for OOK E0 = 0. i.e.:

Pe =12

1 erf

12

EN0

12

(11.5)

These equations can be expressed in received carrier to noise(power) ratios (C/N ) using:

C = E/To (V2) (11.6)N = N0 B (V2) (11.7)E/N0 = To B C/N (11.8)

where C is the carrier power averaged over all symbols, N isthe normalised noise power in the bandwidth B Hz to give:

Pe =12

1 erf

(To B)122

CN

12

(11.9)

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Figure 11.2 Coherent and incoherent bandpass OOK receivers.

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Binary amplitude shift keying (and on-off keying) 5

EXAMPLE 11.1

An OOK IF modulated signal is detected by an ideal matchedfilter receiver. The non-zero symbol at the matched filter inputis a rectangular pulse with an amplitude 100 mV and a durationof 10 ms. The noise at this filter is known to be white andGaussian with RMS of 140 mV in a noise bandwidth of 10 kHz.Calculate the probability of bit error.

Energy per non-zero symbol:E1 = v2rms To

=

100 103

2

2

10 103 = 5. 0 105 (V2s)

Av erage energy per symbol:

E = E1 + 02

= 2. 5 105 (V2s)Noise power spectral density:

N0 =N

BN=

n2rmsBN

=

(140 103)210 103

= 1. 96 106 (V2/Hz)

Thus from equation (11.5):

Pe =12

1 erf

12

EN0

12

=

12

1 erf

12

2. 5 105

1. 96 106

12 =

12

[ 1 erf (2. 525) ] = 1. 778 104

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11.3.3 Binary frequency shift keying (BFSK)

BFSK represents the digital 1s and 0s by carrier pulses withtwo distinct frequencies, f1 and f2:

f (t) = A (t/To) cos 2pi f1 t , for digital 1A (t/To) cos 2pi f2 t , for digital 0

Detection of BFSK can be coherent or incoherent. Incoherentdetection suffers the same CNR penalty compared withcoherent detection as is the case for OOK systems.

The BFSK carrier frequency, fc, is defined by:fc =

f1 + f22

(Hz) (11.19)and BFSK frequency deviation, f , is (Figure 11.6(e)):

f = f2 f12

(11.20)using the zero crossing point in the BFSK voltage spectrum, seeover, yields the bandwidth:

B = 2 f + 2 fo (11.21)

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Binary frequency shift keying (BFSK) 7

A cos 1t ~

A cos 2t

1

2

(c) BFSK modulator

(a) Baseband data

1 1 1 1 0000

1 1 1 10 0 0A

A

A

0

0

0

=

+

(d) BFSK signal and two, component, bandpass OOK signals

fV(f )V(f )

1To01To

f1 f0 f2

fo

f1 + f22

f

(b) Baseband voltage spectrum of a single symbol (e) BFSK voltage spectrum (of two symbols, 0 and 1). Note overlapping OOK spectra

~

Figure 11.6 Binary frequency shift keying (BFSK) modulators, waveforms and spectra.

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Figure 11.7 Coherent and incoherent BFSK receivers.

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Binary frequency shift keying (BFSK) 9

Here fo = 1/To is both the nominal bandwidth and the baud rateof the baseband data stream. If the symbols are orthogonal i.e.:

To

0 cos(2pi f1t) cos(2pi f2t) dt = 0 (11.22)

then when the output of one channel is a maximum the otheroutput will be zero.

If the one sided NPSD at the BFSK receiver input is N0(V2Hz1) then the noise power, 21 = k2E N0/2, received viachannel 1 and the noise power, 22 = k2E N0/2, received viachannel 2 will add power-wise. The total noise power at thereceiver will be:

2 = 21 + 22 = k2E N0 (V2) (11.24)

i.e. 2 the OOK case .

The probability of error for coherently detected orthogonalBFSK is given by:

Pe =12

1 erf

12

EN0

12

(11.25)

i.e. as OOK but E is now twice as large.

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11.3.2 Binary phase shift keying (BPSK)

BPSK codes baseband information onto a carrier, by changingthe carriers phase in sympathy with the baseband digital data:

f (t) = A (t/To) cos 2pi fc t , for digital 1A (t/To) cos(2pi fc t + ) , for digital 0

Usually antipodal states are chosen, i.e. = 180.The post-filtered decision instant voltages are E (V) where E(V2s) is the normalised energy residing in either symbol.The normalised output noise power, 2 (V2) is as in BASK i.e.:

Pe =12

1 erf

EN0

12

(11.11)

The PRK probability of symbol error can be expressed as:

Pe =12

1 erf (To B)12

CN

12

(11.12)

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Binary phase shift keying (BPSK) 11

Figure 11.3 Phase reversal keying (PRK) modulator waveforms, spectra and phasor states.

oscillatorIF/RF

PSK signal

Baseband data

Baseband data

oscillatorReference

PSK signal

Figure a Double balanced mixer connections for modulator and demodulator

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PSK modulation and demodulation both occur in a doublebalanced mixer, Figure b. In the transmitter this switches theoscillator signal, fc, through a 180 phase shift depending onthe digital bipolar (+1/ 1) drive signal, ei(t).The input voltage, ei(t), biases on either diodes D1 and D3 ordiodes D2 and D4 to switch the oscillator connections to theoutput transformer, eo(t).When used as a phase detector, Figure a, the balancedmodulator or mixer is input with the PSK signal and a referenceoscillator from the carrier recovery (CR) circuit.

e

D D

DD 34

eo(t)

cos wt

(t)21

i

c

Figure b Double balanced mixer circuit diagram

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Relative phase between input and reference

Output voltage

0pi2

2pi3pi

2pi

Figure c Mixer output for differing phase between input and reference ports

It outputs a sinusoidal voltage from the baseband portdependent on the relative phase of the two inputs, Figure c.

With a reference synchronised to one of the two input phaseswe can detect the phase transitions by the positive or neg ativeoutput voltages at 0, pi .

Also note, for later use with QPSK, that if the input signalphase is pi /2 or 3pi /2 then the output is zero!

In PSK one must use coherent detection as the information isnow in the phase.

Figure 11.4 PRK correlation detector.

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For a bi-phase (0, pi) modulated signal the normal phase lockedloop (PLL) used to demodulate FM cannot be used forreference oscillator synchronisation.

We need to use a multiplication loop to obtain the synchronisedreference oscillator. After the squaring or self multiplicationoperation both sin c t and sin ( c t + pi ) representing the +1 and1 data symbols both become cos 2 c t with the same phaserelationship:

(sin c t)2 = 12 [1 cos 2 c t](sin( c t + pi ))2 = 12 [1 cos(2 c t + 2pi )]

=12 [1 cos 2 c t]

For bi-phase modulation a squaring loop (based on anotherbiphase modulator) is employed.

Figure 11.11 Squaring loop for suppressed carrier recovery.

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EXAMPLE 11.2

A 140 Mbit/s PRK signalling system uses pulse shaping toconstrain its transmission to the double sideband Nyquistbandwidth. The received signal power is 10 mW and the noisepower spectral density is 6.0 pW/Hz. Find the BER expected atthe output of an ideal correlation receiver.

The double sided Nyquist bandwidth is given by:

B =1

To

N = N0 B = N01

To= N0 Rs (using Rs for the information rate)= 6. 0 1012 140 106 = 8. 4 104 W

Now from equation (11.12):

Pe =12

1 erf (To B)12

CN

12

=

12

1 erf

10 103

8. 4 104

12 =

12

[ 1 erf (3. 450) ]= 5. 33 107

BER = Pe Rs= 5. 33 107 140 106 = 74. 62 errors/s

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Pe

OOK 12 erfc E2N0 12 erfc 14 SNBaseband

Signalling Polar 12 erfc EN0 12 erfc 12 SNOOK 12 erfc 12 EN0 12 erfc To B2 CN

IF/RF

Signalling Orthog BFSK 12 erfc 12 EN0 12 erfc To B2 CNPRK 12 erfc EN0 12 erfc To B CN

Table 11.1. Comparison of the Pe for coherent detection ofbaseband and IF binary signals.

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Figure 11.10 Comparison of binary ASK/PSK/FSK systems performance.

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Multi-phase modulation

We increase the capacity by deploying more phase states e.g. 4,8, 16, 32 etc. 4-phase modulation (pi /4, 3pi /4, 5pi /4, 7pi /4) isobtained in the parallel modulator by summing the outputs oftwo bi-phase modulators.

One modulates 0, pi and the other uses a quadrature carrier togive pi /2, 3pi /2.

Figure 11.22 Quadrature phase shift keying (QPSK) signal constellation showing allowed statetransitions.

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19

Figure shows the parallel QPSK modulator where an input databit stream at 2N bit/s is demultiplexed into two parallel separatebit streams at N bit/s.

Each of these is converted into bipolar signals which driveseparate PSK modulators which are fed with 90 phasedisplaced oscillator drive signals.

Finally the outputs are summed to give the 4-phase(pi /4, 3pi /4, 5pi /4, 7pi /4) signal.

Figure 11.23 Schematic for QPSK (OQPSK) modulator.

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11.4.4 Quadrature phase shift keying (QPSK)

Figure shows the QPSK demodulator. The input isdemodulated with inphase and quadrature versions of theoscillator fed from the carrier recovery (CR) circuit.There is no interference from the two parallel branches due tothe 90 phase difference between them.

These outputs are low-pass filtered and, after the symbol timingrecovery (STR) circuit, sampled to recover the data bits andcombined in the parallel to serial converter.

Figure 11.24 Schematic for QPSK (OQPSK) demodulator.

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Quadrature phase shift keying (QPSK) 21

Constant amplitude BPSK and QPSK are ideal for satellitesystems where a non-linear device is often used as the outputamplifier.

In QPSK the phase change of 90 or 180 or 270 occurs everybit pair on the input data stream, or symbol period.

If these sharp phase transitions can be avoided then the spectralproperties can be improved to reduce adjacent channelinterference. [Offset keyed QPSK (OK-QPSK) staggers the bitstreams by one input bit period (T ), i.e. half the symbol period,To, to avoid simultaneous bit changes.]

Figure 11.25 Unfiltered (constant envelope) QPSK signal (To = 2Tb).

Figure 11.29 Input bit stream and I and Q channel bit streams for QPSK and OQPSK systems.

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4-phase occupies no more bandwidth than 2-phase but as thereare 2 bits defining each of the 4-phase symbols the effectivecapacity in bit/s/Hz, i.e. per unit of bandwidth, is doubled.

Required Eb/N0 Minimum channel Max spectral Requiredfor Pb = 106 bandwidth for ISI free efficiency CNR in

signalling bit/s/Hz min channel( Rb = bit rate) bandwidth

PRK 10.6 dB Rb 1 10.6 dB

QPSK 10.6 dB 0. 5Rb 2 13.6 dB

Table 11.4 Comparison of BPSK & QPSK modulation techniques

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Quadrature phase shift keying (QPSK) 23

11.4.2 M -symbol phase shift keying

The probability of symbol error for MPSK systems is:

Pe = 1 erf

sin pi

M

EN0

12

(11.39(a))

This approximation improves as M and E/N0 increase.Rewriting with CNR using C = E/To and N = N0 B:

Pe = 1 erf

(To B)12 sin pi

M

CN

12

(11.39(b))

Figure 11.16 Phasor states (i.e. constellation diagram) for 16-PSK.

Figure 11.17 Error region (unhatched) for = 0 state of a 16-PSK signal.

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Bits and symbols

Multi-symbol signalling is a coding process in which n binarysymbols (bits) are mapped into a single M-ary symbol. Adetection error in a single symbol can therefore translate intoseveral errors in the corresponding decoded bit sequence.

The probability of bit error, Pb, therefore depends not only onthe probability of symbol error, Pe, and the symbol entropy,H = log2 M , but also on bit mapping and the error types. ForGray coded symbols:

Pb =Pe

log2 M(11.40(a))

We express error rates in terms of Pb as a function of averageenergy per information bit, Eb. The energy, E, of all symbolsin MPSK are:Eb = E/ log2 M (11.40(b))and:

Pb =1

log2 M

1 erf

sin pi

M log2 M

EbN0

12

In terms of CNR this becomes:

Pb =1

log2 M

1 erf

(To B)12 sin pi

M

CN

12

Figure 11.18 shows the probability of bit error, Pb, against C/Nfor MPSK.

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25

The maximum possible ISI free spectral efficiency occurs whenpulse shaping is such that signalling takes place in the doublesided Nyquist bandwidth B = 1/To Hz:

s = log2 M (bit/s/Hz) (11.42(b))Thus we usually say BPSK has an efficiency of 1 bit/s/Hz and16-PSK has 4 bit/s/Hz.

Figure 11.18 MPSK symbol and bit error probabilities: (a) probability of symbol error, Pe,against Eb/N0; and (b) probability of bit error, Pb, against Eb/N0.

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26 IF bandpass (carrier) modulationRequired Eb/N0 Minimum channel Max spectral Required

for Pb = 106 bandwidth for ISI free efficiency CNR insignalling bit/s/Hz min channel

( Rb = bit rate) bandwidth

PRK 10.6 dB Rb 1 10.6 dB

QPSK 10.6 dB 0. 5Rb 2 13.6 dB

8-PSK 14.0 dB 0. 33Rb 3 18.8 dB

16-PSK 18.3 dB 0. 25Rb 4 24.3 dB

Table 11.4 Comparison of several PSK modulation techniques

Figure 11.17 Error region (unhatched) for = 0 state of a 16-PSK signal.

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Example 11.4 - An MPSK system is to operate with 2N PSKsymbols over a 120 kHz channel. The minimum required bitrate is 900 kbit/s. What minimum CNR is required to maintainISI free reception with a Pb no worse that 106?

Maximum (ISI free) baud rate:

Rs =1

To= B

Rs 100 kbaud (k symbol/s). Minimum required entropy is:

H RbRs

=

900 103

120 103= 7. 5 bit/symbol

Minimum number of symbols required is:H log2 M M 2H = 27.5

Since M must be an integer power of 2, M = 28 = 256.For Gray coding:

Pe = Pb log2 M = 106 log2 256 = 8 106

Pe = 1 erf

(To B)12 sin pi

M

CN

12

To find minimum ISI free CNR assumng that To B = 1:

CN

=

erf1 (1 Pe)sin

pi

M

2

=

erf1 (1 8 106)sin

pi

256

2

=

erf1 (0. 999 992)sin

pi

256

2

=

3. 1570. 01227

2

= 66200 = 48. 2 dB

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11.4.3 Multi-phase multi-amplitude modulation

In an unsaturated transmitter operating over a linear channel itis possible to introduce amplitude and phase modulation to givea better distribution of the signal states in the constellation.

Figure shows the three possible 16-state signal constellations.The square quadrature amplitude modulation (QAM)constellation is in (c).

Figure 11.19 Three possible 16-state QAM signal constellations.

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Multi-phase multi-amplitude modulation 29

16-QAM is obtained by modifying the QPSK modulator toaccept bit pairs in the I and Q arms. By feeding the bit pairsinto a DAC a 4-level signal is applied to the phase modulator tocontrol the output amplitude and give 4 distinct amplitudes ineach of the I and Q parallel paths. For the 16-states we encodebinary data in 4-bit sequences and the symbol duration is fourtimes the bit duration.

R s = R /4b

binary data

Pulse shaping

Series to parallelconverter

Baseband Powersplitter

Delaypi2

16-state

2-bitDAC

I channelfilter

Rs= R

Pulse shaping

2-bitDAC

Q channel filter

1cos 2 fpi t

2

1sin 2 fpi t

2

Rs= R /2

= + 1, + 3nb

R s = R /4b

QAM, Rs

= R /4 = 1/Tob

RTb

=1b

(bit/s)

cos 2 pi fc T~

c

cb/2

b

= + 1, + 3na

Figure 11.20(g) Schematic for 16-QAM modulator, asextension of Figure 11.23.

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Figure shows the corresponding time domain wav eform for thesquare 16-state QAM constellation, noting the changes in timescale in (f).

Figure 11.20 The 16-state QAM signal: (a) four-level baseband signals in the inphase and (b)quadrature branches; (c), (d) corresponding four-level modulated complex signals;(e) resulting combined complex (three-level) QAM signal; (f) demodulated signaleye diagram over two symbols (in the I or Q channels).

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The probability of symbol error for M-QAM (M ev en)signalling in Gaussian noise is:

Pe = 2

M 12 1M 12

1 erf 32(M 1) EN0 12 Where E is the average energy per QAM symbol. Forequiprobable rectangular pulse symbols E is given by:

E = 13

V2

2

(M 1) To (11.44)

V is the voltage separation between adjacent inphase orquadrature MASK levels. Using C = E/To and N = N0 B, topequation can be rewritten as:

Pe = 2

M 12 1M 12

1 erf 3 To B2(M 1) CN 12 Denoting average energy per information bit byEb = E/ log2 M , top equation becomes:

Pb =2

log2 M

M 12 1M 12

1 erf 3 log2 M2(M 1) EbN0 12 and in terms of CNR:

Pb =2

log2 M

M 12 1M 12

1 erf 3 To B2(M 1) CN 12

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In IF modulation schemes there are two measures of SNR. Onemeasures the ratio of the carrier power, C, to the noise N ,which is given by the product of Boltzmans constant, k, theoperating temperature, T , with the signal bandwidth, B.

CN

=

CkTB

=

CN0 B

(11.6)

The other measure is the ratio of the energy per information bit,Eb, to the noise power density, N0 = kT . For a bit rate, fb:

EbN0

=

CfbNB

=

CN

Bfb

=

CN

BTo (11.8)

whereBfb

is the ratio of the noise bandwidth to data bit rate.

Thus for bi-phase modulationBfb

= 1 andCN

=

EbN0

Figure 11.21 shows how the spectral efficiency in bit/s/Hzvaries with carrier to noise ratio at a Pe of 106 in thebandwidth limited channel.

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Figure 11.21 Pe and spectral efficiency for multiphase PSK and M-QAM modulation: (a) biterror probability against CNR with PSK shown as dashed and QAM as solidcurves; (b) comparison of the spectral efficiency of these modulation schemes.

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Modulation C/N ratio Eb/N0technique (dB) (dB)

PRK 10.6 10.6

QPSK 13.6 10.6

4-QAM 13.6 10.6

8-PSK 18.8 14.0

16-PSK 24.3 18.3

16-QAM 20.5 14.5

32-QAM 24.4 17.4

64-QAM 26.6 18.8

Table 11.5 Performance comparison of various digitalmodulation schemes (Pb = 106 )

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11.6 Data modems

The ITU-T V series of recommendations covers voicebandmodems.

16-QAM is used in V.22 data modems. In low bit rate datamodems a number of multi-amplitude, multi-phase schemes areused. The 9600 bit/s V.29 modem (b) needs a four-wireconnection.

V.32 has modulation options at 4800 bit/s and 9600 bit/s. V.32modems have major cost savings. Only a two-wire circuit isrequired and a single PSTN connection per end is needed.

Figure 11.53 Examples of signal constellations used in speech band data modems: (a) and (c) asused on switched lines; and (b) on leased lines.

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With launch of V.33 modems (14.4 kbit/s TCM full two-wayduplex), voiceband data rates increased considerably.With the V.fast modems 28.8 kbit/s is possible over voiceband(3 kHz) telephone circuits.Hartley-Shannon equation states that the theoretical maximumbit rate is 30 kbit/s in a 3 kHz speech channel.

The major recent advances have been in the use of digital signalprocessing (DSP) to implement the equaliser combined withpersonal computer use spurring the widespread modem use on2-wire domestic telephone circits.

Figure 11.53 Examples of signal constellations used in speech band data modems: (a) and (c) asused on switched lines; and (b) on leased lines.

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Data modems 37

Todays modems have reduced to a single printed circuit cardimplementation and they hav e combined the improved SNR andreduced noise of modern telephone circuits with VLSI circuitdensity advances to achieve 56 kbit/s rates over the standardtwisted pair 2-wire telephone connection.

This represents over a 40 fold increase in rate from early 1200bit/s designs and more than a 40 fold reduction in sizeemphasising the power of advanced DSP techniques.

Higher rates beyond 56 kbit/s are achieved over a shorterconnection lengths by the digital subscriber line (DSL)techniques.

Table 6.3 Overview of xDSL, compared to ISDN

Standard ISDN HDSL ADSL VDSLIntroduction 1987 1993 1995-9 2000?Spectral allocation (MHz) DC - 0.16 DC - 0.78 0.025 - 1.1 0.3 - 10/30Transmissions symmetric symmetric asymmetric asymmetricUpstream rate (Mbit/s) 0.144 < 2 < 1 < 2Downstream rate (Mbit/s) 0.144 < 2 < 8 < 52Line codes 4B3T, 2B1Q 2B1Q, CAP DMT, CAP/QAM DMT, QAM

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11.5.2 Orthogonal frequency division multiplex (OFDM)In OFDM the modulator comprises a serial to parallel dataconverter followed by a set of mutually orthogonal parallelmodulators.

k sequential high speed (short Tb duration) data bits aremapped into k parallel and simultaneous low speed transmitteddata bits, Figure 11.49(a).

The OFDM symbol duration is thus extended to k Tb s.The individual carriers are spaced by f = 1/(k Tb) to ensureorthogonality between the individual frequencies, i.e. realise anOFDM signal.

In place of a single wideband modulated signal there are nowk simultaneous narrowband FDM signals. Due to multipatheffects some of the FDM channels suffer fading and loss ofreceived data. This is mitigated by wrapping the OFDMmodulator and demodulator in a convolutional coder/decoder toform a COFDM system.

In many practical implementations of COFDM the numberof parallel channels ranges from 1000 to 8000 and so thediscrete modulators in the transmitter are replaced by an inversefast Fourier transform (IFFT) processor, Figure 11.49(b), andthe receiver demodulator by a forward FFT processor.

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Data modems 39

OFDM

registerHolding

Bit rate, R b (bit/s)

fc

fc + f

fc + (k - 1) f

Parallel Shift register modulation

(a)

output

Serial toparallelconverter

I F F ToutputOFDM

(b)

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Further sophistication is achieved by selecting appropriateQPSK, 16-QAM or 64-QAM modulators on each individualchannel to increase the overall data throughput rate.(Remember that 16-QAM gives 4 bit/s/Hz.) COFDM is beingadopted widely in audio (DAB), video broadcast applicationsand WLANs, see last lecture.

Of the 52 carriers used in the HIPERLAN 2 OFDM signal 48carry data and 4 are pilots (known sequences used for frequencyand phase offset correction synchronisation, etc.). The OFDMsubcarrier spacing is 0.3125 MHz giving a nominal OFDMsignal bandwidth of 16.25 MHz.

Table 21.4 HIPERLAN 2 OFDM rates and modulation orders.

Modulation Code rate Bit rateBPSK 1/2 6 Mbit/sQPSK 1/2 12 Mbit/s

16-QAM 9/16 27 Mbit/s64-QAM 3/4 54 Mbit/s

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Data modems 41

The old European local loop connection from exchange tosubscribers is shown.

Exchange

4 km

Overhead distribution

Cross-connect1.5 km

Possible 2nd cross-connect

Overhead drop

Junction box

300 m

Underground drop

50 m

Underground distribution

Figure 6.31 Local loop connections with typical customer distances.

The overall distance from the subscriber to the exchange istypically 4 - 5 km with bundles of 50 twisted pairs with crossconnect points.

DSL is an overlay technology that adds a broadband highspeed network connection on top of the existing copper cablesusing advanced DSP, modulation, coding and equalisationtechniques so that DSL and analogue telephony can co-exist.

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NB (POTS)network

narrowbandPOTS

NB BB

BB

frequency

Existing telephone wiring

equipmenttermination Network

Ethernet or ATMNew home wiring carrying

NB + xDSL

HomeExchange or cabinet

equipmentLine terminatingnetwork

Broadband

modemxDSL

POTS line card

POTS splitter

Figure 6.32 xDSL as a broadband (BB) overlay technology on the narrowband (NB) POTS localloop.

Broadband e.g. DSL has evolved from high-speed digitalsubscriber line (HDSL) in 1993 through assymetric DSL(ADSL) to very-high-speed digital subscriber line (VDSL) withxDSL being the generic term.

The assymetric DSL service provides the high rate downstreamchannel which is required for users to access rapidly internetdata pages.

The key problem with xDSL is that the transmitted spectrumcan extend to 10 MHz to give 10 Mbit/s transmission capability.

As in the previous data modems xDSL relies on multi-symbolmodulation techniques to reduce the signalling rate andoccupied bandwidth. Equalisation then lets a wideband signaloccupy this narrowband channel.

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Data modems 43

Figure 6.18 (a) Input and (b) output 2 Mbit/s pulse for a 2 km length of cable.

Figure 6.20 Line equaliser frequency responses.

Figure 6.25 Combined equaliser response for ISI and crosstalk.

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Conclusion

For simple binary modulation PSK has the best bandwidthefficiency but it requires coherent demodulation.

For improved bandwidth efficiency we need to use multi-symbol alphabets, i.e. MPSK to improve the bit/s/Hz.

MPSK has a constant envelope and can tolerate nonlinearitiesand is a bandwidth efficient modulation technique.

QAM covers same bandwidth efficiency for lower receivedCNR or SNR but it has envelope amplitude modulation andhence needs linear repeaters.

If power efficient modulation is desired and bandwidthoccupancy is not of concern then we must extend binary FSKinto MFSK.

(The GSM system uses Gaussian filtered 2 frequency MFSK-GMSK to optimise its performance.)OFDM is becoming the favoured modulation for mobile,broadcast and other e.g. DSL data transmission channels.

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