1

J.1

Passband Data Transmission IReferences

Phase-shift keyingChapter 4.1-4.3, S. Haykin, Communication Systems, Wiley.

J.2

Introduction

In digital passband transmission, the incoming datastream is modulated onto a carrier with fixedfrequency and then transmitted over a band-passchannel.

In baseband pulse transmission, a data streamrepresented in the form of a discrete pulse-amplitudemodulated (PAM) signal is transmitted over a low-pass channel.

2

J.3

Passband digital transmission allows more efficient use of theallocated RF bandwidth, and flexibility in accommodating differentbaseband signal formats.

ExampleMobile Telephone Systems

GSM: GMSK modulation is used (a variation of FSK)

IS-54: π/4-DQPSK modulation is used (a variation of PSK)

J.4

The modulation process making the transmissionpossible involves switching (keying) the amplitude,frequency, or phase of a sinusoidal carrier inaccordance with the incoming data.

There are three basic signaling schemes:Amplitude-shift keying (ASK)Frequency-shift keying (FSK)Phase-shift keying (PSK)

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J.5

ASK

PSK

FSK

J.6

Unlike ASK signals, both PSK and FSK signals have aconstant envelope.

PSK and FSK are preferred to ASK signals forpassband data transmission over nonlinear channel(amplitude nonlinearities) such as micorwave link andsatellite channels.

4

J.7

Classification of digital modulation techniques

Coherent and Noncoherent

Digital modulation techniques are classified intocoherent and noncoherent techniques, depending onwhether the receiver is equipped with a phase-recovery circuit or not.

The phase-recovery circuit ensures that the localoscillator in the receiver is synchronized to theincoming carrier wave (in both frequency and phase).

J.8

Phase Recovery (Carrier Synchronization)

Two ways in which a local oscillator can be synchronized with anincoming carrier wave

transmit a pilot carrier

use a carrier-recovery circuit such as a phase-locked loop (PPL)

5

J.9

M-ary signaling

In an M-ary signaling scheme, there are M possible signalsduring each signaling interval of duration T. Usually,

nM 2= and bnTT = where bT is the bit duration.

In passband transmission, we have M-ary ASK, M-aryPSK, and M-ary FSK digital modulation schemes.We can also combine different methods:

M-ary amplitude-phase keying (APK)M-ary quadrature-amplitude modulation (QAM)

In baseband transmission, we have M-ary PAM

J.10

M-ary signaling schemes are preferred over binarysignaling schemes for transmitting digital information overband-pass channels when the requirement is to conservebandwidth at the expense of increased power.

The use of M-ary signaling enables a reduction intransmission bandwidth by the factor Mn 2log= overbinary signaling.

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J.11

Coherent PSKThe functional model of passband data transmission systemis

ModulatorSignal

transmission encoder

im

Carrier signal

)(tsi Channel)(tx

Detectoris x Signal transmission

decoder m̂

• im is a sequence of symbol emitted from a messagesource.

• The channel is linear, with a bandwidth that is wideenough to transmit the modulated signal and thechannel noise is Gaussian distributed with zeromean and power spectral density 2/oN .

J.12

The following parameters are considered for a signalingscheme:

Probability of errorA major goal of passband data transmission systems isthe optimum design of the receiver so as to minimizethe average probability of symbol error in the presenceof additive white Gaussian noise (AWGN)

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J.13

Power spectraUse to determine the signal bandwidth and co-channelinterference in multiplexed systems.In practice, the signalings are linear operation,therefore, it is sufficient to evaluate the basebandpower spectral density.

Bandwidth Efficiency

Bandwidth efficiency BRb=ρ bits/s/Hz

where bR is the data rate and B is the usedchannel bandwidth.

J.14

In a coherent binary PSK system, the pair of signals )(1 tsand )(2 ts used to represent binary symbols 1 and 0,respectively, is defined by

)2cos(2)(1 tfTEts cb

b π=

)2cos(2)2cos(2)(2 tfTEtf

TEts c

b

bc

b

b πππ −=+=

where bTt ≤≤0 , and bE is the transmitted signal energyper bit.

8

J.15

For example,

[ ] bb

b

bT

cb

bT

ETTEdttf

TEdttsE

bb=⋅=== ∫∫ 2

2)2(cos2)(0

2

0

21 π

To ensure that each transmitted bit contains an integralnumber of cycles of the carrier wave, the carrier frequency

cf is chosen equal to bTn / for some fixed integer n.

J.16

The transmitted signal can be written as)()(1 tEts bφ= and

)()(2 tEts bφ−=

where bcb

TttfT

t <≤= 0 )2cos(2)( πφ

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J.17

Generation of coherent binary PSK signals

To generate a binary PSK signal, we have to represent theinput binary sequence in polar form with symbols 1 and 0represented by constant amplitude levels of bE+ and

bE− , respectively.

ProductModulator

Signal transmission

encoder

10101 is

)2cos(2)( tfT

t cb

πφ =

)(tsi

J.18

• This signal transmission encoder is performed by apolar nonreturn-to-zero (NRZ) encoder.

−+

=0 is symbolinput 1 is symbolinput

b

bi E

Es

• The carrier frequency bc Tnf /= where n is a fixedinteger.

−=−=

===

bicb

b

bicb

b

i

EstfTEts

EstfTEts

ts if)2cos(2)(

if)2cos(2)()(

2

1

π

π

10

J.19

Detection of coherent binary PSK signals

To detect the original binary sequence of 1s and 0s, weapply the noisy PSK signal to a correlator. The correlatoroutput is compared with a threshold of zero volts.

∫bT

0)(tx

)(tφ

Correlator

X1x Decision

device

0

0 if 00 if 1

1

1

<>

xx

J.20

Example: If the transmitted symbol is 1,

)2cos(2)( tfTEtx cb

b π=

and the correlator output is

b

T

cb

b

T

cb

cb

b

T

E

dttfT

E

dttfT

tfTE

dtttxx

b

b

b

=

⋅=

⋅=

=

∫

∫

∫

0

2

0

01

)2(cos2

)2cos(2)2cos(2

)()(

π

ππ

φ

Similarly, If the transmitted symbol is 0, bEx −=1 .

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J.21

Error probability of binary PSK

We can represent a coherent binary system with a signalconstellation consisting of two message points.

• The coordinates of the message points are all thepossible correlator output under a noiselesscondition.

• The coordinates for BPSK are bb EE − and .

)(tφ

Decisionboundary

bEbE−

J.22

There are two possible kinds of erroneous decision:• Signal )(2 ts is transmitted, but the noise is such that

the received signal point inside region with 01 >xand so the receiver decides in favor of signal )(1 ts .

• Signal )(1 ts is transmitted, but the noise is such thatthe received signal point inside region with 01 <xand so the receiver decides in favor of signal )(2 ts .

∫bT

0

)(tφ

X 1x Decisiondevice

0

0 if 00 if 1

1

1

<>

xx

)()( twtsi +

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J.23

For the first case, the observable element 1x is related to thereceived signal )(tx by

[ ]

∫∫∫

+−=

+=

=

b

b

b

T

b

T

i

T

dtttwE

dtttwts

dtttxx

0

0

01

)()(

)()()(

)()(

φ

φ

φ

1x is a Gaussian process with mean 1x :

b

T

b

ii

E

dtttwEE

xExb

−=

+−=

=

∫ ])()([

][

0φ

J.24

and variance σ :

2

)(2

)()()(2

)()()]()([

)()()()(

)()(

])[(

0

2

0 0

0 0

0 0

2

0

22

o

To

T To

T T

T T

T

ii

N

dttN

dtduututN

dtduutuwtwE

dtduutuwtwE

dtttwE

xxE

b

b b

b b

b b

b

=

=

−=

=

=

=

−=

∫

∫ ∫

∫ ∫

∫ ∫

∫

φ

φφδ

φφ

φφ

φ

σ

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J.25

Therefore, the conditional probability density function of1x , given that symbol 0 was transmitted is

+−=

−−=

o

b

o NEx

N

xxxf

21

2

211

1

)(exp1

2)(exp

21)0|(

π

σσπ

J.26

and the probability of error is

∫

∫∞

∞

+−=

=

0 1

21

0 1110

)(exp1

)0|(

dxN

ExN

dxxfp

o

b

oπ

Putting )(1b

o

ExN

z += , we have

[ ]

=

−= ∫∞

o

b

NE

NE

dzzpob

erfc21

exp10/

210 π

∫∞

−=u

dzzu )exp(2)erfc( 2

π

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J.27

Similarly, the error of the second kind

==

o

b

NEpp erfc

21

1001 and hence

=

o

be N

Ep erfc21

J.28

Quadriphase-shift keying (QPSK)

QPSK has twice the bandwith efficiency of BPSK,since 2 bits are transmitted in a single modulationsymbol. The data input )(tdk is devided into an in-phase stream )(td I , and a quadrature stream )(tdQ .

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J.29

t

1 0 0 1

t

1 0

t

0 1

)(tdk

)(tdI

)(tdQ

J.30

The phase of the carrier takes on one of four equallyspaced values, such as π/4, 3π/4, 5π/4, and 7π/4.

≤≤−+=elsewhere0

0]4/)12(2cos[2)( Ttitf

TE

ts ci

ππ

where .4,3,2,1=iE is the transmitted signal energy per symbol;T is the symbol duration;

Tnfc /= ;

)2 :(Note bTT =

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J.31

Each possible value of the phase corresponds to aunique dibit.

For example, 10 for i=1, 00 for i=2, 01 for i=3and 11 for i=4.(only a single bit is change from one dibit to thenext)

J.32

The transmitted signal can be written as

)()(

]4/)12sin[(]2sin[2

]4/)12cos[(]2cos[2

]4/)12(2cos[2)(

2211 tsts

itfTE

itfTE

itfTEts

ii

c

c

ci

φφ

ππ

ππ

ππ

+=

−−

−=

−+=

where

]2sin[2)(;]2cos[2)( 21 tfT

ttfT

t cc πφπφ ==

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J.33

Input dibit Phase of QPSK 1is 2is10 π/4 2/E 2/E−00 3π/4 2/E− 2/E−01 5π/4 2/E− 2/E11 7π/4 2/E 2/E

J.34

18

J.35

The constellation of QPSK is

2/E

)(1 tφ

)(2 tφ

)10(

)11(

)00(

)01(

2/E

2/E−

2/E−

J.36

Generation of coherent QPSK signals

The incoming binary data sequence is first transformed intopolar form by a nonreturn-to-zero level encoder. The binarywave is next divided by means of a demultiplexer into twoseparate binary sequences.

The result can be regarded as a pair of binary PSKsignals, which may be detected independently due tothe orthogonality of )(1 tφ and )(2 tφ .

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J.37

Demulti-plexerPolar NRZ

10101 is

)2cos(2)(1 tfT

t cb

πφ =

)(ts

is1

is2

X

X

+

)2sin(2)(1 tfT

t cb

πφ =

J.38

Detection of coherent QPSK signals

∫T

0)(tx

)(1 tφIn-phase channel

X1x Decision

device

0

multiplexer

∫T

0

)(2 tφ

X2x Decision

device

0

0 if 00 if 1

1

1

<>

xx

0 if 00 if 1

2

2

<>

xx

Quadrature channel

20

J.39

Error probability of QPSK

The received signal is)()()( twtstx i +=

and the observation elements are

∫∫

+±=

=

b

b

T

b

T

dtttwE

dtttxx

0 1

0 11

)()(

)()(

φ

φ

∫∫

+±=

=

b

b

T

b

T

dtttwE

dtttxx

0 2

0 22

)()(

)()(

φ

φ

J.40

As a coherent QPSK is equivalent to two coherent binaryPSK systems working in parallel and using two carriers thatare in phase quadrature.

Hence, the average probability of bit error in each channelof the coherent QPSK system is

=

=

oo NE

NEp

2erfc

212/erfc

21

21

J.41

Error probability of QPSK

As the bit error in the in-phase and quadrature channels ofthe coherent QPSK system are statistically independent, theaverage probability of a correct decision resulting from thecombined action of the two channels is

+

−=

−=

−=

oo

o

c

NE

NE

NE

pp

2erfc

41

2erfc1

2erfc

211

)1(

2

2

2

J.42

The average probability of symbol error for coherent QPSKis therefore

12/ if2

erfc

2erfc

41

2erfc

1

2

>>

≈

−

=

−=

oo

oo

ce

NENE

NE

NE

pp

22

J.43

In a QPSK system, since there are two bits per symbol, thetransmitted signal energy per symbol is twice the signalenergy per bit,

bEE 2=

and then

≈

o

be N

Ep2

erfct

1 0 0 1

t

1 0

t

0 1

)(tdk

)(tdI

)(tdQ

J.44

With Gray encoding, the bit error rate of QPSK is

Therefore, a coherent QPSK system achieves the sameaverage probability of bit error as a coherent binary PSKsystem for the same bit rate and the same ob NE / but usesonly half the channel bandwidth.

=

o

b

NE

2erfc

21BER

23

J.45

M-ary PSK

During each signaling interval of duration T, one ofthe M possible signals

iiM

tfTEts ci ,...,2,1)1(22cos2)( =

−+=

ππ

is sent.

J.46

M-ary PSK

The signal constellation of M-ary PSK consists of Mmessage points which are equally spaced on a circle ofradius E . For example, the constellation ofoctaphase-shift keying is

≈

MNEP

oe

πsinerfc 4≥M

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J.47

Power spectra of M-ary PSK signals

The symbol function is

≤≤=otherwise0

02)( Tt

TE

tg

where MTT b 2log= and bT is the bit duration.

As the energy spectral density is the magnitude of thesignal’s Fourier transform, the baseband powerspectral density is

)log(sinclog2)(

)(sin2)(

22

2

2

2

MfTMETf

TfEfS

bb=

=ππ

J.48(Normalized to bfT )

25

J.49

Bandwidth efficiencyThe bandwidth required to pass M-ary signal (mainlobe) is given by

MR

MT

TB

b

b

2

2

log2log

2

0)2sinc(2

=

=

== Q

Therefore, the bandwidth efficiency is

2log2 MBRb

=

=ρ