QPSK: Change of phase is ±90º or ±180º

1

PB.51

Offset QPSK

QPSK: Change of phase is ±90º or ±180º

)(1 tφ

)(2 tφ

PB.52

Offset QPSK

Offset QPSK: Change of phase is ±90º only• Envelope variation is reduced.

)(1 tφ

)(2 tφ2/30 2sin2)(

0 2cos2)(

2

1

TttfT

t

TttfT

t

c

c

≤≤=

≤≤=

πφ

πφ

2

PB.53

π/4-shifted QPSK

The carrier phase used for the transmission of successivesymbols is alternately picked from one of the two QPSKconstellation shown below.

)(1 tφ

)(2 tφ

)(1 tφ

)(2 tφ

PB.54

π/4-shifted QPSK

Eight possible phase states for the π/4-shifted QPSK– phase change: ±45º and ±135º– envelope variation is reduced.– Unlike offset QPSK, it can be noncoherently detected.

)(1 tφ

)(2 tφ

3

PB.55

π/4-shifted QPSK

– Can be differentially encoded.– Example:

• π/4-Differential Quaternary Phase Shift Keying(DQPSK) is used in IS-95 (standard for CDMA mobilecommunication)

PB.56

Frequency shifting keying (FSK)

Similar to other passband data transmission system,the function model of FSK is

ModulatorSignal

transmission encoder

im

Carrier signal

)(tsi Channel)(tx

Detectoris x Signal transmission

decoder m̂

4

PB.57

Binary FSK

Symbols 1 and 0 are distinguished from each other bytransmitting one of two sinusoidal wave that differ infrequency.

≤≤=elsewhere0

0)2cos(2)( bi

b

b

iTttf

TE

ts π

where b

i Tinf +

= for some fixed integer and 2,1=i

Note: the two sinusoidal waves must be orthogonal

PB.58

It is a continuous-phase signal as phase continuity isalways maintained, including the inter-bit switchingtimes.

Called continuous-phase frequency-shift keying(CPFSK)

Example:If sTb µ1= , 9=n then MHzf 101 = and MHzf 111 = .

For sending a symbol 1, the sinusoidal wave with 1fcompletes ten cycles.

5

PB.59

The transmitted signal can be written as)()( 11 tEts bφ= and

)()( 22 tEts bφ=

where bib

i TttfT

t <≤= 0 )2cos(2)( πφ

PB.60

Therefore, the constellation of binary FSK is

)(1 tφ

)(2 tφ

bE

bEDecision boundary

bE2

Constellation of binary PSK

)(tφ

Decisionboundary

bEbE−

6

PB.61

GenerationThe binary data sequence is first applied to an on-offlevel encoder, at the output of which symbol 1 isrepresented by a constant amplitude of bE andsymbol 0 is represented by zero volts.

)(1 tφ

X

)(2 tφ

X

On-offlevel

encoder

Inverter

+

PB.62

DetectionThe detector consists of two correlators with acommon input, which are supplied with locallygenerated coherent signals. The correlator outputs arethen subtracted.

∫T

0)(tx

)(1 tφ

X1x

Decisiondevice

0

∫T

0

)(2 tφ

X2x

0 if 00 if 1

<>

yy

+

-

y

7

PB.63

As the decision boundary is )()( 21 tt φφ = ,

choose 1 if 0>ychoose 0 if 0<y

)(1 tφ

)(2 tφ

bE

bEDecision boundary

0=y 0>y

PB.64

Error ProbabilityThe received signal is

∫∫ −=

−=bb TT

dtttxdtttx

xxy

0 20 1

21

)()()()( φφ

The mean of y is

−=

sent was0 Symbolsent was1 Symbol

b

b

EE

y

The variance of y is

o

oo

N

NNxx

=

+=

+=

22

)var()var( 212σ

8

PB.65

Therefore, if symbol 0 was sent,

+−=

−−=

o

b

o NEy

N

yyyf

2)(

exp2

1

2)(exp

21)0|(

2

2

2

π

σσπ

and the error probability is

2)(

exp2

1

)sent was0 symbol0(

0

2

10

dyNEy

N

yPp

o

b

o∫∞

+−=

>=

π

PB.66

Let o

b

NEy

z2+

= ,

[ ]

( )

2/erfc21

exp22/

210

ob

NE

NE

dzzpob

=

−= ∫∞

π

As 1001 pp = ,

( )obe NEP 2/erfc21

=The error probability of binary PSK is

=

o

be N

Ep erfc21 .

i.e., we have to double the bit energy-to-noisedensity ratio, ob NE / , to maintain the samebit error rate as in a binary PSK system.

9

PB.67

Power SpectraThe general form of binary FSK is

≤≤=elsewhere0

0)2cos(2)( bi

b

b

iTttf

TE

tsπ

)/sin()2sin(2)/cos()2cos(2

)/sin()2sin(2)/cos()2cos(2

0 symbolfor 1;- symbolfor )/2cos(2)(

bcb

bbc

b

b

bcb

bbc

b

b

bcb

b

TttfTETttf

TE

TttfTETttf

TE

TttfTEts

ππππ

ππππ

ππ

±=

±−±=

+±=

Consider bTff /121 =− and their arithmetic meanequals to cf

PB.68

• The in-phase component is independent of theinput binary wave. The power spectral densityof this component consists of two deltafunctions. (one delta function if basebandspectrum is considered)

)/sin()2sin(2)/cos()2cos(2)( bcb

bbc

b

b TttfTETttf

TEts ππππ ±=

• The quadrature component is related to theinput binary wave. The power spectral densityis

)14()(cos8

222

2

−fTfTE

b

bb

ππ

10

PB.69

)/sin()2sin(2)/cos()2cos(2)( bcb

bbc

b

b TttfTETttf

TEts ππππ ±=

• The average powers of the delta functionadding up to one-half the total power of thebinary FSK signal.

• The presence of these two discrete frequencycomponents provides a means ofsynchronizing the receiver with the transmitter.

PB.70

11

PB.71

M-ary Quadrature Amplitude Modulation (QAM)

Introduction– PSK is usually limited to BPSK, QPSK and 8-PSK.– For further reducing the transmission bandwidth, QAM is

used.

– Example:Consider a voice signal with 3kHz bandwidth,• If analog amplitude modulation (AM) is used, the

transmission bandwidth is 2x3kHz = 6kHz• If analog single sideband modulation (SSB) is used,

the transmission bandwidth is 3kHz.

PB.72

If digital method is used, the minimum sampling rate is2x3kHz = 6kHz. If there are 256 levels for encoding, thedata rate is 48kbps

Referring to the diagram of PB.70, the transmissionbandwidth for BPSK is

Transmission bandwidth for 8-PSK is

Transmission bandwidth for 1024-QAM is

kHz962/2 == bRT

kHz32log/2/2 2 == MRT b

kHz6.9log/2/2 2 == MRT b

12

PB.73

Generation

4-PAM

4-PAMModulator 16-QAM

32-PAMModulator 1024-QAM

32-PAM

I

Q

I

Q

2-PAM

2-PAMModulator

I

Q QPSK

PB.74

M-ary Quadrature Amplitude Modulation (QAM)The M-ary QAM signal is defined by

[ ]

=

±±=≤≤

−=

)()(

,...2.1,00

)2sin(2)2cos(2)(

2

121 t

tss

kTt

tfbTEtfa

TEts ck

bck

bk

φφ

ππ

where

TttfT

t

TttfT

t

c

c

≤≤=

≤≤=

0 )2sin(2)(

0 )2cos(2)(

2

1

πφ

πφ

[ ] [ ]okokkk EbEass =21

13

PB.75

If there are M symbols and ML = , the M-ary squareconstellation can always be viewed as the Cartesian producta one-dimensional L-ary PAM constellation with itself.

Therefore, we have

{ }

+−−+−+−+−+−

−−−+−−+−−−−+−−+−

=

)1,1()1,3()1,1(

)3,1()3,3()3,1()1,1()1,3()1,1(

LLLLLL

LLLLLLLLLLLL

ba ii

L

MLMM

L

L

PB.76

{ }

−−−−−−−−−−−−

−−−−

=

+−−+−+−+−+−

−−−+−−+−−−−+−−+−

=

)3,3()3,1()3,1()3,3()1,3()1,1()1,1()1,3(

)1,3()1,1()1,1()1,3()3,3()3,1()3,1()3,3(

)1,1()1,3()1,1(

)3,1()3,3()3,1()1,1()1,3()1,1(

LLLLLL

LLLLLLLLLLLL

ba ii

L

MLMM

L

L

Example: Consider a 16-QAM, L=4

14

PB.77

The signal constellation is

oE oE3 1φ

2φ

PB.78

The encoding of the message is as follows:• Two of the four bits, namely, the left-most two bits,

specify the quadrant in the constellation plane inwhich a message point lies. Thus, starting from thefirst quadrant and proceeding counterclockwise, thefour quadrant are represented by 11, 10, 00, and 01.

• The remaining two bits are used to represent one ofthe four possible lying within each quadrant of theplane.

15

PB.79

oE oE3 1φ

2φ

1111

1101

1110

1100

PB.80

Error Probability

The probability of correct detection for M-ary QAM is2)'1( ec PP −=

where 'eP is the probability of symbol error for thecorresponding L-ary PAM.

The probability of symbol error for M-ary QAM is

−=

≈−−=

−=

o

o

e

e

ce

NE

M

PP

PP

erfc112

'2)'1(1

12

16

PB.81

As the transmitted energy in M-ary QAM depends on theparticular symbol transmitted, the probability of symbolerror is expressed in terms of the average value of thetransmitted energy.

Assuming that the L amplitude level levels of the in-phaseor quadrature component are equally likely, we have

3)1(2 o

avEME −

=

Therefore,

−

−=o

ave NM

EM

P)1(2

3erfc112

PB.82

Noncoherent systems

Introduction– For coherent systems, the receiver is perfectly

synchronized to the transmitter, and the only channelimpairment is noise.

– In certain situations, the receiver cannot follow thechange of the received signal phase.• Noncoherent detection is used.

17

PB.83

Noncoherent binary FSK

The transmitted signal is

bib

bi Tttf

TEts ≤≤= 0)2cos(2)( π

For the nonchoherent detection of this frequency-modulated wave, the receiver consists of a pair of matchedfilters followed by envelope detectors.

PB.84

Filtermatched to

bTttf

≤≤0)2cos( 1π

)(txEnvelopedetector

Comparisondevice

1l

21

21

if 0 if 1

llll

<>

Filtermatched to

bTttf

≤≤0)2cos( 2π

Envelopedetector

2l

Sample atbTt =

Sample at bTt =

18

PB.85

[ ]

[ ] [ ] [ ] [ ] ττπτπττπτπ

ττπτ

dfxtTfdfxtTf

dtTfxty

bb

b

TT

T

∫∫∫

−−−=

+−=

0 110 11

0 1

2sin)()(2sin2cos)()(2cos

)(2cos)()(

For example, if the received signal is)2cos()( 1 θπ += tftx

The upper matched filter output is

convolution

The envelope of the matched filter output is

[ ][ ] [ ][ ] 2/12

0 1

2

0 1 2sin)(2cos)(

+ ∫∫ ττπτττπτ dfxdfx bb TT

PB.86

For example,bT

t

The bit error rate for noncoherent binary FSK is

−=

o

be N

EP2

exp21

19

PB.87

Differential Phase-shift keying

Introduction– the noncoherent version of PSK– differential encoding of the input binary wave

– Provided that the unknown phase contained in thereceived wave varies slowly, the phase different betweenwaveforms received in two successive bit intervals will beindependent of

• To send symbol 0, advance the phase of the currentsignal waveform by 180°.

• To send symbol 1, the phase of the current signalwaveform is unchanged.

θ

θ

PB.88

Suppose the transmitted DPSK signal equals to

bcb

b TttfTE

≤≤0)2cos(2

π ,

the transmitted signal for bTt 20 ≤≤ is

≤≤

≤≤=

bbcb

b

bcb

b

TtTtfTE

TttfTE

ts2)2cos(

2

0)2cos(2)(1

π

π

if symbol 1 was sent

≤≤+

≤≤=

bbcb

b

bcb

b

TtTtfTE

TttfTE

ts2)2cos(

2

0)2cos(2)(2

ππ

π

if symbol 0 was sent

No phase change

Change in phase

20

PB.89

GenerationThe differential encoding process starts with anarbitrary first bit, serving as reference.

Delay

Logicnetwork

{ }kb

bT

DPSKsignal

Productmodulator

Amplitudelevel shifter

{ }kd

{ }1−kd)2cos(2 tf

T cb

π

If the incoming binary symbol kb is 1, leave thesymbol kd unchanged with respect to the previous bit.

PB.90

If the incoming binary symbol kb is 0, change thesymbol kd with respect to the previous bit.

Example,

{ }kb 1 0 0 1 0 0 1 1{ }1−kd 1 1 0 1 1 0 1 1

{ }kd 1 1 0 1 1 0 1 1 1( 1−⊕= kkk dbd )

Output 0 0 π 0 0 π 0 0 0phase

21

PB.91

Detection

)(tx

∫bT

0

)2cos( tfcπ

X

)2sin( tfcπ

∫bT

0X

DelaybT

DelaybT

X

X

+

yDecisiondevice

00 if 00 if 1

<>

yy

Suppose the carrier phase is unknown and the initial phaseis 0, the received signal is

)2cos()( θπ += tfAtx c

PB.92

The output of the In-phase correlator is

θθπθπ cos'cos2

)2(cos)2cos(0

AATdttftfA bc

T

cb

==+∫

and the output of the Quadrature-phase correlator is

θθπθπ sin'sin2

)2(sin)2cos(0

AATdttftfA bc

T

cb

==+∫ .

If the initial phase is 180°, the outputs are θcos'A− andθsin'A− .

22

PB.93

Therefore, the two received signal points are)sin',cos'( θθ AA and )sin',cos'( θθ AA −−

)(1 tφ

)(2 tφ

PB.94

The receiver measures the coordinates ),( 00 QI xx at time

bTt = and ),( 11 QI xx at time bTt 2= .

Therefore,

sent was0 symbol if 0

sent was1 symbol if 0

1011

1011

<+

>+

QQII

QQII

xxxx

xxxx

Error probabilityThe bit error rate for DPSK is

−=

0

exp21

NEP b

e

23

PB.95

Comparison of Digital Modulation Schemes

Probability of Error– Signaling Scheme Bit Error Rate

Coherent BPSKCoherent QPSKCoherent MSK

Coherent FSK

DPSK

Noncoherent binary FSK

( )ob NE /erfc21

( )ob NE 2/erfc21

( )ob NE /exp21

−

( )ob NE 2/exp21

−

PB.96

Comparison of DigitalModulation Schemes

The bit error rate for allthe systems decreasemonotonically withincreasing value of

ob NE /

24

PB.97

Bandwidth Efficiency– Example: M-ary PSK for probability of symbol error

=0.0001

M

4 0.5 0.34 dB8 0.333 3.91 dB16 0.25 8.52 dB32 0.2 13.52 dB

Binary

aryM

BandwidthBandwidth

)()( −

Binary

aryM

PowerAveragePowerAverage

) () ( −