Digital communication

H V KUMARASWAMY

Topics to be covered in this session

I Minimum shift keying

II M-ary FSK

III M-ary PSK

Minimum shift keying

Proper utilization of phase during detection, for improving noise performance

Complexity increasesCPFSK (Continuous-phase frequency-shift keying)

.

0])0(2[2

1])0(2[2

)(

2

1

SymbolfortfCosT

E

SymbolfortfCosT

E

ts

b

b

b

b

θ(0) denotes the value of the phase at time t=0

])(2[2

)( ttfCosT

Ets c

b

b

An angle-modulated wave

θ(t) is the phase of s(t), continuous function of time.

)(2

121 fff c Carrier frequency

Phase bb

TttT

ht 0)0()(

)ff(Th 21b Deviation ratio

Measured with respect to bit rate 1/Tb

At time t=Tb

0

1)0()(

Symbolforh

SymbolforhTb

Phase Tree

Phase Trellis, for sequence 1101000

)2()]([2

)2()]([2

)( tfSintSinT

EtfCostCos

T

Ets c

b

bc

b

b

In terms of In phase and Quadrature Component

bb

TttT

t 02

)0()(

+ Sign corresponds to symbol 1

- Sign corresponds to symbol 0

h=1/2

bbbb

b

bb

b

b

b

TtTtT

CosT

E

tT

CosCosT

E

tCosT

Ets

2

2

2])0([

2

])([2

)(1

For the interval of bb TtT

Half cosine pulse

In phase components

+ Sign corresponds to θ(0) =0- Sign corresponds to θ(0) = п

bbb

b

bb

b

b

b

bQ

TttT

CosT

E

tT

CosTSinT

E

tSinT

Ets

202

2

2])([

2

])([2

)(

Quadrature components

+ Sign corresponds to θ(Tb) =п/2- Sign corresponds to θ(Tb) = -п/2

Half sine pulse

Four possibilities

bbcbb

TtTtfCostT

CosT

t

)2(

2

2)(1

bcbb

TttfSintT

SinT

t 20)2(2

2)(2

bTttststs 0)()()( 2211

Basic functions

bbb

T

T

TtTCosE

dtttssb

b

)0(

)()( 11

bbb

T

TtTSinE

dtttssb

b

20)(

)()(2

0

22

coefficients

Signal Space Characterization of MSK

bb

T

T

TtTws

dtttxxb

b

11

11 )()(

b

T

Ttws

dtttxxb

20

)()(

22

2

0

22

0

2

0 4

1

N

Eerfc

N

EerfcP bb

e

0N

EerfcP b

e

MSK receiver

Q-channel

Sketch the waveform of the MSK signal for the sequence for the 101101.Assume that the carrier frequencya) Is 1.25 times the bit rate. b) Equal to the bit

Solution (a) fc =(f1+f2)/2 =1.25/ Tb OR f1+f2=2.5/Tb

Also f1-f2=1/(2Tb)

Solving f1=1.5/Tb f2=1/Tb

(b) fc=1/Tb f1+f2=2/Tb f1-f2=1/(2Tb)\

Solving f1=1.25/Tb f2=0.75/Tb

conclusion