OFFSET QPSK

2

1

s4

s1

s3

s2

(10)(00)

(01) (11)

90 degree shift in phase

180 degree shift in phase

1

OFFSET QPSK

2

OFFSET QPSK

Whenever both bits are changed simultaneously, 180 degree phase-shift occurs.

At 180 phase-shift, the amplitude of the transmitted signal changes very rapidly costing amplitude fluctuation.

This signal may be distorted when is passed through the filter or nonlinear amplifier.

3

OFFSET QPSK

0 1 2 3 4 5 6 7 8-2

-1

0

1

2

0 1 2 3 4 5 6 7 8-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Original Signal

Filtered signal

4

OFFSET QPSK

To solve the amplitude fluctuation

problem, we propose the offset QPSK.

Offset QPSK delay the data in quadrature

component by T/2 seconds (half of

symbol).

Now, no way that both bits can change at

the same time.

5

OFFSET QPSK

In the offset QPSK, the phase of the signal

can change by 90 or 0 degree only while

in the QPSK the phase of the signal can

change by 180 90 or 0 degree.

0 1 2 3 4 5 6 7 8-2

-1

0

1

2

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

OFFSET QPSK

0 1 2 3 4 5 6 7 8-2

-1

0

1

2

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

0 1 10

1 0 0 0

01 10 10 00

Inphase QPSK

Q phase QPSK

QPSK

0 1 1 0

1 0 0

01 10 1000

Inphase Offset QPSK

Q phase Offset QPSK

Offset QPSK

7

Offset QPSK

Possible paths for switching between themessage points in (a) QPSK and (b) offset QPSK.

8

OFFSET QPSK

Bandwidths of the offset QPSK and the

regular QPSK is the same.

From signal constellation we have that

Which is exactly the same as the regular

QPSK.

02erfc

N

EPe

9

/4-shifted QPSK

Try to reduce amplitude fluctuation by

switching between 2 signal constellation

10

/4-shifted QPSK

As the result,

the phase of the

signal can be

changed in

order of /4 or

3/4

11

/4-shifted QPSK

Since the phase of the next will be varied in

order of /4 and 3/4, we can designed the

differential /4-shifted QPSK as given below

Gray-Encoded Input Data Phase Change in radians

00 +/4

01 +3/4

11 -3/4

10 -/4

12

/4-shifted QPSK:00101001

Step Initial phase

Input Dibit Phase change

Transmitted phase

1 /4 00 /4 /2

2 /2 10 -/4 /4

3 /4 10 -/4 0

4 0 01 3/4 3/4

/4-shifted QPSK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2

0

2

QPSK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2

0

2

OFFSET QPSK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1

0

1

D OFFSET QPSK

01 10 10 01

14

/4-shifted QPSK

Since we only measure the phase different

between adjacent symbols, no phase

information is necessary. Hence, non-

coherent receiver can be used.

15

Block diagram of the /4-shifted DQPSK

detector.

16

/4-shifted QPSK

Illustrating the

possibility of phase

angles wrapping

around the positive

real axis.

17

M-array PSK

At a moment, there are M possible symbol

values being sent for M different phase

values, Mii /12

MiiM

tfT

Ets ci ,,2,1,1

22cos

2

18

M-array PSK

Signal-space diagram for

octa phase-shift keying (i.e.,

M 8). The decision

boundaries are shown as

dashed lines.

Signal-space diagram

illustrating the application of

the union bound for octa

phase shift keying.

19

M-array PSK

Probability of errors

MEdd /sin21812

4 ;/sinerfc0

MM

N

EPe

20

M-ary PSK

0 5 10 15 20 25 3010

-50

10-40

10-30

10-20

10-10

100

Eb/N

0 dB

Pro

babili

ty o

f S

ym

bol err

ors

QPSK

8-ary PSK

16-ary PSK

21

M-array PSK

Power Spectra (M-array)

M=2, we have

MfTME

TfEfS

bb

PSK

22

2

2

logsinclog2

sinc2)(

fTEfS bbBPSK2sinc2)(

22

M-array PSK

Power spectra of M-ary PSK

signals for M 2, 4, 8.

Tbf

23

M-array PSK

Bandwidth efficiency:

We only consider the bandwidth of the main lobe (or

null-to-null bandwidth)

Where B is the bandwidth of the transmitted data

Bandwidth efficiency of M-ary PSK is given by

M

R

MTTB b

b 22 log

2

log

22

MMR

R

B

R

b

bb22 log5.0log

2

24

M-ary QAM

QAM = Quadrature Amplitude Modulation

Both Amplitude and phase of carrier change

according to the transmitted symbol, mi

where ai and bi are integers, E0 is the

energy of the signal with the lowest

amplitude

TttfbT

Etfa

T

Ets cicii 0;2sin

22cos

2 00

25

M-ary QAM

Again, we have

as the basis functions

There are two QAM constellations, square

constellation and rectangular

constellation

TttfT

t c 0;2cos2

1

TttfT

t c 0;2sin2

2

26

M-ary QAM

QAM square Constellation

Having even number of bits per symbol,

denoted by 2n

M=L x L possible values

Denoting ML

27

16-QAM

)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(

, ii ba

28

16-QAM

L-ary, 4-PAM

29

QAM Probability of error

Calculation of Probability of errors

Since both basis functions are orthogonal, we

can treat the 16-QAM as combination of two

4-ary PAM systems.

For each system, the probability of error is

given by

0

0

0

11

2

11

N

Eerfc

MN

derfc

LPe

30

16-QAM Probability of error

A symbol will be received correctly if data

transmitted on both 4-ary PAM systems are

received correctly. Hence, we have

Probability of symbol error is given by

21 ec PsymbolP

eee

ece

PPP

PsymbolPsymbolP

2211

111

2

2

31

16-QAM Probability of error

Hence, we have

But because average energy is given by

We have

0

0112

N

Eerfc

MsymbolPe

3

1212

22 0

2/

1

20 EMi

L

EE

L

iav

012

3112

NM

Eerfc

MsymbolP av

e

32