Pass-band Data Transmission Dr. Teerasit Kasetkasem

Pass-band Data Transmission

Dr. Teerasit Kasetkasem

Block Diagram

Functional model of pass-band data transmission system.

Signaling

Illustrative waveforms for the three basic forms of signaling binary information. (a) Amplitude-shift keying. (b) Phase-shift keying. (c) Frequency-shift keying with continuous phase.

What do we want to study?

We are going to study and compare different modulation techniques in terms of Probability of errors Power Spectrum Bandwidth efficiency

B

Rb Bits/s/Hz

Coherent PSK

Binary Phase Shift Keying (BPSK) Consider the system with 2 basis functions

and

tfT

t cb

2cos2

1

tfT

t cb

2sin2

2

BPSK

If we want to fix that for both symbols (0 and 1) the

transmitted energies are equal, we have

1

2

s1

We place s0 to minimize probability of error

s0

s0

BPSK

We found that phase of s1 and s0 are 180 degree difference. We can rotate s1 and s0

1

2

s1

s0

Rotate

BPSK

We observe that 2 has nothing to do with signals. Hence, only one basis function is sufficient to represent the signals

2

s1s0

1

BPSK

Finally, we have

tfT

EtEts c

b

bb 2cos

2)(11

tfT

EtEts c

b

bb 2cos

2)(10

BPSK

Signal-space diagram for coherent binary PSK system. The waveforms depicting the transmitted signals s1(t) and s2(t), displayed in the inserts, assume nc 2.

BPSK

Probability of error calculation. In the case of equally likely (Pr(m0)=Pr(m1)), we have

0

0

erfc2

1

2erfc

2

1

N

E

N

dP

b

ike

BPSK

Block diagrams for (a) binary PSK transmitter and (b) coherent binary PSK receiver.

Quadriphase-Shift Keying (QPSK)

TtitfT

Ets ci

0;

4122cos

2

T is symbol duration E is signal energy per symbol There are 4 symbols for i = 1, 2, 3, and 4

QPSK

TttiEtiE

tfT

iEtfT

iEts cci

0;4

12sin4

12cos

2sin2

412sin2cos

2

412cos

21

Which we can write in vector format as

412sin

412cos

iE

iEis

QPSK

i Input Dibit Phase of QPSK

signaling

Coordinate of Message point

si1 si2

1 10

2 00

3 01

4 11

4/

4/3

4/5

4/7

2/E

2/E

2/E

2/E

2/E

2/E2/E

2/E

QPSK

2

1

s4

s1

s3

s2

(10)(00)

(01) (11)

QPSK signals

QPSK

Block diagrams of (a) QPSK transmitter and (b) coherent QPSK receiver.

QPSK: Error Probability QPSK

Consider signal constellation given in the figure

2

1

s4

s1

s3

s2

(10)(00)

(10) (11)

Z4

Z1

Z3

Z2

2/E

2/E

2/E

2/E

QPSK

We can treat QPSK as the combination of 2 independent BPSK over the interval T=2Tb

since the first bit is transmitted by 1 and the second bit is transmitted by 2.

Probability of error for each channel is given by

00

12

2erf

2

1

2erfc

2

1

N

Ec

N

dP

QPSK

If symbol is to be received correctly both bits must be received correctly.

Hence, the average probability of correct decision is given by

Which gives the probability of errors equal to

21 PPc

0

0

2

0

2erfc

2erfc

4

1

2erfc1

N

E

N

E

N

EPP Ce

QPSK

Since one symbol of QPSK consists of two bits, we have E = 2Eb.

The above probability is the error probability per symbol. The avg. probability of error per bit

Which is exactly the same as BPSK .

0erfcsymbolper

N

EPe b

0erfc

2

1symbolper

2

1bitper

N

EPePe b

BPSK vs QPSK

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Power spectrum density of BPSK vs. QPSK

Normalized frequency,fTb

Nor

mal

ized

PS

D,

Sf/2

Eb

BPSKQPSK

QPSK

ConclusionQPSK is capable of transmitting data twice

as faster as BPSK with the same energy per bit.

We will also learn in the future that QPSK has half of the bandwidth of BPSK.

OFFSET QPSK

2

1

s4

s1

s3

s2

(10)(00)

(01) (11)

90 degree shift in phase

180 degree shift in phase

OFFSET QPSK

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.

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

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.

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 1 0

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

Offset QPSK

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

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

/4-shifted QPSK

Try to reduce amplitude fluctuation by switching between 2 signal constellation

/4-shifted QPSK

As the result, the phase of the signal can be changed in order of /4 or 3/4

/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 +3/411 -3/410 -/4

/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

2QPSK

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

0

2OFFSET QPSK

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

0

1D OFFSET QPSK

01 10 10 01

/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.

Block diagram of the /4-shifted DQPSK detector.

/4-shifted QPSK

Illustrating the possibility of phase angles wrapping around the positive real axis.

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

M-array PSK

Signal-space diagram for octaphase-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 octaphase-shift keying.

M-array PSK

Probability of errors

MEdd /sin21812

4 ;/sinerfc0

MM

N

EPe

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

babi

lity

of S

ymbo

l err

ors

QPSK 8-ary PSK 16-ary PSK

M-array PSK

Power Spectra (M-array)

M=2, we have

MfTME

TfEfS

bb

PSK

22

2

2

logsinclog2

sinc2)(

fTEfS bbBPSK2sinc2)(

M-array PSK

Power spectra of M-ary PSK signals for M 2, 4, 8.

Tbf

M-array PSK

Bandwidth efficiency: We only consider the bandwidth of the main lobe

(or null-to-null bandwidth)

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

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.

TttfbT

Etfa

T

Ets cicii 0;2sin

22cos

2 00

M-ary QAM

Again, we have

as the basis functions

TttfT

t c 0;2cos2

1

TttfT

t cb

02sin2

2

M-ary QAM

QAM square ConstellationHaving even number of bits per symbol,

denoted by 2n. M=L x L possible valuesDenoting ML

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

16-QAM

L-ary, 4-PAM

16-QAM

Calculation of Probability of errorsSince 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

16-QAM

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

16-QAM

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

Coherent FSK

FSK = frequency shift keyingCoherent = receiver have information on

where the zero phase of carrier. We can treat it as non-linear modulation

since information is put into the frequency.

Binary FSK

Transmitted signals are

where

elsewhere ,0

0,2cos2

bib

b

iTttf

T

E

ts

2,1;

iT

inf

b

ci

Binary FSK

S1(t) represented symbol “1”.

S2(t) represented symbol “0”.

This FSK is also known as Sunde’s FSK.It is continuous phase frequency-shift

keying (CPFSK).

Binary FSK

There are two basis functions written as

As a result, the signal vectors are

elsewhere ,0

0,2cos2

bibi

TttfTt

b

bE

E 0 and

021 ss

BFSK

From the figure, we have In case of Pr(0)=Pr(1), the probability of

error is given by

We observe that at a given value of Pe, the BFSK system requires twice as much power as the BPSK system.

bEd 212

022

1

N

EerfcP b

e

TRANSMITTER

RECEIVER

Power Spectral density of BFSK

Consider the Sunde’s FSK where f1 and f2 are different by 1/Tb. We can write

We observe that in-phase component does not depend on mi since

tfT

t

T

Etf

T

t

T

E

T

ttf

T

Ets

cbb

bc

bb

b

bc

b

bi

2sinsin2

2coscos2

2cos2

bb

b

bb

b

T

t

T

E

T

t

T

E cos

2cos

2


We have

For the quadrature component

bbb

b

bb

bBI T

fT

fT

E

T

t

T

EFfS

2

1

2

1

2cos

22

Half of the symbol power

bb

b

T

t

T

Etg

sin

2 2222

2

14

cos8

fT

fTTES

b

bbbBQ


Finally, we obtain )()()( fSfSfS BQBIB

Phase Tree of BFSK

FSK signal is given by

At t = 0, we have

The phase of Signal is zero.

bc

b

b

T

ttf

T

Ets

2cos2

0cos20

02cos2

0b

b

bc

b

b

T

E

Tf

T

Es

Phase Tree of BFSK

At t = Tb, we have

We observe that phase changes by after one symbol (Tb seconds). - for symbol “1” and + for symbol “0”

We can draw the phase trellis as

cos

22cos

2

b

b

b

bbc

b

bb T

E

T

TTf

T

ETs

Minimum-Shift keying (MSK)

MSK tries to shift the phase after one symbol to just half of Sunde’s FSK system. The transmitted signal is given by

"0"for 02cos2

"1"for 02cos2

2cos2

2

1

tfT

E

tfT

E

ttfT

Ets

b

b

b

b

cb

b

MSK

Where

Observe that

tT

ht

b

0

bc

bc T

hff

T

hff

2and

2 21

212

1fffc

MSK

h = Tb(f1-f2) is called “deviation ratio.”

For Sunde’s FSK, h = 1. For MSK, h = 0.5. h cannot be any smaller because the

orthogonality between cos(2f1t) and cos(2f2t) is still held for h < 0.5.

Orthogonality guarantees that both signal will not interfere each other in detection process.

MSK

Phase trellis diagram for MSK signal 1101000

MSK

Signal s(t) of MSK can be decomposed into

where

tftstfts

tftT

Etft

T

E

ttfT

Ets

cQcI

cb

bc

b

b

cb

b

2sin2cos

2sinsin2

2coscos2

2cos2

bb

TttT

t 0;2

0

MSK

Symbol (0) (Tb)

10 /2

-/2

0

0 -/2

/2

MSK

For the interval –Tb < t 0, we have

Let’s note here that the for the interval -Tb<t 0 and 0< tTb may not be the same.

We know that

0;2

0 tTtT

t bb

bbb T

t

T

t

T

t

2sin0sin

2cos0cos

20cos

MSK

Since (0) can be either 0 or depending on the past history. We have

“+” for (0) = 0 and “-” for (0) = Hence, we have

bbb T

t

T

t

T

t

2cos

2cos0cos

20cos

bbbb

bI TtT

T

t

T

Ets

;

2cos

2)(

MSK

Similarly we can write

for 0< tTb and Tb < t2Tb. Note the “+” and “-” may be different between these intervals.

Furthermore, we have that (Tb) can be /2 depending on the past history.

bb

b TtT

Tt 2

MSK

Hence, we have

we have that (Tb) can be /2 depending on the past history.

22sincos

22cossin

2sincos

2cossin

2sin

bb

bb

b

bb

b

bb

b

bb

T

tT

T

tT

T

TtT

T

TtT

T

TtT

bbb

bb T

t

T

t

T

TtT

2sin

22cos

2sin

MSK

Hence, we have

“+” for (Tb) = +/2 and “-” for (Tb) = -/2

The basis functions change to

bbb

bQ Tt

T

t

T

Ets 20;

2sin

2)(

bcbb

TttfT

t

Tt

0;2cos

2cos

21

bcbb

TttfT

t

Tt

0;2sin

2sin

22

MSK

We write MSK signal as

Where and

)()(

)(sinE)(0cosE

2sin2

sinsin2

2cos2

cos0cos2

2sinsin2

2coscos2

2211

2b1b

tsts

tTt

tfT

tT

T

Etf

T

t

T

E

tftT

Etft

T

Ets

b

cb

bb

bc

bb

b

cb

bc

b

b

0cos1 bEs bb TEs sin2

MSK

Symbol (0) s1 (Tb) s2

10 /2

-/2

0

0 -/2

/2

bE

bE

bE

bE

bE

bE

bE

bE

02

1

N

EerfcP b

e

0Phase: /2 /2 0 -/2/2

MSK

We observe that MSK is in fact the QPSK having the pulse shape

Block diagrams for transmitter and receiver are given in the next two slides.

bT

t

2cos

b

b

T

T

dtttxx )()( 11

bT

dtttxx2

022 )()(

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

4

Normalized Frequency, fTb

Nor

mal

ized

PS

D,

S(f

)/E

bMSK BPSKQPSK

MSK

Probability of error of MSK system is equal to BPSK and QPSK

This due to the fact that MSK observes the signal for two symbol intervals whereas FSK only observes for single signal interval.

Bandwidth of MSK system is 50% larger than QPSK.

2

222 116

2cos32)(

fT

fTEfS

b

bbMSK

Noncoherent Orthogonal Modulation

Noncoherent implies that phase information is not available to the receiver.

As a result, zero phase of the receiver can mean any phase of the transmitter.

Any modulation techniques that transmits information through the phase cannot be used in noncoherent receivers.


cos(2ft)

sin(2ft)

Receiver

cos(2ft)sin(2ft)

Transmitter


It is impossible to draw the signal constellation since we do not know where the axes are.

However, we can still determine the distance of the each signal constellation from the origin.

As a result, the modulation techniques that put information in the amplitude can be detected.

FSK uses the amplitude of signals in two different frequencies. Hence non-coherent receivers can be employed.


Consider the BFSK system where two frequencies f1 and f2 are used to represented two “1” and “0”.

The transmitted signal is given by

Problem is that is unknown to the receiver. For the coherent receiver, is precisely known by receiver.

bi TtitfT

Ets 0,2,1;2cos

2)(


Furthermore, we have

To get rid of the phase information (), we use the amplitude

)()(

2sinsin2

2coscos2

2cos2

)(

2211 tsts

tfT

Etf

T

E

tfT

Ets

ii

ii

i

EEEssts ii 2222

21 sincos


Where

The amplitude of the received signal

TT

i dtttxxdtttss0

110

11 )()()()(

TT

i dtttxxdtttss0

220

22 )()()()(

2/12

0

2

0

2sin)(2cos)(

T

i

T

ii dttftxdttftxl

Quadrature Receiver using correlators

Quadrature Receiver using Matched Filter


Decision rule: Let if li > lk for all k. For examples, decide if l1 > l2

This decision rule suggests that if the envelop (amplitude) of the received signal described in term of cos(2f1t) is greater than the envelop of the received signal described in term of cos(2f2t), we say s1(t) was sent.

imm ˆ

1ˆ mm

Noncoherent Matched Filter


Consider the output of matched filter of cos(2fit).

T

ii dtTfxtTftxty0

2cos)(2cos)(

2sin)()(2in-

2cos)()(2cos)(

0

0

T

ii

T

ii

dfxtTfs

dfxtTfty


Envelope at t=T is

Which is exactly the same as in correlator receiver

2/12

0

2

0

2cos)(2cos)(

T

i

T

ii dfxdfxl

Generalized binary receiver for noncoherent orthogonal modulation.

Quadrature receiver equivalent to either one of the two matched filters in part


Probability of Errors

02exp

2

1

N

EPe

Noncoherent: BFSK

For BFSK, we have

elsewhere ; 0

0;2cos2

bib

b

iTttf

T

E

ts

Noncoherent: BFSK

Noncoherent: BFSK

Probability of Errors

02exp

2

1

N

EP b

e

DPSK

Differential PSK Instead of finding the phase of the signal on

the interval 0<tTb. This receiver determines the phase difference between adjacent time intervals.

If “1” is sent, the phase will remain the same If “0” is sent, the phase will change 180 degree.

DPSK

Or we have

and

bbcb

b

bcb

b

TtTtfT

E

TttfT

E

ts

2;2cos2

20;2cos2

)(1

bbcb

b

bcb

b

TtTtfT

E

TttfT

E

ts

2;2cos2

20;2cos2

)(2

DPSK

In this case, we have T=2Tb and E=2Eb

Hence, the probability of error is given by

0exp

2

1

N

EP b

e

DPSK: Transmitter

11 kkkkk dbdbd

DPSK

{bk} 1 0 0 1 0 0 0 1 1

{dk-1} 1 1 0 1 1 0 1 0 0

Differential encoded {dk}

1 1 0 1 1 0 1 0 0 0

Transmitted Phase

0 0 0 0 0

DPSK: Receiver

DPSK: Receiver

From the block diagram, we have that the decision rule as

If the phase of signal is unchanged (send “1”) the sign (“+” or “-”) of both xi and xQ should not change. Hence, the l(x) should be positive.

If the phase of signal is unchanged (send “0”) the sign (“+” or “-1”) of both xi and xQ should change. Hence, the l(x) should be negative.

0

1say

0say

1010

QQII xxxxl x

Signal-space diagram of received DPSK signal.

Documents

Pass-band Data Transmission Dr. Teerasit Kasetkasem