Comm Ch08 Digmod En

8/12/2019 Comm Ch08 Digmod En

1/80

2013/2014 Meixia Tao @ SJTU

Principles of ommunicationsMeixia Tao

Shanghai Jiao Tong University

Chapter 8: Digital Modulation TechniquesTextbook: Ch 8.4 8.5, Ch 10.1-10.5

1


2/80


Topics to be Covered

Binary digital modulation

M-ary digital modulation

Tradeoff study

data basebandDigital

modulator

Bandpass

channel

Digital

demodulatorBPFdetector

Noise

2


3/80


Digital Modulation

The message signal is transmitted by a sinusoidal carrierwave

In digital communications, the modulation process

corresponds to switching orkeying the amplitude, frequency,

orphase of the carrier in accordance with the incomingdigital data

Three basic digital modulation techniques Amplitude-shift keying (ASK) - special case of AM

Frequency-shift keying (FSK) - special case of FM Phase-shift keying (PSK) - special case of PM

Will use signal space approach in receiver design and

performance analysis

3


4/80


8.1 Binary Modulation Types

In binary signaling, the modulator produces one of twodistinct signals in response to 1 bit of source data at atime.

Binary modulation types

Binary PSK (BPSK)

Binary FSK

Binary ASK

4


5/80


Binary Phase-Shift Keying (BPSK)

Modulation

, , bit duration

: carrier frequency, chosen to be for some fixedinteger or

: transmitted signal energy per bit, i.e.

The pair of signals differ only in a relative phase shift of 180degrees

0 1 1 0 1 0 0 1

1

0

1 /c bf T>>

5


6/80


Clearly, there is one basis function of unit energy

Then

A binary PSK system is therefore characterized by having

a signal space that is one-dimensional (i.e. N=1), and with

two message points (i.e. M = 2)

Signal Space Representation for

BPSK

s1s20

6


7/80


Decision Rule of BPSK

Assume that the two signals are equally likely, i.e. P(s1) =P(s2) = 0.5. Then the optimum decision boundary is themidpoint of the line joining these two message points

Decision rule:

Guess signal s1(t) (or binary 1) was transmitted if thereceived signal point r falls in region R1

Guess signal s2(t) (or binary 0) was transmitted otherwise

s1s2

Region R1Region R2

0

7


8/80


Proof of the Decision Rule

Observation scalar (output of the demodulator) r is

where n represents the AWGN component, which hasmean zero and variance

Thus, the likelihood function of r is

If s1 is transmitted

If s2 is transmitted

8


9/80


Recall ML decision criterion:

Thus

And

Finally

Choose s1> r

2

When r1< r

2, r falls inside region R2 and the receiver decides in

favor of s2

Message point

Message

point

Decision boundary

R1

R2

17


18/80


Probabil ity of Error for Binary FSK

Given that s1 is transmitted,

Since the condition r1< r

2corresponds to the receiver

making a decision in favor of symbol s2, the conditional

probability of error given s1 is transmitted is given by

Define a new random variable

Since n1 and n2 are i.i.d with

Thus, n is also Gaussian with

and

18


19/80


By symmetry

Since the two signals are equally likely to be

transmitted, the average probability of errorfor

coherent binary FSK is

3 dB worse than BPSK

i.e. to achieve the same Pe, BFSK needs 3dB more

transmission power than BPSK

19


20/80


Binary FSK Transmitter

On-off signalling form

0

1

20


21/80


Coherent Binary FSK Receiver

bT

dt0

bT

dt0

Decision

Device+

Choose 1 if l>0

Choose 0 otherwise

+

-

21


22/80


Binary ASK

Modulation

Average energy per bit

0 1 1 0 1 0 0 1

(On-off signalling)

1

0

s1s2

Region R1Region R2

0

22


23/80


Probability of Error for Binary ASK

Average probability of error is

Exercise: Prove Pe

Identical to that of coherent binary FSK

23


24/80


Probability of Error and the Distance

Between Signals

These expressions illustrate the dependence of the error

probability on the distance between two signal points. Ingeneral,

BPSK BFSK BASK

24


25/80


0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Probability

ofB

itError

PSK

ASK/FSK

3dB

Probability of Error Curve for BPSK and FSK/ASK

e.g. 25


26/80


Example #1

Binary data are transmitted over a microwave linkat the rate of 106 bits/sec and the PSD of the

noise at the receiver input is 10-10 watts/Hz.

a) Find the average carrier powerrequired tomaintain an average probability of error

for coherent binary FSK.

b) Repeat the calculation in a) for noncoherent

binary FSK

26


27/80

2013/2014 Meixia Tao @ SJTU 27

We have discussed

Coherent modulation schemes, .e.g.

BPSK, BFSK, BASK They needs coherent detection,

assuming that the receiver is able to

detect and track the carrier waves

phase

Update

We now consider:

Non-coherent detection on binary FSK

Differential phase-shift keying (DPSK)

In many practical situations, strict phase

synchronization is not possible. In these

situations, non-coherent reception is required.

27


28/80


8.2: Non-coherent scheme: BFSK

Consider a binary FSK system, the two signals are

Where and are unknown random phases with

uniform distribution

28


29/80


Signal Space Representation

No matter what the two phases are, the signalscan be expressed as a linear combination of the

four basis functions

Signal space representation

29


30/80


Correlating the received signal r(t) with the four basis

functions produces the vector representation of thereceived signal

Detector

30


31/80


Decision Rule for Non-coherent FSK

ML criterion, assume P(s1) = P(s2):

Conditional pdf

Similarly,

Choose s1> envelop detector

Carrier phase is irrelevant in decision making

34


35/80


Structure of Non-Coherent Receiver for

Binary FSK

It can be shown that

Comparator

(selectthe

largest)

(For detailed proof, see Section 10.4.2 in the textbook )

35


36/80


37/80


Differential PSK (DPSK)

DPSK can be viewed as the non-coherent versionof PSK.

Phase synchronization is eliminated using

differential encoding Encoding the information in phase difference

between successive signal transmission

In effect:

to send 0, we phase advance the current signal

waveform by 1800 ;

to send 1, we leave the phase unchanged

37


38/80


DPSK (contd)

Provided that the unknown phase contained inthe received wave varies slowly (constant over two

bit intervals), the phase difference between

waveforms received in two successive bit interval

will be independent of

38


39/80


Generation of DPSK signal

We can generate DPSK signals by combining two basicoperations

Differential encoding of the information binary bits

Phase shift keying

The differential encoding process starts with an arbitraryfirst bit, serving as reference

Let {mi} be input information binary bit sequence, {di} bethe differentially encoded bit sequence

If the incoming bit miis 1, leave the symbol d

iunchanged

with respect to the previous bit di-1

If the incoming bit mi is 0, change the symbol di with respectto the previous bit di-1

39


40/80


Illustration

The reference bit is chosen arbitrary, here taken as 1

DPSK transmitter diagram

1 0 0 1 0 0 1 1Binary data

1 1 0 1 1 0 1 1 1

0 0 0 0 0 0 0

Initial bit

Differentially

encoded

binary data

TransmittedPhase

mi

di___________

1 iii mdd =

40


41/80


42/80


Error Probability of DPSK

The differential detector is suboptimal in the senseof error performance

It can be shown that

42


43/80


Summary of Pe for Different Binary

Modulations

Coherent PSK

Coherent ASK

Coherent FSK

Non-Coherent FSK

DPSK

43


44/80


0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10

-3

10-2

10-1

100

Eb/No in [dB]

Probability

of

BitError

BPSK(QPSK)

ASK/FSK

NC FSK

DPSK

Pe Plots for Different Binary Modulations

44


45/80


We have discussed binary case

Coherent modulation techniques:

BPSK, BFSK, BASK

Noncoherent modulation techniques:

Non-coherent FSK, DPSK

Update

We now consider:

M-ary modulation techniques

MPSK

MQAM

MFSK

45


46/80


8.3 M-ary Modulation Techniques

In binary data transmission, send only one of two possiblesignals during each bit interval Tb

In M-ary data transmission, send one of M possible signalsduring each signaling interval T

In almost all applications, M = 2

n

and T = nTb, where n isan integer

Each of the M signals is called a symbol

These signals are generated by changing the amplitude,

phase or frequency of a carrier in M discrete steps. Thus, we have M-ary ASK, M-ary PSK, and M-ary FSK

digital modulation schemes

46


47/80


Binary is a special case of M-ary

Another way of generating M-ary signals is to

combine different methods of modulation into

hybrid forms

For example, we may combine discrete changes

in both the amplitude and phase of a carrier to

produce M-ary amplitude phase keying.A specialform of this hybrid modulation is M-ary QAM

(MQAM)

47


48/80


M-ary Phase-Shift Keying (MPSK)

The phase of the carrier takes on M possible values:

Signal set:

= Energy per symbol

Basis functions

1c

fT

>>

48


49/80


MPSK (contd)


49


50/80


MPSK Signal Constellations

BPSK QPSK 8PSK 16PSK

50


51/80


The Euclidean distance between any two signal points in the

constellation is

The minimum Euclidean distance is

plays an important role in determining error performance as

discussed previously (union bound)

In the case of PSK modulation, the error probability is dominated by

the erroneous selection of either one of the two signal points adjacent

to the transmitted signal point. Consequently, an approximation to the symbol error probability is

2 ( )

2 1 cosmn m n sm n

d E M

= =

s s

min

22 1 cos 2 sin

s sd E EM M

= =

min

0

/ 22 2 2 sin

/ 2MPSK s

dP Q Q E

MN

=

51


52/80


Exercise

Consider the M=2, 4, 8 PSK signal constellations.All have the same transmitted signal energy Es.

Determine the minimum distance between

adjacent signal points For M=8, determine by how many dB the

transmitted signal energy Es must be increased to

achieve the same as M =4.

mind

mind

52


53/80


Error Performance of MPSK

For large M, doubling the

number of phases requires an

additional 6dB/bit to achieve

the same performance

4dB 5dB 6dB

53

M ary Quadrature Amplitude Modulation


54/80


M-ary Quadrature Amplitude Modulation

(MQAM)

In an M-ary PSK system, in-phase and quadraturecomponents are interrelated in such a way that the

envelope is constant (circular constellation). If we

relax this constraint, we get M-ary QAM.

Signal set:

E0 is the energy of the signal with the lowest amplitude

ai, bi are a pair of independent integers

54


55/80


MQAM (contd)

Basis functions:


55


56/80


MQAM Signal Constellation

Square lattice

Can be related with two L-ary ASK in in-phase and

quadrature components, respectively, where M = L2

1 3 5 7

56


57/80


Error Performance of MQAM

It can be shown that the symbol error probability ofMQAM is tightly upper bounded as

Exercise: From the above expression, determine the increasein the average energy per bit Eb required to maintain the same

error performance if the number of bits per symbol is increased

from k to k+1, where k is large.

0

34

( 1)

b

e

kEP Q

M N

(for )2kM=

57

M ary Frequency Shift Keying (MFSK) or


58/80


M-ary Frequency-Shift Keying (MFSK) or

Multitone Signaling

Signal set:

where

As a measure of similarity between a pair of signal

waveforms, we define the correlation coefficients

58


59/80


MFSK (contd)

For orthogonality, minimum frequency separationbetween successive frequencies is 1/(2T)

The minimum correlation is at

-0.217

59


60/80


M-ary orthogonal FSK has a geometric presenation as MM-dim orthogonal vectors, given as

The basis functions are

( )0 , 0, 0, , 0sE=s

( )1 0, , 0, , 0sE=s

( )1 0, 0, , 0,M sE =s

( )2

cos2m c

f m f tT

= +

60


61/80


62/80


Notes on Error Probabil ity Calculations

Pe is found by integrating conditional probability oferror over the decision region

Difficult for multi-dimensions

Can be simplified using union bound (see ch07)

Pe depends only on the distance profile of signal

constellation

62


63/80


Example #2

The 16-QAM signal constellation shown below is aninternational standard for telephone-line modems (called

V.29).

a) Determine the optimum decision

boundaries for the detectorb) Derive the union bound of the

probability of symbol error

assuming that the SNR is

sufficiently high so that errors

only occur between adjacentpoints

c) Specify a Gray code for this 16-

QAM V.29 signal constellation

63


64/80


Symbol Error versus Bit Error

Symbol errors are different from bit errors When a symbol error occurs, all bits could

be in error

In general, we can find BER using

is the number bits which differ between and

64


65/80


66/80


Example: Gray Code for QPSK

0001

11 10

66


67/80


Bit Error Rate for MPSK and MFSK

For MPSK with gray coding An error between adjacent symbol will most likely occur

Thus, bit error probability can be approximated by

For MFSK When an error occurs anyone of the other symbols may result

equally likely.

On average, therefore, half of the bits will be incorrect. That is k/2bits every k bits will on average be in error when there is a symbol

error Thus, the probability of bit error is approximately half the symbol

error

eb PP2

1

67

8 4 Comparison of M-ary Modulation


68/80


8.4 Comparison of M ary Modulation

Techniques

Channel bandwidth and transmit powerare twoprimary communication resources and have to beused as efficient as possible

Power utilization efficiency (energy efficiency):

measured by the required Eb/No to achieve acertain bit error probability

Spectrum utilization efficiency (bandwidthefficiency): measured by the achievable data rate

per unit bandwidth Rb/B It is always desired to maximize bandwidth

efficiency at a minimal required Eb/No

68

Example # 3


69/80


Example # 3

Suppose you are a system engineerin Huawei, designing a part of thecommunication systems. You are required to design three systems as follow:

I. An ultra-wideband system. This system can use a large of amount ofbandwidth to communicate. But the band it uses is overlaying with the othercommunication system. The main purpose of deploying this system is toprovide high data rates.

II. A wireless remote control system designated for controlling devicesremotely under unlicensed band.

III. A fixed wireless system. The transmitters and receivers are mounted in afixed position with power supply. This system is to support voice and dataconnections in the rural areas. This system works under licensed band.

You are only required to design a modulation scheme for each of the abovesystems. You are allowed to use MFSK, MPSK and MQAM only. You alsoneed to state the modulation level. For simplicity, the modulation level shouldbe chosen from M=[Low, Medium, High]. Justify your answers.

(Hints: Federal Communications Commission (FCC) has a power spectraldensity limit in unlicensed band. It is meant that if your system works underunlicensed band, the power cannot be larger than a limit.)

69

Energy Efficiency Comparison


70/80


e gy c e cy Co pa so

MFSK MPSK

70


71/80


Energy Efficiency Comparison (contd)

MFSK: At fixed Eb/No, increase M can provide an improvement

on Pb

At fixed Pb increase M can provide a reduction in theE

b/N

orequirement

MPSK

BPSK and QPSK have the same energy efficiency

At fixed Eb/No, increase M degrades Pb

At fixed Pb, increase M increases the Eb/Norequirement

MFSK is more energy efficient than MPSK

71


72/80

B d idth Effi i C i ( td)


73/80


Bandwidth Efficiency Comparison (contd)

In general, bandwidth required to pass MPSK/MQAM signalis approximately given by

But

Then bandwidth efficiency may be expressed as

= bit rate

(bits/sec/Hz)

73

B d idth Effi i C i ( t d)


74/80


MFSK:

Bandwidth required to transmit MFSK signal is

Bandwidth efficiency of MFSK signal

Bandwidth Efficiency Comparison (contd)

(Adjacent frequencies need to be separated

by 1/2T to maintain orthogonality)

(bits/s/Hz)

M 2 4 8 16 32 64

1 1 0.75 0.5 0.3125 0.1875(bits/s/Hz)

As M increases, bandwidth efficiency of MPSK/MQAM increases, but

bandwidth efficiency of MFSK decreases.

74

Fundamental Tradeoff :


75/80


Fundamental Tradeoff :Bandwidth Efficiency and Energy Efficiency

To see the ultimate power-bandwidth tradeoff, we need touse Shannons channel capacity theorem:

Channel Capacity is the theoretical upper bound for the maximum

rate at which information could be transmitted without error

(Shannon 1948)

For a bandlimited channel corrupted by AWGN, the maximum rateachievable is given by

Note that

Thus

)1(log)1(log0

22BN

PBSNRBCR s+=+=

R

BSNR

BRN

BP

RN

P

N

TP

N

Esssb

====0000

)12( /

0

= BRbR

B

N

E

75


76/80


77/80

S t D i T d ff


78/80


System Design Tradeoff

Power Limited Systems:Power scarce

but bandwidth available

Bandwidth Limited Systems:

Bandwidth scarce

Power available

Which

Modulation

to Use?

78

Example # 3


79/80


p

Suppose you are a system engineerin Huawei, designing a part of thecommunication systems. You are required to design three systems as follow:

I. An ultra-wideband system. This system can use a large of amount ofbandwidth to communicate. But the band it uses is overlaying with the othercommunication system. The main purpose of deploying this system is toprovide high data rates.

II. A wireless remote control system designated for controlling devicesremotely under unlicensed band.

III. A fixed wireless system. The transmitters and receivers are mounted in a

fixed position with power supply. This system is to support voice and dataconnections in the rural areas. This system works under licensed band.

You are only required to design a modulation scheme for each of the abovesystems. You are allowed to use MFSK, MPSK and MQAM only. You alsoneed to state the modulation level. For simplicity, the modulation level shouldbe chosen from M=[Low, Medium, High]. Justify your answers.

(Hints: Federal Communications Commission (FCC) has a power spectraldensity limit in unlicensed band. It is meant that if your system works underunlicensed band, the power cannot be larger than a limit.)

79

Practical Applications


80/80

Practical Applications

BPSK:

WLAN IEEE802.11b (1 Mbps)

QPSK:

WLAN IEEE802.11b (2 Mbps, 5.5 Mbps, 11 Mbps)

3G WDMA

DVB-T (with OFDM)

QAM

Telephone modem (16QAM)

Downstream of Cable modem (64QAM, 256QAM)

WLAN IEEE802.11a/g (16QAM for 24Mbps, 36Mbps; 64QAM for 38Mbps

and 54 Mbps)

LTE Cellular Systems

FSK:

Cordless telephone

Paging system

Comm Ch08 Digmod En

Documents

Transcript of Comm Ch08 Digmod En