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Differential PSK (DPSK) and Its Performance

( ) ( )cos(2 ( ) )cs t A t f t tπ θ θ= + +

A(t) = amplitude information

θ(t) = phase information

fc = the carrier frequency

θ = the phase shift introduced by the channel

• In coherent communications, θ is known to receivers

• In non-coherent communications, θ is unknown to receivers and

assumed to be a random variable distributed uniformly over (-π, π)

Coherent Detection of Differentially Encoded

M-PSK Signals

• In coherent communications, phase estimation is required.

• The receiver usually derives its frequency and phase demodulation references

from a carrier synchronization loop.

• The synchronization loop may introduce a phase ambiguity

where is the estimate acquired by the receiver through the synchronization

loop.

• For M-PSK signals, the phase ambiguity Φmay take any value in

Thus any value of Φ other than 0 will cause a correctly detected phase to be

erroneously mistaken for one of the other possible phases, even in the

absence of noise.

• Solution: differential encoding while maintaining coherent detection

ˆφ θ θ= −θ̂

2 2 ( 1){0, , }

M

M M

π π −⋯

Differentially Encoded M-PSK Signals

• Instead of encoding information into absolute phases, the

information is now encoded using phase differences between

successive signal transmission.

Example: 0 mapped to phase 0, 1 mapped to phase π

π 0 0 0 π 00Differentially encoded

Phase sequence

π π 0 0 π π0Absolute phase

sequence {∆θn}

1 1 0 0 1 10Binary sequence {an}

The actually transmitted

Phase sequenceInitial value

� The information sequence {an} is now carried by phase difference

in {θn}.

π 0 0 0 π 00Differentially encoded

Phase sequence

π π 0 0 π π0Absolute phase

sequence {∆θn}

1 1 0 0 1 10Binary sequence {an}

The actually transmitted

Phase sequenceInitial value


•Assume the phase ambiguity Φ introduced by the synchronization

loop is constant during successive signal intervals. In the absence

of noise, the receiver would convert the received phase sequence

into 0+Φ, π+Φ, 0+Φ, 0+Φ, 0+Φ, π+Φ, 0+Φ.

•The received sequence would be then differentially decoded into π,

π, 0, 0, π, π.

•Then we get the original binary sequence (110011) back.


Example: M=4. Assume that the following Gray mapping is used

Binary sequence 00 10 11 01

absolute phase 0 π/2 π 3π/2

Binary sequence {an} 10 11 00 10 01 11

Absolute phase sequence {∆θn} π/2 π 0 π/2 3π/2 π

Differentially encoded π/2 3π/2 3π/2 0 3π/2 π/2

phase sequence {θn}

Estimated phase sequence π/2+Φ 3π/2+Φ 3π/2+Φ 0+Φ 3π/2+Φ π/2+Φ

(in the absence of noise)

Differentially decoded sequence π 0 π/2 3π/2 π

Decoded binary sequence 11 00 10 01 11

ˆnθ

ˆna

Φ є { 0, π/2, π, 3π/2} is the phase ambiguity

The Structure of Differential Encoder and Decoder

ˆ{ }nθ

Delay T

{∆θn}{θn}

Modulo 2π addition

Delay T

Figure a. A differential encoder

Figure b. A differential decoder

{∆θn} = the absolute phase

sequence

{θn} = the transmitted phase

sequence

ˆ{ }nθ∆

A digital comm. system that

employ differential encoding of the

inform. Symbols and coherent

detection of successive

differentially encoded M-PSK

signals is called a differentially

encoded coherent M-PSK system.

-

Optimal Coherent Receiver for Differentially

Encoded M-PSK

,1

n

,

tann s

n c

r

rη −=

ˆ2 cos(2 )cf tπ θ+

ˆ2 sin(2 )cf tπ θ− +

r(t)

X

X

( 1 )n T

n Td t

+

∫

( 1)n T

nTdt

+

∫

rn,c

rn,s

900

Carrier

Sync. loop

Sub-

tractor

Select

the

smallest

ˆ{ }nθ

Delay T

Modulo 2π addition

ˆ{ }nθ∆

-

|ηn-β0|

|ηn-β1|

|ηn-βM-1|

1

2( ) cos(2 )

( 1) , 0, 1, 2,

phase shift introduced by the channel

ˆ estimate of θ provided by carrier

synch. loop

ˆ the phase ambiguity

,

{ } represents t

sc n

n n n

n

Er t f t

T

nT t n T n

π θ θ

θ

θ

φ θ θ

θ θ θ

θ

−

= + +

≤ ≤ + = ± ±

=

=

= − =

= −

⋯

△

△ he inform. sequence

Performance

0ˆLet denote Pr{ | }i iP m m m m= =

2

0 20

0

s in1

= 1 - e x p { }s in

e n e rg y / s y m b o l.

sM

s

EMP d

N

E

ππ

π

απ α

−−

=

∫

in the optimal coherent

receiver fro M-PSK.

One can show that the prob. of correct decision in the optimal

coherent receiver for differentially encoded M-PSK is

12

0

12

0

i 0

1

0 coherent M-PSK

0

1 1

Since P for i 0, it follows that

|

M

c i

i

M

e c i

i

M

c o i C

i

P P

P P P

P

P P P P P

−

=

−

=

−

=

=

⇒ = − = −

< ≠

< = =

∑

∑

∑

The symbol error

Pe > Pe|coherent M-PSK

12 2

0

1

12 2

| |

1

1

= 2 ( )

M

e i

i

M

e M PSK e M PSK i

i

P P P

P P P

−

=

−

− −=

= − −

− −

∑

∑

Differential PSK (DPSK)

• The phase shift is treated as a random variable distributed uniformly over

(-π, π), and no phase estimation is provided at the receiver.

• We are now concerned with non-coherent communications

• The phase shift θ is the same during two consecutive signal transmission

intervals [(n-1)T, nT) and [nT, (n+1)T).

• The information phase sequence {∆θn} is still differentially encoded as

before.

• The transmitted signal s(t) in the interval [(n-1)T, (n+1)T] is

• ∆θn= θn- θn-1. θn and θn-1 are independent. Both of them take values

uniformly over

• {∆θn} and {θn} are i.i.d. sequences

1

2cos(2 ), (n-1)T t<nT

( )2

cos(2 ), nT t<(n+1)T

sc n

sc n

Ef t

Ts t

Ef t

T

π θ θ

π θ θ

−

+ + ≤= + + ≤

2 2 ( 1)0, ,

M

M M

π π − ⋯


• The received waveform is then r(t)=s(t)+n(t)

• To derive the optimal receiver, the observation interval should be

taken as (n-1)T ≤ t <(n+1)T since θis the same in this interval.

• We have four orthonormal basis functions:

1

2

3

2cos 2 ( -1)

(t)= T

0 ( 1)

2sin 2 ( -1)

(t)= T

0 ( 1)

0 ( -1)

(t)=

c

c

f t n T t nT

nT t n T

f t n T t nT

nT t n T

n T t

πφ

πφ

φ

≤ ≤ < ≤ +− ≤ ≤ < ≤ +

≤ ≤

4

2cos 2 ( 1)

T

0 ( -1)

(t)= 2sin 2 ( 1)

T

c

c

nT

f t nT t n T

n T t nT

f t nT t n T

π

φπ

< ≤ + ≤ ≤− < ≤ +

2cos 2 cf t

Tπ

2sin 2 cf t

Tπ−

r(t)

X

X

( 1 )n T

n Td t

+

∫

( 1)n T

nTdt

+

∫

rn,c

rn,s

900

Local

Oscillator

Delay

T

Delay

T

rn-1,c

rn-1,s


Waveform

channelVector

channel

, , , ,

, , 1, 1,

0

cos( ) , sin( )

, , , are i.i.d

Each of them N(0, N /2)

n c s n n c n s s n n s

n c n s n c n s

r E r Eθ θ η θ θ η

η η η η− −

= + + = + +

∼

We estimate ∆θn from the new observation (rn,c, rn,s, rn-1,c, rn-1,s)

i

1

, , 1 1, 1,

2 2

, ,

1

2

0 0

1

0

2Let ,0 1.

Given , ,and

( , , , , | , )

[( cos( )] [( sin( )]1

exp( )

2exp | | cos( )n

n i n

n c n s n n s n i n

n

j c s j j s s j

j n

s j

n n

ii M

M

P r r r c r

r E r E

N N

Ec r e r

N

θ

πβ

θ β θ θ

θ β θ θ

θ θ θ θ

π

θ α

−

− − −

= −

− ∆−

= ≤ ≤ −

∆ =

∆ =

− + + − +

= −

= − + −

∑

1) c does not depend on θ, θn-1 and ∆θn

2) rn = rn,c + jrn,s and rn-1=rn-1,c +jrn-1,s

3) αis given by rne-jθn + rn-1e

-jθn-1=| rne-jθn + rn-1|e

jα


θ is distributed uniformly over (-π, π) and θn-1 tales values uniformly

over 2 2 ( 1)

{0, , , }M

M M

π π −⋯

, , 1, 1, 1,

1

0

0 1 0 1

0 0

cos

0

( , , , | , )

21exp{ | | cos( )}

2

2 2( | |) ( | |)

1where I (x)= is modified Bessel function

2

The above

n

n i

n c n s n c n s n i n

s j

n n

s j s j

n n n n

x y

P r r r r

Ec r e r d

N

E EcI r e r cI r e r

N N

e dy

π θ

π

θ β

π

π

θ β θ

θ α θπ

π

− − −

− ∆−−

− ∆ − ∆− −

−

∆ =

= + −

= + = +

∫

∫n-1

, , 1, 1, 0 1

0

formula is independent of

2( , , , | ) ( | |)is j

n c n s n c n s n i n n

EP r r r r cI r e r

N

β

θ

θ β − ∆− − −⇒ ∆ = = +


n k

, , 1, 1, , , 1, 1,

0 1 0 1

0 0

0

1

ˆchoose as iff

( , , , | ) max ( , , , | )

2 2( | |) max ( | |)

( 0, ( ) is an increasing function)

| |

k i

k

n c n s n c n s n k n c n s n c n s n ii

s j s j

n n n ni

j

n n

P r r r r P r r r r

E EI r e r I r e r

N N

x I x

r e r

β β

β

θ β

θ β θ β− − − −

− ∆ − ∆− −

− ∆−

∆

∆ = = ∆ =

⇔ + = +

>

⇔ + =

1

1

1 1 1 1

1 1

1 1

max | |

| || | cos( ) max[| || | cos( )]

| | min | |

where | | and | |

i

n n

j

n ni

n n n n k n n n n ii

n n k n n ii

j j

n n n n

r e r

r r r r

r r e r r e

β

η η

η η β η η β

η η β η η β

−

− ∆−

− − − −

− −

− −

+

⇔ − − = − −

⇔ − − = − −

= =


Applying the MAP rule, we now

Optimal Receiver for M-DPSK

,1

n

,

tann s

n c

r

rη −=

2cos 2 cf t

Tπ

2sin 2 cf t

Tπ−

r(t)

X

X

( 1 )n T

n Td t

+

∫

( 1)n T

nTdt

+

∫

rn,c

rn,s

900

Local

Oscillator

Sub-

tractor

Select

the

smallest

Delay T

ˆnθ∆

-

| ψ-β0|

| ψ -β1|

| ψ -βM-1|

ψ

Difference from the receiver for differentially encoded M-PSK?

+

Performance of M-DPSK

)2

N N(0,~ ,,

)sin(

)cos(

)sin(

)cos()cos(

440

,1,1,,,

,11,1

,11,1

,1,

,1,,

xsncnsncn

snnssn

cnnscn

snnnssn

cnnnscnnscn

I

Er

Er

Er

EEr

−−

−−−

−−−

−

−

++=

++=

++∆+=

++∆+=++=

ηηηη

ηθθ

ηθθ

ηθθθ

ηθθθηθθ

One way to calculate the performance of M-DPSK is to find first the pdf of

phase difference

cn

sn

n

cn

sn

n

nn

r

r

r

r

,1

,11

1

,

,1

1

tan and ,tan−

−−−

−

−

==

−=

ηη

ηηψ

(rn,c, rn,s)

(rn-1,c, rn-1,s)

noise vector (ηn,c, ηn,s)

noise vector

(ηn-1,c, ηn-1,s)

It follows that the pdf of ψ is

independent of the angle θn-1+ θ.

It can be shown that the

conditional pdf of ψ given ∆θn

satisfies the following equation

2

1

1 2

2 1 1 2

2 1 1 2

2 0

2

( | ) ( | )

( ) ( ) 1

( ) ( ) or

exp{ [1 cos( ) cos ]}sin( )

( )4 1 cos( ) cos

n n

n n

n

n n

n

n n

sn

n

n

P f d

F F

F F

Et

NF dt

t

ψ

ψψ

θ θ

θ θ

π

θ π

ψ ψ ψ θ ψ θ ψ

ψ ψ ψ θ ψ

ψ ψ ψ θ θ ψ

θ ψθ ψ

ψπ θ ψ

∆ ∆

∆ ∆

∆−

≤ ≤ ∆ = ∆

− + <∆ <= − >∆ ∆ >

− − ∆ −−∆ −

=− ∆ −

∫

∫

θn-1 + θ

ψ

∆θn


∫

∫

∫

−

−

−

−

−−=⇒

−

−−−=⇒

−

−−−=

+−−+=

=∆+≤≤−=

=∆=∆

2

2

0

2

2

0

2

2

0

coscos1

]}cos4

cos1(exp{

4

sin

coscos1

]}cos4

cos1(exp{

4

sin

1

coscos1

]}cos4

cos1(exp{

4

sin1

1)()(

)|(

)|ˆ(

π

π

π

π

π

π

ββ

π

π

π

π

π

π

π

π

π

π

π

π

πβ

πβ

βθπ

βψπ

β

βθβθ

dt

tM

tN

E

MP

dt

tM

tN

E

MP

dt

tM

tN

E

M

MF

MF

MMP

P

s

e

s

c

s

ii

inii

inin

ii


We then get

Special case

• M=2

Pb=Pe=1/2e-Eb/N0

• For large M, M-DPSK requires 3dB more Eb/N0 than coherent M-PSK

Comparison of Digital Modulation Methods

• To make a fair comparison among different digital modulation

methods, one must investigate the tradeoff among the bandwidth

efficiency, bit error probability, and bit SNR.

• The bandwidth efficiency of a modulation method is defined as the

bit rate (Rb) to bandwidth (W) ratio Rb/W

• The bandwidth is calculated from the power spectral density function

of the transmitted waveform, which in turn depends on the signal set

used.

Assume that the optimal signal set is used.

Theorem 3.7.1: The maximum number N of dimension of the signal

source spanned by a set of signals of duration of T and “bandwidth”

W grows linearly with time T and bandwidth W, respectively:

N=KWT

where K is a constant around 2.

The value of K depends on specific definitions of bandwidth. We

normally choose K=2.


Examples

PAM signals: N=1 => W=1/(2T) (single sideband)

The bandwidth efficiency of M-ary PAM is

M-PSK: N=2 => W=1/T

log /2 log bits/second/Hz

1/ 2

bR M TM

W T= =

log /log bits/second/Hz

1/

bR M TM

W T= =

2log /2 log bits/second/Hz

1/ 2

bR M TM

W T= =

log / 2 log bits/second/Hz

/ 2 M

bR M T M

W M T= =

M2-QAM: N=2 => W=1/T

M-FSK: N=M => W=M/(2T)

As M approaches infinity, Rb/W goes to 0


• In the case of PAM, QAM, and PSK, increasing M results in a higher

bit rate-to-bandwidth ratio R/W.

• However, the cost of achieving the higher data rate is an increase in

the SNR per bit.

• Therefore, these modulation methods are appropriate for channels

that are bandlimited, where we desire R/W>1 and where there is

sufficient high SNR to support increases in M.

• Example: Telephone channels

• M-ary orthogonal signals yields R/W≤1. As M increases, R/W

decreases.

• However, the SNR/bit required to achieve a given error probability

(10-5) decreases as M increases.

• Consequently, M-ary orthogonal signals are appropriate for power-

limited channels that have sufficiently large bandwidth to

accommodate a large number of signals.

• As M goes to infinity, Pe approaches zero provided that SNR/bit≥-

1.6 dB


Information

source

Source

encoderModulator

Channel

DemodulatorSource

decoderUser

Channel

encoder

Channel

decoder

Text, speech,

Audio, image,

video

Natural

redundancy

removal

Controlled

redundancy

insertion

noise

Source coding Channel coding Discrete channel

Digital Communication Block Diagram

数字通信系统框图

Fig 1.1 数字通信系统框图

Channel Capacity and Coded Modulation

Channel Models and Channel Capacity

Binary Symmetric Channels

AWGN

Channel

{an} BPSK

Modulator

s(t) ˆ{ }naDemodulator

and

detector

Assume that an=0 or 1, i.i.d, and symbol 0 and 1 are equally likely.

Then

0

2ˆ( ) b

b n n

EP P a a Q

N

= ≠ =

0

1

0

1

1-q

q

q

1-q

Equivalent diagram

Binary Symmetric Channels

1 2 1 2ˆ ˆ ˆ( ) 1 [1 ]nn n bP a a a a a a p≠ = − −⋯ ⋯

For any fixed bit SNR Eb/N0, the block error prob. goes to 1 as n

approaches infinity.

In order to transmit information reliably over the BSC, one has to

employ channel encoders and decoders.

BSC

{an} Channel

encoder

ˆ{ }naChannel

decoder

An important question: with the use of channel encoders and

decoders, how many number of bits of information can be reliably

transmitted over the BSC?

The answer is the channel capacity of the BSC.

Channel Capacity of the BSC

max ( ; )X

C I X Y=

X is the input random variable to the BSC, Y is the corresponding output.

The maximization is taken over all possible input random variable X.

For BSC, C = 1 – H(q) with the maximum achieved by the equally likely

input X.

AWGN

Channel

{an} BPSK

Modulator

s(t) ˆ{ }naDemodulator

and

detector

Channel

encoder

Channel

decoder

•The concatenation of the channel encoder and BPSK modulator is a

binary coded modulation scheme and gives binary coded signals.

•The BPSK detector is not optimal any more since coded signals are

correlated and the BPSK makes a hard decision (0 or 1) bit by bit.

•To improve the performance, we may consider a detector that outputs

Q>2 outputs. That is, a detector quantizes the demodulator output into

Q>2 outputs (or no quantization). In this case, we say that the detector

has made a soft decision.

•With a detector making a soft decision, we get an equivalent discrete

channel with two inputs and Q>2 possible outputs.

Channel Capacity of the BSC

Discrete Memoryless Channels

m0

m1

mM-1

Y0

Y1

YQ-1

P(y0|m0)

P(yQ-1|mM-1)

•With any M-ary modulator and a soft (or hard) decision detector, we get

an equivalent discrete channel with M inputs and Q≥M possible outputs.

•The maximum number of bits of information that can be reliably

transmitted over the channel is given by channel capacity

A general discrete channel

max ( ; )X

C I X Y=

Discrete Input, Continuous-Output Channels

AWGN

Channel

{an} PAM

Modulator

s(t) Demodulator {yn}

•The demodulator converts the received waveform into scalar vectors,

no estimation is made at this point. The corresponding composite

channel is then equivalent to the scalar channel

Y = X + n

•X takes values in the amplitude set of the PAM modulator

•n is a Gaussian random variable with n ~ N(0, N0/2)

•Channel capacity

max ( ; )X

C I X Y=

•The modulator is given, but coded PAM signal and their receiver

including the detector and channel decoder are left open.

Band-limited, power-limited AWGN channels

Assume the channel is band limited and power limited

r(t) = x(t) + n(t)

x(t) is the input waveform band-limited to W and power-limited to P

n(t) = the AWGN with Sn(f) = N0/2.

To compute the channel capacity, we first convert the waveform

channel into discrete time Gaussian channel.

( ) ( )2

sin 2( ), ( ) sin 22 2

n

n

n

nx t x g t

W

n wtx x g t c Wt

W wt

π

π

+∞

=−∞

= −

= = =

∑

This is equivalent to projecting x(t) into the space spanned by the

orthonormal basis { 2 ( )} :2

2 ( ) ( - )2

n

nW g t

W

nx w x t g t dt

W

+∞−∞

+∞

−∞

−

= ∫

( ) ( )2

n

n

nx t x g t

W

+∞

=−∞

= −∑nx

, n=0, 1, 2,2

nt

W= ± ± ⋯

n(t)

-W W -W W

1/2W 1/2W

Sample at

n n nr x n= +

{rn}

{ } in+∞−∞ is i.i.d and each ni is Gaussian, ni~N(0, N0W)

Since x(t) is power-limited to P, each sample xn is then energy limited

to P. The channel capacity CN per use of the discrete-time Gaussian

channel is

CN = max {I(Xn; rn: Xn is an input with2

0

1[ ] } log(1 )

2n

PE x P

N W≤ = +

During T seconds, we have 2WT samples, and the discrete-time channel

is used 2WT times. Thus, the channel capacity C in bits per second is

0

2 log(1 ) bits/secondN

PC W C W

N W= × = +

Band-limited, power-limited AWGN channels

Bandwidth Efficiency versus Bit SNR

0 0

/

0

log(1 ) log(1 )

Energy/second

2 1

/

b

b

C W

b

E CC P

W N W N W

P E C

E

N C W

= + = +

= =

−⇒ =

0-1.61

0.1

10C/W

Shannon limit

Eb/N0

dB

Ideal diagram of bandwidth efficiency

versus bit SNR

Eb/N0 bit SNR

C/W bandwidth efficiency

0

/

b

/ 00

Let , we have

log bits/second

E 2 1 lim ln 2 1.

/

1

6N

)

C W

C W

W

PC e

N

dBC W

∞

−>

→∞

=

−= = ≈−

2) In order to achieve essentially error free transmission, Eb/N0 must

be ≥-1.6 dB (Shannon limit)

3) An transmission rate Rb< C is achievable, any rate Rb>C is not

achievable (Shannon noisy channel coding theorem)

4) The proof of noise channel coding theorem involves the well-

known random coding argument

Bandwidth Efficiency versus Bit SNR

Achieving Channel Capacity with Orthogonal Signals

• The error probability can be made arbitrarily small as T goes to infinity (M

goes to infinity for a fixed bit rate Rb = log M/T ).

• For a power-limited AWGN channel with unbounded bandwidth orthogonal

signaling (or M-FSK) is asymptotically optimal in the sense that the

corresponding (symbol) error probability goes to o exponentially as M goes

to infinity if the bit rate Rb = log M/T is less than the channel capacity.

2

( ) 2

b

( )

b

2 2 if 0 R C /4

2 2 if C /4 R C

b

b

CT R

eT C R

P

∞

∞

− −

∞

− −∞ ∞

≤ ≤= ≤ ≤

i

i

Coded Modulation-A Probabilistic Approach

• Although orthogonal signaling is asymptotically optimal, there is a

considerable gap between the performance of practical uncoded

communication systems and the optimal performance theoretically

achievable given by the noisy channel coding theorem.

• To reduce the gap, one must resort to coded modulation.

� Algebraic approach (specific coding design techniques)

� Probabilistic approach (analysis of the performance of a general

class of coded signals)

Random Coding Based on M-ary Binary Coded Signals

• Objective

– Design a binary code C consisting of M binary codewords

Ci=(Ci1,Ci2,···, Cin), 0≤ i ≤M-1

– The corresponding coded signals {si(t)} gives good error prob.

performance

• The encoding bit rate is Rb bits/second

• Blocks of k bits are encoded at a time into one of the M waveforms

, M = 2k = 2RbT

• Assume Rb<1/Tc, T=nTc. This implies that k < n, and

1

2( ) (2 1) cos 2 , 1 , [( -1) , ]

[(2 1), , (2 1)]

ci ij c

b

i i in c

s t C f t j n t j Tc jTcT

s C C

επ

ε

= − ≤ ≤ ∈

= − −⋯

1

0{ ( )}M

i is t −=

( )( )22 2 , 1/

2 2b

kT D Rn k

cn n

MD T

− −− −= = = =

• We consider the ensemble of (2n)M distinct ways in which we can

select M codewords from the total 2n possible binary sequences.

• Associated with each of the (2n)M binary codes, there is a coded

communication system.

AWGN

Channel

{an} BPSK

Modulator

ˆ{ }naOptimal

Receiver

Channel

Code C1

1{ ( )}is t

AWGN

Channel

{an} BPSK

Modulator

ˆ{ }naOptimal

Receiver

Channel

Code C2

2{ ( )}is t

AWGN

Channel

{an} BPSK

Modulator

ˆ{ }naOptimal

Receiver

Channel

Code C2nM

2{ ( )}nM

is t


• Instead of evaluating the error prob. of each coded system

individually, we compute the average error prob. of the 2nM coded

communication systems:

• This is equivalent to choosing a binary code from the set of all possible

code Cj, 1≤j ≤2nM, and evaluate the average error prob. , where

{{ ( )}}j

e iP s t

2

1

1{{ ( )}}

2

nM

j

e e inMj

P P s t=

= ∑

[ ( )]eE P C

1( ) , 1 j 2

2

Mn

MnP

jC C= = ≤ ≤

C

2

1

1 1

0 0, ,0 0

[ ( )] {{ ( )}} ( )

Applying the union bound, we get

2 ( , )|| ( ) ( ) ||1 1{{ ( )}} ( )

2 2

where denotes the codeword

nM

j

e e e i

j

j jj jM Mc i lj i l

e i

i l l i i li l

j

i

P E P P s t P

d C Cs t s tP S t Q Q

M MN N

C ith

jC C

ε

=

− −

= = ≠≠

= =

− ≤ =

∑

∑ ∑ ∑

of the codebook .jthjC


2

2

, 0

2

1 , 0

, 0

, 0

( , )1Since ( ) for any 0, {{ ( )}} exp{ }

( , )1 1exp{ }

2

( , )1 1[ exp{ }]2

( , )1[exp{ }]

nM

x j jj c i l

e i

i li l

j j

c i le nM

j i li l

j j

c i l

nMi li l

c i l

i li l

d C CQ x e x P s t

M N

d C CP

M N

d C C

M N

d C CE

M N

ε

ε

ε

ε

−

≠

=≠

≠

≠

≤ > ≤ −

≤ −

= −

= −

∑

∑ ∑

∑

∑

where Ci and Cl are the ith and lth codeword of the random codebook

respectively.

C


0

0 0

00

( ( , ) ) 2

( , )[exp{ }] 2

2 1 2 [1 log(1 )]

c

c c

n

i l

dnNnc i l

d

n

N Nn n

nP d C C d

d

nd C CE e

dN

e e

ε

ε ε

ε

−

−−

=

− −− −

= =

⇒ − =

= + = − +

∑

Ci and Cl are independent, each taking values uniformly over the

set (0,1)

0 b 0 0 b

0

R ( R )

0

( 1)2 [1 log(1 )] < 2 2

1where 1 log(1 ), ,

C

C

N T TDR T DRn

e

N

c

P M e

R e D n TDT

ε

ε

−− − −−

−

⇒

≤ − − + =

= − + = =


Since is the average error prob. of the 2nM coded comm. Systems,

There exists at least one binary code Cj such that the corresponding

error prob.

Pe(Cj)<2-T(DR0-Rb)< 2-n(R0-Rc)

where Rc = k/n = Rb/D is referred to as the code rate.

eP

When Rb<DR0 (or Rc<R0), one can always design a binary coded

system so that the block error prob. goes to 0 exponentially fast as

the block length n approached infinity.


• If one selects a code at random, then this code is good with a very

high probability.

• The probability that the error probability of the randomly chosen code

is less than 1/α.

• As the dimension n is large enough, good codes are abundant. But the

implementation complexity is high.

• The rate R0 is called the cutoff rate.

e eP Pα≥

R0

dB

bits/d

imen

sio

n

-5 0 5 10 Ec/N0—SNR per dimension

0.20.4

0.6

0.8

1.0


Review of Finite Fields

Group Structures

Definition 1 A group is a set G together with a binary operation *

on G such that the following three properties hold:

1) * is associative, that is,

for any a,b,c G, (a*b)*c=a*(b*c)∈

2) There is an identity or (unity)

element e in G such that for all a G,

a*e = e*a = a

∈

3) For each a G, there exists an

inverse element a-1 G such that

a*a-1 = e

Sometimes, we denote the group

as a triple (G, *, e). If the group

also satisfies

4) for all a, b

a * b = b * a

Then the group is called abelian

or commutative.

∈

∈

Example 1 Let

• Z, the set consisting of all integers

• Q, the set of all rational numbers

+ and · are ordinary addition and multiplication.

Then (Z, +, 0), (Q, +, 0), (Q*, ·, 1)

are all groups where Q* is the all nonzero rational

numbers.

Furthermore, they are abelian

How about (Z*, ·, 1) ?

Let n be a positive integer (n>1)

and Zn represent the set of

remainder of all integers on

division n, i.e.,

Zn = {0, 1, 2, ···, n-1}

We define a+b and ab the

ordinary sum and product of a

and b reduced by modulo n

respectively.

Let Zn* = { a Zn| a 0}∈ ≠

Proposition 1

(a) (Zn, +, 0) forms a group

(b) (Zp*, ., 1) forms a group

for any prime p

Definition 2 A multiplicative group

is said to be cyclic if there is an

element a G such that for any b G

there is some integer i with b = ai.

Such and element is called the

generator of the cyclic group, and we

write

G = <a>

∈ ∈

G

1

a

a2

b=ai

Examples

(Z3*,·,1) is a cyclic group with generator 2.

Z3 = {1, 2} = <2> , 20 = 1, 22 = 1 mod 3


31 = 3, 32 = 2, 33 = 6, 34 = 4, 35=5, 36 = 1 mod 7


21 = 2, 22 = 4, 23 = 3, 24 = 1 mod 7

Every element in Z5* can be written in a power of 2

Finite Groups

Definition 3 A group G is called finite if it contains a finite

number of elements. The number of elements in G is called

the order of G, denoted as |G|

Definition 4 A ring (R,+, ·) is a set R, together with two binary

operations, denoted by + and · such that

1) R is an abelian group with respective to +

2) · is associative, that is

( a · b) · c = a · (b · c ) for all a, b, c R.

3) The distributive law holds:

a · (b + c ) = a · b + a · c

(b + c ) · a = b · a + c · a

(Z, +, ·) and (Q, +, ·) are rings

(Zn, +, ·) forms a ring, called residue class ring modulo n

(Z4, +, ·) is a ring.

Rings

∈

Let (F,+, ·) be a ring, and let

F* = { a F| a 0}

The set of nonzero elements in F.

∈ ≠

Definition 5 A field is a ring (F ,+, ·) such that F* together

with the multiplication · forms an abelian group.

Fields

•(Q ,+, ·),

•(R ,+, ·), and

•(C ,+, ·) are fields

where R is the set of all real numbers,

and C is the set of all complex numbers

Definition 6 A finite field F is a field that contains a finite

number q of elements. This number is called the order of the

field. F is also called a Galois field, denoted by GF(q).

Finite Fields

Proposition 2 Let P be a prime, then (Zp ,+, ·) is finite field with

order p. This field is denoted as GF(p).

The addition and multiplication are carried out modulo p.

Examples: GF(2) and GF(3) on page 266

Polynomials

Let R be an arbitrary ring. A polynomial over R is an expression of

the form

f(x) = a0 + a1x + ···anxn

where n is an nonnegative integer, the coefficients ai are elements

of R. x is a symbol not belonging to R, called an indeterminant

over R.

f(x) is said to to be irreducible if f(x) cannot be factored into a

product of lower degree polynomials over R

x2+x+1 is irreducible in GF(2)

Construction of GF(2n) and GF(qn)

Step 1 Selcet n, a positive integer and p a prime.

Step 2 Choose that f(x) is an irreducible polynomial over GF(p) of

degree n.

Step 3 We agree that αis an element that satisfies f(α) = 0.

Let GF(pn) = {a0 + a1 α + ···an αn| ai GF(p) }∈

For two elements g(α) and h(α) in GF(Pn),

g(α)=a0 + a1α+ ··· + an-1αn-1

h(α)=b0 + b1α+ ··· + bn-1αn-1

Define two operations:

a) Addition

g(α)+ h(α)= a0+b0+(a1+b1)α+···+(an-1+bn-1) αn-1

b) Multiplication

g(α) h(α)= r(α)

where r(x) is the remainder of g(x)h(x) divided by f(x)

Theorem: The set GF(pn) together with the two operations above forms

a finite field, and the order of the field is pn

Construction of GF(2n) and GF(pn)

Example: Let p=2 and f(x) = 1+x+x3. Then f(x) is irreducible

over GF(2). Let α be a root of f(x). That is, f(α)=0. The finite

field GF(23) is defined by

GF(23)={a0 + a1 α + ···a2 α2| ai GF(2) }∈

α7 = 1

=α61+ α2101=

=α51+ α + α2111=

=α4α + α2110=

=α31+ α011=

=α2α2100=

=αα010=

=11001=

=00000=

As a power of αAs a polynomialAs a 3-tupe

Primitive Elements and Primitive Polynomials

Fact: For any finite field F, its mutliplcative field F*, the set of

nonzero elements in F is cyclic.

Definition: A generator of the cyclic group GF(pn)* is called a

primitive element of GF(pn). A polynomial has a primitive element

as zero is called a primitive polynomial

Example: x3+x+1 is primitive polynormial over GF(2)

x7-1 = (x+1)(x3+x+1)(x3+x2+1)

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