optimum optimum receivers for the additive...

Optimum Optimum Receivers for the Additive Receivers for the Additive White White Gaussian Noise ChannelGaussian Noise Channel

Wireless Information Transmission System Lab.Wireless Information Transmission System Lab.Institute of Communications EngineeringInstitute of Communications Engineeringg gg gNational Sun National Sun YatYat--sensen UniversityUniversity

Table of ContentsTable of Contents

◊ Optimum Receiver for Signals Corrupted by Additive White Gaussian NoiseWhite Gaussian Noise◊ Correlation Demodulator◊ Matched Filter Demodulator◊ Matched-Filter Demodulator◊ The Optimum Detector

P f f th O ti R i f M l◊ Performance of the Optimum Receiver for MemorylessModulation◊ Probability of Error for Binary Modulation◊ Probability of Error for M-ary Orthogonal Signals◊ Probability of Error for M-ary Biorthogonal Signals◊ Probability of Error for M-ary PAM◊ Probability of Error for M-ary PSK

2

Table of ContentsTable of Contents◊ Differential PSK (DPSK) and Its Performance◊ Probability of Error for QAM◊ Probability of Error for QAM◊ Comparison of Digital Modulation Methods

◊ Optimum Receiver for Signals with Random Phase in◊ Optimum Receiver for Signals with Random Phase in AWGN Channel

O ti R i f Bi Si l◊ Optimum Receiver for Binary Signals◊ Optimum Receiver for M-ary Orthogonal Signals

P b bilit f E f E l D t ti f M O th l◊ Probability of Error for Envelope Detection of M-ary Orthogonal Signals

3

Optimum Receiver for Signals Corrupted by Optimum Receiver for Signals Corrupted by Additive White Gaussian NoiseAdditive White Gaussian Noise

◊ We assume that the transmitter sends digital information by use of M signals waveforms {sm(t)=1,2,…,M }. Each waveform is g { m( ) , , , }transmitted within the symbol interval of duration T, i.e. 0≤t≤T.

◊ The channel is assumed to corrupt the signal by the addition of white Gaussian noise, as shown in the following figure:

( ) ( ) ( ) , 0mr t s t n t t T= + ≤ ≤

where n(t) denotes a sample function of AWGN process with power spectral density Φnn( f )=½N0 W/Hz.

4


◊ Our object is to design a receiver that is optimum in the sense that it minimizes the probability of making an error.p y g

◊ It is convenient to subdivide the receiver into two parts—the signal demodulator and the detector.

◊ The function of the signal demodulator is to convert the received waveform r(t) into an N-dimensional vector r=[r1 r2 ..…rN ] where N is the dimension of theinto an N dimensional vector r [r1 r2 .. rN ] where N is the dimension of the transmitted signal waveform.

◊ The function of the detector is to decide which of the M possible signal f t itt d b d th twaveforms was transmitted based on the vector r.

5


◊ Two realizations of the signal demodulator are described in the following section:g◊ One is based on the use of signal correlators.◊ The second is based on the use of matched filters. f

◊ The optimum detector that follows the signal demodulator is designed to minimize the probability of error.g p y

6

Correlation DemodulatorCorrelation Demodulator◊ We describe a correlation demodulation that decomposes the

receiver signal and the noise into N-dimensional vectors.g

◊ In other words, the signal and the noise are expanded into a series of linearly weighted orthonormal basis functions {f (t)}linearly weighted orthonormal basis functions {fn(t)}.

◊ It is assumed that the N basis function {f n(t)} span the signal space, so every one of the possible transmitted signals of the setso every one of the possible transmitted signals of the set {sm(t)=1≤m≤M } can be represented as a linear combination of {f n(t)}.

◊ In case of the noise, the function {f n(t)} do not span the noise space. However we show below that the noise terms that fall outside the signal space are irrelevant to the detection of the signal.

7

Correlation DemodulatorCorrelation Demodulator◊ Suppose the receiver signal r(t) is passed through a parallel bank of

N basis functions {f n(t)}, as shown in the following figure:

[ ]( ) ( ) ( ) ( ) ( )T T

t f t dt t t f t dt+∫ ∫ [ ]0 0

( ) ( ) ( ) ( ) ( )

, 1, 2,.....k m k

k mk k

r t f t dt s t n t f t dt

r s n k N

= +

⇒ = + =∫ ∫

0( ) ( ) , 1 2

T

mk m k

T

s s t f t dt k , ,......N= =∫

0 ( ) ( ) , 1, 2,.......

T

k kn n t f t dt k N= =∫

◊ The signal is now represented by the vector sm with components smk, k=1,2,…N. Their values depend on which of the M signals was transmitted.transmitted.

8

Correlation DemodulatorCorrelation Demodulator◊ In fact, we can express the receiver signal r(t) in the interval 0 ≤ t ≤

T as: ∑ ∑ ′N N

ttftft )()()()(

∑

∑ ∑= =

′

′++=

Nk k

kkkmk

f

tntfntfstr1 1

)()(

)()()()(

◊ The term n'(t), defined as

∑=

′+=k

kk tntfr1

)()(

◊ The term n (t), defined as

1( ) ( ) ( )

N

k kk

n t n t n f t′ = − ∑is a zero-mean Gaussian noise process that represents the difference between original noise process n(t) and the part corresponding to the

1k =

g p ( ) p p gprojection of n(t) onto the basis functions {fk(t)}.

9

Correlation DemodulatorCorrelation Demodulator◊ We shall show below that n'(t) is irrelevant to the decision as to

which signal was transmitted. Consequently, the decision may be g q y, ybased entirely on the correlator output signal and noise components rk=smk+nk, k=1,2,…,N.

◊ The noise components {nk} are Gaussian and mean values are:

[ ]0

( ) ( ) ( ) 0 for all .T

k kE n E n t f t dt n= =∫ Power spectral densityand their covariances are:

[ ]0∫

[ ]( ) ( ) ( ) ( ) ( )T T

E n n E n t n f t f dtdτ τ τ∫ ∫ Conclusion: The N noise

Power spectral densityis Φnn(f)=½N0 W/Hz

Autocorrelation[ ]

0 0

0

( ) ( ) ( ) ( ) ( )

1 ( ) ( ) ( )

k m k m

T T

k m

E n n E n t n f t f dtd

N t f t f dtd

τ τ τ

δ τ τ τ

=

= −

∫ ∫

∫ ∫

Conclusion: The N noiseComponents {nk} arezero-mean uncorrelatedG i d0 0 0

0 00

( ) ( ) ( )21 1 ( ) ( )2 2

k m

T

k m mk

f f

N f t f t dt N δ= =

∫ ∫

∫

Gaussian randomvariables with a commonvarianceσn

2=½N0.

10

02 2∫

Correlation DemodulatorCorrelation Demodulator◊ From the above development, it follows that the correlator output {rk}

conditioned on the mth signal being transmitted are Gaussian random variables with mean

mkkmkk snsErE =+= )()( mkkmkk )()(

22 1 Nσσand equal variance

◊ Since the noise components {n } are uncorrelated Gaussian random

02Nnr == σσ

◊ Since the noise components {nk} are uncorrelated Gaussian random variables, they are also statistically independent. As a consequence, the correlator outputs {rk} conditioned on the mth signal being

i d i i ll i d d G i i bltransmitted are statistically independent Gaussian variables.

11

Correlation DemodulatorCorrelation Demodulator◊ The conditional probability density functions of the random

variables r=[r1 r2 … rN] are:[ 1 2 N]

MmsrppN

kmkkm ,....,2,1)|()(

1

== ∏=

, s|r ----(A)

NkN

srN

srp mkkmkk ,.....,2,1 , )(exp1)|(

0

2

0

=⎥⎦

⎤⎢⎣

⎡ −−=

π----(B)

NN 00 ⎦⎣π

By substituting Equation (A) into Equation (B), we obtain the jointdi i l PDF

2( )1( | ) 1 2N

k mkr s M⎡ ⎤−⎢ ⎥∑

conditional PDFs

210 0

( )( | ) exp , 1, 2,......,( )

k mkm N

kp m M

N Nπ =

= − =⎢ ⎥⎣ ⎦

∑r s

12

Correlation DemodulatorCorrelation Demodulator◊ The correlator outputs (r1 r2 … rN) are sufficient statistics for

reaching a decision on which of the M signals was transmitted, i.e., g g , ,no additional relevant information can be extracted from the remaining noise process n'(t).

◊ Indeed, n'(t) is uncorrelated with the N correlator outputs {rk}:

[ ] [ ] [ ] [ ]( ) ( ) ( ) ( ) ( ) ( )N

k k k k j j kE n t r E n t s E n t n E n t n E n t n f t n⎧ ⎫⎡ ⎤⎪ ⎪′ ′ ′ ′= + = = −⎨ ⎬⎢ ⎥∑[ ] [ ] [ ] [ ]

[ ]

1( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

k mk k k j j kj

NT

k j k j

E n t r E n t s E n t n E n t n E n t n f t n

E n t n f d E n n f tτ τ τ

=

+ ⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

= −

∑

∑∫ ( ) ( )T

k kn n t f t dt= ∫[ ]

( )

01

0 0

( ) ( ) ( ) ( ) ( )

1 1( ) ( )

k j k jj

NT

k jk j

E n t n f d E n n f t

N t f d N f t

τ τ τ

δ τ τ τ δ

=

= − −

∑∫

∑∫

0( ) ( )k kn n t f t dt∫

( )0 001

( ) ( )2 2

1 2

k jk jj

N t f d N f t

N

δ τ τ τ δ=

=

∑∫

0 01( ) ( ) 02k kf t N f t− = Q.E.D.

13

2 0 0( ) ( )2k kf f Q.E.D.

Correlation DemodulatorCorrelation Demodulator◊ Since n' (t) and {rk} are Gaussian and uncorrelated, they are also

statistically independent.y p◊ Consequently, n'(t) does not contain any information that is relevant

to the decision as to which signal waveform was transmitted.◊ All the relevant information is contained in the correlator outputs {rk}

and, hence, n'(t) can be ignored.

14

Correlation DemodulatorCorrelation Demodulator◊ Example.

◊ Consider an M ary baseband PAM signal set in which the basic◊ Consider an M-ary baseband PAM signal set in which the basic pulse shape g(t) is rectangular as shown in following figure.

◊ The additive noise is a zero-mean white Gaussian noise process◊ The additive noise is a zero mean white Gaussian noise process.◊ Let us determine the basis function f(t) and the output of the

correlation-type demodulator.correlation type demodulator.◊ The energy in the rectangular pulse is

TddTT 222 )( ∫∫ε Tadtadttgg 0

20

2 )( === ∫∫ε

15

Correlation DemodulatorCorrelation Demodulator◊ Example.(cont.)

◊ Since the PAM signal set has dimension N=1 there is only one◊ Since the PAM signal set has dimension N=1, there is only one basis function f(t) given as:

⎨⎧ ≤≤ )0(1)(1)( TtTttf

◊ The output of the correlation-type demodulator is:⎩⎨⎧ ≤≤==

)otherwise( 0)0( 1)()(

2

TtTtgTa

tf

◊ The output of the correlation type demodulator is:

Th l t b i l i t t h f(t) i∫∫ ==

TTdttr

Tdttftrr

00)(1)()(

◊ The correlator becomes a simple integrator when f(t) is rectangular: if we substitute for r(t), we obtain:

[ ]{ }1 1T T T⎡ ⎤∫ ∫ ∫[ ]{ }0 0 0

1 1 ( ) ( ) ( ) ( )T T T

m mr s t n t dt s t dt n t dtT T

s n

⎡ ⎤= + = +⎢ ⎥⎣ ⎦= +

∫ ∫ ∫

16

ms n+

Correlation DemodulatorCorrelation Demodulator◊ Example.(cont.)

◊ The noise term E(n)=0 and:◊ The noise term E(n)=0 and:

( ) ( )2

0 0

1 ( ) ( )T T

n E n t n t E n t n dtdT

σ τ τ⎡ ⎤⎡ ⎤= =⎣ ⎦ ⎢ ⎥⎣ ⎦∫ ∫

[ ] 00 0 0 0

1 ( ) ( ) ( )2

T T T T

TNE n t n dtd t dtd

T Tτ τ δ τ τ

⎣ ⎦

= = −∫ ∫ ∫ ∫0

00

21 1

2 2T

T TN d NT

τ= ⋅ =∫

◊ The probability density function for the sampled output is:

2 2T

⎤⎡ 2

⎥⎦

⎤⎢⎣

⎡ −−=

0

2

0

)(exp1)|(N

srN

srp mm π

17

MatchedMatched--Filter DemodulatorFilter Demodulator◊ Instead of using a bank of N correlators to generate the variables

{rk}, we may use a bank of N linear filters. To be specific, let us { k}, y p ,suppose that the impulse responses of the N filters are:

TttTfth ≤≤−= 0)()(

where {fk(t)} are the N basis functions and hk(t)=0 outside of the

TttTfth kk ≤≤−= 0 ,)()(

interval 0 ≤ t ≤ T.◊ The outputs of these filters are :

( ) ( )( )

( ) ( )

k k

t

y t r t h t

r h t dτ τ τ

= ∗

= −∫0

0

( ) ( )

( ) ( ) , 1, 2,......,

k

t

k

r h t d

r f T t d k N

τ τ τ

τ τ τ

=

= − + =

∫∫

18

0∫

MatchedMatched--Filter DemodulatorFilter Demodulator◊ If we sample the outputs of the filters at t = T, we obtain

( ) ( ) ( ) 1 2T

y T r f d r k Nτ τ τ∫◊ A filter whose impulse response h(t) = s(T–t) , where s(t) is

d t b fi d t th ti i t l 0 ≤ t ≤ T i ll d

0( ) ( ) ( ) , 1, 2,.....,k k ky T r f d r k Nτ τ τ= = =∫

assumed to be confined to the time interval 0 ≤ t ≤ T, is called matched filter to the signal s(t).

◊ An example of a signal and its matched filter are shown in the◊ An example of a signal and its matched filter are shown in the following figure.

19

MatchedMatched--Filter DemodulatorFilter Demodulator◊ The response of h(t) = s(T–t) to the signal s(t) is:

( ) ( ) ( )( ) ( ) ( ) ( )t t

y t s t h t s h t d s s T t dτ τ τ τ τ τ∗ +∫ ∫which is the time-autocorrelation function of the signal s(t).

( ) ( ) ( )0 0

( ) ( ) ( ) ( )y t s t h t s h t d s s T t dτ τ τ τ τ τ= ∗ = − = − +∫ ∫

◊ Note that the autocorrelation function y(t) is an even function of t , which attains a peak at t=T.

20

MatchedMatched--Filter DemodulatorFilter Demodulator◊ Matched filter demodulator that generates the observed variables {rk}

21

MatchedMatched--Filter DemodulatorFilter Demodulator

◊ Properties of the matched filter.◊ If a signal s(t) is corrupted by AWGN the filter with an impulse◊ If a signal s(t) is corrupted by AWGN, the filter with an impulse

response matched to s(t) maximizes the output signal-to-noise ratio (SNR).( )

◊ Proof: let us assume the receiver signal r(t) consists of the signal s(t) and AWGN n(t) which has zero-mean and

1W/Hz.

◊ Suppose the signal r(t) is passed through a filter with impulse ( ) 0

12nm f NΦ =

response h(t), 0≤t≤T, and its output is sampled at time t=T. The output signal of the filter is:

0( ) ( ) ( ) ( ) ( )

t

t t

y t s t h t r h t dτ τ τ= ∗ = −∫∫ ∫

22

0 0 ( ) ( ) ( ) ( )

t ts h t d n h t dτ τ τ τ τ τ= − + −∫ ∫

MatchedMatched--Filter DemodulatorFilter Demodulator◊ Proof: (cont.)

◊ At the sampling instant t=T:◊ At the sampling instant t T:

0 0( ) ( ) ( ) ( ) ( )

T Ty T s h T d n h T dτ τ τ τ τ τ= − + −∫ ∫

◊ This problem is to select the filter impulse response that i i th t t SNR d fi d

( ) ( )s ny T y T= +

maximizes the output SNR0 defined as:

( )2

SNR sy T

T T⎡ ⎤ ∫ ∫

( )( )0 2

SNR s

n

yE y T

=⎡ ⎤⎣ ⎦

[ ]2

0 0

2

( ) ( ) ( ) ( ) ( )

1 1( ) ( ) ( ) ( )

T T

n

T T T

E y T E n n t h T h T t dtd

N t h T h T t dtd N h T t dt

τ τ τ

δ

⎡ ⎤ = − −⎣ ⎦ ∫ ∫

∫ ∫ ∫23

20 00 0 0

( ) ( ) ( ) ( )2 2

N t h T h T t dtd N h T t dtδ τ τ τ= − − − = −∫ ∫ ∫


◊ By substituting for y (T) and into SNR0( )2E y T⎡ ⎤⎣ ⎦◊ By substituting for ys (T) and into SNR0.( )nE y T⎡ ⎤⎣ ⎦

2 2

' Tτ τ= −2 2

0 00

2 2

( ) ( ) ( ') ( ') 'SNR 1 1

T T

T T

s h T d h s T dτ τ τ τ τ τ⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= =∫ ∫

∫ ∫◊ Denominator of the SNR depends on the energy in h(t)

2 20 00 0

1 1( ) ( )2 2

T TN h T t dt N h T t dt− −∫ ∫

◊ Denominator of the SNR depends on the energy in h(t).◊ The maximum output SNR over h(t) is obtained by

maximizing the numerator subject to the constraint that themaximizing the numerator subject to the constraint that the denominator is held constant.

24


◊ Cauchy-Schwarz inequality: if g1(t) and g2(t) are finite-energy◊ Cauchy Schwarz inequality: if g1(t) and g2(t) are finite energy signals, then

∫ ∫∫∞ ∞∞

≤⎤⎡ dttgdttgdttgtg )()()()( 222

with equality when g1(t)=Cg2(t) for any arbitrary constant C.

∫ ∫∫ ∞− ∞−∞−≤⎥⎦

⎤⎢⎣⎡ dttgdttgdttgtg )()()()( 2121

with equality when g1(t) Cg2(t) for any arbitrary constant C.◊ If we set g1(t)=h1(t) and g2(t)=s(T−t), it is clear that the SNR is

maximized when h(t)=Cs(T–t).( ) ( )

25


◊ The output (maximum) SNR obtained with the matched filter◊ The output (maximum) SNR obtained with the matched filter is:

( )( )2 2T T⎡ ⎤ ⎡ ⎤∫ ∫ ( )( )

( )( )0 0

02 2 20

( ) ( ) ( )2SNR 1 1( )T T

s h T d s Cs T T d

NN h T t dt N C s T T t dt

τ τ τ τ τ τ⎡ ⎤ ⎡ ⎤− − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= =− − −

∫ ∫

∫ ∫ ( )( )00 00 0

2

( )2 2

2 2 ( )T

N h T t dt N C s T T t dt

s t dt ε= =

∫ ∫

∫

◊ Note that the output SNR from the matched filter depends on

00 0

( )s t dtN N∫

p pthe energy of the waveform s(t) but not on the detailed characteristics of s(t).

26

MatchedMatched--Filter DemodulatorFilter Demodulator◊ Frequency-domain interpretation of the matched filter

◊ Since h(t)=s(T–t), the Fourier transform of this relationship is:◊ Since h(t) s(T t), the Fourier transform of this relationship is:2

0( ) ( ) let -

T j ftH f s T t e dt T tπ τ−= − =∫

Th h d fil h f h i h l

2 2 * 2

0 ( ) ( )

T j f j fT j fTs e d e S f eπ τ π πτ τ − −⎡ ⎤= =⎢ ⎥⎣ ⎦∫◊ The matched filter has a frequency response that is the complex

conjugate of the transmitted signal spectrum multiplied by the phase factor e-j2πfT (sampling delay of T).p ( p g y )

◊ In other worlds, |H( f )|=|S( f )|, so that the magnitude response of the matched filter is identical to the transmitted signal spectrum.

◊ On the other hand, the phase of H( f ) is the negative of the phase of S( f ).

27


◊ Frequency-domain interpretation of the matched filter◊ If the signal s(t) with spectrum S( f ) is passed through the◊ If the signal s(t) with spectrum S( f ) is passed through the

matched filter, the filter output has a spectrumY( f )=|S( f )|2e-j2πfT ( ) * 2( ) j fTS f S f e π−⋅Y( f ) |S( f )| e j f .

◊ The output waveform is:22 2 2( ) ( ) ( )j ft j fT j ftt Y f df S f dfπ π π∞ ∞ −∫ ∫

( ) ( )S f S f e

◊ By sampling the output of the matched filter at t = T, we obtain (f P l’ l ti )

2 2 2( ) ( ) ( )j ft j fT j ftsy t Y f e df S f e e dfπ π π

−∞ −∞= =∫ ∫

(from Parseval’s relation):ε=== ∫∫

∞

∞−

T

s dttsdffSTy0

22 )()()(

◊ The noise at the output of the matched filter has a power spectral density (See Chapter 2, Page 109)

1

28

02

0 )(21)( NfHf =Φ


◊ Frequency-domain interpretation of the matched filter◊ The total noise power at the output of the matched filter is◊ The total noise power at the output of the matched filter is

∫∞

∞−Φ= 0 )( dffPn

∫∫

∫∞

∞−

∞

∞−

∞−

=== 02

02

0 21)(

21)(

21 NdffSNdffHN ε

◊ The output SNR is simply the ratio of the signal power Ps , given by to the noise power P

222

)(2 TyP =by , to the noise power Pn.

2 2SNR Ps εε)(TyP ss =

00

0

21SNR

NNPn

s

ε===

29


◊ Example:◊ M=4 biorthogonal signals are constructed from the two◊ M=4 biorthogonal signals are constructed from the two

orthogonal signals shown in the following figure for transmitting information over an AWGN channel. The noise is assumed to have a zero-mean and power spectral density ½ N0.

30


◊ Example: (cont.)◊ The M=4 biorthogonal signals have dimensions N=2 (shown in◊ The M=4 biorthogonal signals have dimensions N=2 (shown in

figure a): ⎪⎨⎧ ≤≤= )

210( 2)(1

TtTtf

⎪⎨⎧ ≤≤=

⎪⎩⎨

)21( 2)(

)(otherwise 02)(1

TtTTtf

f

◊ The impulse responses of the two matched filters are (figure b):⎪⎩⎨=

)(otherwise 0)

2()(2 tf

◊ The impulse responses of the two matched filters are (figure b):

⎪

⎪⎨⎧ ≤≤=−= )

21( 2)()( 11

TtTTtTfth

⎪⎨⎧ ≤≤=−=

⎪⎩⎨

)210( 2)()(

)(otherwise 02)()( 11

TtTtTfth

f

31

⎪⎩⎨==

)(otherwise 02)()( 22 tTfth


◊ Example: (cont.)◊ If s (t) is transmitted the responses of the two matched filters are◊ If s1(t) is transmitted, the responses of the two matched filters are

as shown in figure c, where the signal amplitude is A.◊ Since y1(t) and y2(t) are sampled at t=T we observe that◊ Since y1(t) and y2(t) are sampled at t T, we observe that

y1S (T)= , y2S(T)=0.

½ 2

212

A T

◊ From equation Page 28, we have ½A2T=ε and the received vector is:

[ ] ⎥⎤

⎢⎡ +== nnrr εr

where n1(t)=y1n(T) and n2=y2n(T) are the noise components at the

[ ] ⎥⎦⎢⎣+== 2121 nnrr εr

1( ) y1n( ) 2 y2n( ) poutputs of the matched filters, given by

21)()()( ∫ kdttftTT

32

2,1 , )()()(0

== ∫ kdttftnTy kkn


◊ Example: (cont.)◊ Clearly E(n )=E[y (T)] and their variance is◊ Clearly, E(nk)=E[ykn(T)] and their variance is

[ ] [ ]0 0

22 )()()()()( dtdftfntnETyET T

kkknn == ∫ ∫ τττσ [ ]

0 00

0 0

)()()(21 dtdtfftN

T T

kk−= ∫ ∫

∫ ∫τττδ

00

20 2

1)(21

2

NdttfNT

k == ∫◊ Observe that the SNR0 for the first matched filter is

0

21

)(2

NN

εε==

SNR0

33

002

N


◊ Example: (cont.)This result agrees with our previous result◊ This result agrees with our previous result.

◊ We can also note that the four possible outputs of the two matched filters corresponding to the fo r possiblematched filters, corresponding to the four possible transmitted signals are:

( ) ( ) ( ) ( ) ( )1 2 1 2 1 2 1 2 1 2, , , , , , and ,r r n n n n n n n nε ε ε ε= + + − + − +

34

The Optimum DetectorThe Optimum Detector◊ Our goal is to design a signal detector that makes a decision on the

transmitted signal in each signal interval based on the observation of the vector r in each interval such that the probability of a correct decision is maximized.

◊ We assume that there is no memory in signals transmitted in◊ We assume that there is no memory in signals transmitted in successive signal intervals.

◊ We consider a decision rule based on the computation of the pposterior probabilities defined as

P(sm|r)≡P(signal sm was transmitted|r), m=1,2,…,M.◊ The decision criterion is based on selecting the signal corresponding

to the maximum of the set of posterior probabilities { P(sm|r)}. This decision criterion is called the maximum a posterior probabilitydecision criterion is called the maximum a posterior probability (MAP) criterion.

35

The Optimum DetectorThe Optimum Detector◊ Using Bayes’ rule, the posterior probabilities may be expressed as

( | ) ( )( | ) p Pr s s

where P(s ) is the a priori probability of the mth signal being

( | ) ( )( | )( )

m mm

p PPp

=r s ss r

r---(A)

where P(sm) is the a priori probability of the mth signal being transmitted.

◊ The denominator of (A), which is independent of which signal is ( ), p gtransmitted, may be expressed as

∑=M

Ppp )()|(( ssrr)

◊ Some simplification occurs in the MAP criterion when the M signal ll b bl i i i P( ) 1/M

∑=m

mm Ppp1

)()|(( ssrr)

are equally probable a priori, i.e., P(sm)=1/M.◊ The decision rule based on finding the signal that maximizes P(sm|r)

i i l t t fi di th i l th t i i P( | )is equivalent to finding the signal that maximizes P(r|sm).

36

The Optimum DetectorThe Optimum Detector◊ The conditional PDF P(r|sm) or any monotonic function of it is

usually called the likelihood function.y f◊ The decision criterion based on the maximum of P(r|sm) over the M

signals is called maximum-likelihood (ML) criterion.◊ We observe that a detector based on the MAP criterion and one that

is based on the ML criterion make the same decisions as long as a priori probabilities P(sm) are all equal.

◊ In the case of an AWGN channel, the likelihood function p(r|sm) is i bgiven by:

2

2

( )1( | ) exp , 1, 2,......,( )

Nk mk

m N

r sp m MN N

⎡ ⎤−= − =⎢ ⎥∑r s 2

10 0

( | ) p , , , ,( )m N

kp

N Nπ =⎢ ⎥⎣ ⎦

∑

∑ −−−=N

mkkm srN

NNp 20 )(1)ln(

21)|(ln πsr

constant

37

∑=kN 102

The Optimum DetectorThe Optimum Detector◊ The maximum of ln p(r|sm) over sm is equivalent to finding the signal

sm that minimizes the Euclidean distance:m

∑ −=N

kmkkm srD

1

2)(),( sr

◊ We called D(r,sm), m=1,2,…,M, the distance metrics.◊ Hence, for the AWGN channel, the decision rule based on the ML

=k 1

◊ Hence, for the AWGN channel, the decision rule based on the ML criterion reduces to finding the signal sm that is closest in distance to the receiver signal vector r. We shall refer to this decision rule as minimum distance detection.

38

The Optimum DetectorThe Optimum Detector◊ Expanding the distance metrics:

2 2( ) 2N N N

D ∑ ∑ ∑2 2

1 1 12 2

( , ) 2

2 1 2

m n n mn mnn n n

D r r s s

m M= = =

= − +

= − ⋅ + =

∑ ∑ ∑r s

r r s s◊ The term || r||2 is common to all distance metrics, and, hence, it may

be ignored in the computations of the metrics

2 , 1, 2,.......,m m m M= − ⋅ + =r r s s

be ignored in the computations of the metrics.◊ The result is a set of modified distance metrics.

22)(D ssrsr +⋅−=′

◊ Note that selecting the signal sm that minimizes D'(r, sm ) is equivalent to selecting the signal that maximizes the metrics

2),( mmmD ssrsr +=

equivalent to selecting the signal that maximizes the metrics C(r, sm)= –D'(r, sm ),

22)(C ssrsr −⋅=

39

2),( mmmC ssrsr

The Optimum DetectorThe Optimum Detector◊ The term r⋅sm represents the projection of the signal vector onto each

of the M possible transmitted signal vectors.p g◊ The value of each of these projection is a measure of the correlation

between the receiver vector and the mth signal. For this reason, we call C(r, sm), m=1,2,…,M, the correlation metrics for deciding which of the M signals was transmitted.

◊ Finally, the terms || sm||2 =εm, m=1,2,…,M, may be viewed as bias terms that serve as compensation for signal sets that have unequal energiesenergies.

◊ If all signals have the same energy, || sm||2 may also be ignored.◊ Correlation metrics can be expressed as:◊ Correlation metrics can be expressed as:

(r, s ) 2 ( ) ( ) , 0,1,.......,T

C r t s t dt m Mε= − =∫

40

0(r, s ) 2 ( ) ( ) , 0,1,.......,m mC r t s t dt m Mmε∫

The Optimum DetectorThe Optimum Detector◊ These metrics can be generated by a demodulator that cross-

correlates the received signal r(t) with each of the M possible g ( ) ptransmitted signals and adjusts each correlator output for the bias in the case of unequal signal energies.

41

The Optimum DetectorThe Optimum Detector◊ We have demonstrated that the optimum ML detector computes a set

of M distances D(r,sm) or D'(r,sm) and selects the signal ( , m) ( , m) gcorresponding to the smallest (distance) metric.

◊ Equivalently, the optimum ML detector computes a set of Mcorrelation metrics C(r, sm) and selects the signal corresponding to the largest correlation metric.

◊ The above development for the optimum detector treated the important case in which all signals are equal probable. In this case, the MAP criterion is equivalent to the ML criterionthe MAP criterion is equivalent to the ML criterion.

◊ When the signals are not equally probable, the optimum MAP detector bases its decision on the probabilities given by:detector bases its decision on the probabilities given by:

( | ) ( )( | )( )

m mm

p PP =r s ss r ( ) ( ) ( )or , |m m mPM p P=r s r s s

42

( | )( )m p r

( ) ( ) ( ), |m m mp

The Optimum DetectorThe Optimum Detector

◊ Example:◊ Consider the case of binary PAM signals in which the two◊ Consider the case of binary PAM signals in which the two

possible signal points are s1 = −s2 = ,, where εb is the energy per bit. The priori probabilities are P(s1)=p and P(s2)=1-p. Let

bεp p p ( 1) p ( 2) pus determine the metrics for the optimum MAP detector when the transmitted signal is corrupted with AWGN.

◊ The receiver signal vector for binary PAM is: )(Tyr nb +±= ε

where yn(T) is a zero mean Gaussian random variable with variance .

nb

20

12n Nσ =2

43


◊ Example: (cont.)◊ The conditional PDF P(r|s ) for two signals are◊ The conditional PDF P(r|sm) for two signals are

2

1 2

( )1( | ) exp22

brp r s

σπσε⎡ ⎤−

= −⎢ ⎥⎢ ⎥⎣ ⎦

2

22

( )1( | ) exp

nn

nrp r s

σπσ

ε⎢ ⎥⎣ ⎦⎡ ⎤+

= ⎢ ⎥

◊ Then the metrics PM(r,s1) and PM(r,s2) are

2 2( | ) exp22 nn

p r sσπσ

= −⎢ ⎥⎢ ⎥⎣ ⎦

( 1) ( 2)2

1 1 2

( )( , ) ( | ) exp

22b

nn

rpPM p p r sσπσε⎡ ⎤−

= ⋅ = −⎢ ⎥⎢ ⎥⎣ ⎦

r snn ⎢ ⎥⎣ ⎦

( )2

2 2 2

( )1( , ) 1 ( | ) exp22

brpPM p p r sσπσε⎡ ⎤+−

= − ⋅ = −⎢ ⎥⎢ ⎥⎣ ⎦

r s

44

22 nn σπσ ⎢ ⎥⎣ ⎦


◊ Example: (cont.)◊ If PM(r s ) > PM(r s ) we select s as the transmitted signal:◊ If PM(r,s1) > PM(r,s2), we select s1 as the transmitted signal:

otherwise, we select s2. This decision rule may be expressed as:)( 1sPM sr 1),(),(

22

1

sPMPM

srsr

⎤⎡ 22 )()( εε⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−+−

= 2

22

2

1

2)()(

exp1),(

),(

n

bb rrp

pPMPM

σεε

srsr

pprr s

bb −−−+ 1ln2

)()( 1

2

22

σ

εεpsn2

2σ

pNprs −

=− 1ln11ln1 2

1

σε45

pN

pr n

sb = ln

4ln

2 0

2

σε


◊ Example: (cont.)◊ The threshold is denoted by τ divides the real line

1 1ln pN −◊ The threshold is , denoted by τh, divides the real line

into two regions, say R1 and R2, where R1 consists of the set of points that are greater than τh and R2 consists of the set of points

0 ln4

Np

p g h 2 pthat are less than τh.

◊ If , the decision is made that s1 was transmitted.b hr ε τ> 1

◊ If , the decision is made that s2 was transmitted.b hr ε τ<

46


◊ Example: (cont.)◊ The threshold τ depends on N and p If p=1/2 τ =0◊ The threshold τh depends on N0 and p. If p=1/2, τh=0.◊ If p>1/2, the signal point s1 is more probable and, hence, τh<0.

In this case the region R1 is larger than R2 so that s1 is moreIn this case, the region R1 is larger than R2 , so that s1 is more likely to be selected than s2.

◊ The average probability of error is minimized◊ The average probability of error is minimized◊ It is interesting to note that in the case of unequal priori

probabilities, it is necessary to know not only the values of the p y ypriori probabilities but also the value of the power spectral density N0 , or equivalently, the noise-to-signal ratio, in order to

h h h ldcompute the threshold.◊ When p=1/2, the threshold is zero, and knowledge of N0 is not required

by the detectorby the detector.

47

The Optimum DetectorThe Optimum Detector◊ Proof of “the decision rule based on the maximum-likelihood

criterion minimizes the probability of error when the M signals are equally probable a priori”.◊ Let us denote by Rm the region in the N-dimensional space for

which we decide that signal s (t) was transmitted when the vectorwhich we decide that signal sm(t) was transmitted when the vector r=[r1r2….. rN] is received.

◊ The probability of a correct decision given that sm(t) was p y g m( )transmitted is:

( ) ( ) ( ) ( )| s s | s sm m m mR

P c p p p d⋅ = ⋅∫ r r1

◊ The average probability of a correct decision is:mR

1 1( ) ( | ) ( | )M M

m mP c P c p dM M

= =∑ ∑ ∫s r s r

1M

◊ Note that P(c) is maximized by selecting the signal sm if p(r|sm) is larger than p(r|sk) for all m≠k. Q.E.D.

1 1( ) ( | ) ( | )

m

m mm m R

pM M= =

∑ ∑ ∫

larger than p(r|sk) for all m≠k. Q.E.D.

48

The Optimum DetectorThe Optimum Detector◊ Similarly for the MAP criterion, when the M signals are not equally

probable, the average probability of a correct decision is p , g p y

1( ) ( | ) ( )

m

M

mm R

P c P p d=

= ∑ ∫ s r r rsame for all s

◊ In order for P (c) to be as large as possible, the points that are to be included in each particular region Rm are those for which P(sm|r)

same for all sm.

p g m ( m| )exceeds all the other posterior probabilities. Q.E.D.

◊ We conclude that MAP criterion maximize the probability of correct detection.

49

Probability of Error for Binary ModulationProbability of Error for Binary Modulation

◊ Let us consider binary PAM signals where the two signal waveforms are s1(t)=g(t) and s1(t)= −g(t), and g(t) is an arbitrary pulse that is 1( ) g( ) 1( ) g( ), g( ) y pnonzero in the interval 0≤ t ≤Tb and zero elsewhere.

◊ Since s1(t)= −s2(t), these signals are said to be antipodal.◊ The energy in the pulse g(t) is εb.◊ PAM signals are one-dimensional, and, their geometric g g

representation is simply the one-dimensional vector:

1 2 , b bs sε ε= = −1 2b b

Fi (A)

50

Figure (A)


◊ Let us assume that the two signals are equally likely and that signal s1(t) was transmitted. Then, the received signal from the (matched 1( ) , g (filter or correlation) demodulator is

nnsr +=+= ε

where n represents the additive Gaussian noise component, which

nnsr b +=+= ε1

1p p

has zero mean and variance . 20

12n Nσ =

◊ In this case, the decision rule based on the correlation metric given by (Page 39) compares r with the threshold

If 0 h d i i i d i f f ( ) d if 0 h

22),( mmmC ssrsr −⋅=zero. If r >0, the decision is made in favor of s1(t), and if r<0, the decision is made that s2(t) was transmitted.

51


◊ The two conditional PDFs of r are:

( ) ( )20/

10

1| br Np r s eN

ε

π− −

=

( ) ( )20/

20

1| br Np r s eN

ε

π− +

=

52


◊ Given that s1(t) was transmitted, the probability of error is simply the probability that r<0. p y

( ) ( )( )2

0 0

1 11| | exp

brP e s p r s dr dr

ε⎡ ⎤−⎢ ⎥= = −⎢ ⎥∫ ∫( ) ( )

( )

1 100

| | exp P e s p r s dr drNNπ−∞ −∞ ⎢ ⎥

⎢ ⎥⎣ ⎦∫ ∫

( )202 / / 2

0

1 2

2

b N bx re dx x

Nε ε

π− −

−∞

−= =∫

( )2

0

/ 2

2 /

21 2 b

x

Ne dx x x

επ∞ −= = −∫

02

2

b

bQN

π

ε⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

( ) 2 / 21 02

x

tQ t e dx t

π∞ −= ≥∫

53

0N⎜ ⎟⎝ ⎠


◊ If we assume that s2(t) was transmitted, and the probability that r>0 is also . Since the signal

br nε= − +( ) ( )2 0| 2 /bP e s Q Nε=p y g

s1(t) and s2(t) are equally likely to be transmitted, the average probability of error is

( ) ( )2 0| bQ

( ) ( )1 20

21 1| |2 2

bbP P e s P e s Q

Nε⎛ ⎞

= + = ⎜ ⎟⎜ ⎟⎝ ⎠

--(A)

◊ Two important characteristics of this performance measure:◊ First we note that the probability of error depends only on the

0⎝ ⎠

◊ First, we note that the probability of error depends only on the ratio εb /N0.

◊ Secondly we note that 2εb /N0 is also the output SNR0 from the◊ Secondly, we note that. 2εb /N0 is also the output SNR0 from the matched-filter (and correlation) demodulator.

◊ The ratio εb /N0 is usually called the signal-to-noise ratio per bit.◊ The ratio εb /N0 is usually called the signal to noise ratio per bit.

54


◊ We also observe that the probability of error may be expressed in terms of the distance between that the two signals s1and s2 .g 1 2

◊ From Page 52, we observe that the two signals are separated by the distance d12=2 . By substituting into Equation (A),we bε 2

1214b dε =

obtain124b

⎟⎞

⎜⎛ 2d

⎟⎟⎠

⎞⎜⎜⎝

⎛=

0

12

2NdQPb

◊ This expression illustrates the dependence of the error probability on the distance between the two signals points.

55


◊ Error probability for binary orthogonal signals◊ The signal vectors s and s are two dimensional◊ The signal vectors s1and s2 are two-dimensional.

1 [ 0]bs ε=1

2 b

[ ]

[0 ]b

s ε=

where εb denote the energy for each of the waveforms. Note that the distance between these signal points is . 12 2 bd ε=

56


◊ Error probability for binary orthogonal signals◊ To evaluate the probability of error let us assume that s was◊ To evaluate the probability of error, let us assume that s1 was

transmitted. Then, the received vector at the output of the demodulator is .1 2[ ]br n nε= +

◊ We can now substitute for r into the correlation metrics given by to obtain C(r, s1) and C(r, s2).

1 2[ ]b

22),( mmmC ssrsr −⋅= to obtain C(r, s1) and C(r, s2).◊ The probability of error is the probability that C(r, s2) > C(r, s1).

),( mmm

( ) ( )2 12 2 2b b b b bn nε ε ε ε ε⇒ − > + − 2 1 bn n ε> +

( ) ( ) ( )1 2 1 2 2 1| [ , , ] [ ]bP e s P C r s C r s P n n ε= > = − >

57


◊ Error probability for binary orthogonal signals◊ Since n and n are zero mean statistically independent Gaussian◊ Since n1 and n2 are zero-mean statistically independent Gaussian

random variables each with variance ½ N0, the random variable x= n2 – n1 is zero-mean Gaussian with variance N0. Hence,2 1 0 ,

( ) 202

2 10

12 b

x NbP n n e dx

N εε

π∞ −− > = ∫

2

0

2

0

1 2 b

x bN

e dx QNε

επ

∞ − ⎛ ⎞= = ⎜ ⎟⎜ ⎟

⎝ ⎠∫

◊ The same error probability is obtained when we assume that s2 is transmitted:

⎛ ⎞

0⎝ ⎠

( )0

where is the SNR per bit.bb b bP Q Q

Nε γ γ

⎛ ⎞= =⎜ ⎟⎜ ⎟

⎝ ⎠

58


◊ If we compare the probability of error for binary antipodal signals with that for binary orthogonal signals, we find that orthogonal y g g , gsignals required a factor of 2 increase in energy to achieve the same error probability as antipodal signals.

◊ Since 10 log10 2=3 dB, we say that orthogonal signals are 3dB poorer than antipodal signals. The difference of 3dB is simply due to the di t b t th t i l i t hi h i f2 2ddistance between the two signal points, which is for orthogonal signals, whereas for antipodal signals.

◊ The error probability versus 10 log ε /N for these two types of

212 2 bd ε=

212 4 bd ε=

◊ The error probability versus 10 log10 εb/N0 for these two types of signals is shown in the following figure (B). As observed from this figure, at any given error probability, the εb/N0 required for g , y g p y, b 0 qorthogonal signals is 3dB more than that for antipodal signals.

59


◊ Probability of error for binary signals

60

Figure (B)

Probability of Error for MProbability of Error for M--aryary Orthogonal SignalsOrthogonal Signals

◊ For equal-energy orthogonal signals, the optimum detector selects the signal resulting in the largest cross correlation between the g g greceived vector r and each of the M possible transmitted signals vectors {sm}, i.e.,

( ) ∑=

==⋅=M

kmkkmm MmsrsrsrC

1,,2,1 ,, …

◊ To evaluate the probability of error, let us suppose that the signal s1is transmitted. Then the received signal vector is

⎡ ⎤

where ε denotes the symbol energy and n1 n2 … nM are zero-mean

1 2 3 b Mn n n nε⎡ ⎤= +⎣ ⎦r

where εs denotes the symbol energy and n1,n2, ,nM are zero mean, mutually statistically independent Gaussian random variable with equal variance . 2

012n Nσ =

61

2


◊ In this case, the outputs from the bank of M correlations are

( ) ( )C( ) ( )( ) s

ss

nsrC

nsrC

ε

εε

=

+=

,

,

22

11

( ) nsrC ε=

◊ Note that the scale factor εs may be eliminated from the correlatoroutputs dividing each output by

( ) MsM nsrC ε=,

εoutputs dividing each output by .◊ With this normalization, the PDF of the first correlator output is

sε

( ) ⎤⎡ 2

( ) ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=

0

2

1

01 exp1

1 Nx

Nxp s

r

επ

62

⎦⎣


◊ And the PDFs of the other M−1 correlator outputs are

( ) N1 2

◊ It is mathematically convenient to first derive the probability that the

( ) MmeN

xp Nxmr

m

m,,3,2 ,1

02

0

…== −

π◊ It is mathematically convenient to first derive the probability that the

detector makes a correct decision. This is the probability that r1 is larger than each of the other M−1 correlator outputs n2,n3,…, nM. g p 2 3 MThis probability may be expressed as

( ) ( )| drrprrnrnrnPP <<<= ∫∞

where denotes the joint probability ( )2 1 3 1 1 1, , , |MP n r n r n r r< < <…

( ) ( ) 11111312 |,,, drrprrnrnrnPP Mc <<<= ∫ ∞−…

that n2,n3,…, nM are all less than r1, conditioned on any given r1. Then this joint probability is averaged over all r1.

( )

63


◊ Since the {rm} are statistically independent, the joint probability factors into a product of M−1 marginal probabilities of the form:p g p

( ) ( ) MmdxxprrnP mm

r

rm m∫ ∞−==< 1

11 ,,3,2 | … --(B)

dxeNr x∫ ∞−

−= 01 22 2

21 π

02 mx N x=

◊ This probabilities are identical for m=2,3,…,M, and, the joint probability under consideration is simply the result in Equation (B) p y p y q ( )raised to the (M−1)th power. Thus, the probability of a correct decision is

1M

( )21 0

12 2

1 112

Mr N x

cP e dx p r drπ

−∞ −

−∞ −∞

⎛ ⎞= ⎜ ⎟

⎝ ⎠∫ ∫

64


◊ The probability of a (k-bit) symbol error is1

2y r=

cM PP −=1

21M⎧ ⎫⎡ ⎤⎛ ⎞

10

yN

2

212

0

21 1 11 exp22 2

My x s

MP e dx y dyNε

π π

−∞ −

−∞ −∞

⎧ ⎫⎡ ⎤⎛ ⎞⎛ ⎞⎪ ⎪⎢ ⎥= − − −⎜ ⎟⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎪ ⎪⎝ ⎠⎣ ⎦⎩ ⎭∫ ∫

◊ The same expression for the probability of error is obtained when

⎪ ⎪⎝ ⎠⎣ ⎦⎩ ⎭--- (C)

◊ The same expression for the probability of error is obtained when any one of the other M−1 signals is transmitted. Since all the Msignals are equally likely, the expression for PM given above is the average probability of a symbol error.

◊ This expression can be evaluated numerically.

65


◊ In comparing the performance of various digital modulation methods, it is desirable to have the probability of error expressed in terms of p y pthe SNR per bit, εb/N0, instead of the SNR per symbol, εs/N0.

◊ With M=2k, each symbol conveys k bits of information, and hence εs= k εb. Thus, Equation (C) may be expressed in terms of εb/N0 by substituting for εs.

◊ It is also desirable to convert the probability of a symbol error into an equivalent probability of a binary digit error. For equiprobableorthogonal signals all symbol errors are equiprobable and occurorthogonal signals, all symbol errors are equiprobable and occur with probability

MM PP121 −

=− k

MM

M

66


◊ Furthermore, there are ways in which n bits out of k may be in error. Hence, the average number of bit errors per k-bit symbol is

kn

⎛ ⎞⎜ ⎟⎝ ⎠

d h bi b bili i j h l i E i (D)

1

1

22 1 2 1

kkM

Mk kn

k Pn k Pn

−

=

⎛ ⎞=⎜ ⎟ − −⎝ ⎠

∑ --(D)and the average bit error probability is just the result in Equation (D) divided by k, the number of bits per symbol. Thus,

12k P−

◊ The graphs of the probability of a binary digit error as a function of

2 12 1 2

Mb Mk

PP P k= ≈ >>−

g p p y y gthe SNR per bit, εb/N0, are shown in Figure (C) for M=2,4,8,16,32, and 64. This figure illustrates that, by increasing the number M of waveforms one can reduce the SNR per bit required to achieve awaveforms, one can reduce the SNR per bit required to achieve a given probability of a bit error.

67


◊ For example, to achieve a Pb=10–5, the required SNR b , qper bit is a little more than 12dB for M=2, but if M is increased to 64 signal waveforms, the required SNR per bit is approximately 6dBper bit is approximately 6dB. Thus, a savings of over 6dB is realized in transmitter power required to achieve a Pb=10–5 by increasing Mf M 2 M 64from M=2 to M=64.

68 Figure (C)


◊ What is the minimum required εb/N0 to achieve an arbitrarily small probability of error as M→∞?p y

◊ A union bound on the probability of error.◊ Let us investigate the effect of increasing M on the probability of g g p y

error for orthogonal signals. To simplify the mathematical development, we first derive an upper bound on the probability of a symbol error that is much simpler than the exact form given in the following equation:

dyN

ydxeP sM

y xM ⎥

⎥⎤

⎢⎢⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−= ∫ ∫∞

∞−

−

∞−

−

212 2

21exp

211

21 2 ε

ππ N ⎥⎦⎢⎣⎟⎠

⎜⎝⎥⎦⎢⎣ ⎠⎝∫ ∫∞ ∞

0222 ππ

69


◊ A union bound on the probability of error (cont.)◊ Recall that the probability of error for binary orthogonal signals◊ Recall that the probability of error for binary orthogonal signals

is given by: ( )bb

b QN

QP γε=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

◊ Now, if we view the detector for M orthogonal signals as one that makes M −1 binary decisions between the correlator outputs

N ⎟⎠

⎜⎝ 0

y pC(r,s1) that contains the signal and the other M−1 correlatoroutputs C(r,sm), m=2,3,…,M, the probability of error is upper-b d d b i b d f h M 1 Th i if Ebounded by union bound of the M −1 events. That is, if Eirepresents the event that C(r,si)> C(r,s1) for i≠1, then we have Hence( ) ( )

MMP P E P E≤ ∑∪have . Hence, ( ) ( )22

M i iii

P P E P E=

=

= ≤ ∑∪( ) ( ) ( ) ( )0 01 1 / /M b S SP M P M Q N MQ Nε ε≤ − = − <

70

( ) ( )


◊ A union bound on the probability of error(cont.)◊ This bound can be simplified further by upper-bounding◊ This bound can be simplified further by upper bounding

◊ We have( )0/SQ Nε

◊ We have

thus( ) 0

0

2Ns

seNQ εε −< --(E)thus,

( )

0 0

0

2 2

2ln 2 2

2s b

b

N k NkM

k N

P Me e

P e

ε ε

ε

− −

− −

< =

<--(F)

◊ As k→∞ or equivalently, as M→∞, the probability of error approaches zero exponentially provided that εb/N0 is greater than

( )0bMP e<

approaches zero exponentially, provided that εb/N0 is greater than 2ln 2,

( )dB4213912ln2 =>bε

71

( )dB42.1 39.12ln20

=>N


◊ A union bound on the probability of error(cont.)◊ The simple upper bound on the probability of error given by◊ The simple upper bound on the probability of error given by

Equation (F) implies that, as long as SNR>1.42 dB, we can achieve an arbitrarily low PM..

◊ However, this union bound is not a very tight upper bound as a sufficiently low SNR due to the fact that upper bound for the Q function in Equation (E) is loose.

◊ In fact, by more elaborate bounding techniques, it can be shown th t th b d i E ti (F) i ffi i tl ti ht fthat the upper bound in Equation (F) is sufficiently tight for εb/N0>4 ln2.For ε /N <4 ln2 a tighter upper bound on P is◊ For εb/N0<4 ln2, a tighter upper bound on PM is

( )200 2ln2 −−< NkeP ε

72

2<M eP


◊ A union bound on the probability of error(cont.)◊ Consequently PM →0 as k→∞ provided that◊ Consequently, PM →0 as k→∞, provided that

( )dB6.1 693.02ln −=>N

bε

◊ Hence, −1.6 dB is the minimum required SNR per bit to achieve an arbitrarily small probability of error in the limit as

0N

an arbitrarily small probability of error in the limit as k→∞(M→∞). This minimum SNR per bit is called the Shannon limit for an additive Gaussian noise channel.

73

Probability of Error for MProbability of Error for M--aryary BiorthogonalBiorthogonal SignalsSignals

◊ As indicated in Chapter 3, a set of M=2k biorthogonal signals are constructed from ½ M orthogonal signals by including the negatives of the orthogonal signals. Thus, we achieve a reduction in the complexity of the demodulator for the biorthogonal signals relative to that for orthogonal signals, since the former is implemented withto that for orthogonal signals, since the former is implemented with ½ M cross correlation or matched filters, whereas the latter required M matched filters or cross correlators.

◊ Let us assume that the signal s1(t) corresponding to the vector s1=[ 0 0…0] was transmitted. The received signal vector is sε

[ ]n n nε= +rwhere the {nm} are zero-mean, mutually statistically independent and identically distributed Gaussian random variables with

1 2 2[ ]s Mn n nε= +r

yvariance .2

012n Nσ =

74


◊ The optimum detector decides in favor of the signal corresponding to the largest in magnitude of the cross correlatorsg g

( )2

1

1C , , 1, 2,...,2

M

m m k mkk

r s m M= ⋅ = =∑r s r s

while the sign of this largest term is used to decide whether sm(t) or −sm(t) was transmitted.

1 2k =

m( )

◊ According to this decision rule, the probability of a correct decision g , p yis equal to the probability that and r1 exceeds |rm|=|nm| for m=2,3,… ½ M. But

1 1 0sr nε= + >

( ) 2 21 1 00

1 1 0

2 21 1 2

0

1 1| 02

r r Nx N ym r r N

P n r r e dx e dyNπ π

− −

− −< > = =∫ ∫

75

02

Ny x=


◊ Then, the probability of a correct decision is 2

2 121

Mr N

−∞ ⎛ ⎞

∫ ∫from which upon substitution for p(r ) we obtain

( )21 0

1 0

2 21 10 2

12

r N xc r N

P e p r drπ

∞ −

−

⎛ ⎞= ⎜ ⎟⎝ ⎠∫ ∫

from which, upon substitution for p(r1), we obtain

( )2 20

2 12 2 2

2 2

1 12 2

s

Mv N x v

c N v NP e dx e dv

ε

ε ε

−∞ + − −

+

⎛ ⎞= ⎜ ⎟

⎝ ⎠∫ ∫ --(G)

( ) 01

Nv r ε= −

( )0 02 22 2s sN v Nε επ π− − +⎜ ⎟⎝ ⎠∫ ∫

( ) ( )⎥⎤

⎢⎡ −−=

2

1exp1 xxp sε

where we have used the PDF of r given in Page 62

( )1 2sv r ε( )⎥⎥⎦⎢

⎢⎣

=00

1 exp1 NN

xpr π

where we have used the PDF of r1 given in Page 62.◊ The probability of a symbol error PM=1−Pc

76


◊ Pc, and PM may be evaluated numerically for different values yof M. The graph shown in the following figure (D) illustrates PM as a function of εb/N0, where εs=k εb, for M=2,4,8,16 and 32and 32.

77 Fig. (D)


◊ In this case, the probability of error for M=4 is greater than that for M=2. This is due to the fact that we have plotted the symbol error p yprobability PM in Figure(D).

◊ If we plotted the equivalent bit error probability, we should find that the graphs for M=2 and M=4 coincide.

◊ As in the case of orthogonal signals, as M→∞ (k→∞), the minimum required εb/N0 to achieve an arbitrarily small probability of error is

1 6 dB h Sh li i−1.6 dB, the Shannon limit.

78

Probability of Error for Probability of Error for MM--aryary PAMPAM◊ Recall that M-ary PAM signals are represent geometrically as M

one-dimensional signal points with value :g p1 1, 2, ,2 mm gs A m Mε= =

where◊ The Euclidean distance between adjacent signal points is .

( ) MmdMmAm ,,2,1 ,12 =−−=

gd ε2j g p◊ Assuming equally probable signals, the average energy is :

g

( )2

22

1 1 1

1 1 2 12

M M Mg

av m mm m m

ds m M

M M Mεε ε

= = =

= = = − −∑ ∑ ∑

( ) ( )1 1 1

22 2 21 1 1 1

2 3 6

m m m

gg

dM M M d

Mε ε

= = =

⎡ ⎤= − = −⎢ ⎥⎣ ⎦

( )

( )( )1

12

M

m

M Mm

=

+=∑

79

2 3 6M ⎣ ⎦ ( )( )2

1

1 2 16

M

m

M M Mm

=

+ +=∑

Probability of Error for Probability of Error for MM--aryary PAMPAM◊ Equivalently, we may characterize these signals in terms of their

average power, where is :d εε 21

◊ The average probability of error for M-ary PAM :

( )T

dM

TP gav

av

εε 22 1

61

−== (*)

◊ The average probability of error for M ary PAM :◊ The detector compares the demodulator output r with a set of M-1 thresholds,

which are placed at the midpoints of successive amplitude level and decision is made in favor of the amplitude level that is close to rmade in favor of the amplitude level that is close to r.

122 gd ε

We note that if the th amplit de le el is transmitted the demod lator o tp t◊ We note that if the mth amplitude level is transmitted, the demodulator output is 1

2m g mr s n A nε= + = +1where the noise variable n has zero-mean and variance

80

20

2

21 Nn =σ

Probability of Error for Probability of Error for MM--aryary PAMPAM◊ Assuming all amplitude levels are equally likely a priori, the

average probability of a symbol error is the probability that the noise variable n exceeds in magnitude one-half of the distance between levels.

◊ However when either one of the two outside levels ±(M 1) is◊ However, when either one of the two outside levels ±(M-1) is transmitted, an error can occur in one direction only.

◊ As a result, we have the error probability:, p y2

0

20

1 1 1 22 2 2 g

x NM m g d

M MP P r s d e dxM M N επ

ε ∞ −⎛ ⎞− −= − > =⎜ ⎟⎜ ⎟

⎝ ⎠∫

2

20

0

20

1 2 where y 22 g

y

d N

M e dy x NM επ

∞ −

⎝ ⎠−

= =∫( ) 2 / 21 0xQ t d t

∞ − ≥∫( ) 2

0

2 1 gdM

QM N

ε⎛ ⎞−⎜ ⎟=⎜ ⎟⎝ ⎠

( )1 1 12 22

MM M

⋅ − + ⋅ ⋅

( ) / 2 02

x

tQ t e dx t

π= ≥∫

81

0M N⎜ ⎟⎝ ⎠ 2M M

Probability of Error for Probability of Error for MM--aryary PAMPAM◊ From (*) in page 80, we note that TP

Md avg 1

62

2

−=ε

◊ By substituting for , we obtain the average probability of a symbol error for PAM in terms of the average power :

gd ε2

( )( )

( )( )2 2

2 1 2 16 61 1

av avM

M MP TP Q QM MM N M N

ε⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟

◊ It is customary for us to use the SNR per bit as the basic

( ) ( )2 20 01 1M M MM N M N⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

y pparameter, and since T=kTb and k =log2 M:

( ) ( )⎟⎟⎞

⎜⎜⎛−

= 2log612 MQMP avbε

where is the average bit energy and is the average SNR

( ) ⎟⎟⎠

⎜⎜⎝ −

=0

2 1 NMQ

MPM

av avb bP Tε = avb oNεper bit.

82

av avb b avb o

Probability of Error for Probability of Error for MM--aryary PAMPAM◊ The case M=2 corresponds

to the error probability for p ybinary antipodal signals.

◊ The SNR per bit increase◊ The SNR per bit increase by over 4 dB for every factor-of-2 increase in M.

◊ For large M, the additional SNR per bit required to p qincrease M by a factor of 2 approaches 6 dB.

Probability of a symbol error for PAM

83

Probability of a symbol error for PAM

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ Recall from that digital phase-modulated signal waveforms may be

expressed as: p

( ) ( ) ( ) TtMmmM

tftgts cm ≤≤≤≤⎥⎦⎤

⎢⎣⎡ −+= 0 ,1 ,122cos ππ

and have the vector representation:⎦⎣

( ) ( ) ( )ππ 122 ⎤⎡

Energy in each waveform.

◊ Since the signal waveforms have equal energy the optimum detector

( ) ( ) ( ) gsssm mM

mM

ts εεεε ππ21 ,12sin 12cos =⎥⎦

⎤⎢⎣⎡ −−=

◊ Since the signal waveforms have equal energy, the optimum detector for the AWGN channel computes the correlation metrics

( ) MmC mm ,,2,1,, =⋅= srsr

84

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ In other word, the received signal vector r=[r1 r2] is projected onto

each of the M possible signal vectors and a decision is made in favor p gof the signal with the largest projection.

◊ This correlation detector is equivalent to a phase detector that computes the phase of the received signal from r. We selects the signal vector sm whose phase is closer to r.

r◊ The phase of r is

1

21tanrr

r−=Θ

◊ We will determine the PDF of , and compute the probability of error from it.C id h i hi h h i d i l h i

rΘ

◊ Consider the case in which the transmitted signal phase is , corresponding to the signal s1(t).0=Θr

85

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ The transmitted signal vector is , and the received signal

vector has components: 1 0ss ε⎡ ⎤= ⎣ ⎦

p

◊ Because n1 and n2 are jointly Gaussian random variable, it follow 1 1 2 2 sr n r nε= + =

1 2 j y ,that r1 and r2 are jointly Gaussian random variable variables with

2 2 21( ) 0 dE N)(E ε1 2

2 2 22 0

1( ) 0, and .2r r rE r Nσ σ σ= = = =,)( 1 srE ε=

( ) ⎤⎡ 22

( ) ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡ +−−= 2

221

221 2exp

21,

r

s

rr

rrrrp

σπσε

◊ The PDF of the phase Θr is obtained by a change in variables from (r1,r2) to :

⎦⎣

2 2 1 2t rV −+ Θ1 2

86

2 2 1 21 2

1

tanrV r rr

= + Θ =

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ The joint PDF of V and Θr :

1 2dr dr VdVd= Θ

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ Θ−+−=ΘΘ 2

2

2, 2cos2

exp2

, rssrrV

VVVVpσπσ

εε

◊ Integration of over the range of V yields( )rrV Vp ΘΘ ,,( )

r rpΘ Θ⎠⎝ 22 rr σπσ

( ) ( )∞

∫( ) ( )2

202

,0

2 cos 2

,

1

r r

s r r

r V r

V N

p p V dV

V ε σ

Θ Θ

⎛ ⎞− − Θ⎜ ⎟∞

Θ = Θ∫

∫( )

0200

2 2

2 cos 22sin

2 00

/

1 2 2

1

s r rs r

NNN

r

Ve e dVN

ε σε

πσ

Θ⎜ ⎟∞− Θ ⎝ ⎠

′

= ∫V

h d fi th b l SNR

( ) 22 2 cos / 2sin2 0

1 2

s r rs rV

r

e V e dVγ σγ

πσ∞ ′− − Θ− Θ ′ ′= ∫

Nγ ε=0 2VV

N′ =

where we define the symbol SNR as

87

0 .s s Nγ ε=

Probability of Error for Probability of Error for MM--aryary PSKPSK

becomes )( rf ΘΘ

narrower and more peaked about as the SNR increases

0=Θr

)( rrfΘ

γthe SNR increases. sγ

88

Probability density function ( ) for 1,2,4, and 10.r r Sp γΘ Θ =

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ When s1(t) is transmitted, a decision error is made if the noise causes

the phase to fall outside the range . Hence, the MM r ππ ≤Θ≤−p g ,probability of a symbol error is

( ) r

M

rM dpP ΘΘ−= ∫ Θ1π

r

◊ In general, the integral of doesn’t reduced to a simple form

( ) rM rM pr∫− Θπ

( )rrp ΘΘg g p

and must be evaluated numerically, except for M =2 and M =4.◊ For binary phase modulation, the two signals s1(t) and s2(t) are

r

antipodal. Hence, the error probability is

⎟⎟⎞

⎜⎜⎛

=2QP bε

⎟⎟⎠

⎜⎜⎝

=0

2 NQP

89

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ When M = 4, we have in effect two binary phase-modulation signals

in phase quadrature. Since there is no crosstalk or interference p qbetween the signals on the two quadrature carriers, the bit error probability is identical to that of M = 2.

◊ Then the symbol error probability for M=4 is determined by noting that

( )2

2 211 ⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

−=−= QPP bε

where Pc is the probability of a correct decision for the 2-bit symbol.

( )0

2 11⎥⎥⎦⎢

⎢⎣

⎟⎟⎠

⎜⎜⎝

−=−=N

QPPc

c

◊ There, the symbol error probability for M = 4 is

⎡ ⎤⎛ ⎞ ⎛ ⎞4

0 0

2 211 2 12

b bcP P Q Q

N Nε ε⎡ ⎤⎛ ⎞ ⎛ ⎞

= − = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

90

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ For M > 4, the symbol error probability PM is obtained by

numerically integrating Equation ( )MdpP ΘΘ−= ∫1

πnumerically integrating Equation ( ) rM rM dpP

rΘΘ= ∫− Θ1

π

For large values of M, doubling the

b f hnumber of phases requires an additional 6 dB/bit to achieve the same performance

91

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ An approximation to the error probability for large M and for large

SNR may be obtained by first approximating . ( )ΘΘpy y pp g◊ For , is well approximated

as :

( )ΘΘ rp

0 1 and 0.5s rN πε >> Θ ≤ ( )ΘΘrp

( ) 2γ

◊ Performing the change in variable from ,

( ) rs

reP r

sr

Θ−Θ Θ≈Θ

2sincos γ

πγ

rsr u Θ=Θ sin to νg g ,rsr u ΘΘ sto ν

ΘΘ−≈ ∫−

Θ− dePM

M rrs

Mrs

πγπ

πγcos1 sin 2

Mk log

( )≈ ∫

∫∞ − due

Mu

M

sπ

π

πγ

π

2 sin2

2bs kMk

νν == 2log

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

MkQ

MQ bs

πγπγ

π

sin22sin22

92

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ When a Gray Code is used in the mapping of k-bits symbols into the

corresponding signal phases, two k-bit symbols corresponding to p g g p , y p gadjacent signal phases differ in only a signal bit.

◊ The most probable error result in the erroneous selection of an adjacent phase to the true one.

◊ Most k-bit symbol error contain only a single-bit error. The equivalent bit error probability for M-ary PSK is well approximated as :

Mb PP 1≈

◊ In practice, the carrier phase is extracted from the received signal by performing some nonlinear operation that introduces a phase

Mb k

p g p pambiguity.

93

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ For binary PSK, the signal is often squared in order to remove the

modulation, and the double-frequency component that is generated is , q y p gfiltered and divided by 2 in frequency in order to extract an estimate of the carrier frequency and phase φ.◊ This operation result in a phase ambiguity of 180°in the carrier phase.

◊ For four-phase PSK, the received signal is raised to the fourth power t th di it l d l ti d th lti f th h ito remove the digital modulation, and the resulting fourth harmonic of the carrier frequency is filtered and divided by 4 in order to extract the carrier componentextract the carrier component.◊ These operations yield a carrier frequency component containing φ, but there

are phase ambiguities of ±90° and 180° in the phase estimate.

◊ Consequently, we do not have an absolute estimate of the carrier phase for demodulation.

94

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ The phase ambiguity problem can be overcome by encoding the

information in phase differences between successive signal p ff gtransmissions as opposed to absolute phase encoding.◊ For example, in binary PSK, the information bit 1 may be transmitted

by shifting the phase of carrier by 180° relative to the previous carrier phase. Bit 0 is transmitted by a zero phase shift relative to the phase in the previous signaling intervalthe previous signaling interval.

◊ In four-phase PSK, the relative phase shifts between successive intervals are 0, 90°, 180°, and -90°,corresponding to the information bits 00, 01, 11, and 10, respectively.

◊ The PSK signals resulting from the encoding process are said to be d ff ll d ddifferentially encoded.

95

Probability of Error for Probability of Error for MM--aryary PSKPSK◊ The detector is a relatively simple phase comparator that compares

the phase of the demodulated signal over two consecutive interval to extract the information.

◊ Coherent demodulation of differentially encoded PSK results in a higher probability of error than that derived for absolute phasehigher probability of error than that derived for absolute phase encoding.

◊ With differentially encoded PSK, an error in the demodulated phase y , pof the signal in any given interval will usually result in decoding errors over two consecutive signaling intervals.Th b bilit f i diff ti ll d d M PSK i◊ The probability of error in differentially encoded M-ary PSK is approximately twice the probability of error for M-ary PSK with absolute phase encoding.p g

96

Differential PSK (DPSK) and Its PerformanceDifferential PSK (DPSK) and Its Performance

◊ The received signal of a differentially encoded phase-modulated signal in any given signaling interval is compared to the phase of the received signal from the preceding signaling interval.

◊ We demodulate the differentially encoded signal by multiplying r(t) by cos2πf t and sin2πf t integrating the two products over theby cos2πfct and sin2πfct integrating the two products over the interval T.

◊ At the kth signaling interval, the demodulator output : g g , p

i l l( ) ( )

1 2cos sink s k k s k kn nθ φ θ φε ε⎡ ⎤= − + − +⎣ ⎦r

or equivalently,

h θ i h h l f h i d i l h k h

( )k

jsk ner k += −φθε

where θk is the phase angle of the transmitted signal at the kthsignaling interval, φ is the carrier phase, and nk=nk1+jnk2 is the noise vector.

97

vector.


◊ Similarly, the received signal vector at the output of the demodulator in the preceding signaling interval is :p g g g

◊ The decision variable for the phase detector is the phase difference

( )11

1−

−− += −

kj

sk ner k φθεp p

between these two complex numbers. Equivalently, we can project rk onto rk-1 and use the phase of the resulting complex number :

( ) ( ) ( ) *1

*1

*1

11−

−−−

−−− +++= −−

kkkj

skj

sj

skk nnneneerr kkkk φθφθθθ εεε

which, in the absence of noise, yields the phase differenceθk-θk-1.◊ Differentially encoded PSK signaling that is demodulated and

detected as described above is called differential PSK (DPSK).

98


If the pulse g(t) is rectangular, the matched filter may be replaced by integrate-and-dump filtermay be replaced by integrate and dump filter

Block diagram of DPSK demodulator

99


◊ The error probability performance of a DPSK demodulator and detector ◊ The derivation of the exact value of the probability of error for

M-ary DPSK is extremely difficult, except for M = 2.◊ Without loss of generality, suppose the phase difference θk-θk-1=0. Furthermore, the exponential factor e-j(θk-1-φ)

k k 1 pand ej(θk-φ) can be absorbed into Gaussian noise components nk-1and nk, without changing their statistical properties.

◊ The complication in determining the PDF of the phase is the term

( ) *1

*1

*1 −−− +++= kkkksskk nnnnrr εε

◊ The complication in determining the PDF of the phase is the term nkn*k-1.

100


◊ However, at SNRs of practical interest, the term nkn*k-1 is small relative to the dominant noise term (nk+n*k-1).sε ( k k-1)

◊ We neglect the term nkn*k-1 and normalize rkr*k-1 by dividing through by , the new set of decision metrics becomes : sε

s

s

( )( )*

1

*1

ImRe −

+=++=

kk

kks

nnynnx ε

◊ The variables x and y are uncorrelated Gaussian random variable with identical variances σn

2=N0. The phase is

( )1−kky

n 0 p

◊ The noise variance is now twice as large as in the case of PSKxy

r1tan−=Θ

◊ The noise variance is now twice as large as in the case of PSK. Thus we can conclude that the performance of DPSK is 3 dB poorer than that for PSK.

101


◊ This result is relatively good for M ≥4 , but it is pessimistic for M = 2 that the loss in binary DPSK relative to binary PSK is less than 3 y ydB at large SNR.◊ In binary DPSK, the two possible transmitted phase differences

are 0 and π rad. Consequently, only the real part of rkr*k-1 is need for recovering the information.

1

◊ Because the phase difference the two successive signaling

( ) ( )1**

1*

1 21Re −−− += kkkkkk rrrrrr

◊ Because the phase difference the two successive signaling intervals is zero, an error is made if Re(rkr*k-1)<0.

◊ The probability that rkr*k 1+r*krk 1<0 is a special case of a◊ The probability that rkr k-1 r krk-1 0 is a special case of a derivation, given in Appendix B (Proakis 4th Edition).

102


◊ Appendix B concerned with the probability that a general quadratic form in complex-valued Gaussian random Variable is q pless than zero. According to Equation B-21, we find it depend entirely on the first and second moments of the complex-valued Gaussian random variables rk and rk-1.

◊ We obtain the probability of error for binary DPSK in the form

0

21 N

bbeP ε−=

where εb/N0 is the SNR per bit.

103


◊ The probability of a binary digit error for four-phase DPSK with Gray coding can be express in terms of well-known functions, but y g p ,it’s derivation is quite involved.◊ According to Appendix C, it is expressed in the form :

( ) ( ) ( )⎥⎦⎤

⎢⎣⎡ +−−= 22

01 21exp

21, baabIbaQPb

where Q1(a,b) is the Marcum Q function, I0(x) is the modified Bessel function of order zero and the parameters a and b are

⎦⎣ 22

Bessel function of order zero, and the parameters a and b are defined as

⎞⎛⎞⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

2112 and,

2112 bb ba γγ

104


◊ Because binary DPSK is only slightly inferior to binary PSKslightly inferior to binary PSK at large SNR, and DPSK does not require an elaboratenot require an elaborate method for estimate the carrier phase, it is often used in digital communication system.

Probability of bit error for binary and four-phase

105

obab ty o b t e o o b a y a d ou p asePSK and DPSK

Probability of Error for QAMProbability of Error for QAM◊ Recall that QAM signal waveforms may be expressed as

tftAtftAt 2i)(2)()(

where Amc and Ams are the information-bearing signal amplitudes of

tftAtftgAts cmscmcm ππ 2sin)(2cos)()( −=

the quadrature carriers and g(t) is the signal pulse.

◊ The vector representation of these waveform is

⎥⎦

⎤⎢⎣

⎡= gmsgmcm AA εε

21

21 s

◊ To determine the probability of error for QAM, we must specify the i l i ll i

⎦⎣ 22

signal point constellation.

106

Probability of Error for QAMProbability of Error for QAM◊ QAM signal sets that have M = 4 points.

◊ Figure (a) is a four-phase modulated signal and Figure (b) is with two amplitude levels, labeled A1 and A2, and four phases.p 1 2 p

◊ Because the probability of error is dominated by the minimum distance between pairs of signal points let us impose the condition that Ad e 2)( =between pairs of signal points, let us impose the condition that and we evaluate the average transmitter power, based on the premise that all signal points are equally probable.

Ad 2min =

107

Probability of Error for QAMProbability of Error for QAM◊ For the four-phase signal, we have

( ) 22 2241 AAP ==

◊ For the two-amplitude, four-phase QAM, we place the points on circles of radii A and Th s and

( ) 2244

AAPav ==

A3 Ad e 2)(circles of radii A and . Thus, , and

( )[ ] 222 223221 AAAPav =+=

A3 Ad e 2)(min =

which is the same average power as the M = 4-phase signal constellation.

2

◊ Hence, for all practical purposes, the error rate performance of the two signal sets is the same.

Th i d f h li d QAM i l◊ There is no advantage of the two-amplitude QAM signal set over M = 4-phase modulation.

108

Probability of Error for QAMProbability of Error for QAM◊ QAM signal sets that have M = 8 points.

◊ We consider the four signal constellations :◊ We consider the four signal constellations :

Assuming that the signal points are eq all probablepoints are equally probable, the average transmitted signal power is :g p

( )∑=

+=M

mmsmcav AA

MP

1

221

( )∑=

+=M

mmsmc aa

MA

1

222

The coordinates (Amc,Ams) for each signal point are normalized by A

109

signal point are normalized by A

Probability of Error for QAMProbability of Error for QAM◊ The two sets (a) and (c) contain signal points that fall on a

rectangular grid and have Pav = 6A2.g g av

◊ The signal set (b) requires an average transmitted power Pav= 6.83A2, and (d) requires Pav = 4.73A2.

◊ The fourth signal set (d) requires approximately 1 dB less power than the first two and 1.6 dB less power than the third to achieve the same probability of error.

◊ The fourth signal constellation is known to be the best eight-point QAM t ll ti b it i th l t f iQAM constellation because it requires the least power for a given minimum distance between signal points.

110

Probability of Error for QAMProbability of Error for QAM◊ QAM signal sets for M ≥ 16

◊ For 16-QAM, the signal points at a given Q , g p gamplitude level are phase-rotated by relativeto the signal points at adjacent amplitude levels.

◊ However, the circular 16-QAM constellation is not the best 16-point QAM signal constellation for the AWGN channel.

◊ Rectangular M-ary QAM signal are most frequently used in practice. The reasons are :The reasons are :◊ Rectangular QAM signal constellations have the distinct

advantage of being easily generated as two PAM signals

111

g g y g gimpressed on phase-quadrature carriers.

Probability of Error for QAMProbability of Error for QAM◊ The average transmitted power required to achieve a given

minimum distance is only slightly greater than that of the best y g y gM-ary QAM signal constellation.

◊ For rectangular signal constellations in which M = 2k, where k is even, the QAM signal constellation is equivalent to two PAM signals on quadrature carriers, each having signal points.22kM =

◊ The probability of error for QAM is easily determined from the probability of error for PAM.S ifi ll th b bilit f t d i i f th M◊ Specifically, the probability of a correct decision for the M-aryQAM system is

( )21 PP ( )1 Mc PP −=

PAM.ary an oferror ofy probabilit theis where −Mp M

112

Probability of Error for QAMProbability of Error for QAM◊ By appropriately modifying the probability of error for M-ary PAM,

we obtain ⎞⎛⎞⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎠

⎞⎜⎝

⎛−=

013112

NMQ

MP av

M

ε

where εav/No is the average SNR per symbol. ◊ Therefore, the probability of a symbol error for M-ary QAM is

◊ Note that this result is exact for M = 2k when k is even.

( )211 MM PP −−=

◊ Note that this result is exact for M 2 when k is even.

◊ When k is odd, there is no equivalent -ary PAM system. There is still ok, because it is rather easy to determine the error rate for a

My

rectangular signal set.

113

Probability of Error for QAMProbability of Error for QAM◊ We employ the optimum detector

that bases its decisions on the optimum distance metrics, it is relatively straightforward to show that the symbol error probability isthat the symbol error probability is tightly upper-bounded as

⎤⎡⎟⎞

⎜⎛

23ε

( )⎞⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−−≤

0

3

1

321

k

NMQP av

M

ε

ε

f k 1 h /N i h( ) ⎟

⎟⎠

⎞⎜⎜⎝

⎛

−≤

0134

NMkQ bavε

for any k ≥ 1, whereεbav/N0 is the average SNR per bit.

114

Probability of Error for QAMProbability of Error for QAM◊ For nonrectangular QAM signal constellation, we may upper-bound

the error probability by use of a union bound : p y y

( ) ( )[ ] ⎟⎠⎞

⎜⎝⎛−< 0

2min 21 NdQMP e

M

( )where is the minimum Euclidean distance between signal points.◊ This bound may be loose when M is large.

( )edmin

◊ We approximate PM by replacing M-1 by Mn, where Mn is the largest number of neighboring points that are at distance df t ll ti i t

( )emin

from any constellation point.◊ It is interesting to compare the performance of QAM with that of

PSK for any given signal size MPSK for any given signal size M.

115

Probability of Error for QAMProbability of Error for QAM◊ For M-ary PSK, the probability of a symbol error is approximate as

⎞⎛ π

where γ is the SNR per symbol

⎟⎠⎞

⎜⎝⎛≈

MQP sM

πγ sin22

where γs is the SNR per symbol.◊ For M-ary QAM, the error probability is :

⎞⎛⎞⎛ 31 ε⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎠

⎞⎜⎝

⎛−=

013112

NMQ

MP av

M

ε

◊ We simply compare the arguments of Q function for the two signal formats. The ratio of these two arguments is

( )M 13 ( )( )M

MM π2sin2

13 −=R

116

Probability of Error for QAMProbability of Error for QAM◊ We can find out that when M = 4, we have RM = 1.◊ 4-PSK and 4-QAM yield comparable performance for the same SNR◊ 4 PSK and 4 QAM yield comparable performance for the same SNR

per symbol.◊ For M>4, M-ary QAM yields better performance than M-ary PSK., y Q y p y

117

Comparison of Digital Modulation MethodsComparison of Digital Modulation Methods

◊ One can compare the digital modulation methods on the basis of the SNR required to achieve a specified probability of error.q p p y

◊ However, such a comparison would not be very meaningful, unless it were made on the basis of some constraint, such as a fixed data rate of transmission or, on the basis of a fixed bandwidth.

◊ For multiphase signals, the channel bandwidth required is simply the bandwidth of the equivalent low-pass signal pulse g(t) with duration T and bandwidth W, which is approximately equal to the reciprocal of Tof T.

◊ Since T=k/R=(log2M)/R, it follows that M

RW2log

=2g

118


◊ As M is increased, the channel bandwidth required, when the bit rate R is fixed, decreases. The bandwidth efficiency is measured by the , ff y ybit rate to bandwidth ratio, which is

2logR M=

◊ The bandwidth-efficient method for transmitting PAM is single-sideband The channel bandwidth required to transmit the signal is

2gW

sideband. The channel bandwidth required to transmit the signal is approximately equal to 1/2T and,

2 logR M=

this is a factor of 2 better than PSK.

22 log MW

=

◊ For QAM, we have two orthogonal carriers, with each carrier having a PAM signal.

119


◊ Thus, we double the rate relative to PAM. However, the QAM signal must be transmitted via double-sideband. Consequently, QAM and q y, QPAM have the same bandwidth efficiency when the bandwidth is referenced to the band-pass signal.

◊ As for orthogonal signals, if the M = 2k orthogonal signals are constructed by means of orthogonal carriers with minimum f ti f 1/2T th b d idth i d ffrequency separation of 1/2T, the bandwidth required for transmission of k = log2M information bits is

MMM

I h h b d id h i M i( ) R

MM

RkM

TMW

2log222===

In the case, the bandwidth increases as M increases.◊ In the case of biorthogonal signals, the required bandwidth is one-

h lf f th t f th l i l

120

half of that for orthogonal signals.


◊ A compact and meaningful comparison of modulation methods pis one based on the normalized data rate R/W (bits per second per hertz of bandwidth) versus the SNR per bit (εb/N0 ) required to achieve a given error probabilitygiven error probability.

◊ In the case of PAM, QAM, and PSK increasing M results in aPSK, increasing M results in a higher bit-to-bandwidth ratio R/W.

121


◊ However, the cost of achieving the higher data rate is an increase in the SNR per bit.p

◊ Consequently, these modulation methods are appropriate for communication channels that are bandwidth limited, where we desire a R/W >1 and where there is sufficiently high SNR to support increases in M.◊ Telephone channels and digital microwave ratio channels are examples

of such band-limited channels.◊ In contrast M ary orthogonal signals yield a R/W ≤ 1 As M◊ In contrast, M-ary orthogonal signals yield a R/W ≤ 1. As M

increases, R/W decreases due to an increase in the required channel bandwidth.

◊ The SNR per bit required to achieve a given error probability decreases as M increases.

122


◊ 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→∞, the error probability can be made as small as desired, provided that SNR>0 693 ( 1 6dB) This is the minimum SNR perprovided that SNR>0.693 (-1.6dB). This is the minimum SNR per bit required to achieve reliable transmission in the limit as the channel bandwidth W→∞ and the corresponding R/W→0.

◊ The figure above also shown the normalized capacity of the band-limited AWGN channel, which is due to Shannon (1948).Th ti C/W h C ( R) i th it i bit / t th◊ The ratio C/W, where C (=R) is the capacity in bits/s, represents the highest achievable bit rate-to-bandwidth ratio on this channel.

◊ Hence it serves the upper bound on the bandwidth efficiency of any◊ Hence, it serves the upper bound on the bandwidth efficiency of any type of modulation.

123

Optimum Receiver for Signals Optimum Receiver for Signals with Random Phase in AWGN Channelwith Random Phase in AWGN Channel

◊ In this section, we consider the design of the optimum receiver for carrier modulated signals when the carrier phase is unknown and no g pattempt is made to estimate its value.

◊ Uncertainty in the carrier phase of the receiver signal may be due to one or more of the following reasons:◊ The oscillators that are used at the transmitter and the receiver to generate the

carrier signals are generally not phase synchronouscarrier signals are generally not phase synchronous.◊ The time delay in the propagation of the signal from the transmitter to the

receiver is not generally known precisely.

◊ Assuming a transmitted signal of the form( ) ( )[ ]tfj cetgts π2Re=

that propagates through a channel with delay t0 will be received as:( ) ( )[ ]g

( ) ( ) ( ) ( )2 2 2R Rj f t t j f t j fπ π π−⎡ ⎤ ⎡ ⎤

124

( ) ( ) ( ) ( )0 02 2 20 0 0Re Rec c cj f t t j f t j fs t t g t t e g t t e eπ π π− −⎡ ⎤ ⎡ ⎤− = − = −⎣ ⎦⎣ ⎦

Optimum Receiver for Signals Optimum Receiver for Signals with Random Phase in AWGN Channelwith Random Phase in AWGN Channel

◊ The carrier phase shift due to the propagation delay t0 is 02 tfcπφ −=

◊ Note that large changes in the carrier phase can occur due to relatively small changes in the propagation delay.

0fcφ

◊ For example, if the carrier frequency fc=1 MHz, an uncertainty or a change in the propagation delay of 0.5μs will cause a phase uncertainty of π raduncertainty of π rad.

◊ In some channels the time delay in the propagation of the signal from the transmitter to the receiver may change rapidly and in an apparently random fashion.

◊ In the absence of the knowledge of the carrier phase, we may treat this signal parameter as a random variable and determine the form ofthis signal parameter as a random variable and determine the form of the optimum receiver for recovering the transmitted information from the received signal.

125

Optimum Receiver for Binary SignalsOptimum Receiver for Binary Signals◊ We consider a binary communication system that uses the two

carrier modulated signal s1(t) and s2(t) to transmit the information, g 1( ) 2( ) ,where:

( ) ( ) 2Re , 1, 2, 0cj f tm lms t s t e m t Tπ⎡ ⎤= = ≤ ≤⎣ ⎦

and slm(t), m=1,2 are the equivalent low-pass signals.◊ The two signals are assumed to have equal energy g q gy

( ) ( )2

2 1T T

m lms t dt s t dtε = =∫ ∫

◊ The two signals are characterized by the complex-valued correlation

( ) ( )0 02m lm∫ ∫

g y pcoefficient

( ) ( )∫=≡T

ll dttsts0 2

*112 2

1ε

ρρ

126

2ε

Optimum Receiver for Binary SignalsOptimum Receiver for Binary Signals◊ The received signal is assumed to be a phase-shifted version of the

transmitted signal and corrupted by the additive noise g p y

( ) ( ) ( ){ } ( )2 2Re Rec cj f t j f tc sn t n t jn t e z t eπ π⎡ ⎤= + ⋅ = ⋅⎡ ⎤⎣ ⎦ ⎣ ⎦

◊ The received signal may be expressed as

( ) ( ) ( ){ }2Re cj f tjr t s t e z t e πφ⎡ ⎤= +⎣ ⎦

where ( ) ( ) ( ) 0jl lr t s t e z t t Tφ= + ≤ ≤

( ) ( ) ( ){ }Re clmr t s t e z t e⎡ ⎤= + ⋅⎣ ⎦

whererl(t) is the equivalent low-pass received signal.

◊ This received signal is now passed through a demodulator whose

( ) ( ) ( ) , 0l lmr t s t e z t t T+ ≤ ≤

◊ This received signal is now passed through a demodulator whose sampled output at t =T is passed to the detector.

127

Optimum Receiver for Binary SignalsOptimum Receiver for Binary Signals◊ The optimum demodulator

◊ In section 5.1.1, we demonstrated that if the received signals was◊ In section 5.1.1, we demonstrated that if the received signals was correlated with a set of orthogonal functions {fn(t)} that spanned the signal space, the outputs from the bank of correlators provide

t f ffi i t t ti ti f th d t t t k d i ia set of sufficient statistics for the detector to make a decision that minimizes the probability error.

◊ We also demonstrated that a bank of matched filters could be◊ We also demonstrated that a bank of matched filters could be substituted for the bank of correlations.

◊ A similar orthogonal decomposition can be employed for a received signal with an unknown carrier phase.

◊ It is mathematically convenient to deal with the equivalent low-pass signal and to specify the signal correlators or matched filterspass signal and to specify the signal correlators or matched filters in terms of the equivalent low-pass signal waveforms.

128

Optimum Receiver for Binary SignalsOptimum Receiver for Binary Signals

◊ The optimum demodulator◊ The impulse response h (t) of a filter that is matched to the◊ The impulse response hl(t) of a filter that is matched to the

complex-valued equivalent low-pass signal sl(t), 0≤t≤T is given as hl(t)=sl

*(T-t) and the output of such filter at t=T is simplyl( ) l ( ) p p ywhere ε is the signal energy.

◊ A similar result is obtained if the signal sl(t) is correlated with( ) ε2

2

0=∫ dtts

T

l

◊ A similar result is obtained if the signal sl(t) is correlated with sl

*(t) and the correlator is sampled signal sl(t) at t=T.◊ The optimum demodulator for the equivalent low-pass received p q p

signal sl(t) may be realized by two matcher filters in parallel, one matched to sl1(t) and the other to sl2(t).

129


◊ The optimum demodulator◊ Optimum receiver for binary signals◊ Optimum receiver for binary signals

◊ The output of the matched filters or correlators at the sampling instant are the two complex numbers

2,1 , =+= mjrrr msmcm

130


◊ The optimum demodulator◊ Suppose that the transmitted signal is s (t) It can be shown that◊ Suppose that the transmitted signal is s1(t). It can be shown that

( )( ) ( )[ ]

sc

jnjnr 111

i22.sin2cos2 +++=

φφφεφε

where ρ is the complex-valued correlation coefficient of two i l ( ) d ( ) hi h b d

( ) ( )[ ]sc njnr 20202 sin2cos2 +−++−= αφρεαφρε

signals sl1(t) and sl2(t), which may be expressed as ρ=|ρ|exp(jα0).The random noise variables n n n and n are jointly◊ The random noise variables n1c, n1s, n2c, and n2s are jointly Gaussian, with zero-mean and equal variance.

131


◊ The optimum detector◊ The optimum detector observes the random variables [r r r◊ The optimum detector observes the random variables [r1c r1s r2c

r2s]=r, where r1=r1c+jr1s, and r2=r2c+jr2s, and bases its decision on the posterior probabilities P(sm|r), m=1,2.p p ( m| ), ,

◊ These probabilities may be expressed as

( ) ( )| P ssr( ) ( ) ( )( ) 2,1 ,|| == m

pPpP mm

m rssrrs

Likelihood ratio.

◊ The optimum decision rule may be expressed as

( ) ( ) ( ) ( )1 11 2|s sp P> >r s s( ) ( ) ( )

( )( )( )

2 2

1 21 2

2 1

|| | or

|s s

p PP P

p P> >< <

r s ss r s r

r s s

132


◊ The optimum detector◊ The ratio of PDFs on the left hand side is the likelihood ratio◊ The ratio of PDFs on the left-hand side is the likelihood ratio,

which we denote as1( | )( ) p

Λ =r sr

◊ The right-hand side is the ratio of the two prior probabilities,2

( )( | )p

Λ =rr s

◊ The right hand side is the ratio of the two prior probabilities, which takes the value of unity when the two signals are equally probable.

◊ The probability density functions p(r|s1) and p(r|s2) can be obtained by averaging the PDFs p(r|sm,φ) over the PDF of the

d i h irandom carrier phase, i.e.,2

( | ) ( | , ) ( )p p p dπ

φ φ φ= ∫r s r s

133

0( | ) ( | , ) ( )m mp p p dφ φ φ∫r s r s


◊ The optimum detector◊ For the special case in which the two signals are orthogonal i e◊ For the special case in which the two signals are orthogonal, i.e., ρ=0, the outputs of the demodulator are:

r r jr= +1 1 1

1 1 2 cos (2 sin )c s

c s

r r jrn j nε φ ε φ

= +

= + + +

2 2 2

2 2 c s

c s

r r jrn jn

= +

= +

where (n1c, n1s, n2c, n2s) are mutually uncorrelated and, statistically independent zero-mean Gaussian random variable

2 2c sj

statistically independent, zero mean Gaussian random variable.

134


◊ The optimum detector◊ The joint PDF of r=[r r r r ] may be expressed as a product◊ The joint PDF of r=[r1c r1s r2c r2s] may be expressed as a product

of the marginal PDFs.◊ Consequently◊ Consequently

2 21 1

1 1 1 2 2

( 2 cos ) ( 2 sin )1( , | , ) exp2 2

c sc s

r rp r r ε φ ε φφπσ σ

⎡ ⎤− + −= −⎢ ⎥

⎣ ⎦s

2 22 2

2 2 1 2 2

1( , | , ) exp2 2

c sc s

r rp r r φπσ σ

⎣ ⎦⎛ ⎞+

= −⎜ ⎟⎝ ⎠

s

where σ2=2εN0.◊ The uniform PDF for the carrier phase φ (p(φ)=1/2π) represents

2 2πσ σ⎝ ⎠

◊ The uniform PDF for the carrier phase φ (p(φ)=1/2π) represents the most ignorance that can be exhibited by the detector.

◊ This is called the least favorable PDF for φ

135

◊ This is called the least favorable PDF for φ.


◊ The optimum detector◊ With p(φ)=1/2π 0≤φ≤2π substituted into the integral in p(r|s )◊ With p(φ)=1/2π, 0≤φ≤2π, substituted into the integral in p(r|sm),

we obtain:

( )21 π

∫ ( )

( )

2

1 1 10

2 2 2 2 1 11 1

1 , | ,2

2 cos sin41 1

c s

c sc s

p r r d

r rr r d

π

π

φ φπ

ε φ φε φ+⎡ ⎤⎛ ⎞+ +

⎜ ⎟ ⎢ ⎥

∫

∫

s

( )1 11 12 2 20

exp exp2 2 2

c sc s dφ φ

φπσ σ π σ

⎛ ⎞= −⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦∫

⎛ ⎞( ) 2 22 1 11 1

02 20

22 cos sin1 exp2

c sc s r rr rd I

π εε φ φφ

π σ σ

⎛ ⎞++⎡ ⎤⎜ ⎟=⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎝ ⎠

∫

where I0(x) is the modified Bessel function of zeroth order.

⎝ ⎠

136


◊ The optimum detector◊ By performing a similar integration under the assumption that the◊ By performing a similar integration under the assumption that the

signal s2(t) was transmitted, we obtain the result 2 22 2 2 241 r rr r εε ⎛ ⎞+⎛ ⎞+ + ⎜ ⎟

◊ When we substitute these results into the likelihood ratio given

( ) 2 22 22 2 2 02 2 2

241, | exp2 2

c sc sc s

r rr rp r r Iεε

πσ σ σ

⎛ ⎞+⎛ ⎞+ + ⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠s

◊ When we substitute these results into the likelihood ratio given by equation Λ(r), we obtain the result

( )( ) ( ) ( )

( ) ( )( )( )

( )( )

1

2

2 2 20 1 11 1 1 2 2 1 21

2 2 22 1 1 2 2 2 2 10 2 2

2| |( | )( | ) | | 2

sc sc s c s

c s c s sc s

I r rP r r s P r r s Ppp P r r s P r r s PI r r

ε σ

ε σ

+Λ = = =

+><

sr srr s s

◊ The optimum detector computes the two envelopes and and the corresponding values of the Bessel function

( ) 20 2 2c s

21

21 sc rr + 2

22

2 sc rr +

( )222 /2I +

137

and the corresponding values of the Bessel function and to form the likelihood ratio.

( )221

210

/2 σε sc rrI +( )222

220

/2 σε sc rrI +

Optimum Receiver for Binary SignalsOptimum Receiver for Binary Signals◊ The optimum detector

◊ we observe that this computation requires knowledge of the◊ we observe that this computation requires knowledge of the noise variance σ2.

◊ The likelihood ratio is then compared with the threshold pP(s2)/P(s1) to determine which signal was transmitted.

◊ A significant simplification in the g pimplementation of the optimum detector occurs when the two signals are equally probable. In such a case the threshold becomes unity, and, due to the monotonicity of Bessel function shownmonotonicity of Bessel function shown in figure, the optimum detection rule simplifies to

12 2 2 2

s

r r r r>+ +<p

138

2

1 1 2 2c s c ss

r r r r+ +<


◊ The optimum detector◊ The optimum detector bases its decision on the two envelopes◊ The optimum detector bases its decision on the two envelopes

and , and , it is called an envelope detector.◊ We observe that the computation of the envelopes of the received

21

21 sc rr + 2

22

2 sc rr +

◊ We observe that the computation of the envelopes of the received signal samples at the output of the demodulator renders the carrier phase irrelevant in the decision as to which signal was transmitted.

◊ Equivalently, the decision may be based on the computation of the squared envelope r1c

2+r1s2 and r2c

2+r2s2, in which case the

detector is call a square-law detector.

139


◊ Detection of binary FSK signal◊ Recall that in binary FSK we employ two different frequencies◊ Recall that in binary FSK we employ two different frequencies,

say f1 and f2=f1+Δf, to transmit a binary information sequence.◊ The chose of minimum frequency separation Δf = f2 - f1 is◊ The chose of minimum frequency separation Δf f2 f1 is

considered below:( ) ( )1 1 2 22 cos 2 , 2 cos 2 , 0b b b b bs t T f t s t T f t t Tε π ε π= = ≤ ≤

◊ The equivalent low-pass counterparts are:( ) ( ) 22 2 0j fts t T s t T e t Tπε ε Δ ≤ ≤

( ) ( )1 1 2 2, ,b b b b bf f

◊ The received signal may be expressed as:

( ) ( )1 22 , 2 , 0j fl b b l b b bs t T s t T e t Tε ε= = ≤ ≤

( ) ( ) ( )2 cos 2 , 0,1bm m

b

r t f t n t mTε π φ= + + =

140


◊ Detection of binary FSK signal◊ Demodulation and square law detection:◊ Demodulation and square-law detection:

( ) ( )2 2 2 0 1f t f k f t kΔ⎡ ⎤⎣ ⎦( ) ( )

( ) ( )

1

1

cos 2 2 , 0,1

2 sin 2 2 , 0,1

kcb

k

f t f k f t kT

f t f k f t k

π π

π π

= + Δ =⎡ ⎤⎣ ⎦

= + Δ =⎡ ⎤⎣ ⎦

141

( ) ( )1sin 2 2 , 0,1ksb

f t f k f t kT

π π+ Δ⎡ ⎤⎣ ⎦


◊ Detection of binary FSK signal◊ If the mth signal is transmitted the four samples at the detector may be

( ) ( ) ( ) ( ) ( )2 2cos 2 cos 2 2b bT T br r t f t dt f m f t n t f k f t dtε π φ π π⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪+ Δ + + + Δ⎡ ⎤ ⎡ ⎤⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦∫ ∫

◊ If the mth signal is transmitted, the four samples at the detector may be expressed as:

( ) ( ) ( ) ( ) ( )

( ) ( ){ } ( ){ }

1 10 0

1 1 10

cos 2 cos 2 2

2cos 2 cos sin 2 sin cos 2b

bkc kc m

b b

Tbm m kc

r r t f t dt f m f t n t f k f t dtT T

f m f t f m f t f k f t dt nT

π φ π π

επ φ π φ π

= = + Δ + + + Δ⎡ ⎤ ⎡ ⎤⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

= + Δ − + Δ + Δ +⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∫ ∫

∫

i i

i( ) ( ){ } ( ){ }

( )

1 1 10

1cos 2 cos 2 2

m m kcb

b

b

f f f f f fT

m k ft f mT

φ φ

επ π

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= − Δ + +⎡ ⎤⎣ ⎦

∫

( )( ){ }0cosbT

mk f t φ⎡ ⎤+ Δ −⎣ ⎦∫b

( ) ( )( ){ }( ) ( )

1 sin 2 sin 2 2 sin

sin 2 cos 2 b

m kc

T

b

m k ft f m k f t dt n

m k ft m k ft

π π φ

π πε

⎡ ⎤− Δ − + + Δ +⎡ ⎤⎣ ⎦ ⎣ ⎦

⎧ ⎫− Δ − Δ⎡ ⎤ ⎡ ⎤⎪ ⎪⎣ ⎦ ⎣ ⎦( )( )

( )( )

( ) ( )0

cos sin2 2

sin 2 cos 2cos

bm m kc

b

b

f fn

T m k f m k f

m k fT m k

εφ φ

π π

π πε φ

⎡ ⎤ ⎡ ⎤⎪ ⎪⎣ ⎦ ⎣ ⎦= + +⎨ ⎬− Δ − Δ⎪ ⎪⎩ ⎭

− Δ − Δ⎡ ⎤⎣ ⎦ +1

sin 0 1bfTn k mφ

⎧ ⎫−⎡ ⎤⎪ ⎪⎣ ⎦ +⎨ ⎬( )cos

2b mbm k fT

ε φπ⎣ ⎦= +

− Δ ( )sin , 0,1

2 m kcb

n k mm k fT

φπ

⎣ ⎦ + =⎨ ⎬− Δ⎪ ⎪⎩ ⎭

142


◊ Detection of binary FSK signal◊ If the mth signal is transmitted the four samples at the detector may be

( ) ( ) ( ) ( ) ( )2 2cos 2 sin 2 2b bT T br r t f t dt f m f t n t f k f t dtε π φ π π⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪+ Δ + + + Δ⎡ ⎤ ⎡ ⎤⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦∫ ∫

◊ If the mth signal is transmitted, the four samples at the detector may be expressed as (cont.):

( ) ( ) ( ) ( ) ( )

( ) ( ){ } ( ){ }

1 10 0

1 1 10

cos 2 sin 2 2

2cos 2 cos sin 2 sin sin 2b

bks ks m

b b

Tbm m ks

r r t f t dt f m f t n t f k f t dtT T

f m f t f m f t f k f t dt nT

π φ π π

επ φ π φ π

= = + Δ + + + Δ⎡ ⎤ ⎡ ⎤⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

= + Δ − + Δ + Δ +⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∫ ∫

∫

i i

i( ) ( ){ } ( ){ }

( )

1 1 10

1sin 2 sin 2 2

m m ksb

b

b

f f f f f fT

m k ft fT

φ φ

επ π

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= − − Δ + +⎡ ⎤⎣ ⎦

∫

( )( ){ }0cosbT

mm k f t φ⎡ ⎤+ Δ −⎣ ⎦∫b

( ) ( )( ){ }( ) ( )

1 cos 2 cos 2 2 sin

cos 2 sin 2 b

m ks

T

b

m k ft f m k f t dt n

m k ft m k ft

π π φ

π πε

⎡ ⎤− Δ − + + Δ +⎡ ⎤⎣ ⎦ ⎣ ⎦

⎧ ⎫− Δ − Δ⎡ ⎤ ⎡ ⎤⎪ ⎪⎣ ⎦ ⎣ ⎦⎨ ⎬

( )( )

( )( )

( )0

cos sin2 2

cos 2 1 sin 2cos

bm m ks

b

b

f fn

T m k f m k f

m k fT m

εφ φ

π π

π πε φ

⎡ ⎤ ⎡ ⎤⎪ ⎪⎣ ⎦ ⎣ ⎦= − +⎨ ⎬− Δ − Δ⎪ ⎪⎩ ⎭

− Δ −⎡ ⎤⎣ ⎦ ( )sin 0 1bk fT

n k mφ⎧ ⎫− Δ⎡ ⎤⎪ ⎪⎣ ⎦ +⎨ ⎬( )

cos2b m

bm k fTε φ

π⎣ ⎦= −

− Δ ( )sin , 0,1

2 m ksb

n k mm k fT

φπ⎣ ⎦ + =⎨ ⎬− Δ⎪ ⎪⎩ ⎭

143


◊ Detection of binary FSK signal◊ We observe that when k=m the sampled values to the detector◊ We observe that when k=m, the sampled values to the detector

arecosmc b m mcr nε φ= +

◊ We observe that when k≠m the signal components in the samples

sinms b m msr nε φ= − + k=m

◊ We observe that when k≠m, the signal components in the samples rkc and rks will vanish, independently of the values of the phase shifts φk, provided that the frequency separation between φk p q y psuccessive frequency is Δf = 1/T.

◊ In such case, the other two correlator outputs consist of noise only, i.e.,

rkc=nkc, rks=nks, k≠m.

144

Optimum Receiver for Optimum Receiver for MM--aryary Orthogonal SignalsOrthogonal Signals

◊ If the equal energy and equally probable signal waveforms are represented as

where slm(t) are the equivalent low-pass signals.( ) ( ) 2Re s , 1, 2,..., , 0cj f t

m lms t t e m M t Tπ⎡ ⎤= = ≤ ≤⎣ ⎦

lm( ) q p g◊ The optimum correlation-type or matched-filter-type demodulator

produces the M complex-valued random variables( ) ( )*T

∫ ( ) ( )

( ) ( )( ) ( ) ( )

*

0

* * 1 2

T

m mc ms l lm

T T

l l l l

r r jr r t h T t dt

r t s T T t dt r t s t dt m M

= + = −

= − − = =

∫∫ ∫

where rl(t) is the equivalent low-pass signals.◊ The optimum detector based on a random uniformly distributed

( ) ( )( ) ( ) ( )0 0

, 1, 2,...,l lm l lmr t s T T t dt r t s t dt m M∫ ∫

◊ The optimum detector, based on a random, uniformly distributed carrier phase, computes the M envelopes

2 2 , 1, 2,...,m mc msr r r m M= + =

145

or, the squared envelopes |rm|2, and selects the signal with the largest envelope.

Optimum Receiver for Optimum Receiver for MM--aryary Orthogonal SignalsOrthogonal Signals

◊ Optimum receiver for M-ary orthogonal FSK a y o t ogo a Ssignals.◊ There are 2M

correlators: two for each possible transmitted frequencytransmitted frequency.

◊ The minimum frequency separation q y pbetween adjacent frequencies to maintain orthogonalitymaintain orthogonalityis Δf = 1/T.

146

Demodulation of M-ary signals for noncoherent detection

Probability of Error for Envelope Detection Probability of Error for Envelope Detection of of MM--aryary Orthogonal SignalsOrthogonal Signalsyy g gg g

◊ We assume that the M signals are equally probable a priori and that the signal s1(t) is transmitted in the signal interval 0≦t≦T.g 1( ) g

◊ The M decision metrics at the detector are the M envelopes2 2 1 2 M

where

2 2 , 1, 2,...,m mc msr r r m M= + =

1 1 1cosr nε φ= +where

d

1 1 1

1 1 1

cos

sinc s c

s s s

r n

r n

ε φ

ε φ

+

= +2 3 Mand

◊ The additive noise components {nmc} and {nms} are mutually statistically independent zero mean Gaussian variables with equal

, , 2,3,...,mc mc ms msr n r n m M= = =

statistically independent zero-mean Gaussian variables with equal variance σ2=N0/2.

147

Probability of Error for Envelope Detection Probability of Error for Envelope Detection of of MM--aryary Orthogonal SignalsOrthogonal Signals

◊ The PDFs of the random variables at the input to the detector are

( )2 22 2 ⎛ ⎞⎛ ⎞

yy g gg g

( )( )

1

2 22 21 11 1

1 1 02 2 2

1, exp2 2

s c sc s sc s

r rr rp r r Iεε

πσ σ σ

⎛ ⎞+⎛ ⎞+ + ⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠r

( )2 2

2 2

1, exp , 2,3,...,2 2m

mc msmc ms

r rp r r m Mπσ σ

⎝ ⎠⎛ ⎞+

= − =⎜ ⎟⎝ ⎠

r

◊ We define the normalized variables2 2m πσ σ⎝ ⎠

2 2mc ms

m

r rR

σ+

=

1tan msm

mc

rr

−Θ =

148


◊ Clearly, rmc=σRmcosΘm and rms =σRmsinΘm. The Jacobian of this transformation is

yy g gg g

2cos sinJ

sin cosm m

mm m m m

RR R

σ σσ

σ σΘ Θ

= =− Θ Θ

◊ Consequently,21

1 1 1 0 121( , ) exp 2

2 2s sRp R R I R

N Nε ε⎡ ⎤ ⎛ ⎞⎛ ⎞

Θ = − + ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠1 1 1 0 1

0 0

( , ) p2 2

pN Nπ

⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎣ ⎦ ⎝ ⎠

( ) 21, exp , 2,3,...,2 2

mm m m

Rp R R m M⎛ ⎞Θ = − =⎜ ⎟⎝ ⎠

◊ Finally, by averaging p(Rm,Θm) over Θm, the factor of 2π is eliminated.

( ), p , , , ,2 2m m mpπ ⎜ ⎟

⎝ ⎠

eliminated.◊ Thus, we find that R1 has a Rice probability distribution and Rm,

m=2,3,…,M, are each Rayleigh-distribued.

149


◊ The probability of a correct decision is simply the probability that R1>R2, and R1>R3,…,and R1>Rm. Hence,

yy g gg g

1 2, 1 3, , 1 m ,

( )2 1 3 1 1, , ,c MP P R R R R R R∞

= < < <

∫

…

◊ Because the random variables Rm, m=2,3,…,M, are statistically

( ) ( )12 1 3 1 1 10

, , , |M RP R R R R R R R x p x dx∞

= < < < =∫ …

◊ Because the random variables Rm, m 2,3, ,M, are statistically independent and identically distributed, the joint probability conditioned on R1 factors into a product of M-1 identical terms.

where (*)( ) ( )1

1

2 1 10|

M

c RP P R R R x p x dx−∞

= < =⎡ ⎤⎣ ⎦∫0

( ) ( ) 2

2

/22 1 1 2 20

| 1x x

RP R R R x p r dr e−< = = = −∫

150

( ) ( )22 1 1 2 20

| Rp∫


◊ The (M-1)th power may be expressed as

( ) ( )2 211

/ 2 21MM n M−− −⎛ ⎞∑

yy g gg g

◊ Substitution of this result into Equation (*) in Page 150 and

( ) ( )2 2/ 2 2

0

11 1 nx nx

n

Me e

n− −

=

⎛ ⎞− = − ⎜ ⎟

⎝ ⎠∑

◊ Substitution of this result into Equation ( ) in Page 150 and integration over x yield the probability of a correct decision as

( )1 1 1M

n M nε− ⎡ ⎤−⎛ ⎞⎢ ⎥∑

where ε /N0 is the SNR per symbol

( ) ( )0 0

11 exp1 1

n sc

n

nPn n n N

ε=

⎡ ⎤⎛ ⎞= − ⎢ ⎥⎜ ⎟ + +⎝ ⎠ ⎣ ⎦

∑where εs/N0 is the SNR per symbol.

◊ The probability of a symbol error, which is Pm=1-Pc, becomes1 1 1M M nkε− ⎡ ⎤⎛ ⎞

h /N i th SNR bit

( ) ( )1

1

1 0

1 11 exp1 1

Mn b

Mn

M nkPn n n N

ε+

=

⎡ ⎤−⎛ ⎞= − −⎢ ⎥⎜ ⎟ + +⎝ ⎠ ⎣ ⎦

∑

151

where εb/N0 is the SNR per bit.


◊ For binary orthogonal signals (M=2), Equation reduces to the simple form

yy g gg g

q p

◊ For M>2, we may compute the 021

2 2b NP e ε−=

, y pprobability of a bit error by making use of the relationship

122 1

k

b MkP P−

=

◊ Figure shows the bit-error probability

2 1k −

as a function of SNR per bit γb for M=2,4,8,16, and 32.

152


◊ Just as in the case of coherent detection of M-ary orthogonal signals, we observe that for any given bit-error probability, the SNR per bit

yy g gg g

y g p y, pdecreases as M increase.

◊ It can be shown that, in the limit as M→∞, the probability of bit error Pb can be made arbitrarily small provided that the SNR per bit is greater than the Shannon limit of -1.6dB.

◊ The cost for increasing M is the bandwidth required to transmit the signals. F M FSK th f ti b t dj t◊ For M-ary FSK, the frequency separation between adjacent frequencies is Δf=1/T for signal orthogonality.

◊ The bandwidth required for the M signals is W=M Δf=M/T◊ The bandwidth required for the M signals is W=M Δf=M/T.

153


◊ The bit rate is R=k/T, where k=log2M.◊ Therefore the bit rate-to-bandwidth ratio is

yy g gg g

◊ Therefore, the bit rate to bandwidth ratio is

2log MR=

W M=

154

optimum optimum receivers for the additive...

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