EE456 – Digital Communicationsengr.usask.ca/classes/EE/456/slides/Chapter5-Slides-F15.pdf · EE456 – Digital Communications 6. Chapter 5: Optimum Receiver for Binary Data Transmission

Chapter 5: Optimum Receiver for Binary Data Transmission

EE456 – Digital CommunicationsProfessor Ha Nguyen

September 2015

EE456 – Digital Communications 1


Block Diagram of Binary Communication Systems

��

��

( )tm { }kb 10 ( )k s t= ↔b

21 ( )k s t= ↔b

ˆ ( )tm { }ˆkb ( )tr

( )twBits in two different time slots are statistically independent.

a priori probabilities: P [bk = 0] = P1, P [bk = 1] = P2.

Signals s1(t) and s2(t) have a duration of Tb seconds and finite energies:

E1 =∫ Tb0

s21(t)dt, E2 =∫ Tb0

s22(t)dt.

Noise w(t) is stationary Gaussian, zero-mean white noise with two-sided powerspectral density of N0/2 (watts/Hz):

E{w(t)} = 0, E{w(t)w(t + τ)} =N0

2δ(τ), w(t) ∼ N

(

0,N0

2

)

.



�� !" �#$!!�%&��%$'��()�$!*+ ''��,

-�+��%$'��(.�� /��,.��!*'��' �!

( )tm { }kb 10 ( )k s t= ↔b

21 ( )k s t= ↔b

ˆ ( )tm { }ˆkb ( )tr

( )tw

Received signal over [(k − 1)Tb, kTb]:

r(t) = si(t− (k − 1)Tb) +w(t), (k − 1)Tb ≤ t ≤ kTb.

Objective is to design a receiver (or demodulator) such that the probability ofmaking an error is minimized.

Shall reduce the problem from the observation of a time waveform to that ofobserving a set of numbers (which are random variables).



Can you Identify the Signal Sets {s1(t), s2(t)}?

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−1

0

1

t/Tb

NRZ(T

x)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−5

0

5

t/Tb

NRZ(R

x)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−1

0

1

t/Tb

OOK

(Tx)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−5

0

5

t/Tb

OOK

(Rx)



Geometric Representation of Signals s1(t) and s2(t) (I)

Wish to represent two arbitrary signals s1(t) and s2(t) as linear combinations oftwo orthonormal basis functions φ1(t) and φ2(t).

φ1(t) and φ2(t) form a set of orthonormal basis functions if and only if:φ1(t) and φ2(t) are orthonormal if:

∫ Tb

0φ1(t)φ2(t)dt = 0 (orthogonality),

∫ Tb

0φ21(t)dt =

∫ Tb

0φ22(t)dt = 1 (normalized to have unit energy).

If {φ1(t), φ2(t)} can be found, the representations are

s1(t) = s11φ1(t) + s12φ2(t),

s2(t) = s21φ1(t) + s22φ2(t).

It can be checked that the coefficients sij can be calculated as follows:

sij =

∫ Tb

0si(t)φj (t)dt, i, j ∈ {1, 2},

An important question is: Given the signal set s1(t) and s2(t), can one alwaysfind an orthonormal basis functions to represent {s1(t), s2(t)} exactly? If theanswer is YES, is the set of orthonormal basis functions UNIQUE?



Geometric Representation of Signals s1(t) and s2(t) (II)

Provided that φ1(t) and φ2(t) can be found, the signals (which are waveforms) can berepresented as vectors in a vector space (or signal space) spanned (i.e., defined) by theorthonormal basis set {φ1(t), φ2(t)}.

)(1 tφ

)(1 ts

)(2 ts)(2 tφ

0 11s

21s

12s

22s

12 21d d=

2E

1E

s1(t) = s11φ1(t) + s12φ2(t),

s2(t) = s21φ1(t) + s22φ2(t),

sij =

∫ Tb

0si(t)φj(t)dt, i, j ∈ {1, 2},

Ei =

∫ Tb

0s2i (t)dt = s2i1 + s2i2, i ∈ {1, 2},

d12 = d21 =

√∫ Tb

0[s2(t) − s1(t)]2dt

∫ Tb0

si(t)φj (t)dt is the projection of signal si(t) onto basis function φj(t).

The length of a signal vector equals to the square root of its energy.

It is always possible to find orthonormal basis functions φ1(t) and φ2(t) torepresent s1(t) and s2(t) exactly. In fact, there are infinite number of choices!



Gram-Schmidt Procedure

Gram-Schmidt (G-S) procedure is one method to find a set of orthonormal basisfunctions for a given arbitrary set of waveforms.

1 Let φ1(t) ≡ s1(t)√

E1

. Note that s11 =√E1 and s12 = 0.

2 Project s′

2(t) =s2(t)√

E2

onto φ1(t) to obtain the correlation coefficient:

ρ =

∫ Tb

0

s2(t)√E2

φ1(t)dt =1√

E1E2

∫ Tb

0s1(t)s2(t)dt.

3 Subtract ρφ1(t) from s′

2(t) to obtain φ′

2(t) =s2(t)√

E2− ρφ1(t).

4 Finally, normalize φ′

2(t) to obtain:

φ2(t) =φ

′

2(t)√∫ Tb0

[φ

′

2(t)]2

dt=

φ′

2(t)√

1− ρ2

=1

√

1− ρ2

[s2(t)√E2

− ρs1(t)√E1

]

.



Gram-Schmidt Procedure: Summary

)(2 ts

)(1 ts

22s

21s1E0

2E

1( )tφ

2( )s t′

1( )tρφ

2( )tφ

α21d

1 cos( ) 1ρ α− ≤ = ≤

2( )tφ′φ1(t) =

s1(t)√E1

,

φ2(t) =1

√1− ρ2

[s2(t)√E2

− ρs1(t)√E1

]

,

s21 =

∫ Tb

0s2(t)φ1(t)dt = ρ

√

E2,

s22 =(√

1− ρ2)√

E2,

d21 =

√∫ Tb

0[s2(t) − s1(t)]2dt

= E1 − 2ρ√

E1E2 +E2.



Example 1

0

)(2 ts

0bT

V−

0

)(1 ts

V

0bT

123

4

)(1 tφ

bT1

0bT

567

)(1 tφ0

)(1 ts)(2 ts

E− E

89:

(a) Signal set. (b) Orthonormal function. (c) Signal space representation.



Example 2

V−

;

)(1 ts

V

0bT ;

)(2 ts

V

0bT

<=>

?

)(1 tφ

bT1

0bT

bT1−

?

)(2 tφ

0bT

bT1

@AB )(1 tφ0

)(1 ts

)(2 tsE

E

)(2 tφ



Example 3

C

)(2 ts

0bT

V−

C

)(1 ts

V

0bT

V

α

)(1 tφ0

)(1 ts

)(2 ts

),(2 αφ t

0=α

2bT=α

ρα , increasing

( )0,E( )0,E−

E

DDEF

GGHI

−2

3,

2

EE

4bTα =

ρ =1

E

∫Tb

0

s2(t)s1(t)dt

=1

V 2Tb

[

V 2α − V 2(Tb − α)]

=2α

Tb

− 1



Example 4

J

)(2 ts

0bT

V3

J

)(1 ts

V

0bT 2bT

KLM

N

)(1 tφ

bT1

0bT

N

)(2 tφ

0bT

2bT

bT

3−

bT

3

OPQ



2( )tφ

0

1( )s t

2( )s t

32E

1( )tφ

o602E E

ρ =1

E

∫Tb

0

s2(t)s1(t)dt =2

E

∫Tb/2

0

(

2√3

Tb

V t

)

V dt =

√3

2,

φ2(t) =1

(1 − 34 )

12

[s2(t)√

E− ρ

s1(t)√E

]

=2

√E

[

s2(t) −√3

2s1(t)

]

,

s21 =

√3

2

√E, s22 =

1

2

√E.

d21 =

[∫ Tb

0

[s2(t) − s1(t)]2dt

] 12

=

√(

2 −√3)

E.



Example 5

)(1 tφ0

)(1 ts

)(2 ts

E

)(2 tφ

1−==

ρπθ

0

23

==

ρπθ

0

2

==

ρπθ

2locus of ( ) as

varies from 0 to 2 .

s t θπ

θ

s1(t) =√E

√

2

Tb

cos(2πfct),

s2(t) =√E

√

2

Tb

cos(2πfct + θ).

where fc = k2Tb

, k an integer.



Obtaining Different Basis Sets by Rotation

)(1 tφ

)(1 ts

)(2 ts)(2 tφ

)(1 tφ

θ

)(2 tφ

0 11s

21s

12s

22s

)(1 tφ

)(1 ts

)(2 ts

)(2 tφ)(1 tφ

θ

)(2 tφ

0 11s

21s

12s

22s

11 21s s=11

s

21s

22s

12s

11 21s s=

12s

22s

[φ1(t)

φ2(t)

]

=

[cos θ sin θ

− sin θ cos θ

] [φ1(t)φ2(t)

]

.

(a) Show that, regardless of the angle θ, the set {φ1(t), φ2(t)} is also an orthonormalbasis set.

(b) What are the values of θ that make φ1(t) perpendicular to the line joining s1(t)to s2(t)? For these values of θ, mathematically show that the components of

s1(t) and s2(t) along φ1(t), namely s11 and s21, are identical.

Remark: Rotating counter-clockwise for positive θ and clock-wise for negative θ.



Example: Rotating {φ1(t), φ2(t)} by θ = 60◦ to Obtain {φ1(t), φ2(t)}

[φ1(t)

φ2(t)

]

=

[cos θ sin θ

− sin θ cos θ

] [φ1(t)φ2(t)

]

⇒{

φ1(t) = cos θ × φ1(t) + sin θ × φ2(t)

φ2(t) = − sin θ × φ1(t) + cos θ × φ2(t)

00 5.0 5.0 1

1( )tφ

1

2

)31( +

2

)31( −2

)31( −

2

)31( +−

)(2 tφ

0

)(1 tφ

01

1

1 1

1−

5.0

2( )tφ



Gram-Schmidt Procedure for M Waveforms {si(t)}Mi=1

φ1(t) =s1(t)

√∫∞−∞ s21(t)dt

,

φi(t) =φ

′

i(t)√∫∞−∞

[φ

′

i(t)]2

dt, i = 2, 3, . . . , N,

φ′

i(t) =si(t)√Ei

−i−1∑

j=1

ρijφj(t),

ρij =

∫ ∞

−∞

si(t)√Ei

φj(t)dt, j = 1, 2, . . . , i− 1.

If the waveforms {si(t)}Mi=1 form a linearly independent set, then N = M . OtherwiseN < M .



Example: Find a Basis Set for the Following M = 4 Waveforms

)(1 ts

0

1

1

3( )s t

02 2 3

)(2 ts

1

1−

0

1

1

4( )s t

02 2 3

1

1−1−



Answer Found by Applying the G-S Procedure

1( )tφ

0

1

2

1

3( )tφ

02 2 3

2( )tφ

1

1−

0 1 2

1

2

1

2−

s1(t)s2(t)s3(t)s4(t)

=

√2 0 0

0√2 0√

2 0 1√2 0 1

φ1(t)φ2(t)φ3(t)



Representation of Noise with Walsh Functions

−5

0

5x

1(t)

−5

0

5x

2(t)

0

1

2φ

1(t)

−2

0

2φ

2(t)

−2

0

2φ

3(t)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−2

0

2φ

4(t)

t

Exact representation of noise using 4 Walsh functions is not possible.



The First 16 Walsh Functions

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1t

Exact representation might be possible by using many more Walsh functions.



The First 16 Sine and Cosine Functions

Can also use sine and cosine functions (Fourier representation).

0 0.25 0.5 0.75 1−1.5

1.5

t



Representation of Noise

To represent noise w(t), need to use a complete set of orthonormal functions:

w(t) =∞∑

i=1

wiφi(t), where wi =

∫ Tb

0w(t)φi(t)dt.

The coefficients wi’s are random variables and understanding their statisticalproperties is imperative in developing the optimum receiver.

Of course, the statistical properties of random variables wi’s depend on thestatistical properties of the noise w(t), which is a random process.In communications, a major source of noise is thermal noise, which is modelled asAdditive White Gaussian Noise (AWGN):

White: The power spectral density (PSD) is a constant (i.e., flat) over all frequencies.Gaussian: The probability density function (pdf) of the noise amplitude at any giventime follows a Gaussian distribution.

When w(t) is modelled as AWGN, the projection of w(t) on each basis function,

wi =∫ Tb0

w(t)φi(t)dt, is a Gaussian random variable (this can be proved).

For zero-mean and white noise w(t), w1, w2, w3, . . . are zero-mean anduncorrelated random variables:

1 E{wi} = E{∫ Tb

0 w(t)φi(t)dt}

=∫ Tb0 E{w(t)}φi(t)dt = 0.

2 E{wiwj} = E{∫ Tb

0 dλw(λ)φi(λ)∫ Tb0 dτw(τ)φj(τ)

}

=

{N02 , i = j

0, i 6= j.

Since w(t) is not only zero-mean and white, but also Gaussian ⇒ {w1,w2, . . .}are Gaussian and statistically independent!!!



Need to Review Probability Theory &Random Processes – Chapter 3 Slides



Observing a waveform ⇒ Observing a set of numbers

( )ntφ

bt T=

0n n

= +r w

( )0

dbT

t•∫

( )3tφ

bt T=

3 30= +r w

( )0

dbT

t•∫

( )2tφ

bt T=

2 2 2is= +r w

( )0

dbT

t•∫

( )1tφ

bt T=

1 1 1is= +r w

( )0

dbT

t•∫

( )

( ) ( )i

t

s t t

=+

r

w

AWGN

Choose φ1(t) and φ2(t)so that they can be usedto represent the twosignals s1(t) and s2(t)exactly. The remainingorthonormal basisfunctions are simplychosen to complete theset in order to representnoise exactly.

The decision can bebased on theobservationsr1, r2, r3, r4, . . ..

Note that rj , forj = 3, 4, 5, . . ., does notdepend on which signal(s1(t) or s2(t)) wastransmitted.



Optimum Receiver

The criterion is to minimize the bit error probability.

Consider only the first n terms (n can be very very large), ~r = {r1, r2, . . . , rn}⇒ Need to partition the n-dimensional observation space into two decisionregions, ℜ1 and ℜ2.

1ℜ

1ℜ

2ℜ

ℜ space nObservatioDecide a "1" was transmitted

if falls in this region.rR

Decide a "0" was transmitted

if falls in this region.rR



P [error] = P [(“0” decided and “1” transmitted) or

(“1” decided and “0” transmitted)].

= P [0D, 1T ] + P [1D, 0T ]

= P [0D|1T ]P [1T ] + P [1D|0T ]P [0T ]

= P [~r ∈ ℜ1|1T ]P2 + P [~r ∈ ℜ2|0T ]P1

= P2

∫

ℜ1

f(~r|1T )d~r + P1

∫

ℜ2

f(~r|0T )d~r

= P2

∫

ℜ−ℜ2

f(~r|1T )d~r + P1

∫

ℜ2

f(~r|0T )d~r

= P2

∫

ℜf(~r|1T )d~r +

∫

ℜ2

[P1f(~r|0T )− P2f(~r|1T )]d~r

= P2 +

∫

ℜ2

[P1f(~r|0T )− P2f(~r|1T )] d~r

= P1 −∫

ℜ1

[P1f(~r|0T )− P2f(~r|1T )] d~r.

The minimum error probability decision rule is

{P1f(~r|0T )− P2f(~r|1T ) ≥ 0 ⇒ decide “0” (0D)P1f(~r|0T )− P2f(~r|1T ) < 0 ⇒ decide “1” (1D)

.



( )ntφ

bt T=

0n n= +r w

( )0

dbT

t•∫

( )3tφ

bt T=

3 30= +r w

( )0

dbT

t•∫

( )2 tφ

bt T=

2 2 2is= +r w

( )0

dbT

t•∫

( )1tφ

bt T=

1 1 1is= +r w

( )0

dbT

t•∫

( )

( ) ( )i

t

s t t

=+

r

w

0

AWGN,

PSD 2

N

2

1

2 1 2

( )

1 2 2

( )

Pr[error]

[ ( | 0 ) ( |1 )]d

[ ( |1 ) ( | 0 )]d

T T

g r

T T

g r

P P f r P f r r

P P f r P f r r

ℜ

ℜ −

=

+ −

= + −

∫

∫

1 0Dℜ ⇔

21

Dℜ ⇔

( ) 0g r <

( ) 0g r >

( ) 0g r =

nℜ

( ) 0g r><

0

Optimal decision rule: ( ) 0

1

D

D

g r>⇒<



Equivalently,

f(~r|1T )f(~r|0T )

1D

R0D

P1P2

. (1)

The expressionf(~r|1T )f(~r|0T )

is called the likelihood ratio.

The decision rule in (1) was derived without specifying any statistical propertiesof the noise process w(t).

Simplified decision rule when the noise w(t) is zero-mean, white and Gaussian:

(r1 − s11)2 + (r2 − s12)21D

R0D

(r1 − s21)2 + (r2 − s22)2 +N0ln(P1P2

)

.

For the special case of P1 = P2 (signals are equally likely):

(r1 − s11)2 + (r2 − s12)21D

R0D

(r1 − s21)2 + (r2 − s22)2.

⇒ minimum-distance receiver!



Minimum-Distance Receiver

(r1 − s11)2 + (r2 − s12)

2

︸︷︷︸

d21

1D

R0D

(r1 − s21)2 + (r2 − s22)

2

︸︷︷︸

d22

d21

1D

R0D

d22

)(1 tφ

0

)(1 ts

)(2 ts

)(2 tφ

),( 1211 ss

),( 2221 ss

)( Choose 2 ts )( Choose 1 ts 1r

2r

( )r t

( )1 2,r r 1d

2d



Correlation Receiver Implementation

( )S •bT

t0

d

)(2 tφ

( )T •bT

t0

d

)(1 tφ

bTt =

bTt =

Decision

1r

2r

( ) ( ) ( )it s t t= +r w

2 21 1 2 2

0

Compute

( ) ( )

ln( )

for 1, 2

and choose

the smallest

i i

i

r s r s

N P

i

− + −−

=

( )U •bT

t0

d

)(2 tφ

( )V •bT

t0

d

)(1 tφ

bTt =

bTt =

DecisionWXYYZ[\X[

]^_`[Z\2)ln(

21

10 E

PN −

2)ln(

22

20 E

PN −

( )tr

1r

2r

Form

the

dot

product

ir s⋅a a



Receiver Implementation using Matched Filters

( )∫ •bT

t0

d

)(2 tφ

( )∫ •bT

t0

d

)(1 tφ

bTt =

bTt =

)()( 22 tTth b −= φ

)()( 11 tTth b −= φ

bTt =

bTt =Decision

( )tr

1r

2r

( )tr

1r

2r

Decision



Example 5.6

b

)(2 ts

0b

)(1 ts

0 5.0

5.0

1

15.0

5.1

2−cde

f

)(1 tφ

1

0 1f

)(2 tφ

5.00

1−

1

1

ghi



)(1 tφ0

)(1 ts

)(2 ts)(2 tφ

5.0−

5.0

5.0 11−

1

s1(t) = φ1(t) +1

2φ2(t),

s2(t) = −φ1(t) + φ2(t).



For each value of the signal-to-noise ratio (SNR), Matlab simulation was conductedfor transmitting/receiving 500 equally-likely bits.

−3 −2 −1 0 1 2 3−2

−1

0

1

2

3

(E1+E2)/2N0

=0.99 (dB); P[error]=0.1

φ1(t)

φ2(t)

−3 −2 −1 0 1 2 3−2

−1

0

1

2

3

(E1+E2)/2N0

=6.17 (dB); P[error]=0.01

φ1(t)

φ2(t)



)(1 tφ

0

)(1 ts

)(2 ts

)(2 tφ

5.0−

5.0

5.0 11−

1

)( Choose 1 ts)( Choose 2 ts

jkl

1r

2r

)(1 tφ

0

)(1 ts

)(2 ts

)(2 tφ

5.0−

5.0

5.0 11−

1

)( Choose 1 ts)( Choose 2 ts

mno

1r

2r

)(1 tφ

0

)(1 ts

)(2 ts

)(2 tφ

5.0−

5.0

5.0 11−

1

)( Choose 1 ts

)( Choose 2 ts

pqr

1r

2r

(a) P1 = P2 = 0.5, (b) P1 = 0.25, P2 = 0.75. (c) P1 = 0.75, P2 = 0.25.



Example 5.7

s2(t) = φ1(t) + φ2(t),

s1(t) = φ1(t) − φ2(t).

s

)(1 tφ

0

1

s

)(2 tφ

0 1

1−

1

3

2

1



)(1 tφ

0

)(1 ts

)(2 ts1

1−

1

)( Choose 2 ts

)( Choose 1 ts

)(2 tφ

ttuv

wwxy

2

10 ln4 P

PN

2r

1r



z{|}~�~�{�( )� •bT

t0

d

)(2 tφ

bTt =

��

��

=2

10 ln4 P

PNT

�0bT

3

�~�

)(tr 2r 2 2

2 1

choose ( )

choose ( )

r T s t

r T s t

≥ �

< �

��bTt =

��

��

=2

10 ln4 P

PNT

)(2 th

�0bT

3

��

2 2

2 1

choose ( )

choose ( )

r T s t

r T s t

≥ �

< �2r)(tr



Implementation with One Correlator/Matched FilterAlways possible by choosing φ1(t) and φ2(t) such that one of the two basis functionsis perpendicular to the line joining the two signals.

)(1 tφ

)(1 ts

)(2 ts

)(2 tφ)(1 tφ

θ

)(2 tφ

0 11s

21s

12s

22s

11 21s s=

12s

22s

The optimum receiver is still the minimum-distance receiver. However the terms(r1 − s11)2 and (r1 − s21)2 are the same on both sides of the comparison and hencecan be removed. This means that one does not need to compute r1!

(r1 − s11)2 + (r2 − s12)

2

︸︷︷︸

d21

1D

R0D

(r1 − s21)2 + (r2 − s22)

2

︸︷︷︸

d22

⇔ (r2 − s12)2

1D

R0D

(r2 − s22)2

r2

1D

R0D

s22 + s12

2︸︷︷︸

midpoint of two signals

+

(N0/2

s22 − s12

)

ln

(P1

P2

)

︸︷︷︸

equal to 0 if P1=P2

≡ T.



�� ¡¢£¢¤�£( )¥ •bT

t0

d

)(2 tφ

bTt =

T Threshold¦¢§

)(tr 2r 2

2

ˆ 1

ˆ 0

D

D

r T

r T

≥ ¨

< ¨

©ª«¬®¯ª®

bTt =

T Threshold

)(ˆ)( 2 tTth b −= φ

°±²

2r)(tr 2

2

ˆ 1

ˆ 0

D

D

r T

r T

≥ ³

< ³

φ2(t) =s2(t) − s1(t)

(E2 − 2ρ√E1E2 + E1)

12

, T ≡ s22 + s12

2+

(N0/2

s22 − s12

)

ln

(P1

P2

)

.



Example 5.8

)(1 tφ)(1 ts

)(2 ts

)(2 tφ

)(1 tφ)(2 tφ

2111 ˆˆ ss =

4/πθ =

E

E

φ1(t) =1√2[φ1(t) + φ2(t)],

φ2(t) =1√2[−φ1(t) + φ2(t)].



´µ¶·¸¹¸ºµ¹( )» •bT

t0

d

)(2 tφ

bTt =

T Threshold

¼¸½¾0bT

bT

2

2bT

)(tr 2r 2

2

ˆ 1

ˆ 0

D

D

r T

r T

≥ ¿

< ¿

ÀÁÂÃÄÅÄÆÁÅ

bTt =

T Threshold

ÇÈÉ

Ê0

bT

2

2bT

)(th

)(tr 2r 2

2

ˆ 1

ˆ 0

D

D

r T

r T

≥ Ë

< Ë



Receiver Performance

To detect bk , compare r2 =

∫ kTb

(k−1)Tb

r(t)φ2(t)dt to the threshold

T = s12+s222

+ N02(s22−s12)

ln(

P1P2

)

.

12s 22s

( )Trf 02 ( )Trf 12

TT 1 choose 0 choose Ì⇐

2r

ÍÎÏÐÑÐÒÓ ÔÒÕÓÖ×ØÙ

TP [error] = P [(0 transmitted and 1 decided) or (1 transmitted and 0 decided)]

= P [(0T , 1D) or (1T , 0D)].



ÚÛÜÝ Þ

ßàááàâãäßàåæçàèéâêåëìàâíëçàèéâêåëìàâíëîãéïåéâäðññéòàåéâäóôõæöõàä÷ó

ôæøõñéíéáë îãéïåêéñ ùâéøæääôæøõñéíéáë îãéïåêéñ ùâéøæää

óôõæöõàä÷ó

ßàááàâãä ßàåæçàèéâêåë

ìàâíëçàèéâêåëìàâíëîãéïåéâä ðññéòàåéâä

12s 22s

( )Trf 02 ( )Trf 12

TT 1 choose 0 choose ú⇐

2r

ÚÛÜÝ û

T

P [error] = P [0T , 1D] + P [1T , 0D ] = P [1D|0T ]P [0T ] + P [0D |1T ]P [1T ]

= P1

∫∞

T

f(r2|0T )dr2

︸︷︷︸

Area B

+P2

∫ T

−∞

f(r2|1T )dr2

︸︷︷︸

Area A

= P1Q

(

T − s12√

N0/2

)

+ P2

[

1 − Q

(

T − s22√

N0/2

)]

.



Q-function

üýþÿ�� þý��

�üÿý�ÿ ��

� �� ý� ��

� ��

� ��

0λ

x

)(Area xQ=

2

2

e2

1 λ

π−

Q(x) ≡ 1√2π

∫∞

x

exp

(

−λ2

2

)

dλ.

0 1 2 3 4 5 610

−10

10−8

10−6

10−4

10−2

100

x

Q(x

)



Performance when P1 = P2

12s 22s

( )Trf 02 ( )Trf 12

TT 1choose0choose ⇒⇐

2r

12 22

2

s sT

+=

( )22 12

02 /2

s s

NQ

−=

0

P [error] = Q

(

s22 − s12

2√

N0/2

)

= Q

(distance between the signals

2× noise RMS value

)

.

Probability of error decreases as either the two signals become more dissimilar(increasing the distances between them) or the noise power becomes less.

To maximize the distance between the two signals one chooses them so that theyare placed 180◦ from each other ⇒ s2(t) = −s1(t), i.e., antipodal signaling.

The error probability does not depend on the signal shapes but only on thedistance between them.



Example 5.9

)(1 tφ

Tts 0)(1 ⇔

)(1 2 tsT ⇔

)(2 tφ

1−

2

2−

1

1−

1

2

2−

0 1

1

t

01

1

t

1−0

(a) Determine and sketch the two signals s1(t) and s2(t).



(b) The two signals s1(t) and s2(t) are used for the transmission of equally likely bits 0 and 1,respectively, over an additive white Gaussian noise (AWGN) channel. Clearly draw thedecision boundary and the decision regions of the optimum receiver. Write the expression forthe optimum decision rule.

(c) Find and sketch the two orthonormal basis functions φ1(t) and φ2(t) such that the optimumreceiver can be implemented using only the projection r2 of the received signal r(t) onto the

basis function φ2(t). Draw the block diagram of such a receiver that uses a matched filter.

(d) Consider now the following argument put forth by your classmate. She reasons that since the

component of the signals along φ1(t) is not useful at the receiver in determining which bitwas transmitted, one should not even transmit this component of the signal. Thus shemodifies the transmitted signal as follows:

s(M)1 (t) = s1(t) −

(

component of s1(t) along φ1(t))

s(M)2 (t) = s2(t) −

(

component of s2(t) along φ1(t))

Clearly identify the locations of s(M)1 (t) and s

(M)2 (t) in the signal space diagram. What is the

average energy of this signal set? Compare it to the average energy of the original set.Comment.



0t

3−

)(2 ts

1−

1

)(1 ts

01

3

t

1−



)(1 tφ

Tts 0)(1 ⇔

)(1 2 tsT ⇔

)(2 tφ

1−

2

2−

1

1−

1

2

2−

0

2( )tφ

4

πθ = −

�� !"# D0

D1

1( )tφ

2 ( )Ms t

1 ( )Ms t

01

1

t

1−

0 1

1

t



[φ1(t)

φ2(t)

]

=

[cos(−π/4) sin(−π/4)

− sin(−π/4) cos(−π/4)

] [φ1(t)φ2(t)

]

=

[1√2

− 1√2

1√2

1√2

] [φ1(t)φ2(t)

]

.

φ1(t) =1√2[φ1(t) − φ2(t)], φ2(t) =

1√2[φ1(t) + φ2(t)].

0t

01 t

2−

1/ 2

2

1/ 2

1( )tφ 2( )tφ

2( ) (1 )h t tφ= −1t =

2

2

ˆ 0 0

ˆ 0 1D

D

r

r

≥ $

< $

)(tr

0t

2

1/ 2 1



Antipodal Signalling

bkTt =

+

( )tw

( ) ( )( )

k kEκ= ± +y w

( )x tt

0

(1)

( 1)− ( 1)−

bT 2

bT ( )y t

T ( ) ( )h t p t= R ( ) ( )bh t p T tκ= −

0

2PSD

N

0

0

2(0, )

NE

2 1

1 2

( ) ( ) ( )s t s t p t

E E E

= − == =

The pulse shaping filter hT = p(t) defines the power spectrum density of thetransmitted signal, which can be shown to be proportional to |P (f)|2.The error performance, P [error] only depends on the energy E of p(t) and noisePSD level N0. Specifically, the distance between s1(t) and s2(t) is 2

√E (you

should show this for yourself, algebraically or geometrically). Therefore

P [error] = Q

(√

2E

N0

)

.

For antipodal signalling, the optimum decisions are performed by comparing thesamples of the matched filter’s output (sampled at exactly integer multiples ofthe bit duration) with a threshold 0. Of course such an optimum decision ruledoes not change if the impulse response of the matched filter is scaled by apositive constant.



Scaling the matched filter’s impulse response hR(t) does not change the receiverperformance because it scales both signal and noise components by the samefactor, leaving the signal-to-noise ratio (SNR) of the decision variable unchanged!

In the above block diagram, hR(t) = κp(Tb − t). We have been using κ = 1/√E

in order to represent the signals on the signal space diagram (which would be at±√E) and to conclude that the variance of the noise component is exactly N0/2.

For an arbitrary scaling factor κ, the signal component becomes ±κE, while thevariance of the noise component is N0

2κ2E. Thus, the SNR is

SNR =Signal power

Noise power=

(±κE)2

N02κ2E

=2E

N0, (indepedent of κ!)

In terms of the SNR, the error performance of antipodal signalling is

P [error] = Q

(√

2E

N0

)

= Q(√

SNR)

In fact, it can be proved that the receive filter that maximizes the SNR of thedecision variable must be the matched filter. It is important to emphasize thatthe matching property here concerns the shapes of the impulse responses of thetransmit and receive filters.



Outputs of the Matched/Mismatched Filters (No-Noise Scenario)

Clean received signals for rect and half-sine (HS) shaping filters

Output of a matched filter: rect/rect matching

Outputs of HS/HS matched filter (red) and HS/rect mismatched filter (pink)

When the matched filter is used, sampling at exact multiples of the bit duration maximizes the power of the signal component in thedecision variable, hence maximizing the SNR. A timing error (imperfect sampling) would reduce the power of the signal component, hencereducing the SNR, hence degrading the performance, i.e., increasing P [error].

When the receive filter is not matched to the transmit filter, the power of the signal component and the SNR are not maximized, even

under perfect sampling!



Antipodal Baseband Signalling with Rectangular Pulse Shaping

kb

( ) ( )b

h t p T t= −



Antipodal Baseband Signalling with Half-Sine Pulse Shaping

kb

( ) ( )b

h t p T t= −



PSD Derivation of Arbitrary Binary Modulation

Applicable to any binary modulation with arbitrary a priori probabilities, butrestricted to statistically independent bits.

t0

( )T

s t

bT 2

bT 4

bT

1( )s t

1( 2 )

bs t T−

bT−2

bT− 3

bT

2( 3 )

bs t T−

sT (t) =∞∑

k=−∞gk(t), gk(t) =

{s1(t − kTb), with probability P1

s2(t − kTb), with probability P2.

The derivation on the next slide shows that:

SsT (f) =P1P2

Tb

|S1(f)− S2(f)|2 +∞∑

n=−∞

∣∣∣∣∣∣

P1S1

(nTb

)

+ P2S2

(nTb

)

Tb

∣∣∣∣∣∣

2

δ

(

f − n

Tb

)

.



sT (t) = E{sT (t)}︸︷︷︸

DC

+ sT (t) − E{sT (t)}︸︷︷︸

AC

= v(t) + q(t)

v(t) = E{sT (t)} =∞∑

k=−∞

[P1s1(t − kTb) + P2s2(t − kTb)]

Sv(f) =

∞∑

n=−∞

|Dn|2δ(

f − n

Tb

)

, Dn =1

Tb

[

P1S1

(n

Tb

)

+ P2S2

(n

Tb

)]

,

Sv(f) =∞∑

n=−∞

∣∣∣∣∣∣

P1S1

(nTb

)

+ P2S2

(nTb

)

Tb

∣∣∣∣∣∣

2

δ

(

f − n

Tb

)

.

To calculate Sq(f), apply the basic definition of PSD:

Sq(f) = limT→∞

E{|GT (f)|2}T

= · · · = P1P2

Tb

|S1(f) − S2(f)|2.

SsT(f) =

P1P2

Tb

|S1(f) − S2(f)|2 +

∞∑

n=−∞

∣∣∣∣∣∣

P1S1

(nTb

)

+ P2S2

(nTb

)

Tb

∣∣∣∣∣∣

2

δ

(

f − n

Tb

)

.

For the special, but important case of antipodal signalling, s2(t) = −s1(t) = p(t), and equallylikely bits, P1 = P2 = 0.5, the PSD of the transmitted signal is solely determined by theFourier transform of p(t):

SsT(f) =

|P (f)|2

Tb



Baseband Message Signals with Different Pulse Shaping Filters

2 4 6 8 10 12 14−1

−0.5

0

0.5

1

Information bits or amplitude levels

2 4 6 8 10 12 14−2

−1

0

1

2Output of the transmit pulse shaping filter − Rectangular

−2 0 2−60

−40

−20

0

f × Tb

Magnitude(dB)

PSD - Rectangular

2 4 6 8 10 12 14−2

−1

0

1

2Half-sine pulse shaping filter

−2 0 2−60

−40

−20

0

f × Tb

Magnitude(dB)

PSD - Half-sine

2 4 6 8 10 12 14−2

−1

0

1

2

t/Tb

SRRC pulse shaping filter (β = 0.5)

−2 0 2−60

−40

−20

0

Normalized frequency, f × Tb

Magnitude(dB)

PSD - SRRC (30 symbols long)



Building A Binary (Antipodal) Comm. System in Labs #4 and #5

S/Pbits-to-

levels

mapping

pulse

shaping

filter

t

DACN

data

bits

NCO

ˆcf

ˆcos( )cnw

T[ ] ( )h n p nT=symTN

T=

T

sym[ ] ( ) [ ] ( )k

i n i nT a k nT kTd= = -å

sym[ ] [ ] [ ] [ ] ( )k

s n i n h n a k p nT kT= * = -å

( ) ( ) ( ) cos( )s c

k

s t a k p t kT twæ ö

= -ç ÷è øå

low-rate sequence, index k high-rate sequences, index n

...... ...

DAC

correction

filter

ˆc c sFw w=

Figure 1: Block diagram of the transmitter.



∑

cos( )cnω θ−

sin( )cnω θ−

cf

[ ]c

x n

[ ]s

x n

[ ]x n [ ]x n d−

cdω θ=

R[ ] ( )h n p nT= −

T

symTk n

T=

( )s t

θ

Figure 2: Block diagram of the receiver.


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EE456 – Digital Communicationsengr.usask.ca/classes/EE/456/slides/Chapter5-Slides-F15.pdf · EE456 – Digital Communications 6. Chapter 5: Optimum Receiver for Binary Data Transmission