Chapter5-slides-f15.pdf

Chapter 5: Optimum Receiver for Binary Data Transmission

EE456 – Digital Communications Professor Ha Nguyen

September 2015

EE456 – Digital Communications

1

Chapter 5: Optimum Receiver for Binary Data Transmission

Block Diagram of Binary Communication Systems m(t )

ˆ (t ) m


  

 

{bk }

{ } bˆ k

     

    

b k = 0 ↔ s1 (t ) b k = 1 ↔ s2 (t )

 

r( t )

w (t )

Bits 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: R R E1 = 0Tb s21 (t)dt, E2 = 0Tb s22 (t)dt.

Noise w(t) is stationary Gaussian, zero-mean white noise with two-sided power spectral density of N0 /2 (watts/Hz):   N0 N0 E{w(t)} = 0, E{w(t)w(t + τ )} = δ(τ ), w(t) ∼ N 0, . 2 2

EE456 – Digital Communications

2

Chapter 5: Optimum Receiver for Binary Data Transmission

m(t )

ˆ (t ) m

  !"

.!*'' !

{bk }

{ } bˆ k

&%$' ()$!*+ '',

-+%$' (. /,

b k = 0 ↔ s1 (t ) b k = 1 ↔ s2 (t )

#$!!%

r( t )

w (t )

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 of making an error is minimized. Shall reduce the problem from the observation of a time waveform to that of observing a set of numbers (which are random variables).

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Chapter 5: Optimum Receiver for Binary Data Transmission

OOK (Rx)

OOK (Tx)

NRZ (Rx)

NRZ (Tx)

Can you Identify the Signal Sets {s1 (t), s2 (t)}? 1 0 −1 0

0.5

1

1.5

2

2.5 t/Tb

3

3.5

4

4.5

5

0.5

1

1.5

2

2.5 t/Tb

3

3.5

4

4.5

5

0.5

1

1.5

2

2.5 t/Tb

3

3.5

4

4.5

5

0.5

1

1.5

2

2.5 t/Tb

3

3.5

4

4.5

5

5 0 −5 0 1 0 −1 0 5 0 −5 0

EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

Geometric Representation of Signals s1 (t) and s2 (t) (I) Wish to represent two arbitrary signals s1 (t) and s2 (t) as linear combinations of two 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: Z

Tb

φ1 (t)φ2 (t)dt = 0 (orthogonality),

0

Z

0

Tb

φ21 (t)dt =

Z

0

Tb

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

If {φ1 (t), φ2 (t)} can be found, the representations are s1 (t)

=

s11 φ1 (t) + s12 φ2 (t),

s2 (t)

=

s21 φ1 (t) + s22 φ2 (t).

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

Z

0

Tb

si (t)φj (t)dt,

i, j ∈ {1, 2},

An important question is: Given the signal set s1 (t) and s2 (t), can one always find an orthonormal basis functions to represent {s1 (t), s2 (t)} exactly? If the answer is YES, is the set of orthonormal basis functions UNIQUE? EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

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 be represented as vectors in a vector space (or signal space) spanned (i.e., defined) by the orthonormal basis set {φ1 (t), φ2 (t)}.

φ 2 (t ) s2 (t )

s1 (t) = s11 φ1 (t) + s12 φ2 (t),

s22

s12

d12 = d 21 E2

s1 ( t )

s2 (t) = s21 φ1 (t) + s22 φ2 (t), Z Tb sij = si (t)φj (t)dt, i, j ∈ {1, 2}, Ei =

E1 s21

0

s11

φ1 (t )

Z

0 Tb

0

s2i (t)dt = s2i1 + s2i2 , i ∈ {1, 2},

d12 = d21 =

s Z

0

R Tb 0

Tb

[s2 (t) − s1 (t)]2 dt

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) to represent s1 (t) and s2 (t) exactly. In fact, there are infinite number of choices!

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Chapter 5: Optimum Receiver for Binary Data Transmission

Gram-Schmidt Procedure Gram-Schmidt (G-S) procedure is one method to find a set of orthonormal basis functions for a given arbitrary set of waveforms. √ s1 (t) 1 Let φ1 (t) ≡ p . Note that s11 = E1 and s12 = 0. E1 ′ s2 (t) 2 Project s (t) = p onto φ1 (t) to obtain the correlation coefficient: 2 E2 ρ=

Z

Tb

0

s2 (t) 1 √ φ1 (t)dt = √ E2 E1 E2

′

3

′

Subtract ρφ1 (t) from s2 (t) to obtain φ2 (t) =

Z

Tb

s1 (t)s2 (t)dt.

0

s2 (t) √ E2

− ρφ1 (t).

′

4

Finally, normalize φ2 (t) to obtain: ′

φ2 (t)

=

=

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′

φ2 (t) φ (t) = p2 2 Tb  ′ 1 − ρ2 φ (t) dt 2 0   1 ρs1 (t) s2 (t) p √ − √ . E2 E1 1 − ρ2 qR

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Chapter 5: Optimum Receiver for Binary Data Transmission

Gram-Schmidt Procedure: Summary

φ2 (t )

s2′ (t )

φ2′ (t ) s2 (t )

s 22

E2

0

=

φ2 (t)

=

s21

=

s22

=

d21

=

d 21

ρφ1 (t )

α E1

s 21 s1 (t )

−1 ≤ ρ = cos(α ) ≤ 1

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φ1 (t)

φ1 (t )

s1 (t) √ , E1   ρs1 (t) s2 (t) 1 p √ − √ , E2 E1 1 − ρ2 Z Tb p s2 (t)φ1 (t)dt = ρ E2 , 0 p p 1 − ρ2 E2 , s Z Tb [s2 (t) − s1 (t)]2 dt 0

=

p E1 − 2ρ E1 E2 + E2 .

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 1 s1 (t )

s 2 (t )

V

0

Tb

0

Tb

0

0

123

−V

φ1 (t )

1

Tb

s1 (t )

s 2 (t )

0

Tb

567

4 − E

0

φ1 (t )

E

89:

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

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 2 s1 (t )

s 2 (t )

V

V Tb

;

0

0

;

Tb

−V

<=> φ 2 (t )

φ1 (t )

1 Tb

1 Tb Tb 0

φ2 (t )

?

0

s 2 (t )

E

Tb

? s1 (t )

− 1 Tb

@AB

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0

φ1 (t )

E

10

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 3 s1 (t )

s 2 (t )

V

V

0

α

C

Tb

Tb

C

0 −V

φ 2 (t, α )

α= I E 3E G− G 2 , 2 H

α =0

(−

F D D E

Tb 2

s 2 (t )

ρ

=

increasing α , ρ =

α=

E ,0

Tb 4

)

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E =

s1 (t )

0

(

E ,0

)

1 E

Z

Tb

s2 (t)s1 (t)dt

0

1 V 2 Tb 2α −1 Tb

h i V 2 α − V 2 (Tb − α)

φ1 (t )

11

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 4 s1 (t )

s 2 (t )

3V

V

0

Tb

J

0

KLM

φ1 (t )

1

Tb

J

φ2 (t ) 3 Tb

Tb

0

Tb 2

Tb

Tb

N

0

−

Tb 2

N

3 Tb

OPQ EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

φ 2 (t )

s2 (t ) E 2

E

60o 0

s1 (t )

φ1 (t )

3E 2

! √ √ 3 2 3 V t V dt = , Tb 2 0 0 # " √   1 3 s1 (t) 2 s2 (t) s (t) , φ2 (t) = − ρ s (t) − = √ √ √ 1 2 1 2 E E E (1 − 43 ) 2 √ 3√ 1√ E, s22 = E. s21 = 2 2 ρ=

1 E

Z

d21 =

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Tb

Z

s2 (t)s1 (t)dt =

Tb 0

2 E

Z

Tb /2

r 1 √  2 2 = 2 − 3 E. [s2 (t) − s1 (t)] dt 13

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5 φ2 (t )

θ = 3π 2 ρ =0

s1 (t)

=

√

E

s

2 cos(2πfc t), Tb

s2 (t)

=

√

E

s

2 cos(2πfc t + θ). Tb

where fc =

s1 (t ) 0

θ =π ρ = −1

θ

k , k an integer. 2Tb

φ1 (t )

locus of s2 (t ) as θ

E

varies from 0 to 2π . s 2 (t )

θ =π 2 ρ =0

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Chapter 5: Optimum Receiver for Binary Data Transmission

Obtaining Different Basis Sets by Rotation

φ2 (t )

φ 2 (t )

s2 (t )

φ 2 (t )

s22

s22

s11

θ 0

s11 = s21

s12

s1 ( t )

φ1 (t )

s11

s21

s22

φ 2 (t )

s21

s12

φ1 (t )

s2 (t )

φ1 ( t )

s22

s1 (t )

θ s21

0

s12

s11

φ1 (t )

s12 s11 = s21



φˆ1 (t) φˆ2 (t)



=



cos θ − sin θ

sin θ cos θ



φ1 (t) φ2 (t)



.

(a) Show that, regardless of the angle θ, the set {φˆ1 (t), φˆ2 (t)} is also an orthonormal basis set. (b) What are the values of θ that make φˆ1 (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 φˆ1 (t), namely sˆ11 and sˆ21 , are identical. Remark: Rotating counter-clockwise for positive θ and clock-wise for negative θ. EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example: Rotating {φ1 (t), φ2 (t)} by θ = 60◦ to Obtain {φˆ1 (t), φˆ2 (t)} 

ˆ1 (t) φ ˆ2 (t) φ



=



cos θ − sin θ

sin θ cos θ



φ1 (t) φ2 (t)



⇒



φ1 (t )

ˆ1 (t) = cos θ × φ1 (t) + sin θ × φ2 (t) φ ˆ2 (t) = − sin θ × φ1 (t) + cos θ × φ2 (t) φ

φ2 (t )

1

1 0 .5

0

1

0

1

−1 φ1 (t )

φ2 (t )

(1 + 3 ) 2

(1 − 3 ) 2

0

0 .5

1

(1 − 3 ) 2 −

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0

0 .5

1

(1 + 3 ) 2 16

Chapter 5: Optimum Receiver for Binary Data Transmission

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

s1 (t)

φ1 (t)

=

qR

φi (t)

=

qR

′

φi (t)

=

ρij

=

∞ s2 (t)dt −∞ 1

,

′

φi (t) 2 , ∞  ′ φi (t) dt −∞

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

i−1

si (t) X √ − ρij φj (t), Ei j=1 Z ∞ si (t) √ φj (t)dt, j = 1, 2, . . . , i − 1. Ei −∞

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

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example: Find a Basis Set for the Following M = 4 Waveforms s1 (t )

s3 ( t )

1

1

0

1

2

0

2

3

2

3

−1

s4 ( t )

s 2 (t )

1

1 0

1

−1

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2

0 −1

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Chapter 5: Optimum Receiver for Binary Data Transmission

Answer Found by Applying the G-S Procedure φ3 ( t )

φ1 (t ) 1 2 0

1 1

0

2

2

3

−1

φ2 ( t ) 1 2 −

1 2

0

1

2

  √ 2 s1 (t)  s2 (t)     =  √0  s3 (t)   √2 s4 (t) 2 

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√0 2 0 0

   0 φ1 (t) 0    φ2 (t)  1  φ3 (t) 1 19

Chapter 5: Optimum Receiver for Binary Data Transmission

Representation of Noise with Walsh Functions x1(t)

5 0

−5 5 x2(t) 0 −5 2 φ1(t) 1

φ2(t)

0 2 0

−2 2 φ3(t) 0 −2 2

φ4(t)

0

−2 0

0.1

0.2

0.3

0.4

0.5 t

0.6

0.7

0.8

0.9

1

Exact representation of noise using 4 Walsh functions is not possible. EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

The First 16 Walsh Functions

0

0.1

0.2

0.3

0.4

0.5 t

0.6

0.7

0.8

0.9

1

Exact representation might be possible by using many more Walsh functions. EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

The First 16 Sine and Cosine Functions Can also use sine and cosine functions (Fourier representation).

1.5 −1.5 0

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0.25

0.5 t

0.75

1

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Chapter 5: Optimum Receiver for Binary Data Transmission

Representation of Noise To represent noise w(t), need to use a complete set of orthonormal functions: Z Tb ∞ X w(t) = wi φi (t), where wi = w(t)φi (t)dt. 0

i=1

The coefficients wi ’s are random variables and understanding their statistical properties is imperative in developing the optimum receiver. Of course, the statistical properties of random variables wi ’s depend on the statistical properties of the noise w(t), which is a random process. In communications, a major source of noise is thermal noise, which is modelled as Additive 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 given time follows a Gaussian distribution.

When w(t) is modelled as AWGN, the projection of w(t) on each basis function, R wi = 0Tb 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 and uncorrelated random variables: o n 1

2

E{wi } = E

R Tb

E{wi wj } = E

0

w(t)φi (t)dt

nRT 0

b

=

dλw(λ)φi (λ)

R Tb 0

R Tb 0

E{w(t)}φi (t)dt = 0. ( o N0 , 2 dτ w(τ )φj (τ ) = 0,

i=j . i 6= j

Since w(t) is not only zero-mean and white, but also Gaussian ⇒ {w1 , w2 , . . .} are Gaussian and statistically independent!!! EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

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

EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

Observing a waveform ⇒ Observing a set of numbers t = Tb Tb

∫ ( • ) dt 0

r1 = si1 + w1

φ1 ( t ) t = Tb Tb

∫ ( • ) dt 0

r2 = si 2 + w 2

φ2 ( t ) t = Tb

r (t ) =

Tb

∫ ( • ) dt

si (t ) + w (t )

r3 = 0 + w 3

0

AWGN

φ3 ( t )

The decision can be based on the observations r1 , r2 , r3 , r4 , . . ..

t = Tb Tb

∫ ( • ) dt 0

φn ( t )

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Choose φ1 (t) and φ2 (t) so that they can be used to represent the two signals s1 (t) and s2 (t) exactly. The remaining orthonormal basis functions are simply chosen to complete the set in order to represent noise exactly.

rn = 0 + w n

Note that rj , for j = 3, 4, 5, . . ., does not depend on which signal (s1 (t) or s2 (t)) was transmitted.

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Chapter 5: Optimum Receiver for Binary Data Transmission

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 decision regions, ℜ1 and ℜ2 .

Observation space ℜ

Decide a "1" was transmitted R if r falls in this region.

ℜ1

Decide a "0" was transmitted R if r falls in this region.

ℜ2

ℜ1

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Chapter 5: Optimum Receiver for Binary Data Transmission

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 Z Z f (~ r |0T )d~ r f (~ r |1T )d~ r + P1 P2 ℜ2

ℜ1

=

P2

Z

ℜ−ℜ2

=

P2

Z

ℜ

=

P2 +

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

Z

ℜ2

=

P1 −

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

Z

ℜ1

Z

ℜ2

Z

ℜ2

f (~ r |0T )d~ r

[P1 f (~ r |0T ) − P2 f (~ r |1T )]d~ r

[P1 f (~ r |0T ) − P2 f (~ r |1T )] d~ r [P1 f (~ r |0T ) − P2 f (~ r |1T )] d~ r.

The minimum error probability decision rule is  P1 f (~ r |0T ) − P2 f (~ r |1T ) ≥ 0 ⇒ P1 f (~ r |0T ) − P2 f (~ r |1T ) < 0 ⇒

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decide “0” (0D ) . decide “1” (1D )

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Chapter 5: Optimum Receiver for Binary Data Transmission

Pr[error] =

t = Tb

P2 +

Tb

∫ ( • ) dt

∫ [ P f (r | 0 1

ℜ2

) − P2 f ( r |1T )]dr g(r )

= P1 + ∫ [ P2 f ( r |1T ) − P2 f ( r | 0T )]dr

r1 = si1 + w1

0

T

ℜ1

φ1 ( t )

−g(r )

0D

t = Tb

⇒ Optimal decision rule: g ( r )

Tb

∫ ( • ) dt 0

r2 = si 2 + w 2

φ2 ( t ) t = Tb

r (t ) =

Tb

∫ ( • ) dt

si (t ) + w (t )

r3 = 0 + w 3

0

AWGN, PSD

g(r )

ℜn

> 0 < ℜ1 ⇔ 0 D g (r ) > 0

φ3 ( t )

N0 2

> 0 < 1D

t = Tb Tb

∫ ( • ) dt 0

φn ( t )

g(r ) < 0

rn = 0 + w n g (r ) = 0 ℜ 2 ⇔ 1D

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Chapter 5: Optimum Receiver for Binary Data Transmission

Equivalently, 1D

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

R

0D

P1 . P2

(1)

f (~ r |1T ) is called the likelihood ratio. f (~ r |0T )

The decision rule in (1) was derived without specifying any statistical properties of the noise process w(t). Simplified decision rule when the noise w(t) is zero-mean, white and Gaussian: (r1 − s11 )2 + (r2 − s12 )2

1D

R

0D

  (r1 − s21 )2 + (r2 − s22 )2 + N0 ln P1 . P2

For the special case of P1 = P2 (signals are equally likely): (r1 − s11 )2 + (r2 − s12 )2

1D

R

0D

(r1 − s21 )2 + (r2 − s22 )2 .

⇒ minimum-distance receiver!

EE456 – Digital Communications

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Chapter 5: Optimum Receiver for Binary Data Transmission

Minimum-Distance Receiver

1D

(r1 − s11 )2 + (r2 − s12 )2 | {z }

R

0D

d2 1

1D

d21

R

r2 φ 2 (t )

0D

(r1 − s21 )2 + (r2 − s22 )2 {z } | d2 2

d22

s 2 (t )

d2

( s21 , s22 )

r (t )

d1

( r1, r2 )

s1 (t ) ( s11 , s12 )

φ1 (t ) 0

Choose s2 (t )

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Choose s1 ( t )

r1

30

Chapter 5: Optimum Receiver for Binary Data Transmission

Correlation Receiver Implementation t = Tb Tb

T (•)dt

r1

Compute (r1 − si1 )2 + (r2 − si 2 )2

0

r (t ) = si (t ) + w (t )

− N 0 ln( Pi )

φ1 (t )

t = Tb Tb

S (•)dt

r2

Decision

for i = 1, 2 and choose the smallest

0

φ 2 (t ) t = Tb Tb

V (•)dt

r1

Form

0

the

r (t )

φ1 (t )

dot product

t = Tb Tb

U (•)dt

r2

a a

N0 E ln(P1 ) − 1 2 2

WXYYZ[ \X[ ]^_`[Z\

Decision

r ⋅ si

0

φ 2 (t )

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N0 E ln( P2 ) − 2 2 2

31

Chapter 5: Optimum Receiver for Binary Data Transmission

Receiver Implementation using Matched Filters t = Tb Tb

∫ (•)dt

r1

0

r (t )

Decision

φ1 (t )

t = Tb Tb

∫ (•)dt

r2

0

φ 2 (t )

t = Tb r1

h1 (t ) = φ1 (Tb − t ) r (t ) h2 (t ) = φ 2 (Tb − t )

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Decision

t = Tb r2

32

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.6 s1 (t )

s 2 (t )

1.5 0.5 0

0.5

1

b

cde φ1 (t )

1

b

1

f

−2

φ 2 (t )

1

1

0

0.5

0

1

f

0.5

0 −1

ghi

EE456 – Digital Communications

33

Chapter 5: Optimum Receiver for Binary Data Transmission

φ2 (t ) s 2 (t )

1

s1 (t )

0.5

−1

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− 0.5

0

s1 (t)

=

s2 (t)

=

0.5

1

φ1 (t )

1 φ2 (t), 2 −φ1 (t) + φ2 (t).

φ1 (t) +

34

Chapter 5: Optimum Receiver for Binary Data Transmission

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

(E1 +E2 )/2 N0

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

3

3

2

2

1

1

φ2 (t)

φ2 (t)

(E1 +E2 )/2 N0

0 −1 −2 −3

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

0 −1

−2

−1

EE456 – Digital Communications

0 φ1 (t)

1

2

3

−2 −3

−2

−1

0 φ1 (t)

1

2

3

35

Chapter 5: Optimum Receiver for Binary Data Transmission

φ2 (t ) r2

φ2 (t ) r2

s 2 (t )

s 2 (t ) 1

1

jkl

mno

s1 (t )

s1 (t ) 0.5

0.5

Choose s1 (t )

Choose s2 (t ) −1

− 0.5

0

0.5

1

Choose s1 (t )

Choose s2 (t )

φ1 (t ) r1

−1

− 0.5

0

0.5

1

φ1 (t ) r1

φ2 (t ) r2 s 2 (t ) 1

pqr Choose s2 (t ) s1 (t )

0.5 Choose s1 (t )

−1

− 0.5

0

0.5

1

φ1 (t ) r1

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

EE456 – Digital Communications

36

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.7

s2 (t) = φ1 (t) + φ2 (t), s1 (t) = φ1 (t) − φ2 (t).

φ 2 (t )

φ1 (t )

3

1 1 0

−1

EE456 – Digital Communications

1 2

s

0

1

s

37

Chapter 5: Optimum Receiver for Binary Data Transmission

φ2 (t ) r2

1

s 2 (t ) Choose s 2 (t )

N 0 y P1 v lnw t 4 wx P2 tu

φ1 (t ) 0

r1

1 Choose s1 (t )

−1

EE456 – Digital Communications

s1 (t )

38

Chapter 5: Optimum Receiver for Binary Data Transmission

t = Tb Tb

 (•)dt

r (t )

r2

z{|}~~€{

0

φ 2 (t )

T=

3

0

Tb

ˆ

r2

Tb

ŒŽ‘’‘

™ T=

š›œ

EE456 – Digital Communications

N 0 ‡ P1 „ ln… ‚ 4 …† P2 ‚ƒ

t = Tb

3

0

r2 < T ‹ choose s1 (t )

‰~Š

h2 (t )

r (t )

r2 ≥ T ‹ choose s2 (t )

r2 ≥ T  choose s2 (t ) r2 < T  choose s1 (t )

N 0 ˜ P1 • ln – “ 4 –— P2 “”

39

Chapter 5: Optimum Receiver for Binary Data Transmission

Implementation with One Correlator/Matched Filter Always possible by choosing φˆ1 (t) and φˆ2 (t) such that one of the two basis functions is perpendicular to the line joining the two signals. φ2 (t ) φ1 (t ) s2 (t ) s22 s =s 11

φ 2 (t )

21

s12 s22

s1 (t )

θ 0

s21

s11

φ1 (t )

s12

The optimum receiver is still the minimum-distance receiver. However the terms (ˆ r1 − sˆ11 )2 and (ˆ r1 − sˆ21 )2 are the same on both sides of the comparison and hence can be removed. This means that one does not need to compute rˆ1 ! 1D

(ˆ r1 − sˆ11 )2 + (ˆ r2 − sˆ12 )2 {z } |

R

0D

d2 1

1D

rˆ2

R

0D

sˆ22 + sˆ12 2 | {z }

midpoint of two signals EE456 – Digital Communications

+



1D

r2 − sˆ12 )2 R (ˆ r2 − sˆ22 )2 (ˆ r1 − s ˆ21 )2 + (ˆ r2 − sˆ22 )2 ⇔ (ˆ {z } | 0D d2 2



N0 /2 ln sˆ22 − s ˆ12 | {z



P1 P2

equal to 0 if P1 =P2

 }

≡ T.

40

Chapter 5: Optimum Receiver for Binary Data Transmission

t = Tb Tb

¥ (•)dt

r (t )

rˆ2

rˆ2 ≥ T ¨ 1D

žŸ ¡¢£¢¤Ÿ£ rˆ2 < T ¨ 0 D

0

φˆ2 ( t )

Threshold T

¦¢§

t = Tb

r (t )

rˆ2

h (t ) = φˆ2 (Tb − t )

rˆ2 ≥ T ³ 1D

©ª«¬­®­¯ª® rˆ2 < T ³ 0D

°±² ˆ2 (t) = φ

Threshold T

s2 (t) − s1 (t) sˆ22 + sˆ12 , T ≡ + √ 1 2 2 (E2 − 2ρ E1 E2 + E1 )

EE456 – Digital Communications



N0 /2 sˆ22 − sˆ12



ln



P1 P2



.

41

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.8 φ2 (t ) φˆ1 (t )

s2 ( t )

φˆ2 (t )

E

sˆ11 = sˆ21

θ =π /4 E

EE456 – Digital Communications

ˆ1 (t) φ

=

ˆ2 (t) φ

=

s1 (t )

φ1 (t )

1 √ [φ1 (t) + φ2 (t)], 2 1 √ [−φ1 (t) + φ2 (t)]. 2

42

Chapter 5: Optimum Receiver for Binary Data Transmission

t = Tb Tb

r (t )

rˆ2 ≥ T ¿ 1D

rˆ2

» (•)dt

´µ¶·¸¹¸ºµ¹ rˆ2 < T ¿ 0D

0

φˆ2 ( t )

Threshold T

2 Tb

0

Tb 2

Tb

¾

¼¸½ t = Tb

h(t )

r (t )

2 Tb

0

rˆ2

Tb 2

rˆ2 ≥ T Ë 1D

ÀÁÂÃÄÅÄÆÁÅ rˆ2 < T Ë 0D

Ê Threshold T

ÇÈÉ

EE456 – Digital Communications

43

Chapter 5: Optimum Receiver for Binary Data Transmission

Receiver Performance To detect bk , compare ˆ r2 = T =

s ˆ12 +ˆ s22 2

+

N0 2(ˆ s22 −ˆ s12 )

Z

kTb

r(t)φˆ2 (t)dt to the threshold

 b (k−1)T P1 . ln P 2

ÍÎÏÐÑÐÒÓ ÔÒÕÓÖ×ØÙ

f (rˆ2 0T )

f (rˆ2 1T )

sˆ12

choose 0T ⇐ P [error]

sˆ22

rˆ2

Ì choose 1T

=

T P [(0 transmitted and 1 decided) or (1 transmitted and 0 decided)]

=

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

EE456 – Digital Communications

44

Chapter 5: Optimum Receiver for Binary Data Transmission

ßàåæ

f (rˆ2 0T ) ßàááàâãä çàèéâêåë

ìàâíë ìàâíë f (rˆ2 1T ) çàèéâêåë îãéïåéâä ðññéòàåéâä

ìàâíë óôõæ ìàâíë ßàåæ ðññéòàåéâä îãéïåéâä çàèéâêåë çàèéâêåë ßàááàâãä öõàä÷ó óôõæ öõàä÷ó ôæøõñéíéáë îãéïåêéñ ùâéøæää ôæøõñéíéáë îãéïåêéñ ùâéøæää sˆ22

sˆ12

ÚÛÜÝ Þ

rˆ2

ÚÛÜÝ û ú choose 1T

choose 0T ⇐ T P [error]

= =

P [0T , 1D ] + P [1T , 0D ] = P [1D |0T ]P [0T ] + P [0D |1T ]P [1T ] Z T Z ∞ P1 f (ˆ r2 |1T )dˆ r2 f (ˆ r2 |0T )dˆ r2 +P2 −∞ T {z } | {z } | Area B

=

EE456 – Digital Communications

P1 Q

T − sˆ12 p N0 /2

Area A

!

"

+ P2 1 − Q

T − sˆ22 p N0 /2

!#

.

45

Chapter 5: Optimum Receiver for Binary Data Transmission

Q-function 2   

1 − λ2 e 2π

ý

     

     

  

     

üÿý ÿ

  

üýþÿ0
   




x

þý

1 Q(x) ≡ √ 2π

λ Area = Q ( x )

Z

∞

exp x

−

λ2 2

!

dλ.

0

10

−2

10

−4

Q(x)

10

−6

10

−8

10

−10

10

EE456 – Digital Communications

0

1

2

3 x

4

5

6

46

Chapter 5: Optimum Receiver for Binary Data Transmission

Performance when P1 = P2 f (r2 0T )

0

f (r2 1T )

s12

r2

s 22

=Q choose 0T ⇐ T=

P [error] = Q

sˆ22 − sˆ12 p 2 N0 /2

!

(

s22 − s12 2 N 0 /2

)

⇒ choose 1T

s12 + s22 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 they are 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 the distance between them. EE456 – Digital Communications

47

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.9

1

φ2 (t )

1

0

t

2

s1 (t ) ⇔ 0T

1

1 −2

−1

0

1

2

−1

1T ⇔ s 2 (t )

1

φ1 (t )

0

t

−1

−2

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

EE456 – Digital Communications

48

Chapter 5: Optimum Receiver for Binary Data Transmission

(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 the decision boundary and the decision regions of the optimum receiver. Write the expression for the optimum decision rule. ˆ1 (t) and φ ˆ2 (t) such that the optimum (c) Find and sketch the two orthonormal basis functions φ receiver can be implemented using only the projection ˆ r2 of the received signal r(t) onto the ˆ2 (t). Draw the block diagram of such a receiver that uses a matched filter. basis function φ (d) Consider now the following argument put forth by your classmate. She reasons that since the ˆ1 (t) is not useful at the receiver in determining which bit component of the signals along φ was transmitted, one should not even transmit this component of the signal. Thus she modifies the transmitted signal as follows:   (M) ˆ1 (t) s1 (t) = s1 (t) − component of s1 (t) along φ   (M) ˆ1 (t) s2 (t) = s2 (t) − component of s2 (t) along φ (M)

(M)

Clearly identify the locations of s1 (t) and s2 (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.

EE456 – Digital Communications

49

Chapter 5: Optimum Receiver for Binary Data Transmission

s 2 (t )

s1 (t )

3

1

t

1

0

0

−1

−1

t

−3

EE456 – Digital Communications

50

Chapter 5: Optimum Receiver for Binary Data Transmission

1

φ2 (t ) 0   !"#

0D

1D

2

1

t

φˆ2 ( t )

s1M (t ) s1 (t ) ⇔ 0T

1

1 −2

−1

0

1

θ =− s2M (t )

1T ⇔ s 2 (t )

EE456 – Digital Communications

π

2

−1

4

−2

φˆ1 (t )

1

φ1 (t )

0

t

−1

51

Chapter 5: Optimum Receiver for Binary Data Transmission



φˆ1 (t) φˆ2 (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)



.

1 φˆ2 (t) = √ [φ1 (t) + φ2 (t)]. 2

1 φˆ1 (t) = √ [φ1 (t) − φ2 (t)], 2 φˆ1 (t )

φˆ2 (t ) 2

1/ 2

1

t

0

0

1/ 2

t

− 2

h (t ) = φˆ2 (1 − t ) t =1

r (t )

2

0

EE456 – Digital Communications

1/ 2

1

rˆ2 ≥ 0 $ 0D rˆ2 < 0 $ 1D

t

52

Chapter 5: Optimum Receiver for Binary Data Transmission

Antipodal Signalling w (t ) PSD

N0 2

t = kTb

(1)

Tb

2Tb

0

t

( −1) ( −1)

hT (t ) = p( t )

x (t )

s2 (t ) = − s1 ( t ) = p ( t ) E1 = E2 = E

+

hR (t ) = κ p (Tb − t )

y (t ) 0

y ( k ) = ± (κ E ) + w ( k ) (0, N20 E )

The pulse shaping filter hT = p(t) defines the power spectrum density of the transmitted 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 noise PSD level N0 . Specifically, the distance between s1 (t) and s2 (t) is 2 E (you should show this for yourself, algebraically or geometrically). Therefore s ! 2E P [error] = Q . N0 For antipodal signalling, the optimum decisions are performed by comparing the samples of the matched filter’s output (sampled at exactly integer multiples of the bit duration) with a threshold 0. Of course such an optimum decision rule does not change if the impulse response of the matched filter is scaled by a positive constant. EE456 – Digital Communications

53

Chapter 5: Optimum Receiver for Binary Data Transmission

Scaling the matched filter’s impulse response hR (t) does not change the receiver performance because it scales both signal and noise components by the same factor, 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 the variance of the noise component is N20 κ2 E. Thus, the SNR is SNR =

(±κE)2 Signal power 2E = N , = 0 κ2 E Noise power N0 2

(indepedent of κ!)

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

P [error] = Q

s

2E N0

!

=Q

√

 SNR

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

EE456 – Digital Communications

54

Chapter 5: Optimum Receiver for Binary Data Transmission

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 the decision variable, hence maximizing the SNR. A timing error (imperfect sampling) would reduce the power of the signal component, hence reducing 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! EE456 – Digital Communications

55

Chapter 5: Optimum Receiver for Binary Data Transmission

Antipodal Baseband Signalling with Rectangular Pulse Shaping

bk h(t ) = p(Tb − t )

EE456 – Digital Communications

56

Chapter 5: Optimum Receiver for Binary Data Transmission

Antipodal Baseband Signalling with Half-Sine Pulse Shaping

bk h(t ) = p (Tb − t )

EE456 – Digital Communications

57

Chapter 5: Optimum Receiver for Binary Data Transmission

PSD Derivation of Arbitrary Binary Modulation Applicable to any binary modulation with arbitrary a priori probabilities, but restricted to statistically independent bits. sT (t ) s1 ( t − 2Tb )

s1 (t ) −2Tb

−Tb

0

Tb

2Tb

3Tb

4Tb

t

s2 (t − 3Tb )

sT (t) =

∞ X

k=−∞

gk (t),

gk (t) =



s1 (t − kTb ), s2 (t − kTb ),

with probability P1 . with probability P2

The derivation on the next slide shows that:   2   n P S  ∞ + P2 S2 Tn  X 1 1 T P1 P2 b b δ f− n . SsT (f ) = |S1 (f ) − S2 (f )|2 + Tb Tb Tb n=−∞ EE456 – Digital Communications

58

Chapter 5: Optimum Receiver for Binary Data Transmission

sT (t) = E{sT (t)} + sT (t) − E{sT (t)} = v(t) + q(t) | {z } | {z } DC

AC

v(t) = E{sT (t)} =

∞ X

k=−∞

Sv (f ) =

∞ X

n=−∞

Sv (f ) =

|Dn |2 δ



[P1 s1 (t − kTb ) + P2 s2 (t − kTb )]

f−

n Tb



, Dn =

1 Tb



P1 S1



n Tb



+ P2 S2



n Tb



,

  2   n P S n   ∞ X 1 1 Tb + P2 S2 Tb δ f− n . Tb Tb n=−∞

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

T →∞

SsT (f ) =

P1 P2 E{|GT (f )|2 } = ··· = |S1 (f ) − S2 (f )|2 . T Tb

  2   P S n n   ∞ X 1 1 Tb + P2 S2 Tb P1 P2 δ f− n . |S1 (f ) − S2 (f )|2 + Tb Tb Tb n=−∞

For the special, but important case of antipodal signalling, s2 (t) = −s1 (t) = p(t), and equally likely bits, P1 = P2 = 0.5, the PSD of the transmitted signal is solely determined by the Fourier transform of p(t): |P (f )|2 SsT (f ) = Tb EE456 – Digital Communications

59

Chapter 5: Optimum Receiver for Binary Data Transmission

Baseband Message Signals with Different Pulse Shaping Filters Information bits or amplitude levels 1 0.5 0 −0.5 −1 2

4 6 8 10 12 Output of the transmit pulse shaping filter − Rectangular

14 PSD - Rectangular Magnitude (dB)

2 1 0 −1 −2

2

4

6 8 10 Half-sine pulse shaping filter

12

Magnitude (dB)

1 0 −1 2

4

6 8 10 SRRC pulse shaping filter (β = 0.5)

12

Magnitude (dB)

1 0 −1 2

EE456 – Digital Communications

4

6

8 t/T b

10

12

−40 −60

14

−2

0 2 f × Tb PSD - Half-sine

0 −20 −40 −60

14

2

−2

−20

14

2

−2

0

−2 0 2 f × Tb PSD - SRRC (30 symbols long)

0 −20 −40 −60

−2 0 2 Normalized frequency, f × T b

60

Chapter 5: Optimum Receiver for Binary Data Transmission

Building A Binary (Antipodal) Comm. System in Labs #4 and #5 low-rate sequence, index k

high-rate sequences, index n

...

...

... t

s[n ] = i[n ] * h[n ] = å a[k ] p(nT - kTsym ) data bits

k

S/P

bits-tolevels mapping

N N=

Tsym T

pulse shaping filter

hT [n ] = p(nT )

i[n ] = i (nT ) = å a[k ]d (nT - kTsym ) k

DAC correction filter

DAC

cos( wˆ c n )

NCO fˆc

T

wc = wˆ c Fs

æ ö s(t ) = ç å a ( k ) p(t - kTs ) ÷ cos(wct ) è k ø

Figure 1: Block diagram of the transmitter.

EE456 – Digital Communications

61

Chapter 5: Optimum Receiver for Binary Data Transmission

s (t )

∑

x[ n − d ]

x[ n]

hR [n ] = p( − nT ) xc [ n ]

k =n

Tsym T

cos( ω c n − θ )

θ fc

T

sin( ω c n − θ )

ωc d = θ

x s [n ]

Figure 2: Block diagram of the receiver.

EE456 – Digital Communications

62