[PPT] - EE456 Digital Communications Professor Ha Nguyen September 2015 PowerPoint Presentation

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Chapter 5: Optimum Receiver for Binary Data Transmission

EE456 – Digital Communications

Professor Ha Nguyen September 2015

EE456 – Digital Communications 1

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Chapter 5: Optimum Receiver for Binary Data Transmission

Block Diagram of Binary Communication Systems

✁ ✂✄ ☎ ✆ ✝

✞✟

✠ ✡ ✝ ☛ ☞ ✠ ✠ ✆ ✌ ✏

✞

✂ ✌ ☞ ✍

✄

✒ ✓ ✄ ☞ ✠ ✎ ✑ ✟ ✍ ✍✆ ✄ ✖ ✗ ✆ ✑

✞

✂ ✌ ☞ ✍

✄

✒ ✘ ✆ ☎ ✆ ✟ ✔ ✆ ✄ ✖ ✘ ✆ ☎

✠ ✎

✍ ✄ ✂ ☎ ✍ ✟

✠

( ) t m

{ }

k

b

1

( )

k

s t = ↔ b

2

1 ( )

k

s t = ↔ b ˆ ( ) t m

{ }

ˆ

k

b ( ) t r ( ) t w 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: E1 = Tb s2

1(t)dt, E2 = Tb

s2

2(t)dt.

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

0, N0

2

.

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Chapter 5: Optimum Receiver for Binary Data Transmission

✕ ✙ ✚✛ ✜ ✢ ✣ ✙ ✤✥ ✦ ✧ ✣ ★ ✩ ✦ ✦ ✢ ✪ ✫ ✙ ✤ ✚✪ ✩ ✬ ✙ ✛ ✭ ✮ ✛ ✩ ✦ ✯ ✰ ✥ ✬ ✬ ✢ ✛ ✱ ✲ ✢ ✰ ✙ ✤ ✚ ✪ ✩ ✬ ✙ ✛ ✭ ✳ ✢ ✜ ✢ ✥ ✴ ✢ ✛ ✱ ✳ ✢ ✜ ✙ ✦ ✯ ✬ ✛ ✚ ✜ ✬ ✥ ✙ ✦

( ) t m

{ }

k

b

1

( )

k

s t = ↔ b

2

1 ( )

k

s t = ↔ b ˆ ( ) t m

{ }

ˆ

k

b ( ) t r ( ) t w 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

bserving a set of numbers (which are random variables).

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Chapter 5: Optimum Receiver for Binary Data Transmission

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

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −1 1 t/Tb NRZ (Tx) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −5 5 t/Tb NRZ (Rx) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −1 1 t/Tb OOK (Tx) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −5 5 t/Tb OOK (Rx)

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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: Tb φ1(t)φ2(t)dt = 0 (orthogonality), Tb φ2

1(t)dt =

Tb φ2

2(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 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?

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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

rthonormal basis set {φ1(t), φ2(t)}.

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ

11

s

21

s

12

s

22

s

12 21

d d =

2

E

1

E

s1(t) = s11φ1(t) + s12φ2(t), s2(t) = s21φ1(t) + s22φ2(t), sij = Tb si(t)φj(t)dt, i, j ∈ {1, 2}, Ei = Tb s2

i (t)dt = s2 i1 + s2 i2, i ∈ {1, 2},

d12 = d21 = Tb [s2(t) − s1(t)]2dt Tb 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.

1 Let φ1(t) ≡ s1(t)

E1

. Note that s11 = √E1 and s12 = 0.

2 Project s ′

2(t) = s2(t)

E2
nto φ1(t) to obtain the correlation coefficient:

ρ = Tb s2(t) √E2 φ1(t)dt = 1 √E1E2 Tb s1(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)

Tb φ

′

2(t)2 dt

= φ

′

2(t)

1 − ρ2

= 1

1 − ρ2

s2(t) √E2 − ρs1(t) √E1

.

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Chapter 5: Optimum Receiver for Binary Data Transmission

Gram-Schmidt Procedure: Summary ) (

2 t

s ) (

1 t

s

22

s

21

s

1

E

2

E

1( )

t φ

2( )

s t ′

1( )

t ρφ

2( )

t φ α

21

d 1 cos( ) 1 ρ α − ≤ = ≤

2( )

t φ′

φ1(t) = s1(t) √E1 , φ2(t) = 1

1 − ρ2

s2(t) √E2 − ρs1(t) √E1

,

s21 = Tb s2(t)φ1(t)dt = ρ

E2,

s22 =

1 − ρ2

E2, d21 = Tb [s2(t) − s1(t)]2dt = E1 − 2ρ

E1E2 + E2.

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 1

✵

) (

2 t

s

b

T V −

✵

) (

1 t

s V

b

T

✶ ✷ ✸ ✹

) (

1 t

φ

b

T 1

b

T

✺ ✻ ✼

) (

1 t

φ ) (

1 t

s ) (

2 t

s E − E

✽ ✾ ✿

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

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 2

V −

❀

) (

1 t

s V

b

T

❀

) (

2 t

s V

b

T

❁ ❂ ❃ ❄

) (

1 t

φ

b

T 1

b

T

b

T 1 −

❄

) (

2 t

φ

b

T

b

T 1

❅ ❆ ❇

) (

1 t

φ ) (

1 t

s ) (

2 t

s E E ) (

2 t

φ

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 3

❈

) (

2 t

s

b

T V −

❈

) (

1 t

s V

b

T V α ) (

1 t

φ ) (

1 t

s ) (

2 t

s ) , (

2

α φ t = α 2

b

T = α ρ α , increasing

( )

, E

( )

, E − E

❉ ❉ ❊ ❋

❍

■−

2 3 , 2 E E 4

b

T α =

ρ = 1 E Tb s2(t)s1(t)dt = 1 V 2Tb

V 2α − V 2(Tb − α)
=

2α Tb − 1

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 4

❏

) (

2 t

s

b

T V 3

❏

) (

1 t

s V

b

T 2

b

T

❑ ▲ ▼ ◆

) (

1 t

φ

b

T 1

b

T

◆

) (

2 t

φ

b

T 2

b

T

b

T 3 −

b

T 3

❖ P ◗

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Chapter 5: Optimum Receiver for Binary Data Transmission 2( )

t φ

1( )

s t

2( )

s t

3 2 E 1( )

t φ

60

2 E E

ρ = 1 E Tb s2(t)s1(t)dt = 2 E Tb/2

2

√ 3 Tb V t

V dt =

√ 3 2 , φ2(t) = 1 (1 − 3

4 ) 1 2

s2(t) √ E − ρ s1(t) √ E

=

2 √ E

s2(t) −

√ 3 2 s1(t)

,

s21 = √ 3 2 √ E, s22 = 1 2 √ E. d21 = Tb [s2(t) − s1(t)]2dt 1

2 =

2 −

√ 3

E.

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Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5

) (

1 t

φ ) (

1 t

s ) (

2 t

s E ) (

2 t

φ 1 − = = ρ π θ 2 3 = = ρ π θ 2 = = ρ π θ

2

locus of ( ) as varies from 0 to 2 . s t θ π θ

s1(t) = √ E

2

Tb cos(2πfct), s2(t) = √ E

2

Tb cos(2πfct + θ). where fc = k 2Tb , k an integer.

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Chapter 5: Optimum Receiver for Binary Data Transmission

Obtaining Different Basis Sets by Rotation

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) (

1 t

φ θ ) (

2 t

φ

11

s

21

s

12

s

22

s ) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) (

1 t

φ θ ) (

2 t

φ

11

s

21

s

12

s

22

s

11 21

s s =

11

s

21

s

22

s

12

s

11 21

s s =

12

s

22

s

ˆ φ1(t) ˆ φ2(t)

=
cos θ

sin θ − sin θ cos θ φ1(t) φ2(t)

.

(a) Show that, regardless of the angle θ, the set {ˆ φ1(t), ˆ φ2(t)} is also an orthonormal basis set. (b) What are the values of θ that make ˆ 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 ˆ φ1(t), namely ˆ s11 and ˆ s21, are identical. Remark: Rotating counter-clockwise for positive θ and clock-wise for negative θ.

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Chapter 5: Optimum Receiver for Binary Data Transmission

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

ˆ

φ1(t) ˆ φ2(t)

=
cos θ

sin θ − sin θ cos θ φ1(t) φ2(t)

⇒
ˆ

φ1(t) = cos θ × φ1(t) + sin θ × φ2(t) ˆ φ2(t) = − sin θ × φ1(t) + cos θ × φ2(t) 5 . 5 . 1

1( )

t φ 1 2 ) 3 1 ( + 2 ) 3 1 ( − 2 ) 3 1 ( − 2 ) 3 1 ( + − ) (

2 t

φ ) (

1 t

φ

1 1 1 1 1 − 5 .

2( )

t φ

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Chapter 5: Optimum Receiver for Binary Data Transmission

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

i=1 φ1(t) = s1(t) ∞

−∞ s2 1(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)}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

) (

1 t

s 1 1

3( )

s t 2 2 3 ) (

2 t

s 1 1 − 1 1

4( )

s t 2 2 3 1 1 − 1 −

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Chapter 5: Optimum Receiver for Binary Data Transmission

Answer Found by Applying the G-S Procedure

1( )

t φ 1 2 1

3( )

t φ 2 2 3

2( )

t φ 1 1 − 1 2 1 2 1 2 −     s1(t) s2(t) s3(t) s4(t)     =     √ 2 √ 2 √ 2 1 √ 2 1       φ1(t) φ2(t) φ3(t)  

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Chapter 5: Optimum Receiver for Binary Data Transmission

Representation of Noise with Walsh Functions

−5 5 x1(t) −5 5 x2(t) 1 2 φ1(t) −2 2 φ2(t) −2 2 φ3(t) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 −2 2 φ4(t) t

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

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Chapter 5: Optimum Receiver for Binary Data Transmission

The First 16 Walsh Functions

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t

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

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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). 0.25 0.5 0.75 1 −1.5 1.5 t

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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: w(t) =

∞

i=1

wiφi(t), where wi = Tb w(t)φi(t)dt. 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, wi = Tb 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:

1

E{wi} = E Tb w(t)φi(t)dt

=

Tb E{w(t)}φi(t)dt = 0.

2

E{wiwj} = E Tb dλw(λ)φi(λ) Tb dτw(τ)φj(τ)

=
N0

2 , i = j 0, i = j .

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

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Chapter 5: Optimum Receiver for Binary Data Transmission

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

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Chapter 5: Optimum Receiver for Binary Data Transmission

Observing a waveform ⇒ Observing a set of numbers

( )

n t

φ

b

t T =

n n

= + r w

( )

d

b

T

t

∫

( )

3 t

φ

b

t T =

3 3

= + r w

( )

d

b

T

t

∫

( )

2 t

φ

b

t T =

2 2 2 i

s = + r w

( )

d

b

T

t

∫

( )

1 t

φ

b

t T =

1 1 1 i

s = + r w

( )

d

b

T

t

∫

( ) ( ) ( )

i

t s t t = + r w AWGN

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
rthonormal basis

functions are simply chosen to complete the set in order to represent noise exactly. The decision can be based on the

bservations

r1, r2, r3, r4, . . .. 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.

1

ℜ

1

ℜ

2

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

❘

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

❘

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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 = 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

ℜ

f( 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

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Chapter 5: Optimum Receiver for Binary Data Transmission

( )

n t

φ

b

t T =

n n

= + r w

( )

d

b

T

t

∫

( )

3 t

φ

b

t T =

3 3

= + r w

( )

d

b

T

t

∫

( )

2 t

φ

b

t T =

2 2 2 i

s = + r w

( )

d

b

T

t

∫

( )

1 t

φ

b

t T =

1 1 1 i

s = + r w

( )

d

b

T

t

∫

( ) ( ) ( )

i

t s t t = + r w 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

0D ℜ ⇔

2

1D ℜ ⇔ ( ) g r < ( ) g r > ( ) g r =

n

ℜ ( ) g r >

<

Optimal decision rule: ( ) 1

D D

g r > ⇒ <

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Chapter 5: Optimum Receiver for Binary Data Transmission

Equivalently, f( r|1T ) f( r|0T )

1D

0D

P1 P2 . (1) The expression f( r|1T ) f( r|0T ) is called the likelihood ratio. The decision rule in (1) was derived without specifying any statistical properties

f 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

0D

(r1 − s21)2 + (r2 − s22)2 + N0ln P1 P2

.

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

1D

0D

(r1 − s21)2 + (r2 − s22)2. ⇒ minimum-distance receiver!

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Chapter 5: Optimum Receiver for Binary Data Transmission

Minimum-Distance Receiver

(r1 − s11)2 + (r2 − s12)2

d2

1

1D

0D

(r1 − s21)2 + (r2 − s22)2

d2

2

d2

1 1D

0D

d2

2

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) , (

12 11 s

s ) , (

22 21 s

s ) ( Choose

2 t

s ) ( Choose

1 t

s

1

r

2

r ( ) r t

( )

1 2

, r r

1

d

2

d

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Chapter 5: Optimum Receiver for Binary Data Transmission

Correlation Receiver Implementation

( )

❙ •

b

T

t d

) (

2 t

φ

( )

❚ •

b

T

t d

) (

1 t

φ

b

T t =

b

T t = Decision

1

r

2

r ( ) ( ) ( )

i

t s t t = + r w

2 2 1 1 2 2

Compute ( ) ( ) ln( ) for 1, 2 and choose the smallest

i i i

r s r s N P i − + − − =

( )

❯ •

b

T

t d

) (

2 t

φ

( )

❱ •

b

T

t d

) (

1 t

φ

b

T t =

b

T t = Decision

❲ ❳ ❨ ❨ ❩ ❬ ❭ ❳ ❬ ❪ ❫ ❴ ❵ ❬ ❩ ❭

2 ) ln( 2

1 1

E P N − 2 ) ln( 2

2 2

E P N − ( ) t r

1

r

2

r Form the dot product

i

r s ⋅

❛ ❛

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Chapter 5: Optimum Receiver for Binary Data Transmission

Receiver Implementation using Matched Filters

( )

∫ •

b

T

t d

) (

2 t

φ

( )

∫ •

b

T

t d

) (

1 t

φ

b

T t =

b

T t = ) ( ) (

2 2

t T t h

b −

= φ ) ( ) (

1 1

t T t h

b −

= φ

b

T t =

b

T t = Decision ( ) t r

1

r

2

r ( ) t r

1

r

2

r Decision

EE456 – Digital Communications 32

SLIDE 33

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.6

❜

) (

2 t

s

❜

) (

1 t

s 5 . 5 . 1 1 5 . 5 . 1 2 −

❝ ❞ ❡ ❢

) (

1 t

φ 1 1

❢

) (

2 t

φ 5 . 1 − 1 1

❣ ❤ ✐

EE456 – Digital Communications 33

SLIDE 34

Chapter 5: Optimum Receiver for Binary Data Transmission

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ 5 . − 5 . 5 . 1 1 − 1 s1(t) = φ1(t) + 1 2φ2(t), s2(t) = −φ1(t) + φ2(t).

EE456 – Digital Communications 34

SLIDE 35

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. −3 −2 −1 1 2 3 −2 −1 1 2 3

(E1+E2)/2 N0

=0.99 (dB); P[error]=0.1 φ1(t) φ2(t) −3 −2 −1 1 2 3 −2 −1 1 2 3

(E1+E2)/2 N0

=6.17 (dB); P[error]=0.01 φ1(t) φ2(t)

EE456 – Digital Communications 35

SLIDE 36

Chapter 5: Optimum Receiver for Binary Data Transmission ) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ 5 . − 5 . 5 . 1 1 − 1 ) ( Choose

1 t

s ) ( Choose

2 t

s

❥ ❦ ❧

1

r

2

r ) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ 5 . − 5 . 5 . 1 1 − 1 ) ( Choose

1 t

s ) ( Choose

2 t

s

♠ ♥ ♦

1

r

2

r ) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ 5 . − 5 . 5 . 1 1 − 1 ) ( Choose

1 t

s ) ( Choose

2 t

s

♣ q r

1

r

2

r (a) P1 = P2 = 0.5, (b) P1 = 0.25, P2 = 0.75. (c) P1 = 0.75, P2 = 0.25. EE456 – Digital Communications 36

SLIDE 37

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.7

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

s

) (

1 t

φ 1

s

) (

2 t

φ 1 1 − 1 3 2 1

EE456 – Digital Communications 37

SLIDE 38

Chapter 5: Optimum Receiver for Binary Data Transmission

) (

1 t

φ ) (

1 t

s ) (

2 t

s 1 1 − 1 ) ( Choose

2 t

s ) ( Choose

1 t

s ) (

2 t

φ

t t ✉ ✈ ✇ ✇ ① ②

2 1 0 ln

4 P P N

2

r

1

r

EE456 – Digital Communications 38

SLIDE 39

Chapter 5: Optimum Receiver for Binary Data Transmission

③ ④ ⑤ ⑥ ⑦ ⑧ ⑦ ⑨ ④ ⑧

( )

⑩ •

b

T

t d

) (

2 t

φ

b

T t =

❶ ❶ ❷ ❸ ❹ ❹ ❺ ❻

=

2 1 0 ln

4 P P N T

❼

b

T 3

❽ ⑦ ❾

) (t r

2

r

2 2 2 1

choose ( ) choose ( ) r T s t r T s t ≥

❿

<

❿ ➀ ➁ ➂ ➃ ➄ ➅ ➄ ➆ ➁ ➅

b

T t =

➇ ➇ ➈ ➉ ➊ ➊ ➋ ➌

=

2 1 0 ln

4 P P N T ) (

2 t

h

➍

b

T 3

➎ ➏ ➐

2 2 2 1

choose ( ) choose ( ) r T s t r T s t ≥

➑

<

➑

2

r ) (t r

EE456 – Digital Communications 39

SLIDE 40

Chapter 5: Optimum Receiver for Binary Data Transmission

Implementation with One Correlator/Matched Filter

Always possible by choosing ˆ φ1(t) and ˆ φ2(t) such that one of the two basis functions is perpendicular to the line joining the two signals. ) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) (

1 t

φ θ ) (

2 t

φ

11

s

21

s

12

s

22

s

11 21

s s =

12

s

22

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

(ˆ r1 − ˆ s11)2 + (ˆ r2 − ˆ s12)2

d2

1 1D

0D

(ˆ r1 − ˆ s21)2 + (ˆ r2 − ˆ s22)2

d2

2

⇔ (ˆ r2 − ˆ s12)2

1D

0D

(ˆ r2 − ˆ s22)2 ˆ r2

1D

0D

ˆ s22 + ˆ s12 2

midpoint of two signals

+

N0/2

ˆ s22 − ˆ s12

ln

P1 P2

equal to 0 if P1=P2

≡ T.

EE456 – Digital Communications 40

SLIDE 41

Chapter 5: Optimum Receiver for Binary Data Transmission

➒ ➓ ➔ →➣ ↔ ➣ ↕ ➓ ↔

( )

➙ •

b

T

t d

) ( ˆ

2 t

φ

b

T t = T Threshold

➛ ➣ ➜

) (t r

2

ˆ r

2 2

ˆ 1 ˆ

D D

r T r T ≥

➝

<

➝ ➞ ➟ ➠ ➡➢ ➤ ➢ ➥ ➟ ➤

b

T t = T Threshold ) ( ˆ ) (

2

t T t h

b −

= φ

➦ ➧ ➨

2

ˆ r ) (t r

2 2

ˆ 1 ˆ

D D

r T r T ≥

➩

<

➩

ˆ φ2(t) = s2(t) − s1(t) (E2 − 2ρ√E1E2 + E1)

1 2

, T ≡ ˆ s22 + ˆ s12 2 +

N0/2

ˆ s22 − ˆ s12

ln

P1 P2

.

EE456 – Digital Communications 41

SLIDE 42

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.8

) (

1 t

φ ) (

1 t

s ) (

2 t

s ) (

2 t

φ ) ( ˆ

1 t

φ ) ( ˆ

2 t

φ

21 11

ˆ ˆ s s = 4 / π θ = E E

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

EE456 – Digital Communications 42

SLIDE 43

Chapter 5: Optimum Receiver for Binary Data Transmission

➫ ➭ ➯ ➲➳ ➵ ➳ ➸ ➭ ➵

( )

➺ •

b

T

t d

) ( ˆ

2 t

φ

b

T t = T Threshold

➻ ➳ ➼ ➽

b

T

b

T 2 2

b

T

) (t r

2

ˆ r

2 2

ˆ 1 ˆ

D D

r T r T ≥

➾

<

➾ ➚ ➪ ➶ ➹➘ ➴ ➘ ➷ ➪ ➴

b

T t = T Threshold

➬ ➮ ➱ ✃

b

T 2 2

b

T ) (t h

) (t r

2

ˆ r

2 2

ˆ 1 ˆ

D D

r T r T ≥

❐

<

❐

EE456 – Digital Communications 43

SLIDE 44

Chapter 5: Optimum Receiver for Binary Data Transmission

Receiver Performance

To detect bk, compare ˆ r2 = kTb

(k−1)Tb

r(t)ˆ φ2(t)dt to the threshold T = ˆ

s12+ˆ s22 2

+

N0 2(ˆ s22−ˆ s12) ln

P1

P2

.

12

ˆ s

22

ˆ s

( )

T

r f ˆ

2

( )

T

r f 1 ˆ

2 T T

1 choose choose

❒

⇐

2

ˆ r

❮ ❰ Ï Ð Ñ Ð Ò Ó Ô Ò ÕÓ Ö× Ø Ù

T P [error] = P [(0 transmitted and 1 decided) or (1 transmitted and 0 decided)] = P [(0T , 1D) or (1T , 0D)].

EE456 – Digital Communications 44

SLIDE 45

Chapter 5: Optimum Receiver for Binary Data Transmission

Ú Û Ü Ý Þ ß à á á à â ã ä ß à å æ ç à èé â ê å ë ì à â í ë ç à èé â ê å ë ì à â í ë î ã é ï å é â ä ð ñ ñ é ò à å é â ä ó ôõ æ ö õ à ä ÷ ó ô æ ø õ ñ é í é á ë î ã é ï å ê é ñ ù â é ø æ ä ä ô æ ø õ ñ é í é á ë î ã é ï å ê é ñ ù â é ø æ ä ä ó ôõ æ ö õ à ä ÷ ó ß à á á à â ã ä ß à å æ ç à èé â ê å ë ì à â í ë ç à èé â ê å ë ì à â í ë î ã é ï å é â ä ð ñ ñ é ò à å é â ä

12

ˆ s

22

ˆ s

( )

T

r f ˆ

2

( )

T

r f 1 ˆ

2 T T

1 choose choose

ú

⇐

2

ˆ r

Ú Û Ü Ý û

T P [error] = P [0T , 1D] + P [1T , 0D] = P [1D|0T ]P [0T ] + P [0D|1T ]P [1T ] = P1 ∞

T

f(ˆ r2|0T )dˆ r2

Area B

+P2 T

−∞

f(ˆ r2|1T )dˆ r2

Area A

= P1Q

T − ˆ

s12

N0/2
+ P2
1 − Q
T − ˆ

s22

N0/2
.

EE456 – Digital Communications 45

SLIDE 46

Chapter 5: Optimum Receiver for Binary Data Transmission

Q-function

ü ý þ ÿ ✁

✂
✄

☎ ✆✝

✞

✟ ✠

✁

✡☛

þ ý☞

☞ ✓ ü ÿ ý ✔ ÿ ✌ ☞ ✍ ✓ ✗ ✌ ✄ ✄ ✌ ☛ ✝ ☞ ✗ ✌ ✟ ý ✘ ✌ ✎

☛

✠ ✟ ☎ ✚ ✌ ☛ ✂ ☎ ✘ ✌ ✎

☛

✠ ✟ ☎ ✚ ✌ ☛ ✂ ☎ ✆✝

✞

✟

☛

☞ ✛ ✁ ✁

✏

✌ ✟

☛

☞

λ x ) ( Area x Q =

2

e 2 1

λ

π

−

Q(x) ≡ 1 √ 2π ∞

x

exp

− λ2

2

dλ.

1 2 3 4 5 6 10

−10

10

−8

10

−6

10

−4

10

−2

10 x Q(x) EE456 – Digital Communications 46

SLIDE 47

Chapter 5: Optimum Receiver for Binary Data Transmission

Performance when P1 = P2

12

s

22

s

( )

T

r f

2

( )

T

r f 1

2 T T

1 choose choose ⇒ ⇐

2

r

12 22

2 s s T + =

( )

22 12

2 /2 s s N

Q

−

=

P [error] = Q

ˆ

s22 − ˆ 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 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

SLIDE 48

Chapter 5: Optimum Receiver for Binary Data Transmission

Example 5.9

) (

1 t

φ

T

t s ) (

1

⇔ ) ( 1

2 t

s

T ⇔

) (

2 t

φ 1 − 2 2 − 1 1 − 1 2 2 − 1 1 t 1 1 t 1 −

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

EE456 – Digital Communications 48

SLIDE 49

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. (c) Find and sketch the two orthonormal basis functions ˆ φ1(t) and ˆ φ2(t) such that the optimum receiver can be implemented using only the projection ˆ r2 of the received signal r(t) onto the basis function ˆ 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 ˆ φ1(t) is not useful at the receiver in determining which bit was transmitted, one should not even transmit this component of the signal. Thus she modifies the transmitted signal as follows: s(M)

1

(t) = s1(t) −

component of s1(t) along ˆ

φ1(t)

s(M)

2

(t) = s2(t) −

component of s2(t) along ˆ

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.

EE456 – Digital Communications 49

SLIDE 50

Chapter 5: Optimum Receiver for Binary Data Transmission

t 3 − ) (

2 t

s 1 − 1 ) (

1 t

s 1 3 t 1 −

EE456 – Digital Communications 50

SLIDE 51

Chapter 5: Optimum Receiver for Binary Data Transmission

) (

1 t

φ

T

t s ) (

1

⇔ ) ( 1

2 t

s

T ⇔

) (

2 t

φ 1 − 2 2 − 1 1 − 1 2 2 −

2

ˆ ( ) t φ 4 π θ = −

✑ ✒ ✕ ✖ ✙ ✖ ✜ ✢ ✣ ✜ ✤ ✢ ✥✦ ✧ ★

D D

1

ˆ ( ) t φ

2 ( ) M

s t

1 ( ) M

s t 1 1 t 1 − 1 1 t

EE456 – Digital Communications 51

SLIDE 52

Chapter 5: Optimum Receiver for Binary Data Transmission

ˆ φ1(t) ˆ 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)

.

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

t 1 t 2 − 1/2 2 1/2

1

ˆ ( ) t φ

2

ˆ ( ) t φ

2

ˆ ( ) (1 ) h t t φ = − 1 t =

2 2

ˆ ˆ 1

D D

r r ≥

✩

<

✩

) (t r t 2 1/2 1

EE456 – Digital Communications 52

SLIDE 53

Chapter 5: Optimum Receiver for Binary Data Transmission

Antipodal Signalling

b

kT t =

+

( ) t w

( ) ( )

( )

k k

E κ = ± + y w ( ) x t t (1) ( 1) − ( 1) −

b

T 2

b

T ( ) y t

T( )

( ) h t p t =

R( )

( )

b

h t p T t κ = −

2

PSD

N 2

(0, )

N E 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 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 P [error] = Q

2E

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

SLIDE 54

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 N0

2 κ2E. Thus, the SNR is

SNR = Signal power Noise power = (±κE)2

N0 2 κ2E

= 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 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

SLIDE 55

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

SLIDE 56

Chapter 5: Optimum Receiver for Binary Data Transmission

Antipodal Baseband Signalling with Rectangular Pulse Shaping

k

b ( ) ( )

b

h t p T t = −

EE456 – Digital Communications 56

SLIDE 57

Chapter 5: Optimum Receiver for Binary Data Transmission

Antipodal Baseband Signalling with Half-Sine Pulse Shaping

k

b ( ) ( )

b

h t p T t = −

EE456 – Digital Communications 57

SLIDE 58

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. t ( )

T

s t

b

T 2 b T 4 b T

1( )

s t

1(

2 )

b

s t T −

b

T − 2 b T − 3 b T

2(

3 )

b

s 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
n

Tb

+ P2S2
n

Tb

Tb
2

δ

f − n

Tb

.

EE456 – Digital Communications 58

SLIDE 59

Chapter 5: Optimum Receiver for Binary Data Transmission

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
n

Tb

+ P2S2
n

Tb

Tb
2

δ

f − n

Tb

.

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

T →∞

E{|GT (f)|2} T = · · · = P1P2 Tb |S1(f) − S2(f)|2. SsT (f) = P1P2 Tb |S1(f) − S2(f)|2 +

∞

n=−∞
P1S1
n

Tb

+ P2S2
n

Tb

Tb
2

δ

f − n

Tb

.

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): SsT (f) = |P (f)|2 Tb

EE456 – Digital Communications 59

SLIDE 60

Chapter 5: Optimum Receiver for Binary Data Transmission

Baseband Message Signals with Different Pulse Shaping Filters

2 4 6 8 10 12 14 −1 −0.5 0.5 1 Information bits or amplitude levels 2 4 6 8 10 12 14 −2 −1 1 2 Output of the transmit pulse shaping filter − Rectangular −2 2 −60 −40 −20 f × Tb Magnitude (dB) PSD - Rectangular 2 4 6 8 10 12 14 −2 −1 1 2 Half-sine pulse shaping filter −2 2 −60 −40 −20 f × Tb Magnitude (dB) PSD - Half-sine 2 4 6 8 10 12 14 −2 −1 1 2 t/Tb SRRC pulse shaping filter (β = 0.5) −2 2 −60 −40 −20 Normalized frequency, f × Tb Magnitude (dB) PSD - SRRC (30 symbols long) EE456 – Digital Communications 60

SLIDE 61

Chapter 5: Optimum Receiver for Binary Data Transmission

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

S/P

bits-to- levels mapping pulse shaping filter

t DAC N data bits NCO ˆ

c

f

ˆ cos( )

cn

T[ ]

( ) h n p nT

sym

T N T

T

sym

[ ] ( ) [ ] ( )

k

i n i nT a k nT kT

sym

[ ] [ ] [ ] [ ] ( )

k

s n i n h n a k p nT kT

( )

( ) ( ) cos( )

s c k

s t a k p t kT t

low-rate sequence, index k

high-rate sequences, index n

... ... ...

DAC correction filter

ˆ

c c s

F

Figure 1:

Block diagram of the transmitter.

EE456 – Digital Communications 61

SLIDE 62

Chapter 5: Optimum Receiver for Binary Data Transmission

∑

cos( )

cn

ω θ − sin( )

cn

ω θ −

c

f

[ ]

c

x n [ ]

s

x n [ ] x n [ ] x n d −

cd

ω θ =

R[ ]

( ) h n p nT = − T

sym

T k n T = ( ) s t

θ

Figure 2: Block diagram of the receiver.

EE456 – Digital Communications 62