A First Course in Digital Communications Ha H. Nguyen and E. Shwedyk - PowerPoint PPT Presentation

Chapter 5: Optimum Receiver for Binary Data Transmission A First Course in Digital Communications Ha H. Nguyen and E. Shwedyk February 2009 A First Course in Digital Communications 1/58

Chapter 5: Optimum Receiver for Binary Data Transmission Block Diagram of Binary Communication Systems { } b = ↔ s t m b t 0 ( ) ✞ ☞ ( ) ✍ ✎ ✁ ✂ ✌ ✁ ✄ k 1 k ☛ ☞ ✠ ✠ � ✞ ✟ ✠ ✡ ✝ ✆ ✌ ✁ ✂ ✄ ☎ ✆ ✝✁ ✏ ✑ ☞ ✠ ✒ ✟ ✔ ✎ ✎ ✄ ✓ ✆ ✄ b = ↔ s t 1 ( ) k 2 ✞ ☞ ✕ ✎ ✆ ✓ ✁ ✂ ✌ ✁ ✄ ✠ ✒ ✟ ✠ ✖ ✎ ✎ ✆ ☎ ✁ ✄ ✂ ☎ ✁ { } ✏ ✖ ✟ ✗ ✔ m t r ✆ ☎ ✆ ✆ ✄ t ˆ ( ) b ( ) ˆ k w t ( ) Bits in two different time slots are statistically independent . a priori probabilities: P [ b k = 0] = P 1 , P [ b k = 1] = P 2 . Signals s 1 ( t ) and s 2 ( t ) have a duration of T b seconds and � T b � T b s 2 s 2 finite energies: E 1 = 1 ( t )d t , E 2 = 2 ( t )d t . 0 0 Noise w ( t ) is stationary Gaussian , zero-mean white noise with two-sided power spectral density of N 0 / 2 (watts/Hz): E { w ( t ) w ( t + τ ) } = N 0 E { w ( t ) } = 0 , 2 δ ( τ ) . A First Course in Digital Communications 2/58

Chapter 5: Optimum Receiver for Binary Data Transmission { } b = ↔ s t m b t 0 ( ) ✤ ✩ ( ) ✫ ✬ ✙ ✚ ✪ ✙ ✛ k 1 k ★ ✩ ✦ ✦ ✘ ✤ ✥ ✦ ✧ ✣ ✢ ✪ ✙ ✚ ✛ ✜ ✢ ✣✙ ✭ ✮ ✩ ✦ ✯ ✥ ✱ ✬ ✬ ✛ ✰ ✢ ✛ b = ↔ s t 1 ( ) k 2 ✤ ✩ ✲ ✬ ✢ ✰ ✙ ✚ ✪ ✙ ✛ ✦ ✯ ✥ ✦ ✳ ✬ ✬ ✢ ✜ ✙ ✛ ✚ ✜ ✙ { } ✭ ✳ ✥ ✴ ✱ m t r ✢ ✜ ✢ ✢ ✛ t ˆ ( ) b ( ) ˆ k w t ( ) Received signal over [( k − 1) T b , kT b ] : r ( t ) = s i ( t − ( k − 1) T b ) + w ( t ) , ( k − 1) T b ≤ t ≤ kT b . 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). A First Course in Digital Communications 3/58

Chapter 5: Optimum Receiver for Binary Data Transmission Geometric Representation of Signals s 1 ( t ) and s 2 ( t ) (I) Wish to represent two arbitrary signals s 1 ( t ) and s 2 ( t ) as linear combinations of two orthonormal basis functions φ 1 ( t ) and φ 2 ( t ) . φ 1 ( t ) and φ 2 ( t ) are orthonormal if: � T b φ 1 ( t ) φ 2 ( t )d t = 0 ( ortho gonality ) , 0 � T b � T b φ 2 φ 2 1 ( t )d t = 2 ( t )d t = 1 ( normal ized to have unit energy ) . 0 0 The representations are s 1 ( t ) = s 11 φ 1 ( t ) + s 12 φ 2 ( t ) , s 2 ( t ) = s 21 φ 1 ( t ) + s 22 φ 2 ( t ) . � T b where s ij = s i ( t ) φ j ( t )d t, i, j ∈ { 1 , 2 } , 0 A First Course in Digital Communications 4/58

Chapter 5: Optimum Receiver for Binary Data Transmission Geometric Representation of Signals s 1 ( t ) and s 2 ( t ) (II) φ 2 t ( ) s 1 ( t ) = s 11 φ 1 ( t ) + s 12 φ 2 ( t ) , s s 1 t ( ) s 2 ( t ) = s 21 φ 1 ( t ) + s 22 φ 2 ( t ) , 12 � T b s ij = s i ( t ) φ j ( t )d t, i, j ∈ { 1 , 2 } , 0 s s 2 t ( ) 22 φ 1 t ( ) s s 0 11 21 � T b s i ( t ) φ j ( t )d t is the projection of s i ( t ) onto φ j ( t ) . 0 How to choose orthonormal functions φ 1 ( t ) and φ 2 ( t ) to represent s 1 ( t ) and s 2 ( t ) exactly? A First Course in Digital Communications 5/58

Chapter 5: Optimum Receiver for Binary Data Transmission Gram-Schmidt Procedure . Note that s 11 = √ E 1 and s 12 = 0 . Let φ 1 ( t ) ≡ s 1 ( t ) 1 � E 1 2 ( t ) = s 2 ( t ) ′ Project s onto φ 1 ( t ) to obtain the correlation 2 � E 2 coefficient : � T b � T b s 2 ( t ) 1 ρ = √ E 2 φ 1 ( t )d t = √ E 1 E 2 s 1 ( t ) s 2 ( t )d t. 0 0 2 ( t ) = s 2 ( t ) ′ ′ Subtract ρφ 1 ( t ) from s 2 ( t ) to obtain φ √ E 2 − ρφ 1 ( t ) . 3 ′ Finally, normalize φ 2 ( t ) to obtain: 4 ′ ′ φ 2 ( t ) φ 2 ( t ) φ 2 ( t ) = = �� T b � � 2 d t � 1 − ρ 2 ′ φ 2 ( t ) 0 � s 2 ( t ) � 1 − ρs 1 ( t ) = √ E 2 √ E 1 . � 1 − ρ 2 A First Course in Digital Communications 6/58

Chapter 5: Optimum Receiver for Binary Data Transmission Gram-Schmidt Procedure: Summary φ t 2 ( ) s 1 ( t ) φ 1 ( t ) = √ E 1 , � s 2 ( t ) � 1 − ρs 1 ( t ) √ E 2 √ E 1 φ 2 ( t ) = , ′ s t � 2 ( ) φ ′ t 1 − ρ 2 2 ( ) � T b s 2 t � ( ) s s 21 = s 2 ( t ) φ 1 ( t )d t = ρ E 2 , 22 0 E d �� � � 2 s 22 = 1 − ρ 2 E 2 , 21 ρφ α t 1 ( ) �� T b φ t 1 ( ) 0 E s s 1 t d 21 = [ s 2 ( t ) − s 1 ( t )] 2 d t ( ) 21 1 0 � − ≤ ρ = α ≤ 1 cos( ) 1 = E 1 − 2 ρ E 1 E 2 + E 2 . A First Course in Digital Communications 7/58

Chapter 5: Optimum Receiver for Binary Data Transmission Gram-Schmidt Procedure for M Waveforms { s i ( t ) } M i =1 s 1 ( t ) φ 1 ( t ) = , �� ∞ −∞ s 2 1 ( t )d t ′ φ i ( t ) φ i ( t ) = , i = 2 , 3 , . . . , N, �� ∞ � 2 d t � ′ φ i ( t ) −∞ i − 1 s i ( t ) � ′ φ i ( t ) = √ E i − ρ ij φ j ( t ) , j =1 � ∞ s i ( t ) √ E i ρ ij = φ j ( t )d t, j = 1 , 2 , . . . , i − 1 . −∞ If the waveforms { s i ( t ) } M i =1 form a linearly independent set , then N = M . Otherwise N < M . A First Course in Digital Communications 8/58

Chapter 5: Optimum Receiver for Binary Data Transmission Example 1 s 1 t s 2 t ( ) ( ) V T b ✵ ✵ T 0 0 b ✶ ✷ ✸ − V φ 1 t ( ) T 1 b s 2 t s 1 t ( ) ( ) φ 1 t ✹ ( ) T 0 b − E 0 E ✺ ✻ ✼ ✽ ✾ ✿ (a) Signal set. (b) Orthonormal function. (c) Signal space representation. A First Course in Digital Communications 9/58

Chapter 5: Optimum Receiver for Binary Data Transmission Example 2 s 1 t s 2 t ( ) ( ) V V T b ❀ ❀ T 0 0 b − V ❁ ❂ ❃ φ 2 t φ 1 t ( ) ( ) T T 1 1 b b T b ❄ ❄ T 0 0 b − T 1 b ❅ ❆ ❇ A First Course in Digital Communications 10/58 φ ( t )

Chapter 5: Optimum Receiver for Binary Data Transmission Example 3 s 1 t s 2 t ( ) ( ) V V T α b ❈ ❈ T 0 0 b − V φ α t ( , ) 2 T α = b � T b 1 2 ρ = s 2 ( t ) s 1 ( t )d t s 2 t ■ ❋ ( ) E E ● 3 ❉ E − ● ❉ , 0 ❍ ❊ α ρ 2 2 increasing , 1 � � V 2 α − V 2 ( T b − α ) = V 2 T b E T α = b α = 2 α 0 4 = − 1 s 1 t ( ) T b φ 1 t ( ) ( ) ( ) 0 − E E , 0 , 0 A First Course in Digital Communications 11/58

Chapter 5: Optimum Receiver for Binary Data Transmission Example 4 s 1 t s 2 t ( ) ( ) V 3 V ❏ ❏ T T T 0 0 2 b b b ❑ ▼ ▲ φ 1 t φ 2 t ( ) ( ) 3 T b T 1 b T b ◆ ◆ T T 0 0 2 b b 3 − T b ❖ P ◗ A First Course in Digital Communications 12/58

Chapter 5: Optimum Receiver for Binary Data Transmission φ 2 t ( ) s 2 t ( ) ( ) E E 3 2 , 2 s 1 t ( ) φ 1 t ( ) ( ) 0 E , 0 √ √ � � � T b � T b / 2 ρ = 1 s 2 ( t ) s 1 ( t )d t = 2 2 3 3 V t V d t = 2 , E E T b 0 0 √ � � � s 2 ( t ) � 1 − ρs 1 ( t ) 2 3 φ 2 ( t ) = √ √ = √ s 2 ( t ) − 2 s 1 ( t ) , 1 (1 − 3 4 ) E E E 2 √ √ √ 3 s 22 = 1 s 21 = E, E. 2 2 � 1 �� T b �� 2 √ � [ s 2 ( t ) − s 1 ( t )] 2 d t d 21 = = 2 − 3 E. 0 A First Course in Digital Communications 13/58

Chapter 5: Optimum Receiver for Binary Data Transmission Example 5 φ 2 t ( ) � √ 2 s 1 ( t ) = E cos(2 πf c t ) , θ = π 3 2 T b ρ = 0 � √ 2 s 2 ( t ) = E cos(2 πf c t + θ ) . T b k where f c = 2 T b , k an integer. s 1 t ( ) φ 1 t ( ) 0 θ θ = π s t θ locus of ( ) as ρ = − E 1 2 π varies from 0 to 2 . s 2 t ( ) θ = π 2 ρ = 0 A First Course in Digital Communications 14/58

Chapter 5: Optimum Receiver for Binary Data Transmission Representation of Noise with Walsh Functions 5 x 1 ( t ) 0 −5 5 x 2 ( t ) 0 −5 2 φ 1 ( t ) 1 0 2 φ 2 ( t ) 0 −2 2 φ 3 ( t ) 0 −2 2 φ 4 ( t ) 0 −2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t Exact representation of noise with 4 Walsh functions is not possible. A First Course in Digital Communications 15/58

Chapter 5: Optimum Receiver for Binary Data Transmission The First 16 Walsh Functions 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t Exact representations might be possible with many more Walsh functions. A First Course in Digital Communications 16/58

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 0.25 0.5 0.75 1 t A First Course in Digital Communications 17/58

Chapter 5: Optimum Receiver for Binary Data Transmission Representation of the Noise I To represent the random noise signal, w ( t ) , in the time interval [( k − 1) T b , kT b ] , need to use a complete orthonormal set of known deterministic functions: � T b ∞ � w ( t ) = w i φ i ( t ) , where w i = w ( t ) φ i ( t )d t. 0 i =1 The coefficients w i ’s are random variables and understanding their statistical properties is imperative in developing the optimum receiver. A First Course in Digital Communications 18/58