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

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 { } b = ↔ s t b m 0 ( ) t ( ) k k 1 ✏ � ✞ ✂ ✌ ☞ ✍ � ✄ 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 finite energies: E 1 = � T b 1 ( t )d t , E 2 = � T b s 2 s 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 ) } = 0 , E { w ( t ) w ( t + τ ) } = N 0 0 , N 0 2 δ ( τ ) , w ( t ) ∼ N . 2 EE456 – Digital Communications 2

Chapter 5: Optimum Receiver for Binary Data Transmission { } b = ↔ s t b m t 0 ( ) ( ) k 1 k ✫ ✙ ✤ ✚✪ ✩ ✬ ✙ ✛ ✕ ✙ ✚✛ ✜ ✢ ✣ ✙ ✤✥ ✦ ✧ b = ↔ s t ✣ ★ ✩ ✦ ✦ ✢ ✪ 1 ( ) ✭ ✮ ✛ ✩ ✦ ✯ ✰ ✥ ✬ ✬ ✢ ✛ ✱ k 2 ✲ ✢ ✰ ✙ ✤ ✚ ✪ ✩ ✬ ✙ ✛ { } m r t 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). EE456 – Digital Communications 3

Chapter 5: Optimum Receiver for Binary Data Transmission Can you Identify the Signal Sets { s 1 ( t ) , s 2 ( t ) } ? NRZ (Tx) 1 0 −1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t/T b NRZ (Rx) 5 0 −5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t/T b OOK (Tx) 1 0 −1 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t/T b OOK (Rx) 5 0 −5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 t/T b EE456 – Digital Communications 4

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 ) form a set of orthonormal basis functions if and only if: 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 If { φ 1 ( t ) , φ 2 ( t ) } can be found, 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 ) . It can be checked that the coefficients s ij can be calculated as follows: � T b s ij = s i ( t ) φ j ( t )d t, i, j ∈ { 1 , 2 } , 0 An important question is: Given the signal set s 1 ( t ) and s 2 ( t ) , can one always find an orthonormal basis functions to represent { s 1 ( t ) , s 2 ( t ) } exactly? If the answer is YES, is the set of orthonormal basis functions UNIQUE? EE456 – Digital Communications 5

Chapter 5: Optimum Receiver for Binary Data Transmission Geometric Representation of Signals s 1 ( t ) and s 2 ( t ) (II) Provided that φ 1 ( t ) and φ 2 ( 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 { φ 1 ( t ) , φ 2 ( t ) } . φ 2 t ( ) s 2 t ( ) s 1 ( t ) = s 11 φ 1 ( t ) + s 12 φ 2 ( t ) , s 22 = d d s 2 ( t ) = s 21 φ 1 ( t ) + s 22 φ 2 ( t ) , 12 21 � T b E s ij = s i ( t ) φ j ( t )d t, i, j ∈ { 1 , 2 } , 2 s s 1 t ( ) 0 12 � T b s 2 i ( t )d t = s 2 i 1 + s 2 E i = i 2 , i ∈ { 1 , 2 } , E 0 1 φ �� T b 1 t ( ) s s 0 11 [ s 2 ( t ) − s 1 ( t )] 2 d t 21 d 12 = d 21 = 0 � T b s i ( t ) φ j ( t )d t is the projection of signal s i ( t ) onto basis function φ j ( t ) . 0 The length of a signal vector equals to the square root of its energy. It is always possible to find orthonormal basis functions φ 1 ( t ) and φ 2 ( t ) to represent s 1 ( t ) and s 2 ( t ) exactly. In fact, there are infinite number of choices! EE456 – Digital Communications 6

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. . Note that s 11 = √ E 1 and s 12 = 0 . 1 Let φ 1 ( t ) ≡ s 1 ( t ) � E 1 2 ( t ) = s 2 ( t ) ′ 2 Project s onto φ 1 ( t ) to obtain the correlation coefficient : � E 2 � 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 ) 3 Subtract ρφ 1 ( t ) from s 2 ( t ) to obtain φ √ E 2 − ρφ 1 ( t ) . ′ 4 Finally, normalize φ 2 ( t ) to obtain: ′ ′ φ 2 ( t ) φ 2 ( t ) φ 2 ( t ) = = �� T b � 2 ( t ) � 2 d t � φ 1 − ρ 2 ′ 0 � s 2 ( t ) � 1 − ρs 1 ( t ) √ E 2 √ E 1 = . � 1 − ρ 2 EE456 – Digital Communications 7

Chapter 5: Optimum Receiver for Binary Data Transmission Gram-Schmidt Procedure: Summary φ t 2 ( ) s 1 ( t ) ′ s t φ 1 ( t ) = √ E 1 , 2 ( ) φ ′ t 2 ( ) � s 2 ( t ) � 1 − ρs 1 ( t ) s 2 t φ 2 ( t ) = √ E 2 √ E 1 , ( ) � s 1 − ρ 2 22 � T b E � d 2 s 21 = s 2 ( t ) φ 1 ( t )d t = ρ E 2 , 21 0 ρφ α t 1 ( ) �� � � 1 − ρ 2 φ t s 22 = E 2 , 1 ( ) 0 E s s 1 t �� T b ( ) 21 1 [ s 2 ( t ) − s 1 ( t )] 2 d t d 21 = 0 − ≤ ρ = α ≤ 1 cos( ) 1 � E 1 − 2 ρ = E 1 E 2 + E 2 . EE456 – Digital Communications 8

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. EE456 – Digital Communications 9

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 ( ) ( ) φ 2 t ( ) T T 1 1 b b s 2 t E ( ) T b T 0 ❄ 0 ❄ b s 1 t ( ) − T 1 b φ 1 t ( ) 0 E ❅ ❆ ❇ EE456 – Digital Communications 10

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 t α ( , ) 2 T α = b 2 � Tb s 2 t 1 ( ) E E 3 ρ = s 2 ( t ) s 1 ( t )d t ■ − , ❋ E α ρ 0 ● ❉ 2 2 increasing , ● ❉ 1 � � V 2 α − V 2 ( T b − α ) ❍ ❊ = V 2 T b E T α = b 2 α α = 0 = − 1 4 T b s 1 t ( ) φ 1 t ( ) ( ) ( ) 0 − E E , 0 , 0 EE456 – Digital Communications 11

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 ◗ EE456 – Digital Communications 12

Chapter 5: Optimum Receiver for Binary Data Transmission φ 2 ( ) t s t 2 ( ) E E s t 1 ( ) 2 o 60 φ 1 ( ) t 0 3 E 2 √ √ � � � Tb � Tb/ 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 − ρ s 1 ( t ) 2 3 φ 2 ( t ) = √ √ = √ s 2 ( t ) − 2 s 1 ( t ) , 1 (1 − 3 E E E 4 ) 2 √ √ √ 3 s 22 = 1 s 21 = E, E. 2 2 � 1 �� Tb �� √ 2 = � [ s 2 ( t ) − s 1 ( t )] 2 d t d 21 = 2 − 3 E. 0 EE456 – Digital Communications 13

Chapter 5: Optimum Receiver for Binary Data Transmission Example 5 φ 2 t ( ) � √ 2 θ = π 3 2 s 1 ( t ) = E cos(2 πf c t ) , 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 EE456 – Digital Communications 14