Optimal Receiver using Complex Baseband Representation Saravanan - PowerPoint PPT Presentation

Optimal Receiver using Complex Baseband Representation Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay September 26, 2013 1 / 13

Passband Signals in Passband Noise Consider M -ary passband signaling over a channel with passband Gaussian noise H i : y p ( t ) = s i , p ( t ) + n p ( t ) , i = 1 , . . . , M where y p ( t ) Real passband received signal s i , p ( t ) Real passband signals n p ( t ) Real passband GN with PSD N 0 2 Signal Noise N 0 2 − f c f c f Note: A WSS random process is passband if its autocorrelation function is a passband signal 2 / 13

Passband Signals in Passband Noise Consider M -ary passband signaling over a channel with passband Gaussian noise H i : y p ( t ) = s i , p ( t ) + n p ( t ) , i = 1 , . . . , M where y p ( t ) Real passband received signal s i , p ( t ) Real passband signals n p ( t ) Real passband GN with PSD N 0 2 The equivalent problem in complex baseband is H i : y ( t ) = s i ( t ) + n ( t ) , i = 1 , . . . , M where y ( t ) Complex envelope of y p ( t ) s i ( t ) Complex envelope of s i , p ( t ) n ( t ) Complex envelope of n p ( t ) What is the optimal receiver in terms of the complex baseband signals? 3 / 13

Complex Envelope of Passband Gaussian Noise • The complex baseband representation of n p ( t ) is given by 1 n p ( t )] e − j 2 π f c t n ( t ) = n c ( t ) + jn s ( t ) = √ [ n p ( t ) + j ˆ 2 where ˆ n p ( t ) is the Hilbert transform of n p ( t ) • The inphase and quadrature components of n ( t ) are given by 1 [ n p ( t ) cos 2 π f c t + ˆ n c ( t ) = √ n p ( t ) sin 2 π f c t ] 2 1 √ [ˆ n s ( t ) = n p ( t ) cos 2 π f c t − n p ( t ) sin 2 π f c t ] 2 • n c ( t ) and n s ( t ) are jointly Gaussian and independent random processes (Proof in Proakis Section 2.9) 4 / 13

In-Phase and Quadrature Component PSDs N 0 � | f − f c | < W S n p ( f ) = 2 0 otherwise Passband Gaussian Noise PSD N 0 2 − f c f c f N 0 � | f | < W S n c ( f ) = S n s ( f ) = 2 0 otherwise In-Phase and Quadrature Component PSDs N 0 2 − f c f c f 5 / 13

Complex Envelope PSD • By the independence of n c ( t ) and n s ( t ) , we have R n ( τ ) = E [ n ( t + τ ) n ∗ ( t )] = R n c ( τ ) + R n s ( τ ) • S n ( f ) = S n c ( f ) + S n s ( f ) Complex Envelope PSD N 0 − f c f c f • If n c ( t ) and n s ( t ) are approximated by white Gaussian noise, n ( t ) is said to be complex white Gaussian noise 6 / 13

Complex White Gaussian Noise Definition (Complex Gaussian Random Process) A complex random process Z ( t ) = X ( t ) + jY ( t ) is a Gaussian random process if X ( t ) and Y ( t ) are jointly Gaussian random processes. Definition (Complex White Gaussian Noise) A complex Gaussian random process Z ( t ) = X ( t ) + jY ( t ) is complex white Gaussian noise with PSD N 0 if X ( t ) and Y ( t ) are independent white Gaussian noise processes with PSD N 0 2 . 7 / 13

Optimal Detection in Complex Baseband • The continuous time hypothesis testing problem in complex baseband H i : y ( t ) = s i ( t ) + n ( t ) , i = 1 , . . . , M where y ( t ) Complex envelope of y p ( t ) s i ( t ) Complex envelope of s i , p ( t ) n ( t ) Complex white Gaussian noise with PSD N 0 = 2 σ 2 • The equivalent problem in terms of complex random vectors H i : Y = s i + N , i = 1 , . . . , M where Y , s i and N are the projections of y ( t ) , s i ( t ) and n ( t ) respectively onto the signal space spanned by { s i ( t ) } . • m = E [ N ] = 0 , C N = 2 σ 2 I 8 / 13

Complex White Gaussian Noise through Correlators E [ � n , ψ 1 � ( � n , ψ 2 � ) ∗ ] cov ( � n , ψ 1 � , � n , ψ 2 � ) = �� � � n ( t ) ψ ∗ n ∗ ( s ) ψ 2 ( s ) ds = E 1 ( t ) dt � � ψ 2 ( t ) ψ ∗ 2 ( s ) E [ n ( t ) n ∗ ( s )] dt ds = � � ψ 2 ( t ) ψ ∗ 1 ( s ) 2 σ 2 δ ( t − s ) dt ds = � 2 σ 2 ψ 2 ( t ) ψ ∗ = 1 ( t ) dt 2 σ 2 � ψ 2 , ψ 1 � = 9 / 13

MPE and ML Rules in Complex Baseband • N is a circularly symmetric Gaussian vector and the pdf of Y under H i is 1 � � − ( y − s i ) H C − 1 N ( y − s i ) p i ( y ) = π K det ( C N ) exp −� y − s i � 2 1 � � = ( 2 πσ 2 ) K exp 2 σ 2 • The MPE rule is given by Re ( � y , s i � ) − � s i � 2 + σ 2 log π i δ MPE ( y ) = argmax 2 1 ≤ i ≤ M Re ( � y , s i � ) − � s i � 2 + σ 2 log π i = argmax 2 1 ≤ i ≤ M • The ML rule is given by Re ( � y , s i � ) − � s i � 2 � y − s i � 2 = argmax δ ML ( y ) = argmin 2 1 ≤ i ≤ M 1 ≤ i ≤ M Re ( � y , s i � ) − � s i � 2 � y − s i � 2 = argmax = argmin 2 1 ≤ i ≤ M 1 ≤ i ≤ M 10 / 13

ML Receiver for QPSK QPSK signals where p ( t ) is a real baseband pulse, A is a real number and 1 ≤ m ≤ 4 √ � 2 π f c t + π ( 2 m − 1 ) � s p m ( t ) = 2 Ap ( t ) cos 4 � √ � � � 2 π f c t + π ( 2 m − 1 ) j = Re 2 Ap ( t ) e 4 Complex Envelope of QPSK Signals s m ( t ) = Ap ( t ) e j π ( 2 m − 1 ) , 1 ≤ m ≤ 4 4 Orthonormal basis for the complex envelope consists of only φ ( t ) = p ( t ) � E p 11 / 13

ML Receiver for QPSK A √ Let √ E b = E p 2 . The vector representation of the QPSK signals is √ � � s 1 = E b + j E b � � s 2 = − E b + j E b � � s 3 = − E b − j E b � � s 4 = E b − j E b The hypothesis testing problem in terms of vectors is H i : Y = s i + N , i = 1 , . . . , 4 where N ∼ CN ( 0 , 2 σ 2 ) The ML rule is given by � y − s i � 2 δ ML ( y ) = argmin 1 ≤ i ≤ 4 12 / 13

Thanks for your attention 13 / 13