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Random Variables and Stochastic Processes – 0903720 Lecture#21 Dr. Ghazi Al Sukkar Email: [email protected] Office Hours: Refer to the website Course Website: http://www2.ju.edu.jo/sites/academic/ghazi.alsukkar 1 Chapter 9 Systems with Stochastic Inputs Deterministic Systems Memoryless Systems Linear Time Invariant Systems (LTI) Differentiators Upcrossings and Downcrossings of a stationary Gaussian process Vector Processes and Multiterminal Systems Discrete-Time Systems 2 Systems with Stochastic Inputs • A deterministic system1 transforms each input waveform 𝑋 𝑡, 𝜁𝑖 into an output waveform 𝑌 𝑡, 𝜁𝑖 = 𝑇 𝑋 𝑡, 𝜁𝑖 by operating only on the time variable 𝑡. Thus a set of realizations at the input corresponding to a process 𝑋(𝑡) generates a new set of realizations 𝑌 𝑡, 𝜁 at the output associated with a new process 𝑌(𝑡). X (t , i ) X (t ) T [] Y (t , i ) (t ) Y t t • Our goal is to study the output process statistics in terms of the input process statistics and the system function. 1A stochastic system on the other hand operates on both the variables 𝑡 and 𝜁 3 Deterministic Systems Memory Memoryless Time variance Linearity with Memory Time-varying Y (t ) g[ X (t )] Time-Invariant Non-Linear Linear Y (t ) L[ X (t )] Linear-Time Invariant (LTI) systems X (t ) h(t ) LTI system Y (t ) X (t ) h(t ) h(t ) X ( )d Convolution 4 h( ) X (t )d . Memoryless Systems: The output Y(t) in this case depends only on the present value of the input X(t). i.e., Y (t ) g{ X (t )} Strict-sense stationary input Wide-sense stationary input X(t) stationary Gaussian with RXX ( ) Memoryless system Strict-sense stationary output. Memoryless system Need not be stationary in any sense. Memoryless system Y(t) stationary, but not Gaussian with RXY ( ) RXX ( ). 5 𝑌 𝑡 = 𝑔 𝑋(𝑡) Here 𝑔(𝑥) is a function of 𝑥. • Therefore, the first-order density 𝑓𝑌 (𝑦; 𝑡) of 𝑌(𝑡) can be expressed in terms of the corresponding density 𝑓𝑋 (𝑥; 𝑡) of 𝑋 𝑡 . Furthermore: ∞ 𝐸 𝑌(𝑡) = 𝑔 𝑥 𝑓𝑋 𝑥; 𝑡 𝑑𝑥 −∞ • Since 𝑌 𝑡1 = 𝑔 𝑋 𝑡1 and 𝑌 𝑡2 = 𝑔 𝑋 𝑡2 , then the second-order density 𝑓𝑌 𝑦1 , 𝑦2 ; 𝑡1 , 𝑡2 of 𝑌(𝑡) can be determined in terms of the corresponding density 𝑓𝑋 𝑥1 , 𝑥2 ; 𝑡1 , 𝑡2 of 𝑋 𝑡 . Furthermore: ∞ ∞ 𝐸 𝑌 𝑡1 𝑌(𝑡2 ) = 𝑔 𝑥1 𝑔(𝑥2 )𝑓𝑋 𝑥1 , 𝑥2 ; 𝑡1 , 𝑡2 𝑑𝑥1 𝑑𝑥2 −∞ −∞ • In the same way the nth-order density 𝑓𝑌 𝑦1 , … , 𝑦𝑛 ; 𝑡1 , … , 𝑡𝑛 of 𝑌(𝑡) can be determined in terms of the corresponding density 𝑓𝑋 𝑥1 , … , 𝑥𝑛 ; 𝑡1 , … , 𝑡𝑛 of 𝑋 𝑡 6 • If the input 𝑋(𝑡) to a memoryless system is an SSS process, then to determine the nth-order density of the output 𝑌(𝑡) we solve the system: 𝑔 𝑥1 = 𝑦1 , 𝑔 𝑥2 = 𝑦2 , … , 𝑔 𝑥𝑛 = 𝑦𝑛 If this system of equations has a unique solution, then: 𝑓𝑋 𝑥1 , … , 𝑥𝑛 ; 𝑡1 , … , 𝑡𝑛 𝑓𝑌 𝑦1 , … , 𝑦𝑛 ; 𝑡1 , … , 𝑡𝑛 = 𝑔′ 𝑥1 ⋯ 𝑔′ (𝑥𝑛 ) Since 𝑋(𝑡) is SSS, then the numerator is invariant to a time shift, and the denominator does not depend on 𝑡, then 𝑌(𝑡) is also an SSS process. 7 Theorem: If X(t) is a zero mean stationary Gaussian process, and Y(t) = g[X(t)], where g () represents a nonlinear memoryless device, then RXY ( ) RXX ( ), E{g ( X )}. Proof: RXY ( ) E{ X (t )Y (t )} E[ X (t ) g{ X (t )}] x1 g ( x2 ) f X1X 2 (x1 , x2 )dx1dx2 where X 1 X (t ), X 2 X (t ) are jointly Gaussian random variables, and hence x* A1 x / 2 1 𝑋≡𝑋 f X1X 2 ( x1 , x2 ) e 2 | A| X ( X 1 , X 2 )T , x ( x1 , x2 )T R (0) R ( ) * A E{ X X } LL R ( ) R (0) * XX XX XX XX 8 where L is an upper triangular factor matrix with positive diagonal entries. i.e., l11 l12 L . 0 l22 Consider the transformation Z L1 X ( Z1 , Z 2 )T , z L1 x ( z1 , z2 )T so that 1 *1 1 *1 E{Z Z } L E{X X }L L AL I * * and hence Z1, Z2 are zero mean independent Gaussian random variables. Also x L z x1 l11 z1 l12 z2 , x2 l22 z2 and hence x A1 x z L* A1 Lz z z z12 z22 . * * * The Jacobaian of the transformation is given by 9 | J || L1 || A |1/ 2 . RXY ( ) (l 11 z g (l l11 l12 z1 l12 z2 ) g (l22 z2 ) 1 1 | J | 2 | A|1/ 2 22 z2 ) f z1 ( z1 ) f z2 ( z2 )dz1dz2 22 z2 ) f z1 ( z1 ) f z2 ( z2 )dz1dz2 1 z g (l 2 e z12 / 2 z22 / 2 0 e l11 z1 f z1 ( z1 )dz1 g (l22 z2 ) f z2 ( z2 )dz2 l12 z2 g (l22 z2 ) f z2 ( z2 ) dz2 1 2 l12 l222 ug (u) let u l22 z2 . 1 2 e e z / 2 2 2 u 2 / 2 l222 du, 10 fu ( u ) RXY ( ) l12 l22 g (u ) u l2 1 2 l222 22 e u 2 / 2 l222 du df u ( u ) f u ( u ) du RXX ( ) g (u ) f u(u )du, since A LL* gives l12 l22 RXX ( ). Hence 0 RXY ( ) RXX ( ){ g (u ) f u (u ) | g (u ) f u (u )du} RXX ( ) E{g ( X )} RXX ( ), the desired result, where E[ g ( X )]. Thus if the input to a memoryless device is stationary Gaussian, the cross correlation function between the input and the output is proportional to the input autocorrelation function. 11 Linear Systems L[] represents a linear system if L{a1 X (t1 ) a2 X (t2 )} a1 L{X (t1 )} a2 L{X (t2 )}. Let Y (t ) L{ X (t )} represent the output of a linear system. (1) Time-Invariant System: L[] represents a time-invariant system if (2) Y (t ) L{ X (t )} L{ X (t t0 )} Y (t t0 ) i.e., shift in the input results in the same shift in the output also. If L[] satisfies both (1) and (2), then it corresponds to a linear time-invariant (LTI) system. LTI systems can be uniquely represented in terms of their output to h (t ) Impulse a delta function (impulse response) (t ) LTI h (t ) response of the system t Impulse Impulse response 12 Where ℎ 𝑡 = 𝐿 𝛿(𝑡) Y (t ) X (t ) X (t ) t t Y (t ) LTI Y (t ) h(t ) X ( )d arbitrary input h( ) X (t )d Which follows by expressing X(t) as X (t ) X ( ) (t )d and applying (1) and (2) to Y (t ) L{ X (t )}. Thus Y (t ) L{ X (t )} L{ X ( ) (t )d } L{ X ( ) (t )d } By Linearity X ( ) L{ (t )}d By Time-invariance X ( )h (t )d h ( ) X (t )d . 13 Output Statistics: • The mean of the output process is given by: (t ) E{Y (t )} E{ X ( )h(t )d } Y X ( )h(t )d X (t ) h(t ). • Similarly the cross-correlation function between the input and Output processes is given by: R XY (t1 , t2 ) E{ X (t1 )Y * (t2 )} E{ X (t1 ) X * (t2 )h * ( )d } E{ X (t1 ) X * (t2 )}h * ( )d R XX (t1 , t2 )h * ( )d R XX (t1 , t2 ) h * (t2 ). Where 𝐿2 means that the system operates on the variable 𝑡2 . 14 • Finally the output autocorrelation function is given by: RYY (t1 , t 2 ) E{Y (t1 )Y * (t 2 )} E{ X (t1 )h( )d Y * (t 2 )} E{ X (t1 )Y * (t 2 )}h( )d RXY (t1 , t 2 )h( )d RXY (t1 , t 2 ) h(t1 ) L1RXY t1 , t 2 , or RYY (t1 , t2 ) RXX (t1 , t2 ) h* (t2 ) h(t1 ). 𝜂𝑋 (𝑡) 𝑅𝑋𝑋 (𝑡1 , 𝑡2 ) ℎ∗ (𝑡2 ) ℎ(𝑡) 𝑅𝑋𝑌 (𝑡1 , 𝑡2 ) 𝜂𝑌 (𝑡) ℎ(𝑡1 ) 𝑅𝑌𝑌 (𝑡1 , 𝑡2 ) 15 • Also the autocovariance 𝐶𝑌𝑌 (𝑡1 , 𝑡2 ) of 𝑌(𝑡) is the autocorrelation 𝑅𝑌𝑌 (𝑡1 , 𝑡2 ) of the centered process 𝑌 𝑡 = 𝑌 𝑡 − 𝜂𝑌 (𝑡) And 𝑌 𝑡 = ℎ 𝑡 ∗ 𝑋(𝑡) where 𝑋 𝑡 = 𝑋 𝑡 − 𝜂𝑋 (𝑡) 𝑅𝑋𝑋 𝑡1 , 𝑡2 = 𝐸 𝑋 𝑡1 − 𝜂𝑋 (𝑡1 ) 𝑋 𝑡2 − 𝜂𝑋 (𝑡2 ) ∗ = 𝐶𝑋𝑋 𝑡1 , 𝑡2 𝑅𝑌𝑌 𝑡1 , 𝑡2 = 𝐸 𝑌 𝑡1 − 𝜂𝑌 (𝑡1 ) 𝑌 𝑡2 − 𝜂𝑌 (𝑡2 ) ∗ = 𝐶𝑌𝑌 𝑡1 , 𝑡2 𝑅𝑋𝑌 𝑡1 , 𝑡2 = 𝐸 𝑋 𝑡1 − 𝜂𝑋 (𝑡1 ) 𝑌 𝑡2 − 𝜂𝑌 (𝑡2 ) ∗ = 𝐶𝑋𝑌 𝑡1 , 𝑡2 Then: 𝐶𝑋𝑌 𝑡1 , 𝑡2 = 𝐶𝑋𝑋 𝑡1 , 𝑡2 ∗ ℎ∗ (𝑡2 ) 𝐶𝑌𝑌 𝑡1 , 𝑡2 = 𝐶𝑋𝑌 𝑡1 , 𝑡2 ∗ ℎ (𝑡1 ) 16 In particular if 𝑋(𝑡) is wide-sense stationary, then we have (t ) so that: X ∞ 𝜂𝑌 𝑡 = 𝜂𝑋 ℎ 𝜏 𝑑𝜏 = 𝜂𝑋 𝑐, 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 −∞ Also 𝑅𝑋𝑋 𝑡1 , 𝑡2 = 𝑅𝑋𝑋 (𝑡1 − 𝑡2 ) so: ∞∞ 𝑅𝑋𝑋 𝑡1 − 𝑡2 + 𝛼 ℎ∗ 𝛼 𝑑𝛼 = 𝑅𝑋𝑋 𝜏 ∗ ℎ∗ −𝜏 𝑅𝑋𝑌 𝑡1 , 𝑡2 = −∞ = 𝑅𝑋𝑌 𝜏 , 𝜏 = 𝑡1 − 𝑡2 Thus 𝑋(𝑡) and 𝑌(𝑡) are jointly WSS. Further, the output autocorrelation simplifies to: RYY (t1 , t 2 ) R XY (t1 t 2 )h( )d , t1 t 2 R XY ( ) h( ) RYY ( ). We obtain RYY ( ) RXX ( ) h* ( ) h( ). 17 X Thus the output process is also wide-sense stationary. This gives rise to the following representation X (t ) wide-sense stationary process LTI system ℎ(𝑡) X (t ) wide-sense stationary process. X (t ) strict-sense stationary process LTI system ℎ(𝑡) X (t ) strict-sense stationary process X (t ) Linear system Gaussian ℎ(𝑡) process (also stationary) X (t ) Gaussian process (also stationary) 18 White Noise Process: • Recall that 𝑊(𝑡) is said to be a white noise process if RWW (t1 , t2 ) q(t1 ) (t1 t2 ), i.e., 𝐸[𝑊(𝑡1 )𝑊 ∗ (𝑡2 )] = 0 unless 𝑡1 = 𝑡2 . • 𝑊(𝑡) is said to be wide-sense stationary (WSS) white noise if 𝐸[𝑊(𝑡)] = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, and 𝑅𝑊𝑊 𝑡1 , 𝑡2 = 𝑞𝛿 𝑡1 − 𝑡2 = 𝑞𝛿 𝜏 • If 𝑊(𝑡) is also a Gaussian process (white Gaussian process), then all of its samples are independent random variables. White noise 𝑊(𝑡) LTI ℎ(𝑡) Colored Noise 𝑁 𝑡 = ℎ 𝑡 ∗ 𝑊(𝑡) 19 • For WSS white noise input 𝑊(𝑡), we have: E[ N (t )] W h( )d , a constant and RNN ( ) q ( ) h* ( ) h( ) qh* ( ) h( ) q ( ) where ( ) h( ) h ( ) h( )h * ( )d . * • Thus when the input of an LTI system is white noise process then the output represents a (colored) noise process. Note: White noise need not be Gaussian. “White” and “Gaussian” are two different concepts! 20 Case study:Differentiators • A differentiator is a linear system whose output is the derivative on the input: 𝑑𝑋(𝑡) 𝑌 𝑡 = 𝐿 𝑋(𝑡) = = 𝑋 ′ (𝑡) 𝑑𝑡 • The mean of the output: 𝜂𝑋 ′ 𝑡 = 𝐿 𝜂𝑋 (𝑡) = 𝜂𝑋′ (𝑡) • The crosscorrelation between 𝑋(𝑡) and 𝑋 ′ (𝑡) is: 𝜕𝑅𝑋𝑋 (𝑡1 , 𝑡2 ) 𝑅𝑋𝑋 ′ 𝑡1 , 𝑡2 = 𝐿2 𝑅𝑋𝑋 (𝑡1 , 𝑡2 ) = 𝜕𝑡2 • The output autocorrelation: 𝜕𝑅𝑋𝑋 ′ 𝑡1 , 𝑡2 𝑅𝑋 ′ 𝑋 ′ 𝑡1 , 𝑡2 = 𝐿1 𝑅𝑋𝑋 ′ 𝑡1 , 𝑡2 = 𝜕𝑡1 • Combining both: 𝜕 2 𝑅𝑋𝑋 (𝑡1 , 𝑡2 ) 𝑅𝑋 ′𝑋 ′ 𝑡1 , 𝑡2 = 𝜕𝑡1 𝜕𝑡2 21 • If 𝑋(𝑡) is WSS, then 𝜂𝑋 𝑡 = 𝜂𝑋 which is constant, hence: 𝜂𝑋 ′ 𝑡 = 0 Also since 𝑅𝑋𝑋 (𝑡1 , 𝑡2 ) = 𝑅𝑋𝑋 (𝜏), 𝜏 = 𝑡1 − 𝑡2 Then 𝜕𝑅𝑋𝑋 𝑡1 − 𝑡2 𝑑𝑅𝑋𝑋 𝜏 =− 𝜕𝑡2 𝑑𝜏 𝜕 2 𝑅𝑋𝑋 (𝑡1 − 𝑡2 ) 𝑑 2 𝑅𝑋𝑋 𝜏 =− 𝜕𝑡1 𝜕𝑡2 𝑑𝜏 2 Hence: ′ 𝑅𝑋𝑋 ′ 𝜏 = −𝑅𝑋𝑋 𝜏 ′′ 𝑅𝑋 ′ 𝑋 ′ 𝜏 = −𝑅𝑋𝑋 𝜏 22 Upcrossings and Downcrossings of a stationary Gaussian process: Consider a zero mean stationary Gaussian process 𝑋(𝑡) with autocorrelation function 𝑅𝑋𝑋 (𝜏) .An upcrossing over the mean value occurs whenever the realization 𝑋(𝑡) passes through zero with positive slope. Let 𝜌∆𝑡 𝑋(𝑡) represent the probability Upcrossings of such an upcrossing in the interval 𝑡, 𝑡 + ∆𝑡 𝑡 We wish to determine 𝜌. Downcrossing Since 𝑋(𝑡) is a stationary Gaussian process, its derivative process 𝑋 ′ (𝑡) is also zero mean stationary Gaussian with autocorrelation function ′′ 𝑅𝑋 ′𝑋 ′ 𝜏 = −𝑅𝑋𝑋 𝜏 . Further 𝑋(𝑡) and 𝑋 ′ (𝑡) are jointly Gaussian stationary processes, and since ′ 𝑅𝑋𝑋 ′ 𝜏 = −𝑅𝑋𝑋 𝜏 23 we have 𝑅𝑋𝑋 ′ −𝜏 = 𝑑𝑅𝑋𝑋 −𝜏 − 𝑑 −𝜏 = 𝑑𝑅𝑋𝑋 𝜏 𝑑𝜏 = −𝑅𝑋𝑋 ′ 𝜏 (odd function) which for 𝜏 = 0 gives 𝑅𝑋𝑋 ′ 0 = 0 ⟹ 𝐸 𝑋 𝑡 𝑋 ′ 𝑡 =0 i.e., the jointly Gaussian zero mean random variables X 1 X (t ) and X 2 X (t ) are uncorrelated and hence independent with variances ′′ 𝜎12 = 𝑅𝑋𝑋 0 𝑎𝑛𝑑 𝜎22 = 𝑅𝑋 ′𝑋 ′ 0 = −𝑅𝑋𝑋 0 >0 respectively. Thus 𝑥12 𝑥22 − 2 +2𝜎 2 2𝜎 1 2 𝑒 1 𝑓𝑋1 𝑋2 𝑥1 , 𝑥2 = 𝑓𝑋1 𝑥1 𝑓𝑋2 𝑥2 = 2𝜋𝜎1 𝜎2 To determine 𝜌 the probability of upcrossing rate, 24 we argue as follows: In an interval(t , t t ), the realization moves from X(t) = X1 to X (t t ) X (t ) X (t )t X 1 X 2 t, and hence the realization intersects with the zero level somewhere in that interval if X 1 0, X 2 0, and X (t t ) X 1 X 2 t 0 i.e., X 1 X 2 t. Hence the probability of upcrossing in (t , t t ) is given by t x 2 0 0 x x t f 1 2 X1 X 2 ( x1 , x2 )d x1dx2 X (t ) X ( t t ) t t t t X (t ) 0 f X 2 ( x2 )d x2 x t f X1 ( x1 )d x1 . 2 Differentiating both sides of (14-53) with respect to t , we get 0 f ( x2 )x2 f ( x2 t )dx2 X2 X1 and letting t 0, this equation reduce to 25 0 1 x2 f X ( x2 ) f X (0)dx2 x2 f X ( x2 )dx2 0 2R XX (0) 1 1 1 ( 2 2 / ) 2R XX (0) 2 2 (0) R XX R XX (0) [where we have made use of (5-78), Text]. There is an equal probability for downcrossings, and hence the total probability for crossing the zero line in an interval (t , t t ) equals 0 t , where 0 1 (0) / RXX (0) 0. RXX It follows that in a long interval T, there will be approximately 0T (0) is large, then the crossings of the mean value. If RXX autocorrelation function RXX ( ) decays more rapidly as moves away from zero, implying a large random variation around the origin (mean value) for X(t), and the likelihood of zero crossings should increase with increase in RXX (0) 26 Vector Processes and Multiterminal Systems • We consider now a MIMO system which is a system with 𝑛 input Random processes 𝑋𝑖 (𝑡) 𝑛𝑖=1 and 𝑟 output R.P.s 𝑟 𝑌𝑗 (𝑡) 𝑗=1 𝑋1 (𝑡) 𝑋2 (𝑡) ⋮ 𝑋𝑛 (𝑡) MIMO 𝑌1 (𝑡) 𝑌2 (𝑡) ⋮ 𝑌𝑟 (𝑡) • Let 𝑋 𝑡 = 𝑋𝑘 (𝑡) = 𝑋1 𝑡 , 𝑋2 𝑡 , … , 𝑋𝑛 (𝑡) 𝑡 be the input column Random Process vector, whose 𝑛 elements are Random Processes. • Let 𝑌 𝑡 = 𝑌𝑙 (𝑡) = 𝑌1 𝑡 , 𝑌2 𝑡 , … , 𝑌𝑟 (𝑡) 𝑡 be the output column Random Process vector, whose 𝑟 elements are Random Processes. 27 • The mean of 𝑋 𝑡 is: 𝐸𝑋 𝑡 = 𝜂𝑋 𝑡 = 𝜂𝑖 (𝑡) , where 𝜂𝑖 𝑡 = 𝐸 𝑋𝑖 (𝑡) . • The autocorrelation 𝑅𝑋𝑋 𝑡1 , 𝑡2 of the vector process 𝑋 𝑡 is an 𝑛 × 𝑛 matrix: 𝑅𝑋𝑋 𝑡1 , 𝑡2 = 𝐸 𝑋 𝑡1 𝑋 † (𝑡2 ) With elements 𝐸 𝑋𝑘 (𝑡1 )𝑋𝑙∗ 𝑡2 , 𝑘, 𝑙 = 1, … , 𝑛 • The cross-correlation between 𝑋 𝑡 and 𝑌 𝑡 is an 𝑛 × 𝑟 matrix: 𝑅𝑋𝑌 𝑡1 , 𝑡2 = 𝐸 𝑋 𝑡1 𝑌 † (𝑡2 ) With elements 𝐸 𝑋𝑘 (𝑡1 )𝑌𝑙∗ 𝑡2 , where 𝑘 = 1, … , 𝑛 and 𝑙 = 1, … , 𝑟 28 • If the MIMO system is an LTI system, then it is specified in terms of its impulse response matrix ℎ 𝑡 = ℎ𝑙𝑘 (𝑡) , 𝑘 = 1, … , 𝑛 and 𝑙 = 1, … , 𝑟 which is an 𝑟 × 𝑛 matrix Where it component ℎ𝑙𝑘 (𝑡) is the response of the 𝑙th output terminal when the 𝑘th input terminal has input equals 𝛿(𝑡) and all other input terminal has input equal 0. • The response of the 𝑙th output terminal 𝑌𝑙 (𝑡) to an arbitrary input 𝑋 𝑡 = 𝑋𝑘 (𝑡) is: 𝑌𝑙 𝑡 = ∞ ℎ −∞ 𝑙1 𝛼 𝑋1 𝑡 − 𝛼 𝑑𝛼 + ⋯ + Hence ∞ ℎ −∞ 𝑙𝑛 𝛼 𝑋𝑛 𝑡 − 𝛼 𝑑𝛼 ∞ 𝑌 𝑡 = ℎ 𝛼 𝑋 𝑡 − 𝛼 𝑑𝛼 −∞ 29 • By setting 𝑡 = 𝑡2 and take the Hermitian of 𝑌 𝑡 and Premultiplying by 𝑋 𝑡1 ∞ 𝑋 𝑡1 𝑌 † 𝑡2 = 𝑋 𝑡1 𝑋 † 𝑡2 − 𝛼 ℎ† 𝛼 𝑑𝛼 −∞ Then by taking the mean of both sides: ∞ 𝑅𝑋𝑋 𝑡1 , 𝑡2 − 𝛼 ℎ† 𝛼 𝑑𝛼 𝑅𝑋𝑌 𝑡1 , 𝑡2 = −∞ • By setting 𝑡 = 𝑡1 and Postmultiplying 𝑌 𝑡 by 𝑌 † 𝑡2 ∞ 𝑌 𝑡1 𝑌 † 𝑡2 = ℎ 𝛼 𝑋 𝑡1 − 𝛼 𝑌 † 𝑡2 𝑑𝛼 −∞ Then by taking the mean of both sides: ∞ 𝑅𝑌𝑌 𝑡1 , 𝑡2 = ℎ 𝛼 𝑅𝑋𝑌 𝑡1 − 𝛼, 𝑡2 𝑑𝛼 −∞ 30 Discrete-Time Systems • A discrete-time system is a system where the input and the output are discrete-time signals. • An LTI discrete-time system is specified through its impulse response ℎ[𝑛] which is the response of the system when the input is a discrete impulse 1, 𝑛 = 0 𝛿𝑛 = 0, 𝑛 ≠ 0 • The output 𝑌[𝑛] of a discrete-time system when the input is a Stochastic process 𝑋[𝑛] is given as: ∞ 𝑌 𝑛 =𝑋 𝑛 ∗ℎ 𝑛 = ∞ 𝑋 𝑛−𝑘 ℎ 𝑘 = 𝑘=−∞ 𝑋[𝑛] ℎ[𝑛] ℎ 𝑛−𝑘 𝑥 𝑘 𝑘=−∞ 𝑌[𝑛] Output Statistics • The mean of the output process is given by: ∞ 𝜂𝑌 𝑛 = 𝐸 𝑌[𝑛] = 𝐸 ∞ ℎ 𝑛 − 𝑘 𝑋[𝑘] = 𝑘=−∞ ℎ 𝑛 − 𝑘 𝐸 𝑋[𝑘] 𝑘=−∞ = 𝜂𝑋 𝑛 ∗ ℎ[𝑛] • The cross-correlation function between the input and Output processes is given by: ∞ 𝑅𝑋𝑌 𝑛1 , 𝑛2 = 𝐸 𝑋 𝑛1 𝑌 ∗ 𝑛2 ∞ ∞ 𝐸 𝑋 𝑛1 𝑋 ∗ 𝑛2 − 𝑘 ℎ∗ 𝑘 = = 𝑘=−∞ = 𝑅𝑋𝑋 𝑛1 , 𝑛2 ∗ ℎ∗ 𝑘=−∞ 𝑅𝑋𝑋 𝑛1 , 𝑛2 − 𝑘 ℎ∗ 𝑘 𝑘=−∞ 𝑛2 𝑋 ∗ 𝑛2 − 𝑘 ℎ∗ 𝑘 = 𝐸 𝑋 𝑛1 • The output autocorrelation function is given by: 𝑅𝑌𝑌 𝑛1 , 𝑛2 = 𝐸 𝑌 𝑛1 𝑌 ∗ 𝑛2 ∞ 𝑋 𝑛1 − 𝑚 ℎ 𝑚 𝑌 ∗ 𝑛2 =𝐸 𝑚=−∞ ∞ 𝐸 𝑋 𝑛1 − 𝑚 𝑌 ∗ 𝑛2 ℎ 𝑚 = 𝑚=−∞ ∞ = 𝑅𝑋𝑌 𝑛1 − 𝑚, 𝑛2 ℎ 𝑚 = 𝑅𝑋𝑌 𝑛1 , 𝑛2 ∗ ℎ 𝑛1 𝑚=−∞ • Then by combining both relations: 𝑅𝑌𝑌 𝑛1 , 𝑛2 = 𝑅𝑋𝑋 𝑛1 , 𝑛2 ∗ ℎ∗ 𝑛2 ∗ ℎ 𝑛1 Discrete-Time WSS input • If the input process 𝑋[𝑛] is WSS, then 𝑌[𝑛] is also WSS with: • Mean: ∞ 𝜂𝑌 = 𝜂 𝑋 ℎ[𝑘] = 𝜂𝑋 𝐶: constant 𝑘=−∞ • The cross-correlation between 𝑋(𝑡) and 𝑌(𝑡) is: 𝑅𝑋𝑌 𝑚 = 𝑅𝑋𝑋 𝑚 ∗ ℎ∗ [−𝑚] • The output autocorrelation is: 𝑅𝑌𝑌 𝑚 = 𝑅𝑋𝑌 𝑚 ∗ ℎ[𝑚] Or: 𝑅𝑌𝑌 𝑚 = 𝑅𝑋𝑋 𝑚 ∗ ℎ∗ −𝑚 ∗ ℎ 𝑚 = 𝑅𝑋𝑋 𝑚 ∗ 𝜌[𝑚] ∗ Where 𝜌 𝑚 = ∞ ℎ 𝑚 + 𝑘 ℎ 𝑘 𝑘=−∞