By a result we will show later we will find that r x

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By a result we will show later, we will find that R X 0 (0) = - R X (0) Z -∞ ω 2 S X ( ω ) . So the existence of X 0 t for a WSS process X t is equivalent to the condition that the second moment of the PSD of X t is finite. Example 3 Let X t be a process having S X ( ω ) = k ω 2 + β 2 . In this case, - R X 0 (0) = 1 2 π Z -∞ ω 2 k ω 2 + β 2 = , so the process is not mean-square differentiable. 2 Integration For a function f : T R , we define a Riemann integral as lim max | t i - t i - 1 |→ 0 n X i =1 f ( ζ i )( t i - t i - 1 ) = Z b a f ( t ) dt where a = t 0 < t 1 < · · · < t n = b and z i ( t i - 1 , t i ) . Let us define a similar sort of limit for a random process. We will define the limits in the mean-square sense. Then Z b a X t dt is the mean-square integral of X t . Properties of M.S. integrals R b a X t dt : 1. The integral exists if and only if Z b a Z b a R X ( t, s ) dtds < . Proof R b a X t dt exists if and only if lim max | t i - t i - 1 |→ 0 , max | s i - s i - 1 |→ 0 E n X i =1 X ζ i ( t i - t i - 1 ) - m X j =1 X β j ( s j - s j - 1 ) 2 = 0 = lim n X i =1 n X k =1 E [ X ζ i X ζ k ]( t i - t i - 1 )( t k - t k - 1 ) - 2 n X i =1 m X j =1 E [ X ζ i X β j ]( t i - t i - 1 )( s j - s j - 1 )+ m X j =1 m X l =1 E [ X β j X β l ]( s j - s j - 1 )( s l - s l - 1 )
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ECE 6010: Lecture 7 – Analytical Properties of Random Processes 5 The expectations can be replaced as lim n X i =1 n X k =1 R X ( ζ i , ζ k )( t i - t i - 1 )( t k - t k - 1 ) - 2 n X i =1 m X j =1 R X ( ζ i , β j )( t i - t i - 1 )( s j - s j - 1 )+ m X j =1 m X l =1 R X ( β j , β l )( s j - s j - 1 )( s l - s l - 1 ) = 0 The equality to zero is the Cauchy criterion for the Riemann sums defining R b a R b a R X ( t, s ) dtds . Thus, the integral exists implies that R b a R b a R X ( t, s ) dtds exists. 2 2. Assume that R b a X t dt exists. Then E [ R b a X t dt ] = R b a E [ X t ] dt = R b a μ x ( t ) dt .
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  • Fall '08
  • Stites,M
  • Probability theory, Xt

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