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Unformatted text preview: ECE514 Random Process Fall 2010 HW4  Due: November 16, 2010 Contact [email protected] Recommended reading: Please read most of Sections 4.14.8 and Section 4.9 pages 113114. The fol lowing can be skipped: • The proof in Sec. 4.5. • Examples 4.8.1 and 4.8.4 in Sec. 4.8. The rest of the chapter will definitely interest some of you, but is not of highest priority. 1. Problem 4.1 in course notes. Answer : (a) Possible sample functions are x t = 2 + t, x t = 2 t, x t = 2 + t, x t = 2 t. (b) Because Pr( B = +1) = Pr( B = 1) = Pr( A = +1) = Pr( A = 1) = 0 . 5 , we have via symmetry that Pr( X t = α ) = Pr( X t = α ) , ∀ t ∈ R , and so Pr( X t > 0) = Pr( X t < 0) . It may appear at this point that Pr( X t ≥ 0) = 0 . 5 . However, we must also examine the cases where X t = 0 , t = 2 : 2 t = 0 , 2 + t = 0 , t = 2 : 2 + t = 0 , 2 t = 0 . We conclude that Pr( X t ≥ 0) = 0 . 75 for t ∈ { 2 , +2 } , else Pr( X t ≥ 0) = 0 . 5 . (c) For all functions above (Part a), x t < for some t . Therefore, Pr( X t ≥ , ∀ t ∈ R ) = 0 . 2. Problem 4.3 in course notes. Answer : Because θ ∼ U [0 , 2 π ] , phases are equally likely, and so μ X ( t ) = 0 , ∀ t ∈ R . Therefore, the covariance and correlation are identical, and we have C X ( s,t ) = R X ( s,t ) = E [ X s X t ] = E A 2 cos(2 πV t + θ )cos(2 πV s + θ ) = E [ A 2 ] E cos(2 πV ( t + s ) + 2 θ ) + cos(2 πV ( s t )) 2 (1) where (1) relies on independence of A , V , and θ . Because θ ∼ U [0 , 2 π ] , E [cos(2 πV ( t + s ) + 2 θ )] = 0 . We now evaluate the second cosine term in detail, E [cos(2 πV ( s t ))] = Z 5 v =0 cos(2 πv ( s t )) dv = 1 2 π ( s t ) sin(2 πv ( s t )) 5 = sin(10 π ( s t )) 2 π ( s t ) = 5 sinc (10 π ( s t )) , (2) 1 ECE514 Random Process Fall 2010 HW4  Due: November 16, 2010 Contact [email protected] where the sinc function is utilized, sinc ( a ) = sin ( a ) a . Next, E [ A 2 ] = ( E [ A ]) 2 + var ( A ) = 8 . (3) Combining (1)(3), C X ( s,t ) = 20 sinc (10 π ( s t )) . Finally, the process is wide sense stationary (WSS), because C X ( s,t ) = f ( s t ) . 3. Problem 4.7 in course notes. Answer : (a) A sketch of a Brownian bridge would look similar to a Wiener process, where the special features would be that the bridge is only defined over the range [0 , 1] , and that the Wiener process returns to zero at the end of the range....
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This note was uploaded on 01/25/2011 for the course ECE 514 taught by Professor Krim during the Fall '08 term at N.C. State.
 Fall '08
 Krim

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