EE 571 PROBABILITY THEORY AND STOCHASTIC PROCESSSES FINAL EXAM Date: 25 Jan 2005 Duration: 150 min. 1-) The random process X(t) is defined as X (t ) u (t ) , where α is a random variable uniformly distributed in (0, T) and t0 t0 (a) Determine the autocorrelation function RXX (t1 , t2 ) E{ X (t1 ), X (t2 )} of X(t). (b) Sketch the following cross-sections of RXX (t1 , t2 ) 1 u (t ) 0 (i) RXX (t , t ) (ii) RXX (t1 , t2 T / 2) j ωt 2-) Consider the random process X (t ) Ae , where A is a complex constant, ω is a random variable with pdf fω ( ) . (a) Find the autocorrelation function of X(t) in terms of fω ( ) . (b) Show that the power spectral density of X(t) is given by S XX ( ) 2 A f ω ( ) . 2 3-) Consider the binary communication channel shown below, where Xn denotes the input (0 or 1) and Xn+1 denotes the output. The channel delivers the input symbol to the output with a certain error probability: P X n 1 1| X n 0 P X n 1 0 | X n 1 Consider a cascade of N such binary communication channels (n= 0,1, …,N-1). (a) Write the probability transition matrix for one channel P(1). (b) Find the probability transition matrix for the cascade P(N). (c) Find the probability that a digit arriving as 1 through the cascade was in fact transmitted as 1 (i.e the probability P X 0 1| X N 1 . The state probabilities of the input to the cascaded channel are P X 0 1 p , P X 0 0 q . 0 1-α 0 β Xn Xn+1 α 1 1 1-β 4-) Consider a homogeneous Poisson process X(t) such that E{X(9)}=18. (a) Find the mean and variance of X(8). (b) Find P X (2) 3 . (c) Find P X (4) 5 | X (2) 3. (Hint: use the fact that the Poisson process has independent increments) 5-) A LTI continuous-time system has impulse response e4t h(t ) 0 t 0 t 0 The input to the system X(t) is a random process with autocorrelation RXX ( ) e (a) Find the power spectral density SYY ( f ) of the output Y(t) . (b) Find the average power of the output. 2 . SOLUTION 1-) (a) RXX (t1 , t2 ) E u (t1 α ) u (t2 α ) u (t 1 ) u (t2 ) f α ( )d T 1 u(t1 ) u(t2 ) d T 0 0 t1 = tT2 T 1 t1 0 or t2 0 0 t1 T t2 >t1 0 t2 T t1 >t2 t1 , t2 >T (c) (i) 0 RXX (t , t ) Tt 1 t0 0t T t T 1 t T (ii) Tt1 RXX (t1 , t2 T / 2) 12 0 t1 T / 2 t1 T / 2 t1 0 1/2 T/2 2-) (a) RXX ( ) E X * (t ) X (t ) E A*e j ω t Ae j ω (t ) A E e j ω A 2 2 e j f ω ( )d t (b) RXX ( ) in terms of the power spectral density is 1 RXX ( ) 2 S XX ( )e j d Comparing with the expression in part (a), the uniqueness property of Fourier transforms gives S XX ( ) 2 A fω ( ) 2 3-) (a) 1 P(1) 1 (b) From lecture notes P( N ) P(1) N 1 (c) P X 0 1| X N 1 (1 ) N N (1 ) (1 ) N (1 ) N P X N 1| X 0 1 .P{ X 0 1} P11 (0, N ). p1 (0) P{ X N 1} p1 ( N ) where P11 (0, N ) is the transition probability (1 ) N The state probabilities are p1 (0) p p1 ( N ) P01 (0, N ). p0 (0) P11 (0, N ). p1 (0) P11 (0, N ) (1 ) N (1 ) N q p p q ( p q) (1 ) N p1 ( N ) p q 1 P X 0 1| X N 1 [ (1 ) N ] p ( p q)(1 ) N 4-) (a) E{X(t)}= λt = 18 λ= 18/9 =2 t=9 E{X(8)}= 8×2 = 16 Var{X(8)}= λt = 16 (t )k t 3 (4)k 4 4 P X (2) 3 e e e [1 1!4 162! 64 3! ] 0.4335 k 0 k ! k 0 k ! 3 (b) (c) Let Y X (2) , Z X (4) X (2) X (4) Y X (4) 5 Z Y 5 , P Z Y 5 | Y 3 P Y k 4k 4 e k! Y and Z are independent Poisson random variables. P Z Y 5, Y 3 P Y 3 P Z n and 4n 4 e n! P Y k , Z n 4n k k 0 n 0 n !k ! 3 5 k 3 5 k P Z Y 5, Y 3 P Y k , Z n e 8 k 0 n 0 2 5 4n 4 4n 1 3 4n 2 4n 3 e n 0 n ! n 0 n ! n 0 2(n !) n 0 6(n !) 0.1705 8 P Z Y 5 | Y 3 0.1705 0.3933 0.4335 5-) (a) SYY ( f ) S XX ( f ) H ( f ) 2 H ( f ) h(t )e j 2 ft dt e (4 j 2 f )t dt 0 0 S XX ( f ) e 0 2 e j 2 f d 1 4 j 2 f 4 4 (2 f ) 2 4 1 SYY ( f ) 2 2 4 (2 f ) 16 (2 f ) 4n k 8 e n !k ! (b) E{Y (t )} 2 S YY ( f )df s j 2 f E{Y (t )} 2 j c ( s )c ( s ) ds 2 j j d ( s )d ( s ) 1 c( s ) c( s ) 2 2 4 1 SYY ( s ) 2 2 4 s 16 s (2 s)(4 s) (2 s)(4 s) d ( s) d ( s) c( s ) 2 c0 2 c1 0 d ( s) s 2 6s 8 E{Y 2 (t )} d0 8 d1 6 c12 d 0 c02 d 2 4 1 2d 0 d1d 2 2 8 6 24 d2 1