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EE562aFinalSolF09

# EE562aFinalSolF09 - EE 562a Final Solution December 8 2009...

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EE 562a Final Solution December 8, 2009 1 1. (a) Using the fact that t is an integer, g ( t ) = Z 1 2 - 1 2 G ( f ) e i2 πft df = 2 Z 1 2 0 e i2 πft df = 1 i πt ( e i πt - 1) , t 6 = 0 1 , t = 0 = 1 , t = 0 - 2 i πt , t odd 0 , otherwise. h ( t ) = Z 1 2 - 1 2 H ( f ) e i2 πft df = Z 0 - 1 2 e i2 πft df = 1 i2 πt [1 - e - i πt ] , t 6 = 0 1 2 , t = 0 = 1 2 , t = 0 1 i πt , t odd 0 , otherwise. (b) (YES or NO answers) y ( u, t ) z(u,t) v(u,t) wide-sense stationary? YES YES YES mean zero? YES YES NO real? NO NO YES if real, positive? YES Gaussian? YES YES NO strictly stationary? YES YES YES The tricky parts of this answer come from the fact that, since g ( t ) = 2 h * ( t ), v ( u, t ) = 2 | z ( u, t ) | 2 = 1 2 | y ( u, t ) | 2 . Note also that the input’s power spectral density does not have a Dirac delta function at f = 0, and hence must be mean zero. (c) S y ( f ) = | G ( f ) | 2 S x ( f ) = ( 12 , 0 f 1 4 0 , otherwise. S yz * ( f ) = G ( f ) H * ( f ) S x ( f ) = 0 for all f .

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2 EE 562a Final Solution December 8, 2009 (d) E {| y ( u, t ) | 2 } = R yy * (0) = Z 1 2 - 1 2 S y ( f ) df = 3 (e) Using the fact that G ( f ) G ( - f ) = 0 for all f 6 = 0, we have R yy ( t 1 , t 2 ) = E X α g ( α ) x ( u, t 1 - α ) X β g ( β ) x ( u, t 1 - β ) = X α X β g ( α ) g ( β ) R x ( t 1 - t 2 - α + β ) = X α X β g ( α ) g ( β ) Z 1 2 - 1 2 S x ( f ) e i 2 πf ( t 1 - t 2 - α + β ) df = Z 1 2 - 1 2 S x ( f ) G ( f ) G ( - f ) e i 2 πf ( t 1 - t 2 ) df = 0 (f) NO. The general factorization technique for causal filter construction fails because Z 1 2 - 1 2 | ln S
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