EE562aFinalSolF09

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

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Unformatted text preview: 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 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 , otherwise. h ( t ) = Z 1 2- 1 2 H ( f ) e i2 πft df = Z- 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 , 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 , ≤ f ≤ 1 4 , otherwise. S yz * ( f ) = G ( f ) H * ( f ) S x ( f ) = 0 for all f . 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...
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This note was uploaded on 04/03/2010 for the course EE 562a taught by Professor Toddbrun during the Fall '07 term at USC.

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

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