This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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) widesense 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...
View
Full
Document
This note was uploaded on 04/03/2010 for the course EE 562a taught by Professor Toddbrun during the Fall '07 term at USC.
 Fall '07
 ToddBrun

Click to edit the document details