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Unformatted text preview: ECE 302: Homework 7 Prof. Saul Gelfand Fall 2009 Due December 11, 2009 Problem 1 a. Figure ?? shows some examples of X ( t ) with A and B randomly generated with A = 1, 2 A = 1 4 , B = 0, 2 B = 1. 0.2 0.4 0.6 0.8 121 1 2 3 4 t X(t) Figure 1: Problem 1(a) b. Mean function m X ( t ) = E [ At + B ] = tE [ A ] + E [ B ] = t A + B. Correlation function R X ( t 1 ,t 2 ) = E [ X ( t 1 ) X ( t 2 )] = E [( At 1 + B )( At 2 + B )] = E [ A 2 t 1 t 2 + AB ( t 1 + t 2 ) + B 2 ] = ( t 1 t 2 ) E [ A 2 ] + ( t 1 + t 2 ) E [ AB ] + E [ B 2 ] = ( t 1 t 2 )( 2 A + A 2 ) + ( t 1 + t 2 ) A B + 2 B + B 2 1 c. X ( t ) is not widesense stationary, because m X ( t ) is not constant for all t , and further, R X ( t 1 ,t 2 ) cannot be written as a function of = t 1 t 2 . d. Let Z = tA . We have that F Z ( z ) = P [ Z z ] = P [ tA z ] = P [ A z/t ] = F A ( z/t ) f Z ( z ) = d dz F Z ( z ) = d dz F A ( z/t ) = 1 t f A ( z/t ) Since A and B are independent, then Z and B are also indepen dent. Thus f X ( t ) ( x ) = f Z ( z ) * f B ( b ) = integraldisplay  f Z ( x ) f B ( ) d = integraldisplay  1 t f A parenleftbigg x t parenrightbigg f B ( ) d Now, we have the joint distribution f X ( t 1 ) ,X ( t 2 ) ( x 1 ,x 2 ) = f X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ) f X ( t 2 ) ( x 2 ) . To find the conditional PDF f X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ), we first find the conditional CDF F X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ) = P [ X ( t 1 ) x 1  X ( t 2 ) = x 2 ] = P [ At 1 + B x 1  At 2 + B = x 2 ] = P [ At 1 + B x 1  B = x 2 At 2 ] . Since we are given that B = x 2 At 2 , then F X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ) = P [ At 1 + B x 1  B = x 2 At 2 ] = P [ At 1 + x 2 At 2 x 1 ] = P bracketleftbigg A x 1 x 2 t 1 t 2 bracketrightbigg = F A parenleftbigg x 1 x 2 t 1 t 2 parenrightbigg f X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ) = d dx 1 F X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ) = 1 t 1 t 2 f A parenleftbigg x 1 x 2 t 1 t 2 parenrightbigg 2 Therefore f X ( t 1 ) ,X ( t 2 ) ( x 1 ,x 2 ) = f X ( t 1 )  X ( t 2 ) ( x 1  X ( t 2 ) = x 2 ) f X ( t 2 ) ( x 2 ) = 1 t 1 t 2 f A parenleftbigg x 1 x 2 t 1 t 2 parenrightbiggintegraldisplay  1 t 2 f A parenleftbigg x 2 t 2 parenrightbigg f B ( ) d e. Since, for each t , X ( t ) is a linear combination of Guassians, we have that X ( t ) is itself Gaussian. Since all Gaussian random vari ables are jointly Gaussian, then X ( t 1 ) and X ( t 2 ) are indeed jointly Guassian....
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 Spring '08
 GELFAND

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