ECE302HW7Soln_Fa09

# ECE302HW7Soln_Fa09 - ECE 302 Homework 7 Prof Saul Gelfand...

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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, σ 2 A = 1 4 , ¯ B = 0, σ 2 B = 1. 0.2 0.4 0.6 0.8 1-2-1 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 )( σ 2 A + ¯ A 2 ) + ( t 1 + t 2 ) ¯ A ¯ B + σ 2 B + ¯ B 2 1 c. X ( t ) is not wide-sense 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 τ = 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- λ ) f B ( λ ) dλ = integraldisplay ∞-∞ 1 t f A parenleftbigg x- λ t parenrightbigg f B ( λ ) dλ 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- λ t 2 parenrightbigg f B ( λ ) dλ 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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ECE302HW7Soln_Fa09 - ECE 302 Homework 7 Prof Saul Gelfand...

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