ECE302HW7Soln_Fa09

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 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 )( 2 A + A 2 ) + ( t 1 + t 2 ) A B + 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 = 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....
View Full Document

Page1 / 9

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online