sol-exam1.2011

sol-exam1.2011 - 1 EE 7615 Solutions to Exam I Oct. 2011 1)...

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Unformatted text preview: 1 EE 7615 Solutions to Exam I Oct. 2011 1) We have S N ( f ) = S N W ( f ) | H ( f ) | 2 = braceleftBigg 10- 6 | f | < 1 otherwise. a) We have E [ N (0)] = 0 and E [ N (0) 2 ] = R N (0) = integraldisplay - S N ( f ) df = 2 10- 6 Thus var ( N (0)) = 2 = 2 10- 6 . We have f N (0) ( x ) = 1 2 e- x 2 2 2 b) E [ N ( . 5)] = 0 and var ( N ( . 5)) = 2 = 2 10- 6 . Now E [ N (0) N ( . 5)] = R N ( . 5) . We need R N ( ) . R N ( ) = F- 1 [ S N ( f )] = integraldisplay 1- 1 10- 6 e j 2 f df = 2 10- 6 sin(2 ) 2 Thus E [ N (0) N ( . 5)] = R N ( . 5) = 0 . This implies that the two r.v.s are independent. f N (0) N ( . 5) ( x,y ) = 1 2 2 e- x 2 + y 2 2 2 2) a) E [ Y ( t )] = E [ X (2 t + T )] = 0 . R Y ( t + ,t ) = E [ Y ( t + ) Y ( t )] = E [ X (2( t + ) + T ) X (2 t + T )] = R X (2( t + ) + T )- (2 t + T )) = R X (2 ) From the above we see that { Y ( t ) } is also W.S.S....
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sol-exam1.2011 - 1 EE 7615 Solutions to Exam I Oct. 2011 1)...

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