sol-exam1.2011

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

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 | &lt; 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....
View Full Document

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online