solution_set7

# solution_set7 - EE 511 Solutions to Problem Set 7 1(i SX(f...

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

EE 511 Solutions to Problem Set 7 1. (i) S ˆ X ( f ) = | H ( f ) | 2 S X ( f ) where H ( f ) = - j sgn( f ). Therefore, S ˆ X ( f ) = S X ( f ). (ii) Suppose h ( τ ) is the impulse response of the Hilbert transformer, - h ( τ ) = h ( - τ ) is the impulse response of the inverse Hilbert transformer. Therefore, we have R ˆ XX ( τ ) = R X ( τ ) ⋆ h ( τ ) and R X ˆ X ( τ ) = R ˆ X ( τ ) ( - h ( τ )) = R X ( τ ) ( - h ( τ )) = - R ˆ XX ( τ ) , where denotes convolution. From the above result, we have S ˆ XX ( f ) = - S X ˆ X ( f ) . (iii) E [ X t ˆ X t ] = R X ˆ X (0) and E [ X t ˆ X t ] = R ˆ XX (0). Therefore, we have E [ X t ˆ X t ] = 0. (iv) R Z ( τ ) = E [( X t + τ + j ˆ X t + τ )( X t - j ˆ X t )] = R X ( τ ) - jR X ˆ X ( τ ) + jR ˆ XX ( τ ) + R ˆ X ( τ ) = 2 R X ( τ ) + 2 jR ˆ XX ( τ ) Therefore S Z ( f ) = 2 S X ( f ) + 2 jS ˆ XX ( f ) = 2 S X ( f ) + 2 jS X ( f )[ - j sgn( f )] = 2 S X ( f )[1 + sgn( f )] = b 4 S X ( f ) f > 0 0 else

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.

{[ snackBarMessage ]}

### Page1 / 4

solution_set7 - EE 511 Solutions to Problem Set 7 1(i SX(f...

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

View Full Document
Ask a homework question - tutors are online