{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

s16_ps13_sol

# s16_ps13_sol - Unied Engineering II Spring 2004 Problem S16...

This preview shows pages 1–3. Sign up to view the full content.

Unified Engineering II Spring 2004 Problem S16 Solution The Fourier transform of x ( t ) is given by X ( f ). Then the FT of x 1 ( t ) is given by X 1 ( f ) = H ( f ) X ( f ) = jX ( f ) , 0 < f < f M + jX ( f ) , f M < f < 0 0 , | f | > f M The signal x 2 ( t ) is given by x 2 ( t ) = w 1 ( t ) x 1 ( t ) where w 1 ( t ) = cos 2 πf c t . The FT of w 1 ( t ) is 1 W 1 ( f ) = [ δ ( f f c ) + δ ( f + f c )] 2

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

View Full Document
The FT of x 2 ( t ) is then X 2 ( f ) = X 1 ( f ) W 1 ( f ) 1 = [ X 2 1 ( f f c ) + X 1 ( f + f c )] j X ( f f c ) , f c < f < f c + f M 2 + j X ( f f c ) , f c f M < f < f c 2 j X ( f + f c ) , f c < f < f c + f 2 = M + j X ( f + f c ) , f c f M < f < f c 2 0 , else The signal x 3 ( t ) is given by x 3 ( t ) = w 2 ( t ) x ( t ) where w 2 ( t ) = sin 2 πf c t . The FT of w 2 ( t ) is 1 W 2 ( f ) = [ ( f f c ) + ( f + f c )] 2 The FT of x 3 ( t ) is then X 3 ( f ) = X ( f ) W 2 ( f ) 1 [ = 2 jX ( f f c ) + jX ( f + f c )] j X ( f f c ) , f c < f < f c + f M 2 j X ( f f c ) , f c f M < f < f c 2 + j X ( f + f c ) , f c < f < f c + f 2 = M + j X ( f + f c ) , 2 M < f < f c f c f 0 , else Finally, the FT of y ( t ) is given by Y ( f ) X 2 ( f ) + X 3 ( f ) = jX ( f f c ) , f c < f < f c + f M 0 , f c f M < f < f c 0 , f c < f < f c + f = = M + jX ( f + f c ) , f c f M < f < f c 0 , else jX ( f f c ) , f c < f < f c + f M + jX ( f + f c ) , f c f M < f < f c 0 , else First, y ( t ) is guaranteed to be real if x ( t ), because if x ( t ) real, X ( f ) has conjugate symmetry, and then
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

s16_ps13_sol - Unied Engineering II Spring 2004 Problem S16...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online