s16_ps13_sol

s16_ps13_sol - Unified Engineering II Spring 2004 Problem...

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Unformatted text preview: 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 , | 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 f c t . The FT of w 1 ( t ) is 1 W 1 ( f ) = [ ( f f c ) + ( f + f c )] 2 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 , else The signal x 3 ( t ) is given by x 3 ( t ) = w 2 ( t ) x ( t ) where w 2 ( t ) = sin 2 f c t . The FT of w 2 ( t ) is 1 W 2 ( f ) = [ j ( f f c ) + j ( f + f c )] 2 The FT of...
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This note was uploaded on 01/28/2012 for the course AERO 16.01 taught by Professor Markdrela during the Fall '05 term at MIT.

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s16_ps13_sol - Unified Engineering II Spring 2004 Problem...

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