BME 343_Fall 2009_HW 7_solution

BME 343_Fall 2009_HW 7_solution - 7.14(a) T T 0 0 X ( ) = e...

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7.1 4 (a) () 00 1 TT ja T at j t j a t e Xe e d t e d t j a ω ωω −+ −− === + ∫∫ 7.2 4 (a) 11 22 jt j tt x te e d e d ππ == 2s i n [ ( ) ] 2( ) t t j tt ee jt t 0 0 sin[ ( )] sin [ ( )] ct t 7.3 3 (c) () cos [ () ( ) ] cos () () cos () ( ) xt t ut ut tut π =− = Since sin( ) cos( ) 2 −= Î ( ) c o s ( ) ( ) s i n t =+ /2 1 ( ) [ (1 )(1 ) ] { [ ) ] } 21 2 1 j j j πω ωπ ωδ δ + + + + + + ) ) ) jj j δω += + = + ; ) ) ) j = + Î [ ) ] [ ) ] 0 j e j + + + = 1 j j 7.3 6 (a) The signal x(t) in this case is a triangle pulse 2 t Δ multiplied by cos( ) t ( )cos (10) 2 t x From Table 7.1, pair 19 (page 702) 2 s i n t c Δ⇔ and 10 10 1 cos(10 ) [ ] 2 tee So ( 10) ( 10) ( )cos(10 ) {sin ( ) sin ( )} 2 2 t tc c + +
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7.3 7 (a) 44 () ( ) ( ) 22 Xr e c t r e c t ω −+ =+ Since 1 sin ( ) ( ) 2 c t rect π So 11 2 s in () s s in ()cos (4) jt x tc t e c t e c t t ππ = 7.3 11 (a) ( ) X xte d t −∞ = [( ) ] [ ( ) ] [ ( ) ] dX d d x t e dt x t e dt jtx t e dt dd d ωωω ωω ∞∞ −−− −∞ −∞ −∞ === ∫∫ Therefore, dX jtx t d −⇔ (b) -15 -10 -5 0 5 10 15 0 0.5 1 1.5 2 w radian Magnitude -15 -10 -5 0 5 10 15 -1 -0.5 0 1 w radian Phase
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() [ ] at at te u t jt je u t
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This note was uploaded on 12/14/2011 for the course BME 343 taught by Professor Emelianov during the Spring '09 term at University of Texas at Austin.

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BME 343_Fall 2009_HW 7_solution - 7.14(a) T T 0 0 X ( ) = e...

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