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3104442_18_Ch3-作业解答

3104442_18_Ch3-作业解答

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Ch_3 3-1 3-1ê[email protected] êF t 1 3-1 1
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0 0, 2, 4, 1 2 1, 3, n n n F jb jE n n π = ± ± = - = - = ± ± L L 2 T
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3-4 3-4áÅ }+ t 2 3-4 h #w ½ ª*𠹪* 0 0 1 ( ) 2 T E a f t dt T = = 3-4 ( ) f t G 0 2 2 2 0 2 2 2 0 2 2 0 2 ( )cos( ) 2 ( )cos( ) 4 2 cos( ) 8 1 sin( ) sin( ) 8 cos 1 ( ) 2 0 4 ( ) T n T T T T T a f t n t dt T f t n t dt T E t n t dt T T E t n t n t dt T n E n T n T n E n n ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ π - = = = = - = - = - g ° 2 T π ϖ = °° 0 n b = °° 3
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0 1 2 1,3,5, 2 2 2 ( ) [ cos( ) sin( )] 4 ( )cos( ) 2 ( ) 4 1 1 [cos( ) cos(3 ) cos(5 ) ] 2 3 5 n n n n f t a a n t b n t E E n t n E E t t t ϖ ϖ ϖ π ϖ ϖ ϖ π = = = + + = + - = - + + + L L …… 3-4-1 ;ˆ½ª * 3 3-4 t 3 3-4 @@ @ êF% ƒ~ ¹+ p êF f ( t ) E O 2 T T t t 4 3-4 4 p êF ïêF à êFP êF h ¹ª*h ¹ª* @@ @ êFY ƒ~ ¹+ p êF 4
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g ( t ) E O 2 T - t 2 T F 5 3-4 a -40 -30 -20 -10 0 10 20 30 40 0 0.1 0.2 0.3 0.4 0.5 G(w) ( 6 ° 3-4 b 5 2 | | (1 ), (| | ) 2 ( ) 0 , (| | ) 2 t T E t T g t T t - = |{ o ( ) g t { } 2 2 2 8 ( ) ( ) sin 2 4 4 a ET T E T G g t S T ϖ ϖ ϖ ϖ = = = F 6 3 ( ) ( ) ( ) ( ) 2 2 2 T T T f t g t g t g t = + + - + - Ξ Γ…♠ { } 0 0 ( ) ( ) j t f t t F e ϖ ϖ - - = F 5
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{ } 3 2 2 2 3 2 2 2 3 2 2 2 2 2 3 ( ) ( ) ( ) ( ) ( ) 2 2 2 ( ) ( ) ( ) ( ) 8 sin ( ) 4 T T T j j j T T T j j j T T T j j j T T T F f t g t g t g t G e G e G e G e e e E T e e e T ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ - - - - - - = = + + - + - = + + = + + = + + F F 3 2 2 2 2 3 2 2 2 2 2 2 3 2 2 2
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