AME451_HW_Solution_3

AME451_HW_Solution_3 - AME 451 HW Solution #3 1 a) Let g (t...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
AME 451 HW Solution #3 1 a) Let () cos (2)1 () gt t t = i 2 4 s Gs s = + 0.5 2 [ () ] [ ] ( 0 .5 ) 0.5 (0 . 5 )4 t Fs Lft Le gt s s == = + + = ++ b) Let 0.5 cos (3)1 t e t t = i From problem a), we know 22 0.5 . 5 )3 s s + = 2 2 ] [ ( 2 ) ] 0.5 . 5 s s Lgt e Gs s e s = + = c) 0.5 0 0.5 0.5 0.5 0.5 (5 45)1 (5 )1 4 [cos(5 )cos( ) sin(5 )sin( )] 1( ) 44 cos(5 ) 1( ) sin(5 ) 1( ) tt t f te t t t et t t t t π ππ −− =+ =− ii i From problem a) we know 0.5 0.5 0.5 [c o s ( 5 ) 1 ( ) ] . 5 )5 5 [s i n ( 5 ) 1 ( ) ] . 5 t t s Le t t s t t s + = = i i Then 0.5 0.5 ] [ cos(5 ) 1( )] [ sin(5 ) 1( )] 20 . 52 5 2( 0 . 5 ) 5 . 5 24 . 5 . 5 t t t t s ss s s = + = d)
Background image of page 1

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

View Full DocumentRight Arrow Icon
0.5 0 0.5 0.5 0.5 0.5 () cos (3 45)1 (3 )1 4 [cos(3 )cos( ) sin(3 )sin( )] 1( ) 44 22 cos(3 ) 1( ) sin(3 ) 1( ) tt t gt e t t e t t et t t t e t t π ππ −− =+ =− ii i Then, 0.5 0.5 [ () ] [ cos(3 ) 1( )] [ sin(3 ) 1( )] 20 . 52 3 2( 0 . 5 ) 3 . 5 . 5 . 5 Gs Lgt Le t t t t s ss s s = + ++ = Finally, 2 2 ] [ ( 2 ) ] . 5 2( 0 . 5 ) 3 s s Fs Lft e Gs s e s == = = e) 3 () 3s in s in2 2 f t t ω ωω 3 ] [ s in2 ] 2 3 (2 ) L t s = + f) 0.5 0 0.5 0.5 0.5 0.5 0.5 0.5 ( ) 3 sin cos( 45 ) 3 sin [cos cos sin sin ] 32 (sin cos sin sin ) 2 1 1c o s ( 2 ) (s i n ( ) 2 [s i n ( 2 ) c o s ( 2 ) ] 4 t t t ft e t t e t t t t t t t e e t = +
Background image of page 2
0.5 0.5 0.5 0.5 0.5 0.5 22 32 ( ) [ ( )] [ [ sin(2 ) cos(2 )]] 4 ( [ sin(2 )] [ ] [ cos(2 )]) 4 2 1 0 . 5 () 4 ( 0.5) (2 ) 0.5 ( 0.5) (2 ) 0 . 5 2 32 1 4 ( 0.5) (2 ) 4 0.5 tt t t Fs Lft L e t e e t Le t t s ss s s ωω ω −− == + =− + + + ++ + + 2 a) Using partial fraction decomposition, 2 11 61(31 0 ) (31 0 ) 0 ) 0 ) AB +− =+ From multiple poles method, q p s r q k k k r q s F p s ds d k A = + = | )] ( ) [( ! 1 ) ( Then we have 10 1 2 1 | ) 10 3 ( 1 | ] ) 10 3 )( 10 3 ( 1 ) 10 3 [( ! 0 1 10 3 10 3 1 0 0 + = + + = + + + + + + = = = s s s s s s s s ds d A 10 1 2 1 | ) 10 3 ( 1 | ] ) 10 3 )( 10 3 ( 1 ) 10 3 [( ! 0 1 10 3 10 3 1 0 0 = + + + = + + + + + = + = + = s s s s s s s s ds d B So 1 1 () ( ) ( ) 10 ( 3 10) 10 ( 3 1 (3 1 1 [ () ] ( ) [ ] 10 ( 3 10 ( 3 10 10 ft L Fs L L ee −+ + + b) Let 2 1 61 s Gs + =
Background image of page 3

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

View Full DocumentRight Arrow Icon
2 () s Fs e Gs =
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

AME451_HW_Solution_3 - AME 451 HW Solution #3 1 a) Let g (t...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online