solution06 - 1 (see the notes, p 4-27) 2.0833e+000 //...

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1/. (a) )) 66 . 8 5 ( ))( 66 . 8 5 ( ( ) 10 ( 2 . 0 ) ( 02 . 0 ) ( 100 10 100 10 ) ( 2 j s j s s s s Y s s X s s s s H + + = = + + + = poles = 0, -5+8.66j, -5-8.66j ° = = = = + = = = = = = 150 0115 . 0 005773 . 0 01 . 0 005773 . 0 01 . 0 ) ( ) ( 02 . 0 ) ( ) ( 2 2 2 3 2 2 1 1 2 1 c c j c c j s Y p s c s Y p s c p s p s ) radians 618 . 2 66 . 8 cos( 023 . 0 02 . 0 ) 150 66 . 8 cos( ) 0115 . 0 ( 2 02 . 0 ) ( 5 5 + = ° + = t t t y t e t e
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1/. (b)
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1/. (c) 2/. (a) 100 20 5 260 126 20 + + = + + + x x y y y (b) zeros at -2+4j, -2-4j pole at -10 residue = 15.385 pole at -5+1j residue = -5.1923-10.962j pole at -5-1j residue = -5.1923+10.962j magnitude (-5.1923-10.962j) = 12.1295 angle (-5.1923-10.962j) = -115.3452º ) 3452 . 115 cos( 259 . 24 385 . 15 ) ( 5 10 ° + = t e e t h t t See next page for graph.
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Graphical computation of residue for p 2 :
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3/. (a) 2 1 ) ( ) ( s s X t t x = = ) 24 11 ( 50 ) ( ) ( ) ( 2 2 + + = = s s s s X s H s Y >> num = [50]; >> den = [1 11 24 0 0]; >> [c,p] = residue(num, den) c = // residues -1.5625e-001 1.1111e+000 -9.5486e-001 // residue for (s-0)
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Unformatted text preview: 1 (see the notes, p 4-27) 2.0833e+000 // residue for (s-0) 2 p = // poles -8.0000e+000 -3.0000e+000 0 0 t e e t y t t 0833 . 2 95486 . 1111 . 1 15625 . ) ( 3 8 + + = (b) the input is the sum of i) a ramp with slope 3 ii) a ramp with a slope of -9 delayed by 4 seconds iii) a ramp with a slope of 6 delayed by 6 seconds therefore the output is the sum of the responses to these three inputs let r(t) be the ramp response determined in part (a) then ) 6 ( ) 6 ( 6 ) 4 ( ) 4 ( 9 ) ( 3 ) ( + = t u t r t u t r t r t y...
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This note was uploaded on 07/16/2009 for the course SYSC 3600 taught by Professor John bryant during the Winter '08 term at Carleton CA.

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solution06 - 1 (see the notes, p 4-27) 2.0833e+000 //...

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