s11_ps10_sol

s11_ps10_sol - Unified Engineering II Spring 2004 Problem...

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Unformatted text preview: Unified Engineering II Spring 2004 Problem S11 (Signals and Systems) Solution 1. From the problem statement, 9 . 82 m/s 2 n = 2 = 0 . 1077 r/s 129 m/s 1 1 = = = 0 . 0471 2( L /D 0 2 15 Therefore, 1 G ( s ) = s ( s 2 + 0 . 01015 s + 0 . 0116) The roots of the denominator are at s = 0, and . 01915 . 01015 2 4 . 0116 s = 2 = . 005075 . 1075 j So 1 G ( s ) = s ( s [ . 005075 + 0 . 1075 j ]) ( s [ . 005075 . 1075 j ]) Use the coverup method to obtain the partial fraction expansion G ( s ) = 86 . 283 + 43 . 142 + 2 . 036 j s s [ . 005075 + 0 . 1075 j ] 43 . 142 2 . 036 j + [ . 005075 . 1075 j ] s Taking the inverse Laplace transform (assuming that g ( t ) is causal), we have g ( t ) =86 . 283 ( t ) + ( 43 . 142 + 2 . 036 j ) e ( . 005075+0 . 1075 j ) t + ( 43 . 142 2 . 036 j ) e ( . 005075 . 1075 j ) t Therefore, g ( t ) = ( t ) 86 . 283 + 2 e . 005075...
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s11_ps10_sol - Unified Engineering II Spring 2004 Problem...

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