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 ω n = √ 2 = 0 . 1077 r/s 129 m/s 1 1 ζ = = = 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 σ ( 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 ) = σ ( t ) 86 . 283 + 2 e − . 005075...
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s11_ps10_sol - Unified Engineering II Spring 2004 Problem...

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