s10_ps10_sol

s10_ps10_sol - Unified Engineering II Spring 2004 Problem...

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Unformatted text preview: Unified Engineering II Spring 2004 Problem S10 (Signals and Systems) Solution 1. Because the numerator is the same order as the denominator, the partial fraction expansion will have a constant term: 3 s 2 + 3 s − 10 G ( s ) = 2 s − 4 3 s 2 + 3 s − 10 = ( s − 2)( s + 2) b c = a + + s + 2 s − 2 To find a , b , and c , use coverup method: a = G ( s ) | = 3 s = ∞ 3 s 2 + 3 s − 10 b = s + 2 = 2 s =2 3 s 2 + 3 s − 10 1 c = = s − 2 s = − 2 So 2 1 G ( s ) = 3 + s − 2 + s + 2 , Re[ s ] > 2 We can take the inverse LT by simple pattern matching. The result is that g ( t ) = 3 δ ( t ) + 2 e 2 t + e − 2 t σ ( t ) 2. 6 s 2 + 26 s + 26 G ( s ) = ( s + 1)( s + 2)( s + 3) a b c = + + s + 1 s + 2 s + 3 Using partial fraction expansions, 6 s 2 + 26 s + 26 ( s + 2)( s + 3) 3 a = = s = − 1 6 s 2 + 26 s + 26 b = 2 = ( s + 1)( s + 3) s = − 2 6 s 2 + 26 s + 26 c = = 1 ( s + 1)( s + 2) s = − 3 So 3 2 1 G ( s ) = + + , Re[ s ] > − 1 s + 1 s + 2 s + 3 The inverse LT is given by...
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This note was uploaded on 01/28/2012 for the course AERO 16.01 taught by Professor Markdrela during the Fall '05 term at MIT.

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s10_ps10_sol - Unified Engineering II Spring 2004 Problem...

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