HW4_Solution - Home Work 4 Solution 1. We start by noting...

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Home Work 4 Solution 1. We start by noting that 2 3 2 3 5 () ( 1)( 2) ss Gs s s s  has a triple pole at s = 0, and single poles at -1 and -2. Therefore, the partial fraction expansion will have the form 23 ( 1) ( 2) A B C D E s s s s s Now we have many procedures to calculate A, B, C, D, and E. (i) We can bring the partial fraction expansion to a common denominator as 2 3 3 3 4 3 2 3 2 2 4 3 4 3 3 4 3 2 3 ( 2) ( 2) ( 2) ( 2) ( ( 2) ( 3 2 ) ( 3 2 ) ( 3 2) ( 2 ) ( ) ( 2) ( ) (3 2 ) (2 3 ) (2 3 ) 2 ( 2 As s s Bs s s C s s Ds s Es s s s s A s s s B s s s C s s D s s E s s s s s A D E s A B D E s A B C s B C s C s s s ) Now equating coefficients of the numerator gives us the following simultaneous equations: 0 3 2 0 2 3 2 2 3 3 25 A D E A B D E A B C BC C  We can now solve these equations to get: C=5/2, B=-9/4, A=25/8, D=-4, E=7/8 (ii) We can use the reside formulae to first calculate D and E: D=G(s)(s+1)|s=-1 = -4, E = G(s)(s+2)|s=-2 = 7/8, C=G(s)s 3 |s=0 = 2.5. Now we cannot calculate A and B so easily unless we use the more complicate formula for residues for repeated poles involving derivatives. A simple solution at this point is to note that 32 ( ( 2) C D E A B s s s s s
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This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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HW4_Solution - Home Work 4 Solution 1. We start by noting...

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