478 Auxtendix A Solutions to the Exercises Solution 5.6 ( s -$ l)(s - 2) A B +-+F ( s )= sZ(s 2)(s+3)(.s - 1) s s?- c +- D +- E s s -1 + 1/3 F(.v) = ,s2 R -1/3 5/18 -1/6 s-1 +++-+s s+2 4- Determiiic B by evaluating F ( s ) at the zcro s = -1 1 F(-l) =; - - L ? - - + - + - = - B + - = j B = 3 36 12 36 1/3 2/9 5/18 1/G + -+ - _ T l i e ~ : F ( s ) =I s2 s s+2 Ss.3 s-1 Inverse trarisforrnation using Table 4.1 yicltls: 1/3 ~ Solution 5.7 - - 2 - 16s - A F ( s )= ( s 4-1)”s 4-2 ) ( s -I- ) (s 1)2 + A = ( + l)”F(s)( (s+1)2 D s +3 =- =1 hT-1 F ( s ) = _ B c‘ +-+ +-s 41 s J, 2 B + + s s Si-3 s+3 A trick t,o calculate B is putting some value of this casc F’(s) is evaluated at s = R (but not a pole) into b’(%s).lu