Control ENG HW_Part_5

Control ENG HW_Part_5 - 3s 3s 1 1 = 2 −2 2 ( s + 1)( s +...

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Unformatted text preview: 3s 3s 1 1 = 2 −2 2 ( s + 1)( s + 4) 3 s + 1 s + 4 3.3 a) 2 1 1 = 2 2 − 2 s + 22 s +1 1 1 L−1 2 2 − 2 = Cost − Cos 2t s + 22 s +1 b) 1 1 A B+C = = +2 2 2 s ( s − 2 s + 5) s ( s − 1) + 2 s s − 2s + 5 2 [ A+B=0 -2A+C=0 5A=1 A=1/5 ;B=-1/5;C=2/5 We get X ( s) = 1 1 2−s s + s 2 − 2s + 5 5 Inverting,we get = 1 1 1 + e t Sin 2t − e t Cos 2t 5 2 = 1 t1 1 + e 2 Sin 2t − Cos 2t 5 3s 2 − s 2 − 3s + 2 A B C D = + 2+ + c) 2 2 s ( s − 1) ss s − 1 ( s − 1) 2 As( s − 1) 2 + B( s − 1) 2 + Cs 2 ( s − 1) + Ds 2 = 3s 3 − s 2 − 3s + 2 A( s 3 − 2 s + s ) + B ( s 2 − 2 s + 1) + C ( s 3 − s 2 ) + Ds 2 = 3s 3 − s 2 − 3s + 2 A+C=3 -2A+B-C+D=-1 A-2B=-3 B=2; A=2(2)-3=1 C=3-1=2 D=2(1)-2+2-1=1 We get X ( s ) = 12 2 1 + 2+ + ss s + 1 ( s − 1) 2 By inverse L.T L−1 [X (t )] = 1 + 2t + 2e t + te t L−1 [X (t )] = 1 + 2t + e t ( 2 + t ) 3.4 Expand the following function by partial fraction expansion. Do not evaluate co-efficient or invert expressions X ( s) = 2 ( s + 1)( s + 1) 2 ( s + 3) X ( s) = A Bs + C Ds + E F +2 +2 + 2 s + 1 s + 1 ( s + 1) s+3 2 = A( s 2 + 1) 2 ( s + 3) + ( Bs + C )( s + 1)( s + 3)( s 2 + 1) + ( Ds + E )( s + 1)( s + 3) + F ( s + 1)( s 2 + 1) 2 = A( s 4 + 2 s 2 + 1)( s + 3) + ( Bs + C )( s 2 + 4 s + 3)( s 2 + 1) + ( Ds + E )( s 2 + 4 s + 3) + F ( s + 1)( s 2 + 4 s + 1) ...
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