Control ENG HW_Part_5

# Control ENG HW_Part_5 - 3s 3s 1 1 = 2 −2 2 s 1 s 4 3 s 1...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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) ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online