Control ENG HW_Part_3

Control ENG HW_Part_3 - X ( s) = 1 s( s + s + 1) 1 A Bs + C...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: X ( s) = 1 s( s + s + 1) 1 A Bs + C = +2 s ( s + s + 1) s s + 2 s + 1 2 1 = A(s2 + 2s +1) + Bs2 + Cs 0 = A + B(by equating the co − effecient of s 2 ) 0 = 2 A + C (by equating the co − effecients of s ) 1 = A(by equating the co − effecients of const) A+ B = 0 B = −1 C = −2 A A = 1, B = −1, C = −2 1 s+2 −2 s s + 2s + 1 1 (s + 1) + 1 L−1{ X ( s )} = L−1 − (s + 1)2 s X ( s) = 1 1 { X (t )} = 1 − L−1 + 2 s + 1 (s + 1) { X (t )} = 1 − e − t (1 + t ) dx 2 dx 3.1 C + 3 + x = 1 x ( 0) = x ' ( 0) = 0 2 dt dt by Applying laplace transforms, we get = ( s 2 + 3s + 1) X ( s ) = X ( s) = 1 s( s + 3s + 1) 2 1 s 2 X ( s) = A Bs + C +2 s s + 3s + 1 1 = A( s 2 + 3s + 1) + Bs 2 + Cs 0 = A + B(by equating the co − effecient of s 2 ) 0 = 3 A + C (by equating the co − effecients of s) 1 = A(by equating the co − effecients of const) A+ B = 0 B = −1 C = −3 A = −3 A = 1, B = −1, C = −3 s+3 1 L−1{ X ( s )} = L−1 − 2 s s + 3s + 1 −1 −1 1 L { X ( s )} = L − s s + 1 L−1{ X ( s )} = L−1 − s s + X (t ) = 1 − e − 3t 2 (Cos s+3 2 3 − 2 2 5 2 3 2 3 . s+ 2 5 2 − 2 2 2 3 3 5 − s+ − 2 2 2 5t 3 5 + t sinh 2 2 5 3.2(a) dx 4 d 3 x + 3 = Cos t; x (0) = x ' (0) = x ''' (0) = 0 4 dt dt x11 (0) = 1 2 5 2 5 2 ...
View Full Document

This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.

Ask a homework question - tutors are online