CUAD. CONTROL I.pdf - Profe Saul

−1L f ( s )= 2−11LS2−−1L1( S −1)f ( t)= 2t+ etPolos complejos conjugadosLo identificamos con un S 2 multiplicando en los polos.2S + 4 + 2SF s)=2S−13S(2Lf3Ss)=2S2=+ 422( S + 4) S ( S + 4)+ 42( S + 4)2A B= +2S SCS + D A S+ =2S + 4222[ ( S + 4)] + B[ S + 4] + CS + D[ S ]2S ( S + 4)2(2=AS3− 4AS+ BS2S2− 4B+ CS2( S + 4)2+ DS2S=32[ A + C] + S [ B + D]S22( S + 4)+ 4AS+ 4BA + C = 0B – D = 34A = 04B = 4B = 1B + D = 31 + D = 3D = 3 – 1 = 24A = 0A = 0C = 0−1L f ( s )=−10L +S−11LS2+−10S+ 2=2S + 4−11LS2+S22=+ 4−11LS2+S22+( 2) 2Tabla−1L f ( s )=−11LS2+−1LS22+( 2) 2f ( t)= t + sen2t36