CUAD. CONTROL I.pdf - Profe Saul

C(s)=R(s)Wn22( S + LWn + Wn 1−L )( S + LWn −Wn1−L )22C(s)WnR(s)= ( S + LWn + J& Wd )( S + LWn − J&Wd )Forma general 2do orden para la condición0 < L < 1Encontrar la respuesta transitoria (C(s)) de 2do de un sistema con una R(s) = 1/S parauna condición 0 < L < 1C(s)=R(s)1 ⎡C(s)=S⎢⎣Wn( S + LWn + J& Wd )( S + LWn − J&Wd )2Wn( S + LWn + J&Wd )( S + LWn − J&Wd ) ⎥ ⎦[( S + LWn + J& Wd )( S + LWn − J&Wd )]22WnC(s)= S(a+b) (a-b) = a 2 -b 2C(s)C(s)=Sa b a bWn222= sacamos a J − ( J&) = −(− −1)= 12S[ ( S + LWn) − ( J&Wd ) ]2Wn= +[( S + LWn) − ( Wd ) ] 2( S + LWn) 2+ ( Wd ) 222 2S + 2SLWn+ L Wn2 2AS⎤BS + C=A22[( S + LWn) + ( Wd ) ] + S[ BS + C]22S ( S + LWn) + ( Wd )[ ]=22 2 2[ + 2SLWn+ L Wn + Wd ] + S[ BS + C]2S[ ( S + LWn) + ( Wd ) ]2A S=AS2+ A2SLWn+S2 2AL Wn+ AWd2[( S + LWn) + ( Wd ) ]22+ BS2+ CSS=S=222( A + B) + S( A2LWn+ C)+2S ( S + LWn) + ( Wd )2AL Wn[ ]2+ AWd2 2 2[ A + B] + S[ A2LWn+ C] + A[ L Wn + Wd ]2S[ ( S + LWn) + ( Wd ) ]2242