ECE 423Law, J.D.Energy Systems IISpring 2009Session 32aPage 1/2STEADY STATE ERROR TO A UNIT STEP INPUTP(s)H(s)C(s)+ -X(s) E(s) Y(s)F(s)Y(s)=C(s)P(s)E(s) (1)E(s)=X(s) − H(s)Y(s) (2)Y(s)=C(s)P(s)(X(s) − H(s)Y(s)) (3)Y(s)=C(s)P(s)X(s) − C(s)P(s)H(s)Y(s) (4)Y(s)+C(s)P(s)H(s)Y(s)=C(s)P(s)X(s) (5)(1+C(s)P(s)H(s))Y(s)) = C(s)P(s)X(s) (6)Y(s)) =C(s)P(s)1+C(s)P(s)H(s)X(s) (7)E(s)=X(s) − H(s)Y(s) (8)E(s)=X(s) −H(s)C(s)P(s)1+C(s)P(s)H(s)X(s) (9)E(s)=(1−C(s)P(s)H(s)1+C(s)P(s)H(s))X(s) (10)limt⇒∞e(t)=lims⇒0se(s) (11)ECE 423Law, J.D.Energy Systems IISpring 2009Session 32aPage 2/2x(t)=u(t) Unit step function (12)x(s)=1s(13)limt⇒∞e(t)=lims⇒0s (1−C(s)P(s)H(s)1+C(s)P(s)H(s))1s(14)C(s)=CN(s)CD(s)(15)P(s)=PN(s)PD(s)(16)H(s)=HN(s)HD(s)(17)limt⇒∞e(t)=0 = lims⇒0(1−CN(s)CD(s)PN(s)PD(s)HN(s)HD(s)1+CN(s)CD(s)PN(s)PD(s)HN(s)HD(s)) (18)1 = lims⇒0CN(s)CD(s)PN(s)PD(s)HN(s)HD(s)1+CN(s)CD(s)PN(s)PD(s)HN(s)HD(s)(19)1 = lims⇒0CN(s)PN(s)HN(s)CD(s)PD(s)HD(s)+CN(s)PN(s)HN(s)(20)0 = lims⇒0CD(s)PD(s)HD(s) (21)0 = lims⇒0s(sn+ an−1sn−1+ an−2sn−2+ ..... + a1s+ a0) (22)Must have at least one pure
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