《自动控制原理》课程教学资源（PPT课件讲稿）Module 3 First-Order SystemModule3

3.3 Step Response Step input: =H(1)→R= →C(s)= 1+S To obtain the inverse Laplace, we need partial fraction expansion: 1+酉 S+ s+ 0+ S c()=1-e

3.3 Step Response

Step Res ponse 0.9 0.8 0.7 0.6 0.5 0.632 0.4 0.2 0.1 0.5 1.5 2.5 3.5 Time(sec

 0.632 

3.4 Ramp response Ramp input t→R →C(S) 2 Partial fraction expansion C C(s)= 十一十 +sls +sS S a=C(s)s 1: b dC(s)s - 0 C(s)+ 1/r

3.4 Ramp Response

c(1)=t-21- Linear S imulation Results p output Steady state error Q2: What is the magnitude of the steady state error? How can it be minimized Time(sec.)