Module 3 First-Order System (1 hours) First order systems Time response of first order systems . Proportional feedback control of first order systems
Module 3 First – Order System (1 hours) •First order systems •Time response of first order systems •Proportional feedback control of first order systems
First order systems - systems described by a single first order differential equation First order systems contain one energy storage element. lump mass sprin g capacitor inductor
First order systems – systems described by a single first order differential equation First order systems contain one energy storage element: • lump mass • spring • capacitor • inductor
3.1 The model of First-Order system Kirchoff,'s law for node 1 R =O Cν R Take the laplace transform Q1: What assumption have we made for y when Csv taking its Apace transform? L R V: -V=RCsV T=RC 11+RCs1+
3.1 The Model of First-Order system
3.2 Impulse response Time response of first order systems: C I Impulse input: R1+ ( r=0(1)→R=1→C(s) 1+ Take the inverse Laplace of C(s)
3.2 Impulse Response
Impulse response 0.8 0.6 C(0)-7 0.4 0.2 心中书 1中 0 0.5 1.5 2 2.5 3.5 Time(sec.)
1 C(0+ ) = 2 1 (0 ) C + = − 2 1 − ( )
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.)
The steady state error ess in the ramp response of first-order system is e()=r(t)-c(t) =t-(t-t+Te tt→>0 T-te z√→e√ SS
• The steady state error in the ramp response of first-order system is: ss e e(t) = r(t) − c(t) ( ) 1 t t t e − = − − + = − − t e 1 t → es s
