ME369 Modeling,Analysis and System Control-A --Lecture 141027 Week 7# Ch5.3~5. Oct.27(M) Transient,second order system, spec. 6 Transient,higher order system Ch5.45. 0ct.29(W) HW4 Stability,Routh Criterion 7 Week 8# Routh Criterion Ch5.7 Nov.3(M) HW3 due Steady State Error Ch5.8 Practical lecture 1-control Nov.5(W) system modeling Quiz 2
ME369 Modeling, Analysis and System Control-A --Lecture_141027 Week 7# Oct.27(M) Transient, second order system, spec. Ch5.3~5. 6 Oct.29(W) Transient, higher order system Stability, Routh Criterion Ch5.4~5. 7 HW4 Week 8# Nov.3(M) Routh Criterion Steady State Error Ch5.7 Ch5.8 HW3 due Nov.5(W) Practical lecture 1 – control system modeling Quiz 2
Outline of today's PPT file Transient Response of Second-Order Systems ·二阶系统的瞬态响应 Step response Ramp response Transient Response Analysis and Specification
Outline of today’s PPT file • Transient Response of Second-Order Systems • 二阶系统的瞬态响应 – Step response – Ramp response • Transient Response Analysis and Specification
Transient Response of 2nd-Order Systems m m成=F-bx-x mx+bx+kx=F X(s) 1 7777777777777777777777777 F ms2 +bs+k -Forced response,ZSR,particular solution Free response,ZIR,homogeneous solution Total response Z/R+ZSR(for linear system!) 5= 日96
Transient Response of 2nd-Order Systems mx F bx kx mx bx kx F 2 X s 1 F ms bs k – Forced response, ZSR, particular solution – Free response, ZIR, homogeneous solution – Total response = ZIR + ZSR (for linear system!) n k m 2 b km
Free Response )+250n)+oy=0 I.C.y(0),(0) → [s2Y(s)-y(0)-(0)]+250n[sY(s)-y(0)]+o,Y(s)=0 s=-0+j@d Damped natural frequency Y(s)= (s+250n)y(0)+(0) 阻尼自然频率 → s2+250nS+0n o=Cw,:wj=wV1-52 S+σ > Y(s)=a(s+a)+0j +6 (s+o)2+o → Y(s)=0)+0) @a S+O @d +o+a+X0 s+o)2+o7 → 0=g0+0e"sino,1+X0ecosa,1 @a 日96
Free Response 2 2 0 n n y yy I.C. (0), (0) y y 2 2 [ ( ) (0) (0)] 2 [ ( ) (0)] ( ) 0 n n s Y s sy y sY s y Y s 2 2 ( 2 ) (0) (0) ( ) 2n n n s y y Y s s s 22 22 ( ) () () d d d s Y s a s s b 22 22 ( ) ( ) (0) (0 ) ( ) (0 ) d d d d y y s s s y s Y (0) (0) ( ) sin (0) cos t t d d d yt e t e y y y t 2 ; 1 s zw w w z = =- nd n d s j Damped natural frequency 阻尼自然频率
Solutions for 2nd Order Systems x()◆ )+250ny+o元y=u(t) x(0) 01 0 5>1 Overdamped 5=1 Critically damped 0<<1 Underdamped
Solutions for 2nd Order Systems 2 2 () n n y y y ut 1 1 0 1 Overdamped Critically damped Underdamped
Transient response with unit-step input F my+bi+ky=F m b b @.-Vm 2√mk +250n少+o7y= kS 1/m y(o)s+25os+@(
F m y k b m y by ky F n k m 2 b mk 2 2 n n F y yy m 2 2 1/ 2 n n m Ys Fs s s Transient response with unit-step input
Transient response with unit-step input Standard expression +250n立+o7y=ou(t) Roots are S1,2=±j0m c(t) y(t)=1-cos(@,t) 2 Free vibration Critically stable 5=0 Without damping零阻尼
Transient response with unit-step input 2 2 2 () nn n y y y ut 0 Without damping 零阻尼 Roots are n s 1,2 j 2 2 2 () () n n Ys Us s y ( t ) 1 cos( t ) n Free vibration Critically stable Standard expression
Transient response with unit-step input j9+250n少+07y=07u(t) Roots are s.2=-50n±j0nV1-52=-60n±j0a Damped natural frequency 阻尼自然频率 213816412 1 0.2 0.4 Rs)s 0.6 ξ=0.8 c(t)=1- s:0 sin@,1+A)(e0) B=g-5 tp 5 10 15 0<5<1 Underdamped (欠阻尼)
Roots are 0 1 Underdamped (欠阻尼) n n n d s j j 2 1,2 1 sin( ) 1 ( ) 1 2 t e c t d t n (t0) 2 d n 1 Damped natural frequency 阻尼自然频率 s R s 1 ( ) 2 1 1 tg Transient response with unit-step input 2 2 2 () nn n y y y ut
5.2=-g0n±j0nV1-52=-50n±j0a c(0=1- -sno,i+B jo (t≥0) S平面 B 04=0nV1-52 B=你个≤ 日6
2 d n 1 sin( ) 1 ( ) 1 2 t e c t d t n (t0) 2 1 1 tg n n n d s j j 2 1,2 1
Transient response with unit-step input +250n少+oy=ou(t) Roots are 2=-0n±0nV52-1 Rising monotone,without vibration,long transient reponse,can be divided as two first-order systems. 5>1 Overdamped
Roots are 1 Overdamped 1 2 P1、2 n n Rising monotone, without vibration, long transient reponse, can be divided as two first-order systems. Transient response with unit-step input 2 2 2 () nn n y y y ut
