Module 4 Second- Order System (1 hours) · Poles and zeros Second order systems: damping, natural frequency Step response of second order system
Module 4 Second – Order System (1 hours) • Poles and zeros • Second order systems: damping, natural frequency • Step response of second order system
Second order systems- systems described by a second order differential equation Second order systems contain two kinds of energy storage elements Lump mass and spring Capacitor and inductor Others
Second order systems – systems described by a second order differential equation Second order systems contain two kinds of energy storage elements: • Lump mass and spring • Capacitor and inductor • Others
Second order systems: Example - motor positioned load OV Gear box Potentiometer 2 Load 10V-10V+v OE Amplifier Motor Potentiometer 1 van k K K R+Lsc+Js」Ns Potentiometer 1 K Potentiometer 2
va K K K R+Lsc+Js Potentiometer I K Potentiometer 2 Field controlled motor model R = iR+l 丿=R+Lsl 丿R+SL a T orque t.=Ki, K.-torque constant
Model of the gear box: N T=J0+c0
6 K K,○ K a(R+∠se+)1N Potentiometer 1 K Potentiometer 2 Often the motor tF can be simplified K K K K R+Ls)c+ s)rc 1+s1+s L Rcl 1+=s R(c+Js R C C Dominant pole
K R(c+Js) N S K , G KKKINR 8d 1+GH Js"+CS+K,KKm/NR $+250, s+a Natural frequenc Damping ratio K. K C NR NRJ 21KKKJ a p m
4. 1 Second-Order System Transfer Function G(s)= s(IS+ G(S) K R(s)1+G(S)H(S)Ts2+S+K 3+250,s+On Where K-system total gain(actual parameter) T-time constant(actual parameter s-the damping ratio(characteristic parameter) n-the natural frequency (characteristic parameter
4.1 Second-Order System Transfer Function 2 2 2 2 1 ( ) ( ) 2 ( ) ( ) ( ) n n n Ts s K s s K G s H s G s R s C s + + = + + = + = ( 1) ( ) + = s Ts K G s Where, K – system total gain (actual parameter) T – time constant (actual parameter) – the damping ratio (characteristic parameter) – the natural frequency (characteristic parameter) n
It can be drawn into a feedback system block di agra R K C R K TS<+s+K S(7S+1) R R S4+25O,S+On s(s+25O,) K T
• It can be drawn into a feedback system block diagram: Ts s K K + + 2 R C s(Ts +1) K R C R C ( 2 ) 2 n n s s + R C 2 2 2 2 n n n s s + + K T n n = = 2 1 ; 2
The closed-loop characteristic equation G(s) K R(S)1+G(S)H(S) TS'+S+K S+250, 5+0 1+G(s)H(S)=0■ +24aS+ 0 2=-5on±jon1-2=-on±j the damped natural frequency
The closed-loop characteristic equation: 2 2 2 2 1 ( ) ( ) 2 ( ) ( ) ( ) n n n Ts s K s s K G s H s G s R s C s + + = + + = + = 1+ G(s)H(s) = 0 2 0 2 2 s + n s +n = n n n d s s = − j − = − j 2 1 , 2 1 2 d =n 1− – the damped natural frequency
