Module 5 Second-Order System Time-Domain Response (2 hours) Ramp response ● Harmonic response Transient response as a function of the location of the poles
Module 5 Second – Order System : Time-Domain Response (2 hours) • Ramp response • Harmonic response • Transient response as a function of the location of the poles
Ramp response 5。1 C(S) R(S s2+25o,s+ (1)=t→R(S)=- S Assume: +250S+a <1 12525s+50 a,s, (s+so, ,+a2(s+so, )+a2
5. 1
S+4 C(s)= s20,S On (s+5a,+a2(s+5 a, ),+02 c()=1-+e coS(@,t)+ sin(at) O,v1-s 2 Ideal Constant error Transient error
Step Response MATLAB L6. m on the web 16 R amp Step 20 Second order svstem Ime (sec Step response-zero error Ramp response- constant error
5.2 Impulse response 2 R(s)=1,C(s) s2+25n,S+n Assume: O<2<I e s@nt SIn a st t→)00 0 e(∞)=0 c(t) 0<<1 5>1
5.2 Impulse Response 2 2 2 2 ( ) 1, ( ) n n n s s R s C s + + = = Assume: 0 1 sin 1 0 1 ( ) 2 2 − ⎯ ⎯→ − = − t→ n n t c t e t n e() = 0 t c(t) 0 1 1