《自动控制原理》课程教学资源（PPT课件讲稿）Module 5 Second-Order System Time-Domain Response

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: O1

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

Harmonic response: TC+c=r 5.3 Joule Jo+ cne o(+Jat, I+ joT Recall that for the first order system we have found the harmonic response by substituting in the transfer function S=10

5.3

Harmonic response: This substitution can be done for any system! C(jO)=C(s) R(0) (o)+2o,(o)+m2 +2 +1 amplitude ratio Phase change C(ja 250 R(jO) 2(=∠C-∠R=m0

Bode Diagrams MATLAB: Resonance peak L6. m on the web 100 10 Frequency(rad/sec)

Q: Where does the resonance peak occur? C(a) R(jo) 0\+ 45 A Where the denominator has a minimum 0→a=O 2 do

5.4 Relationship between System poles and Transient Response (P79) The characteristic equation and the characteristic roots are s2+25n+Cn2=0 s1,s2=-2On±jonV1-52=-on±jon tg

5.4 Relationship between System Poles and Transient Response (P79) The characteristic equation and the characteristic roots are: 2 0 2 2 s +  n s +n = n n n d s s = −   j − = −  j 2 1 , 2 1       2 1 1 1− = = − − t g t g n d