디지털 제어 Sun Moon University 1 of 19 목 차 9. Frequency response analysis Kyoung-Chul DIGITAL CONTROL.

Presentation on theme: "디지털 제어 Sun Moon University 1 of 19 목 차 9. Frequency response analysis Kyoung-Chul DIGITAL CONTROL."— Presentation transcript:

1 디지털 제어 Sun Moon University 1 of 19 목 차 9. Frequency response analysis Kyoung-Chul Koh(kckoh@sunmoon.ac.kr) DIGITAL CONTROL

2 디지털 제어 Sun Moon University 2 of 19 강의진도 참고문헌 [1] 고경철, 이장명 저, “C 로 구현한 디지털 제어 시스템 ”, 홍릉과학서적, to be published 2004.4 [2] Ogata, Discrete-Time Control System, [3] Dorf, Modern Control Systems, Wesley, 1998 [1 주 ] 1 장 서론 [2 주 ] 2 장 Z 변환 [3 주 ] 3 장 역 Z 변환 [4 주 ] 4 장 이산 제어 시스템의 모델링 [5 주 ] 5 장 시간응답 [6 주 ] 6 장 정상 상태 오차 분석 [7 주 ] 7 장 안정도 해석 [8 주 ] mid-test [9 주 ] 8 장 피드포워드 보상 [10 주 ] 9 장 주파수 응답 특성

3 디지털 제어 Sun Moon University 3 of 19 Design based on the frequency-response method Frequency response - continuous time control systems > conventional frequency response design techniques >> Bode diagram - compensator design - response of the linear time-invariant system to a sinusoidal input > amplitude > phase Response of a linear time-invariant discrete time system to a sinusoidal input - frequency response of G(z) > by substituting

4 디지털 제어 Sun Moon University 4 of 19 9.1 Sinusoidal input Sinusoidal input -> frequency -> period void reference_input_set(int itype,int no_of_samples,double *r) { int k,T=0.01; if (itype==SINE) { double Ts=0.1; for(k=0;k<1000;k++) r[k]=sin(2*PHI*k*T/Ts); } ex) making sinusoidal input withand

5 디지털 제어 Sun Moon University 5 of 19 9.2 Frequency response z-transform of the sinusoidal input sinusoidal signal u(kT) frequency response of the system is given by

6 디지털 제어 Sun Moon University 6 of 19 9.2 Frequency response response of the system is given by partial fraction Multiplying both sides by (9.5) (9.6)

7 디지털 제어 Sun Moon University 7 of 19 9.2 Frequency response since and by same method, we obtain (9.9) (9.10)

8 디지털 제어 Sun Moon University 8 of 19 9.2 Frequency response (9.9),(9.10)->(9.6) where (9.14) by inverse Z transform of (9.14), then (9.15)

9 디지털 제어 Sun Moon University 9 of 19 9.2 Frequency response (9.15) The last term on the right-hand side of Eq.(9.15) represents the transient response. Since the system has been assumed to be stable, all transient response terms will disappear at steady state and we will get the following steady state response : (9.16) By Eq.(9.16) becomes where (9.18)

10 디지털 제어 Sun Moon University 10 of 19 9.2 Frequency response In terms of, Eq.(9.16) can be written as follows: We have shown that indeed gives the magnitude and phase of the frequency response of. Thus, to obtain the frequency response of, we need only to substitutefor z in G(z). (9.21) For cascaded systems as (9.23)

11 디지털 제어 Sun Moon University 11 of 19 9.3 Frequency response of 1 st order system For the 1 st order system given by (9.30)

12 디지털 제어 Sun Moon University 12 of 19 9.3 Frequency response of 1 st order system Ex.1) For a cascaded 1 st order system given by (9.35) From Eq.(9.30)

13 디지털 제어 Sun Moon University 13 of 19 9.3 Frequency response of 1 st order system Ex.2) For a discretised 1 st order system given by Its pulse transfer function is When K=1 and From Eq.(9.35) T=0.01

14 디지털 제어 Sun Moon University 14 of 19 9.3 Frequency response of 1 st order system #include #define SamplingTime 0.01 #define PHI 3.141592 double T; void main() { double M,Theta,Ts,omeg; T=SamplingTime; Ts=1.0; omeg=2.0*PHI/Ts; M=(1-exp(-T))/sqrt(1+exp(-2.0*T)-2.0*exp(-T)*cos(omeg*T)); Theta=-atan(exp(-T)*sin(omeg*T)/(1.0-exp(-T)*cos(omeg*T)))-omeg*T; printf("M, Theta=%f\t%f\n",M,-Theta/2./PHI*Ts); }

15 디지털 제어 Sun Moon University 15 of 19 9.4 Frequency response of 2nd order system For the 2nd order system given by where

16 디지털 제어 Sun Moon University 16 of 19 9.4 Frequency response of 2nd order system 7:void main() 8:{ 9:double M,M1,M2,Theta,Theta1,Theta2,omeg; 10:T=SamplingTime; 11:Ts=10.0; 12:double alpha, beta,K,a,b; 13:printf("alpha,beta,K="); 14:scanf("%lf %lf %lf",&alpha,&beta,&K); 15://G(z)=1/(z-z1)(z-z2), z1=a+bj, z2=a-bj 16:a=-alpha/2.0; 17:b=sqrt(4.0*beta-alpha*alpha)/2.0; 18://Amplitue and phase 19:omeg=2.0*PHI/Ts; 20:M1=1/sqrt(1+a*a+b*b-2.0*(a*cos(omeg*T)+b*sin(omeg*T))); 21:M2=1/sqrt(1+a*a+b*b-2.0*(a*cos(omeg*T)-b*sin(omeg*T))); 22:M=K*M1*M2; 23:Theta1=atan((a*sin(omeg*T)-b*cos(omeg*T))/(1.-a*cos(omeg*T)-b*sin(omeg*T))); 24:Theta2=atan((a*sin(omeg*T)+b*cos(omeg*T))/(1.-a*cos(omeg*T)+b*sin(omeg*T))); 25:Theta=-Theta1-Theta2-omeg*T; 26:printf("M=%f\n",M); 27:printf("theta(in radian),theta(in sec)=%f\t%f\n",Theta,Theta/2./PHI*Ts); 28: }

17 디지털 제어 Sun Moon University 17 of 19 9.4 Frequency response of 2nd order system Ex.3) For the second order system given by Obtain the frequency response when

18 디지털 제어 Sun Moon University 18 of 19 Conclusion 주파수 응답 특성은 제어 시스템의 성능을 판정하는 또 하나의 중요한 요소중의 하나이다. 정현파 입력에 대한 시스템의 정상 상태 주파수 응답은 출력이 입력의 주파수에 에 따라, 어떻게 변화하는 가 알아보는데 유용하다. 이 장에서는 복소수 함수 해석을 통해, 1차 및 2차 선형 이산 시간 시스템의 주파수 응답을 이론적으로 분석하는 방법을 보았다. 그리고 직접 C 프로그램을 이용한 컴퓨터 시뮬레이션을 통해, 주파수 응답 특성을 확인하여 보았다.

19 디지털 제어 Sun Moon University 19 of 19 HW#9 Solve Pr.9.2, 9.4, 9.6, 9.8, 9.10 Due: Next week this time