자동제어 | lag compensator 설계하기 |
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작성자 cemtool 작성일14-08-19 11:44 조회16,695회 댓글0건본문
다음과 같은 3차 시스템을 다시 생각하자.
이때, phase margin이 적어도 25°이고, = 9인 lag compensator를 설계하시오.
- ex6_18p.cem
/* Example 6.18 Lag Compensator design for a Type 0 System */ del *; "****************************************************" "** **" "** example 6.18 **" "** Lag Compensator design for a Type 0 System **" "** **" "****************************************************" " " sleep(2); "Plant" " " "CEMTool>> K = 2;" "CEMTool>> numG = 4.5;" "CEMTool>> denG = convo(convo([2 1],[1 1]),[.5 1]);" " " K = 2; numG = 4.5; denG = convo(convo([2 1],[1 1]),[.5 1]); sleep(2); "Lag compensator" " " "CEMTool>> numD = [5 1];" "CEMTool>> denD = [10 1];" " " numD = [5 1]; denD = [10 1]; sleep(2); "open loop" " " "CEMTool>> num = K*convo(numG,numD);" "CEMTool>> den = convo(denG,denD);" " " num = K*convo(numG,numD); den = convo(denG,denD); sleep(2); "Bode plot for KG(s), KD(s)G(s)" " " // Figure 6.63: Frequency response of lag-compensation design w = logspace(-2, 2, 1000); figure(1); bode(num,den,w); holdon; bode(K*numG,denG,w); "figure1.빨간색: Bode plot for KG(s)" "figure2.파란색: Bode plot for KD(s)G(s)" " " sleep(2); "Step response of lag-compensation" " " "CEMTool>> [numCL,denCL] = feedback(num,den,1,1);" "CEMTool>> t = 0:20:.02;" "CEMTool>> step(numCL,denCL,t);" // Figure 6.64: Step response of lag-compensation design figure(2); [numCL,denCL] = feedback(num,den,1,1); t = 0:20:.02; step(numCL,denCL,t); title("Step Response of lag-compensation design in Example 6.18");
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