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자동제어 | lag compensator 설계하기 |

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작성자 cemtool 작성일14-08-19 11:44 조회13,813회 댓글0건

본문

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");
6_18_1_figure.png
 


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