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#1 |
初级会员
注册日期: 2012-10-10
帖子: 2
声望力: 0 ![]() |
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clear;clc; close all;%初始化
XF=20;CUB0=10;OF=1; WF=3; KD=0.09; KL=0.46; Ka=10^(-6);H=10^(-7);f=8;K=10;B=0.000001;%用f=input('input f=') A=1+H/Ka; %迭代计算eeorg eeaq for i=1:K; XD(i)=XF; XL(i)=XF; PF(i)=B; end %赋初值 TD=0;TL=0;RD=0.1; RL=0.1; while (RD>0.000001)|(RL>0.000001); %迭代停止误差要求 for i=1:K CUB(i)=CUB0/(1+(KD*XD(i)+KL*XL(i))/(A*PF(i))); end for i=1:K kD(i)=KD*CUB(i)/(A*PF(i)); end RD=abs((kD(1)-TD)/kD(1)); TD=kD(1); for i=1:K kL(i)=KL*CUB(i)/(A*PF(i)); end RL=abs((kL(1)-TL)/kL(1)); TL=kD(1); for i=1:f-1 ED(i)=OF/WF*kD(i);EL(i)=OF/WF*kL(i); end for i=f:K ED(i)=(OF)*(WF+1)*kD(i);EL(i)=(OF)*(WF+1)*kL(i); end AD=zeros(K,K);AL=zeros(K,K); for i=1:1:K AD(i,i)=-(1+ED(i));AL(i,i)=-(1+EL(i)); end for i=2:1:K if i==f AD(i,i-1)=(WF)/(WF+1);AL(i,i-1)=(WF)/(WF+1); else AD(i,i-1)=1;AL(i,i-1)=1; end end for i=1:1:K-1 AD(i,i+1)=ED(i+1);AL(i,i+1)=EL(i+1); end b=zeros(K,1); b(f,1)=-XF/(WF+1); XD=AD\b; XL=AL\b; CUBD=(kD.').*XD;CUBL=(kL.').*XL; PF(i)=CUBD(i)+CUBL(i); end eeaq=abs((XL(K)-XD(K))/(XL(K)+XD(K))) eeorg=abs((CUBD(1)-CUBL(1))/(CUBD(1)+CUBL(1))) |
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#2 |
初级会员
注册日期: 2012-10-10
帖子: 2
声望力: 0 ![]() |
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怎么没人解答呀 在线等
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