matlab 解微分方程出错clear all;clc;load 15.mat;t=celiang(:,1);u1=celiang(:,2);syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等vU1=celiang(:,2);k=length(u1);C1=1;h=5e-7;C3=1;R1=1;L1=1;R2=1;L2=1;C2=1;i1(1)=0;i1(2)=0;%在开始的两个点用的
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![matlab 解微分方程出错clear all;clc;load 15.mat;t=celiang(:,1);u1=celiang(:,2);syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等vU1=celiang(:,2);k=length(u1);C1=1;h=5e-7;C3=1;R1=1;L1=1;R2=1;L2=1;C2=1;i1(1)=0;i1(2)=0;%在开始的两个点用的](/uploads/image/z/13304428-52-8.jpg?t=matlab+%E8%A7%A3%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E5%87%BA%E9%94%99clear+all%3Bclc%3Bload+15.mat%3Bt%3Dceliang%28%3A%2C1%29%3Bu1%3Dceliang%28%3A%2C2%29%3Bsyms+u2+i5+ii+x+u5%25%E5%AE%9A%E4%B9%89u1%2CL1%2CL2%2CR1%2CC1%2CC3%2Cii+%E7%AD%89vU1%3Dceliang%28%3A%2C2%29%3Bk%3Dlength%28u1%29%3BC1%3D1%3Bh%3D5e-7%3BC3%3D1%3BR1%3D1%3BL1%3D1%3BR2%3D1%3BL2%3D1%3BC2%3D1%3Bi1%281%29%3D0%3Bi1%282%29%3D0%3B%25%E5%9C%A8%E5%BC%80%E5%A7%8B%E7%9A%84%E4%B8%A4%E4%B8%AA%E7%82%B9%E7%94%A8%E7%9A%84)
matlab 解微分方程出错clear all;clc;load 15.mat;t=celiang(:,1);u1=celiang(:,2);syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等vU1=celiang(:,2);k=length(u1);C1=1;h=5e-7;C3=1;R1=1;L1=1;R2=1;L2=1;C2=1;i1(1)=0;i1(2)=0;%在开始的两个点用的
matlab 解微分方程出错
clear all;clc;
load 15.mat;
t=celiang(:,1);
u1=celiang(:,2);
syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等
vU1=celiang(:,2);
k=length(u1);
C1=1;
h=5e-7;
C3=1;
R1=1;
L1=1;
R2=1;
L2=1;
C2=1;
i1(1)=0;i1(2)=0;%在开始的两个点用的是最简单的使之为零的方法.
for j=3:k-1
i1(j)=(vU1(j+1)-vU1(j))*12*C1/(23*h)+16*i1(j-1)/23-5*i1(j-2)/23;
end
i1(k)=(vU1(k)-vU1(k-1))*C1/h;
i1=i1';
i3=i5+ii-i1;
fai=x;
p1 = 1.367e-014;
p2 =-2.585e-027;
p3 = -8.37e-011;
p4 = 3.585e-023;
p5 = 3.2e-007;
p6 =-1.298e-019;
p7 = -0.0002556;
p8 = 1.143e-016;
ih= p1*x.^7 + p2*x.^6 + p3*x.^5 + p4*x.^4 + p5*x.^3 +p6*x.^2 + p7*x + p8;
i7=i3-ih;
i4=-100*i7;
u4=u5/100;
i2=-i4-i5;
%%%%%%%%%%%%%%%%%%%%%
s=dsolve(diff((u2-u1),t)==i5/C3,diff(i3,t)==(u1-R1*i3-u5)/L1,diff(fai,t)==u5,diff(i4,t)==(u2-u4- R2*i4)/L2);
u2=s.u2
已知u1如何求u2
i1=C1*du1/dt;%%含有微分
i5=C3*d(u2-u1)/dt;%%含有微分
i3=i5+ii-i1;
u5=u1-R1*i3-L1*di3/dt;%%含有微分
u5=dfai/dt;%%含有微分
fai=x;
p1 = 1.367e-014;
p2 = -2.585e-027;
p3 = -8.37e-011;
p4 = 3.585e-023;
p5 = 3.2e-007;
p6 = -1.298e-019;
p7 = -0.0002556;
p8 = 1.143e-016;
ih= p1*x.^7 + p2*x.^6 + p3*x.^5 + p4*x.^4 + p5*x.^3 + p6*x.^2 + p7*x + p8;
i7=i3-ih;
i4=-100*i7;
u4=u5/100;
u2=u4+R2*i4+L2*di4/dt;%%含有微分
i2+i4+i5=0;
matlab 解微分方程出错clear all;clc;load 15.mat;t=celiang(:,1);u1=celiang(:,2);syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等vU1=celiang(:,2);k=length(u1);C1=1;h=5e-7;C3=1;R1=1;L1=1;R2=1;L2=1;C2=1;i1(1)=0;i1(2)=0;%在开始的两个点用的
clear all;clc;
load 15.mat;
t=celiang(:,1);
u1=celiang(:,2);
syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等
vU1=celiang(:,2);
k=length(u1);
C1=1;
h=5e-7;
C3=1;
R1=1;
L1=1;
R2=1;
L2=1;
C2=1;
i1(1)=0;i1(2)=0;%在开始的两个点用的是最简单的使之为零的方法.
for j=3:k-1
i1(j)=(vU1(j+1)-vU1(j))*12*C1/(23*h)+16*i1(j-1)/23-5*i1(j-2)/23;
end
i1(k)=(vU1(k)-vU1(k-1))*C1/h;
i1=i1';
i3=i5+ii-i1;
fai=x;
p1 = 1.367e-014;
p2 =-2.585e-027;
p3 = -8.37e-011;
p4 = 3.585e-023;
p5 = 3.2e-007;
p6 =-1.298e-019;
p7 = -0.0002556;
p8 = 1.143e-016;
ih= p1*x.^7 + p2*x.^6 + p3*x.^5 + p4*x.^4 + p5*x.^3 +p6*x.^2 + p7*x + p8;
i7=i3-ih;
i4=-100*i7;
u4=u5/100;
i2=-i4-i5;
u21=u2-u1;
%%%%%%%%%%%%%%%%%%%%%
s=dsolve(diff(u21,t)==i5/C3,diff(i3,t)==(u1-R1*i3-u5)/L1,diff(fai,t)==u5,diff(i4,t)==(u2-u4- R2*i4)/L2);
u2=s.u2