不是六字ID😂的个人资料

在for循环中如何读取solve函数求出来的解如题，不在for循环中好求，但是到for循环就没法用S.x的方法求解，请问如何破 syms x1a x2a x4a x5a x6a x7a x8a syms x1b x2b x4b x5b x6b x7b x8b u = 5*10^(-26); h = 6.62*10^(-34); T2 = 10^(-6); c = -1; w1 = 10^(15); w = w1; T1 = T2/2; o = 0.001; E1 = h*o/(2*u*T2); E2 =0; E3 = 10^(-6); b = 0; a=w1-w; w3=10^15-2*10^7 disp('o='),disp(o) S=solve((-1/T1)*(x1a - c) == -2*u^2*h^-2*(conj(E1)*(E1*x1a + E2*x2a + E3*x5a)/(i*(w1 + b) - (i*w + 1/T2)) +... conj(E2)*(E2*x1a + E1*x2a + E3*x6a)/(i*(w1 - b) - (i*w + 1/T2)) + conj(E3)*(E1*x5a + E2*x6a + E3*x1a)/(i*w3 - (i*w + 1/T2)) -... E1*(conj(E1)*x1a + conj(E2)*x2a +conj(E3)*x5a)/(i*(w1 + b) - (i*w - 1/T2)) -... E2*(conj(E2)*x1a + conj(E1)*x2a + conj(E3)*x6a)/(i*(w1 - b) - (i*w - 1/T2)) -... E3*(conj(E1)*x5a + conj(E2)*x6a + conj(E3)*x1a)/(i*w3 - (i*w - 1/T2))),... (i*2*b - 1/T1)*x2a == -2*u^2*h^-2*(conj(E1)*(E1*x2a + E2*x4a + E3*x7a)/(i*(w1 + 3*b) - (i*w + 1/T2)) +... conj(E2)*(E1*x1a + E2*x2a + E3*x5a)/(i*(w1 + b) - (i*w + 1/T2)) - E1*(conj(E2)*x1a +conj(E1)*x2a +... conj(E3)*x6a)/(i*(w1 - b) - (i*w - 1/T2)) - E2*(conj(E2)*x2a + conj(E1)*x4a + conj(E3)*x8a)/(i*(w1 - 3*b) - (i*w - 1/T2))),... (i*4*b - 1/T1)*x4a == -2*u^2*h^-2*(conj(E2)*(E1*x2a + E2*x4a + E3*x7a)/(i*(w1 + 3*b) - (i*w + 1/T2)) -... E1*(conj(E2)*x2a + conj(E1)*x4a + conj(E3)*x8a)/(i*(w1 - 3*b) - (i*w - 1/T2))),... (i*(w1 + b -w3) - 1/T1)*x5a == -2*u^2*h^-2*(conj(E1)*(E1*x5a + E2*x7a)/(i*(2*w1 + 2*b - w3) - (i*w + 1/T2)) +... conj(E2)*(E2*x6a + E1*x8a)/(i*(2*w1 - w3) - (i*w + 1/T2)) + conj(E3)*(E1*x1a + E2*x2a + E3*x5a)/(i*(w1 + b) - (i*w + 1/T2)) -... E1*(conj(E3)*x1a + conj(E1)*x5a + conj(E2)*x6a)/(i*w3 - (i*w - 1/T2))),... (i*(w1 - b - w3) - 1/T1)*x6a == -2*u^2*h^-2*(conj(E2)*(E2*x6a + E1*x8a)/(i*(2*w1 - 2*b - w3) - (i*w + 1/T2)) +... conj(E1)*(E1*x6a + E2*x5a)/(i*(2*w1 - w3) - (i*w + 1/T2)) + conj(E3)*(E2*x1a + E1*x2a + E3*x6a)/(i*(w1 - b) - (i*w + 1/T2)) -... E2*(conj(E3)*x1a + conj(E1)*x5a + conj(E2)*x6a)/(i*w3 - (i*w - 1/T2))),... (i*(w1 + 3*b - w3) - 1/T1)*x7a == -2*u^2*h^-2*(conj(E2)*(E1*x5a + E2*x7a)/(i*(2*w1 + 2*b - w3) - (i*w + 1/T2)) +... conj(E3)*(E1*x2a + E2*x4a + E3*x7a)/(i*(w1 + 3*b) - (i*w + 1/T2))),... (i*(w1 - 3*b - w3) - 1/T1)*x8a == -2*u^2*h^-2*(conj(E3)*(E2*x2a + E1*x4a + E3*x8a)/(i*(w1 - 3*b) - (i*w + 1/T2)) +... conj(E1)*(E2*x6a + E1*x8a)/(i*(2*w1 - 2*b - w3) - (i*w + 1/T2))),x1a,x2a,x4a,x5a,x6a,x7a,x8a); % format shortEng disp(S.x1a) x10=S.x1a; x20=S.x2a; x40=S.x4a; x50=S.x5a; x60=S.x6a; x70=S.x7a; x80=S.x8a; x10 = double(x10); x20 = double(x20); x40 = double(x40); x50 = double(x50); x60 = double(x60); x70 = double(x70); x80 = double(x80); x1(1) = x10 x2(1) = x20 x4(1) = x40 x5(1) = x50 x6(1) = x60 x7(1) = x70 x8(1) = x80 for n = 1:1:320 w3(n)=10^(15)-1.6*10^8+1*10^6*n; w3w1T2(n)=(w3(n)-w1)*T2; y11(n) = i*u*h^-1*(E1*x1(n) + E2*x2(n) + E3*x5(n))/(i*(w1 + b) - (i*w+1/T2)); y12(n) = i*u*h^-1*(E2*x1(n) + E1*x2(n) + E3*x6(n))/(i*(w1 - b) - (i*w+1/T2)); y31(n) = i*u*h^-1*(E1*x2(n) + E2*x4(n) + E3*x7(n))/(i*(w1 + 3*b) - (i*w+1/T2)); y32(n) = i*u*h^-1*(E2*x2(n) + E1*x4(n) + E3*x8(n))/(i*(w1 - 3*b) - (i*w+1/T2)); y51(n) = i*u*h^-1*E1*x4(n)/(i*(w1 + 5*b) - (i*w+1/T2)); y52(n) = i*u*h^-1*E2*x4(n)/(i*(w1 - 5*b) - (i*w+1/T2)); y6(n) = i*u*h^-1*(E3*x1(n) + E1*x5(n) + E2*x6(n))/(i*w3(n) - (i*w + 1/T2)); y71(n) = i*u*h^-1*(E1*x5(n) + E2*x7(n))/(i*(2*w1 + 2*b - w3(n)) - (i*w + 1/T2)); y72(n) = i*u*h^-1*(E2*x6(n) + E1*x8(n))/(i*(2*w1 - 2*b - w3(n)) - (i*w + 1/T2)); y73(n) = i*u*h^-1*(E1*x6(n) + E2*x5(n))/(i*(2*w1 - w3(n)) - (i*w + 1/T2)); absy31(n) = abs((h/(u*T2*E3))*y31(n)); solve(x1b(n) == c -T1*i*2*u*h^-1*(conj(E1)*y11(n)+conj(E2)*y12(n)+conj(E3)*y6(n)-E1*conj(y11(n))-E2*conj(y12(n))-E3*conj(y6(n))),... x2b(n) == i*2*u*h^-1*(conj(E1)*y31(n)+conj(E2)*y11(n)-E1*conj(y12(n)))/(i*2*b-1/T1),... x4b(n) == i*2*u*h^-1*(conj(E2)*y31(n)-E1*conj(y32(n)))/(i*4*b-1/T1),... x5b(n) == i*2*u*h^-1*(conj(E1)*y71(n)+conj(E2)*y73(n)+conj(E3)*y11(n)-E1*conj(y6(n)))/(i*(w1+b-w3)-1/T1),... x6b(n) == i*2*u*h^-1*(conj(E2)*y72(n)+conj(E1)*y73(n)+conj(E3)*y12(n)-E2*conj(y6(n)))/(i*(w1-b-w3)-1/T1),... x7b(n) == i*2*u*h^-1*(conj(E2)*y71(n)+conj(E3)*y31(n))/(i*(w1+3*b-w3)-1/T1),... x8b(n) == i*2*u*h^-1*(conj(E3)*y32(n)+conj(E1)*y72(n))/(i*(w1-3*b-w3)-1/T1),... x1b(n),x2b(n),x4b(n),x5b(n),x6b(n),x7b(n),x8b(n)); x10(n) = double(x1b(n)); %不知为何赋值不了，前面都挺好的 x20(n) = double(x2b(n)); x40(n) = double(x4b(n)); x50(n) = double(x5b(n)); x60(n) = double(x6b(n)); x70(n) = double(x7b(n)); x80(n) = double(x8(n)); x1(n+1) = double(x10(n)); x2(n+1) = double(x20(n)); x4(n+1) = double(x40(n)); x5(n+1) = double(x50(n)); x6(n+1) = double(x60(n)); x7(n+1) = double(x70(n)); x8(n+1) = double(x80(n)); imy6(n)=imag((h/(u*T2*E3))*y6(n)); rey6(n)=real((h/(u*T2*E3))*y6(n)); if(mod(n,100)==0) disp(n) end end

求教：mathematica复数和实数的定义与符号计算如何定义虚数啊，我的计算中有实数和复数，但是该如何分别定义代码如下 d = -2*u^2*h^2*T1; a11 = I*h*u*c*T1*E1 - I*w - 1/T2 + d*(E1*Conjugate[E1] + E3*Conjugate[E3]); a12 = d*E1*Conjugate[E3]; a13 = d*E3*Conjugate[E1]; a14 = -d*E1^2; a15 = -2*d*E1*E3; a21 = d*E3*Conjugate[E1]; a22 = I*h*u*c*T1*E3 - I*w - 1/T2 + d*(E1*Conjugate[E1] + E3*Conjugate[E3]); a24 = -2*d*E1*E3; a25 = -d*E3^2; a26 = a14; a31 = d*E3*Conjugate[E1]; a33 = -I*w - 1/T2 + d*(E1*Conjugate[E1] + E3*Conjugate[E3]); a35 = a14; b1 = -I*w1; b2 = -I*w3; b3 = -I*(2*w1 - w3); eq1 = (a11 - b1)*x1 + a12*x2 + a13*x3 + a14*x4 + a15*x5; eq2 = a21*x1 + (a22 - b2)*x2 + a24*x4 + a25*x5 + a26*x6; eq3 = a31*x1 + (a33 - b3)*x3 + a35*x5; eq4 = (Conjugate[a11] + b1)*x4 + Conjugate[a12]*x5 + Conjugate[a13]*x6 + Conjugate[a14]*x1 + Conjugate[a15]*x2; eq5 = Conjugate[a21]*x4 + (Conjugate[a22] + b2)*x5 + Conjugate[a24]*x1 + Conjugate[a25]*x2 + Conjugate[a26]*x3; eq6 = Conjugate[a31]*x4 + (Conjugate[a33] + b3)*x6 + Conjugate[a35]*x2; s1 = Solve[eq1 == 0, eq2 == 0, eq3 == 0, eq4 == 0, eq5 == 0, eq6 == 0] 其实很简单，就是用符号计算算一个线型方程组，h,u,T1,T2，w1,w3,w为实数，E1,E3为复数，x1到x6为所求解的未知数

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