Matlab花费小时来解决方程式

Question

我一直在尝试求解由矩阵组成的方程式。 Matlab现在已经运行了将近两个小时，什么也没有。它应该乘以约8x8矩阵并倒入一个8x8矩阵。应该花这么长时间吗？我使用的是在4GHz，16 GB和RTX 2060的i7第8代2.20。

代码（最后一行是MATLAB花费的时间很长）：

%% Modelagem do conversor SEPIC-Zeta em espaço de estados
%% Declarando variáveis simbólicas:
% Variáveis com o índice "_" são derivadas de primeira ordem.
% RL1, RL2, RL3 e RL4 são as resistências série dos indutores.
% RseC1, RseC2, RseC3 e RseC4 são as resistências série dos capacitores.

Vi = sym('Vi');

C1 = sym('C1');
C2 = sym('C2');
C3 = sym('C3');
C4 = sym('C4');

L1 = sym('L1');
L2 = sym('L2');
L3 = sym('L3');
L4 = sym('L4');

VC1 = sym('VC1');
VC2 = sym('VC2');
VC3 = sym('VC3');
VC4 = sym('VC4');
VC1_ = sym('VC1_');
VC2_ = sym('VC2_');
VC3_ = sym('VC3_');
VC4_ = sym('VC4_');

IL1 = sym('IL1');
IL2 = sym('IL2');
IL3 = sym('IL3');
IL4 = sym('IL4');
IL1_ = sym('IL1_');
IL2_ = sym('IL2_');
IL3_ = sym('IL3_');
IL4_ = sym('IL4_');

RL1 = sym('RL1');
RL2 = sym('RL2');
RL3 = sym('RL3');
RL4 = sym('RL4');
RseC1 = sym('RseC1');
RseC2 = sym('RseC2');
RseC3 = sym('RseC3');
RseC4 = sym('RseC4');
Ro = sym('Ro');

D = sym('D');
s = sym('s');

%% Determinando as matrizes de coeficientes do conversor em t = [0, DTs]:

% Equações LKT no conversor:
eqn11 = IL1_ - (-RL1*IL1 + Vi)/L1 == 0;
eqn21 = IL2_ - ((RseC1 - RL2)*IL2 - VC1)/L2 == 0;
eqn31 = IL3_ - (VC2 + (RseC2 - RL3)*IL3 + RseC2*IL4)/L3 == 0;
eqn41 = IL4_ - ((-VC3 + RseC2*C2*VC2_ - (RseC3 + RL4)*IL4 + Ro*C4*VC4_ + VC2)/L4) == 0;
eqn51 = VC1_ - (Vi - RL1*IL1 + L1*IL1_ + VC1 - L2*IL2_)/C1*(RseC1 - RL2) == 0;
eqn61 = VC2_ - (-RseC1*IL2 - VC1 - RL2*IL2 - L2*IL2_ + VC2 + L3*IL3_ + RL3*IL3)/(C2*RseC2) == 0;
eqn71 = VC3_ - (Ro*IL4 - Ro*C4*VC4_ - VC3 - L4*IL4_ + L3*IL3_ + RL3*IL3 - RL4*IL4)/(C3*RseC3) == 0;
eqn81 = VC4_ - (Ro*IL4 - VC4)/(C4*(RseC4 + Ro)) == 0;

% Solucionando o sistema de equações lineares:
sol1 = solve([eqn11, eqn21, eqn31, eqn41, eqn51, eqn61, eqn71, eqn81], [IL1_, IL2_, IL3_, IL4_, VC1_, VC2_, VC3_, VC4_]);

% Separando as matrizes A e B dos coeficientes:
system_A1 = [sol1.IL1_ == IL1_, sol1.IL2_ == IL2_, sol1.IL3_ == IL3_, sol1.IL4_ == IL4_,...
sol1.VC1_ == VC1_, sol1.VC2_ == VC2_, sol1.VC3_ == VC3_, sol1.VC4_ == VC4_];
vars1 = [ IL1, IL2, IL3, IL4, VC1, VC2, VC3, VC4 ];
vars21 = [ Vi ];

% Definindo as matrizes A1, B1, C1 e D1:
A1 = equationsToMatrix(system_A1,vars1);
B1 = equationsToMatrix(system_A1,vars21);
C1_ = [ 0; 0; 0; RseC4*Ro/(RseC4 + Ro); 0; 0; 0; Ro/(RseC4 + Ro) ];
D1 = 0;

%% Deterninando as matrizes dos coeficientes em t = [DTs, Ts]:
eqn12 = IL1_ - (-Vi + IL1*(RseC1 - RL1) + VC1 + RL2*IL2 + L2*IL2_)/L1 == 0;
eqn22 = IL2_ - (RseC2*IL1 + IL2*(-RL2 + RseC2) + VC2)/L2 == 0;
eqn32 = IL3_ - (-VC3 -RL3*IL3 - RseC3*IL3)/L3 == 0;
eqn42 = IL4_ - ((IL4*RL4 - Ro*IL4 - Ro*C4*VC4_)/L4) == 0;
eqn52 = VC1_ - (Vi - RseC1*IL1 - L1*IL1_ - VC1 + RL2*IL2 + L2*IL2_)/C1*(RseC1 + RseC2 - RL1) == 0;
eqn62 = VC2_ - (RL2*IL2 + L2*IL2_ - VC2)/(C2*RseC2) == 0;
eqn72 = VC3_ - (-Ro*IL4 - Ro*C4*VC4_ - VC3 + L4*IL4_ - L3*IL3_ + RL4*IL4)/(C3*(RL3 + RseC3)) == 0;
eqn82 = VC4_ - (Ro*IL4 - VC4)/(C4*(RseC4 - Ro)) == 0;

% Solucionando o sistema de equações lineares:
sol2 = solve([eqn12, eqn22, eqn32, eqn42, eqn52, eqn62, eqn72, eqn82], [IL1_, IL2_, IL3_, IL4_, VC1_, VC2_, VC3_, VC4_]);

% Separando as matrizes A e B dos coeficientes:
system_A2 = [sol2.IL1_ == IL1_, sol2.IL2_ == IL2_, sol2.IL3_ == IL3_, sol2.IL4_ == IL4_,...
sol2.VC1_ == VC1_, sol2.VC2_ == VC2_, sol2.VC3_ == VC3_, sol2.VC4_ == VC4_];

vars2 = [ IL1, IL2, IL3, IL4, VC1, VC2, VC3, VC4 ];
vars22 = [ Vi ];

% Definindo as matrizes A2, B2, C2 e D2:
A2 = equationsToMatrix(system_A2,vars2);
B2 = equationsToMatrix(system_A2,vars22);
C2_ = [ 0; 0; 0; RseC4*Ro/(RseC4 - Ro); 0; 0; 0; -Ro/(RseC4 - Ro) ];
D2 = 0;

%% Equacionamento para o espaço de estados médio:
A = simplify(A1*D + A2*(1-D));
B = simplify(B1*D + B2*(1-D));
C = simplify(C1_*D + C2_*(1-D));
D = simplify(D1*D + D2*(1-D));
X = simplify(-inv(A)*B*Vi);
I = eye(8);
Tp = C*inv(s*I - A)*((A1-A2)*X + (B1-B2)*Vi)+(C1-C2)*X;

Answer 1

在不详细了解数学的情况下，我可以看到它在第 108 行 X = simple(-inv(A)*B*Vi); 处变慢，而不是最后一行，但我还没有运行这么长的时间<1分钟。尝试使用 A\b 而不是 inv(A)*b ，这会花费更多时间。