# I gave the initial condition correctly still the program not working.

조회 수: 3 (최근 30일)
SAHIL SAHOO 2022년 10월 11일
답변: Walter Roberson 2022년 10월 11일
ti = 0;
tf = 70E-8;
tspan=[ti tf];
k = (0.62).*10^(-5);
% y0= [(10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% ((-3.14).*rand(5,1) + (3.14).*rand(5,1))];
y0 = [ 0.00001; 0.00001; 0.00001; 0.00001; 0.00001;
0.00001; 0.00001; 0.00001; 0.00001; 0.00001; 2.5669; 2.0482; 2.0454; -0.7968; 0.2303];
yita_mn = [
0 1 0 0 1;
1 0 1 0 0;
0 1 0 1 0;
0 0 1 0 1;
1 0 0 1 0;
]*(k);
N = 5;
tp = 1E-12;
[T,Y]= ode45(@(t,y) rate_eq(t,y,yita_mn,N),tspan./tp,y0);
Index exceeds the number of array elements. Index must not exceed 15.

Error in solution>rate_eq (line 81)
dy(n16) = -a.*(Gt(n2)-Gt(n1)) + (k).*(y(j2)./y(j5)).*cos(y(n16)) - (k).*(y( j5)./y(j2)).*cos(y(n16)) + (k).*(y(j8)./y(j5)).*cos(y(n17)) - (k).*(y(j19)./y(j2)).*cos(y(n20));

Error in solution (line 24)
[T,Y]= ode45(@(t,y) rate_eq(t,y,yita_mn,N),tspan./tp,y0);

Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 107)
odearguments(odeIsFuncHandle,odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);
figure(1)
plot(T./t,(Y(:,16)),'linewidth',0.8);
hold on
for m = 16:20
plot(T./t,(Y(:,m)),'linewidth',0.8);
end
hold off
grid on
xlabel("time")
ylabel("phase difference")
set(gca,'fontname','times New Roman','fontsize',18,'linewidth',1.8);
function dy = rate_eq(t,y,yita_mn,N,o)
dy = zeros(4*N,1);
dGdt = zeros(N,1);
dOdt = zeros(N,1);
P = 0.5;
a = 1;
T = 2E3;
Gt = y(1:3:3*N-2);
At = y(2:3:3*N-1);
Ot = y(3:3:3*N-0);
k = (0.62).*10^(-5);
for i = 1:N
dGdt(i) = (P - Gt(i) - (1 + 2.*Gt(i)).*(At(i))^2)./T ;
dOdt(i) = -a.*(Gt(i));
for j = 1:N
dOdt(i) = dOdt(i)+yita_mn(i,j).*((At(j)/At(i)))*cos(Ot(j)-Ot(i));
end
end
dy(1:3:3*N-2) = dGdt;
dy(3:3:3*N-0) = dOdt;
n1 = (1:5)';
n2 = circshift(n1,-1);
n16 = n1 + 15;
n17 = circshift(n16,-1);
n20 = circshift(n16,1);
j2 = 3*(1:5)-1;
j5 = circshift(j2,-1);
j8 = circshift(j2,-2);
j19 = circshift(j2,1);
dy(n16) = -a.*(Gt(n2)-Gt(n1)) + (k).*(y(j2)./y(j5)).*cos(y(n16)) - (k).*(y( j5)./y(j2)).*cos(y(n16)) + (k).*(y(j8)./y(j5)).*cos(y(n17)) - (k).*(y(j19)./y(j2)).*cos(y(n20));
end

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### 채택된 답변

Walter Roberson 2022년 10월 11일
y0 = [ 0.00001; 0.00001; 0.00001; 0.00001; 0.00001;
0.00001; 0.00001; 0.00001; 0.00001; 0.00001; 2.5669; 2.0482; 2.0454; -0.7968; 0.2303];
That is 15 initial values.
for m = 16:20
plot(T./t,(Y(:,m)),'linewidth',0.8);
end
But you are trying to plot assuming 20 results. The only way to get 20 results is to have 20 or more initial values.

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### 추가 답변 (1개)

Benjamin Thompson 2022년 10월 11일
circshift returns a vector of the same length as its input. So, j2, j5, j8, and j19 are vectors and not scalar values as the line having the failure seems to expect. You can use breakpoints in your script in MATLAB to investigate further and debug the problems.

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