is the "for loop" is wrong? what can be the solution?
显示 更早的评论
clc
ti = 0;
tf = 100E-4;
tspan=[ti tf];
o = 1E6;
tc = 70E-9;
tf = 240E-6;
a1 = 0.02;
a2 = 0.02;
P1 = 1;
P2 = 1;
k = 0.033;
l = 0.5;
f = @(t,y) [ ((y(2)-a1).*y(1)) + k.*(y(1).*cos(y(3))).*(2/tc);
(P1 - y(2).*(1+y(1)))./tf;
o - (k / tc) * 2 * sin(y(3));
];
[T,Y] = ode45(f,tspan,[1;1;0].*10E-3);
% this is the for loop, maybe this is wrong
Y(:,3) = -3:0.01:3;
U = zeros(length(k),1) ;
for i = 1:length(k)
U(i) = -o.*(Y(:,3)) - 2.*(k./tc).*cos(Y(:,3) - pi/2)
end
%plotting the graphs
plot(T,Y(:,3));
xlim([0 10E-5])
xlabel('t')
ylabel('phase difference')
legend('k = 0.033')
plot(Y(:,3),U)
6 个评论
Abderrahim. B
2022-7-20
Hi!
The below lines of code are not correct. Read the comments below and try to rereview your code, and maybe you share what you are trying to do with this for loop.
Y(:,3) = -3:0.01:3; % this assignement is not correct size(-3:0.01:3) is different from size(Y(:,3))
U = zeros(length(k),1) ; % k is a scalar thus U is a scalar
for i = 1:length(k) % length(k) is one >> one iteratio, what is the point to use a for loop!!
U(i) = -o.*(Y(:,3)) - 2.*(k./tc).*cos(Y(:,3) - pi/2) % This assignment is not possible size(U(i)) does not match the size of the right side.
end
SAHIL SAHOO
2022-7-20
I calculated the Y(:,3) which represent Φ, i want to calculate the U for different value of Φ, where the U is
then I want to plot (Y(:,3), Φ)
Abderrahim. B
2022-7-20
Okay now is clear what you re trying to do. You do not need a loop. One of MATLAB advantages is vectorization. Check out @Torsten answer.
VBBV
2022-7-20
You can also use for loop as shown in your code. But plot you mentioned plot (Y(:,3), Φ) is strange. It will be a straight line since Y(:,3) and Φ are same according to your equation and code written. Do you mean plot (Y(:,3), U) ?
VBBV
2022-7-20
The for loop code works well as shown in my answer below. which can also be done without a loop.
SAHIL SAHOO
2022-7-20
采纳的回答
clc
ti = 0;
tf = 100E-4;
tspan=[ti tf];
o = 1E6;
tc = 70E-9;
tf = 240E-6;
a1 = 0.02;
a2 = 0.02;
P1 = 1;
P2 = 1;
k = 0.033;
l = 0.5;
f = @(t,y) [ ((y(2)-a1).*y(1)) + k.*(y(1).*cos(y(3))).*(2/tc);
(P1 - y(2).*(1+y(1)))./tf;
o - (k / tc) * 2 * sin(y(3));
];
[T,Y] = ode45(f,tspan,[1;1;0].*10E-3);
% this is the for loop, maybe this is wrong ... yes this assignment is
% incorrect
Y(:,3) = linspace(-3,3,length(Y)); % change this
U = zeros(length(Y),1) ;
for i = 1:length(Y)
U(i) = -o.*(Y(i,3)) - 2.*(k./tc).*cos(Y(i,3) - pi/2); % also this
end
%plotting the graphs
figure(1)
plot(T,Y(:,2));
figure(2)
plot(T,U)
xlim([0 10E-5])
xlabel('t')
ylabel('phase difference')
legend('','k = 0.033')
plot(T,U)
You can plot both U and T variables , but for loop needs modififcation
2 个评论
clc
ti = 0;
tf = 100E-4;
tspan=[ti tf];
o = 1E6;
tc = 70E-9;
tf = 240E-6;
a1 = 0.02;
a2 = 0.02;
P1 = 1;
P2 = 1;
k = 0.033;
l = 0.5;
f = @(t,y) [ ((y(2)-a1).*y(1)) + k.*(y(1).*cos(y(3))).*(2/tc);
(P1 - y(2).*(1+y(1)))./tf;
o - (k / tc) * 2 * sin(y(3));
];
[T,Y] = ode45(f,tspan,[1;1;0].*10E-3);
% this is the for loop, maybe this is wrong ... yes this assignment is
% incorrect
Y(:,3) = linspace(-3,3,length(Y)); % change this
U = zeros(length(Y),1) ;
for i = 1:length(Y)
U(i) = -o.*(Y(i,3)) - 2.*(k./tc).*cos(Y(i,3) - pi/2); % also this
end
%plotting the graphs
figure(1)
plot(Y(:,3),U); % i guess you are looking for this
i guess you are looking for this
SAHIL SAHOO
2022-7-20
更多回答（1 个）
ti = 0;
tf = 100E-4;
tspan=[ti tf];
o = 1E6;
tc = 70E-9;
tf = 240E-6;
a1 = 0.02;
a2 = 0.02;
P1 = 1;
P2 = 1;
k = 0.033;
l = 0.5;
f = @(t,y) [ ((y(2)-a1).*y(1)) + k.*(y(1).*cos(y(3))).*(2/tc);
(P1 - y(2).*(1+y(1)))./tf;
o - (k / tc) * 2 * sin(y(3));
];
[T,Y] = ode45(f,tspan,[1;1;0].*10E-3);
% this is the for loop, maybe this is wrong
U = -o.*Y(:,3) - 2.*(k./tc).*cos(Y(:,3) - pi/2);
%plotting the graphs
plot(T,Y(:,3));
xlim([0 10E-5])
xlabel('t')
ylabel('phase difference')
legend('k = 0.033')
plot(Y(:,3),U)
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