is the "for loop" is wrong? what can be the solution?

조회 수: 1 (최근 30일)
SAHIL SAHOO
SAHIL SAHOO 2022년 7월 20일
댓글: SAHIL SAHOO 2022년 7월 20일
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)
  댓글 수: 6
VBBV
VBBV 2022년 7월 20일
The for loop code works well as shown in my answer below. which can also be done without a loop.
SAHIL SAHOO
SAHIL SAHOO 2022년 7월 20일
yeah I mean plot (Y(:,3), U) and the graph is correct too.

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채택된 답변

VBBV
VBBV 2022년 7월 20일
편집: VBBV 2022년 7월 20일
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')
Warning: Ignoring extra legend entries.
plot(T,U)
You can plot both U and T variables , but for loop needs modififcation
  댓글 수: 2
VBBV
VBBV 2022년 7월 20일
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
SAHIL SAHOO 2022년 7월 20일
yes, this graph is correct that what i expected, without for loop I get a straight line.

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추가 답변 (1개)

Torsten
Torsten 2022년 7월 20일
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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