How to solve the differential equation say ode45 with multiple inputs(vector)
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I have a a differential equation swing100
function dydt= swing100(t,y)
global Tm Te D H wB ;
dydt=[wB.*y(2);(0.5/H).*(Tm-Te-(D.*y(2)))];
This is called from a call_swing100.m file:
clc
clear all
global Tm Te D H wB ;
Tm=[1.0];
Te=[0.99];
D=[0];
H=[4];
wB=2*pi*50; %constant
y1=[0];
y2=[0];
y0=[y1; y2]; % intial conditions matrix
%dydt = odefun(t,y);
[t,y] = ode45(@swing100,[0 20],y0);
plot(t,y(:,1),'-',t,y(:,2),'--')
title('Solution of Swing Equation, ');
xlabel('time t');
ylabel('solution y');
legend('y_1','y_2')
This works Fine.
Now I want to solve the same with not one Tm but with say three values of Tm,Te,D,H. That is now Tm etc are vector inputs. So how do I do that? I have tried many 'wrong 'things and keep getting errors.
In simulink for example it works when I give an input vector. How do I do that in .m? Should I have to write a for loop and call each one once?
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Star Strider
2014-4-16
This works:
swing100 = @(t,y, Tm,Te,D,H,wB) [wB.*y(2); (0.5/H).*(Tm-Te-(D.*y(2)))];
Tm = 1:3:9;
Te = .99*(1:3);
D = (0:2);
H = 4:6;
wB=2*pi*50; %constant
y0 = [0 0];
ts = linspace(0,20,20);
tic
for k1 = 1:3
for k2 = 1:3
for k3 = 1:3
for k4 = 1:3
f = @(t,y) swing100(t,y, Tm(k1),Te(k2),D(k3),H(k4),wB);
[t,y] = ode45(f, ts, y0);
ymtx(k1,k2,k3,k4,:,:) = y;
end
end
end
end
toc
figure(1)
plot(ts, squeeze(ymtx(1,2,3,3,:,:)) )
legend('y_1', 'y_2', 'Location', 'SE')
grid
I specified a range of integration times in variable ts so all the y vectors would have the same number of rows. (The entire set of integrations took about 90 seconds.)
I created the vectors for the variables Tm, Te, D, H you want to iterate over, so you will have to substitute the correct ones for mine.
Plotting it is not straightforward, so I included an example of that in case you want to plot them.
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Satyanarayan
2014-4-16
3 个评论
Star Strider
2014-4-16
My pleasure!
You can use the array reference I used in the plot command to get the values you want from your ODE integrations.
And thanks!
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