Differential equation not working

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Reji G
Reji G on 17 Oct 2022
Commented: Walter Roberson on 21 Oct 2022
I wanted to implement the equation below:
I wrote the program as follows. But nothing showing up in the plot.
clc; close all; clear all;
t=0.01:0.001:10;
V=575*sin(2*pi*t);
I=100*sin(2*pi*t);
P0=250;
syms V(t)
syms I(t)
ode = diff(V) == (V*I)*diff(I)-((V/0.01)*(((V*I)/P0)-1));
VSol(t) = dsolve(ode);
fplot(VSol(t),[0 10]);
  2 Comments
Torsten
Torsten on 17 Oct 2022
Edited: Torsten on 17 Oct 2022
Why do you specify I and V and then try to solve a differential equation for V ? If you know that V =575*sin(2*pi*t), you don't need to solve a differential equation.
Further, in order to plot a possible solution for V, you will have to specify an initial condition, e.g. V at t=0.
Reji G
Reji G on 18 Oct 2022
What should I modify in the code ?
If V(0)=0.1, then ?
I'm not knowing much of these. That's why came to help center

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Answers (1)

Torsten
Torsten on 18 Oct 2022
I = @(t)100*sin(2*pi*t);;
dIdt = @(t) 2*pi*100*cos(2*pi*t);
V0 = 0.1;
fun = @(t,V,I,dIdt) V/I(t)*dIdt(t)-V/0.01*(V*I(t)/250-1.0);
[T,V] = ode45(@(t,V) fun(t,V,I,dIdt),0.01:0.001:10,V0);
plot(T,V)
  19 Comments
Reji G
Reji G on 21 Oct 2022
Thank you Torsten.
For I = @(t)100*sin(2*pi*t); your first code itself is working fine.
My concern is I = @(t)100*sin(100*pi*t); (in place of 1Hz if I give 50 HZ(2->100)), the program is showing 5 errors. I want a plot for I = @(t)100*sin(100*pi*t).
Walter Roberson
Walter Roberson on 21 Oct 2022
f = 50; %Hz
I = @(t)100*sin(2*pi*f*t);
dIdt = @(t) 2*pi*f*cos(2*pi*t); %guessing here
V0 = 0.1;
fun = @(t,V,I,dIdt) V/I(t)*dIdt(t)-V/0.01*(V*I(t)/250-1.0);
delta = 1/(10*f);
tspan = delta:delta:1/(2*f)-delta;
[T,V] = ode45(@(t,V) fun(t,V,I,dIdt), tspan, V0);
fun_inter = @(t)interp1(T,V,delta+mod(t-delta, 1/f-delta));
x = 0:delta:10;
plot(x,fun_inter(x))
Heavy aliasing on the drawing.
x = 0:delta:1;
plot(x,fun_inter(x))

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