How to plot a current based on active voltage?

I am to plot a current line in a graphs based on a active voltage. When I plot it, 200+ graphs appear and i wonder if maybe it is a mistake to not include it as an array, but cant quite figure it out. Can anybode please help?:)

tspan = [0 0.5];

R = 8.314;

F = 9.648e4;

T = 80 + 273.15; %Conductivity reference temperature

ipp_0 = 0.67e-4;

A_m = 232;

b_ca = 0.55 ;

b_an = 1-b_ca ;

C_dl = 3.5e-2*A_m ;

u0 = 0;

ipp_1 =0;

[T,U] = ode15s(@(t,u) u_act(t,u,A_m,ipp_0,b_an,F,R,T,b_ca,C_dl),tspan,u0);

plot(T,U)

%im = A_m*ipp_0*(exp(b_an*F/R/T*U) - exp(-b_ca*F/R/T*U));

%ipp = (idl+im)/A_m

function i_ret = i(t)

if t< 0.2

i_ret = 100;

elseif t < 0.4

i_ret = 10;

else

i_ret = 100;

end

end

function dudt = u_act(t,u,A_m,ipp_0,b_an,F,R,T,b_ca,C_dl)

im = A_m*ipp_0*(exp(b_an*F/R/T*u) - exp(-b_ca*F/R/T*u));

idl = i(t) - im;

dudt = idl/C_dl;

ipp = (idl+im)/A_m;

figure

plot(T, ipp)

end

It is supposed to look something like this, but we are to find i_pp which is i/A_m.

[Torsten]

If differential equations have discontinuous parameters (like "i" in the above case), one should restart the optimizer with the solution obtained so far at the points of discontinuity instead of integrating over. That's why I arranged the code the way I did.

Answer 1

[Torsten]

i = [100,10,100];
t = [0,0.2,0.4,0.6];
R = 8.314;
F = 9.648e4;
Temp = 80 + 273.15; %Conductivity reference temperature
ipp_0 = 0.67e-4;
A_m = 232;
b_ca = 0.55 ;
b_an = 1-b_ca ;
C_dl = 3.5e-2*A_m ;
T = [];
U = [];
u0 = 0;
for j = 1:numel(i)
  fun= @(t,u) (i(j)-A_m * ipp_0 * (exp(b_an*F/(R*Temp)*u)-exp(-b_ca*F/(R*Temp)*u)))/C_dl;
  tspan = [t(j) t(j+1)];
  [Tstep,Ustep] = ode15s(fun,tspan,u0);
  T = [T;Tstep];
  U = [U;Ustep];
  u0 = Ustep(end,:);
end
figure(1)
plot(T,U)
grid on

ifun = @(time) 0;
for j = 1:numel(t)-1
   ifun = @(time)ifun(time)+i(j)*(time>=t(j))*(time<t(j+1));
end
for j = 1:numel(T)
   iplot(j) = ifun(T(j));
   im(j) = A_m*ipp_0*(exp(b_an*F/R/Temp*U(j)) - exp(-b_ca*F/R/Temp*U(j)));
   idl(j) = iplot(j) - im(j);   
end
figure(2)
plot(T,[iplot;idl;im].')
grid on

[Anniken]

This is really clever and i feel like its above my level of expertise in matlab and will try to understand the complex code you have provided! I am truly grateful for your help:))

Answer 2

[Florian Bidaud]

Well you have a plot inside your u_act function, which is gonna be called at each calculation step in ode15s.

If you remove them you get the plot you want

tspan = [0 0.5];
R = 8.314;
F = 9.648e4;
T = 80 + 273.15; %Conductivity reference temperature
ipp_0 = 0.67e-4;
A_m = 232;
b_ca = 0.55 ;
b_an = 1-b_ca ;
C_dl = 3.5e-2*A_m ;
u0 = 0;
ipp_1 =0;
[T,U] = ode15s(@(t,u) u_act(t,u,A_m,ipp_0,b_an,F,R,T,b_ca,C_dl),tspan,u0);
plot(T,U)

%im = A_m*ipp_0*(exp(b_an*F/R/T*U) - exp(-b_ca*F/R/T*U));
%ipp = (idl+im)/A_m
function i_ret = i(t)
    if t< 0.2
        i_ret = 100;
    elseif t < 0.4
        i_ret = 10;
    else
        i_ret = 100;
    end
end
function dudt = u_act(t,u,A_m,ipp_0,b_an,F,R,T,b_ca,C_dl)
    im = A_m*ipp_0*(exp(b_an*F/R/T*u) - exp(-b_ca*F/R/T*u));
    idl = i(t) - im;
    dudt = idl/C_dl;
    ipp = (idl+im)/A_m; 
end

[Anniken]

But we need to use the solution gived from the plot in dudt function? So we kind of need to collect the results from function dudt and implement them in maybe antoher function which plots the current ipp in regards to the time(x-axis) while take into consideration the function values.

[Florian Bidaud]

The plots you have inside your function are not of any use for the solving

[Anniken]

Yeah i should remove it. But do you know how i can plot the ipp based on the u_act and time?