How to plot a current based on active voltage?
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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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1514274/image.png)
1 件のコメント
Torsten
2023 年 10 月 18 日
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.
採用された回答
Torsten
2023 年 10 月 18 日
編集済み: Torsten
2023 年 10 月 18 日
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
その他の回答 (1 件)
Florian Bidaud
2023 年 10 月 18 日
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
3 件のコメント
Florian Bidaud
2023 年 10 月 18 日
編集済み: Florian Bidaud
2023 年 10 月 18 日
The plots you have inside your function are not of any use for the solving
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